url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://learn.saylor.org/mod/book/tool/print/index.php?id=37127&chapterid=20666 | 1,721,052,105,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514707.52/warc/CC-MAIN-20240715132531-20240715162531-00759.warc.gz | 319,336,781 | 12,692 | # The Definition of a Derivative
The graphical idea of a slope of a tangent line is very useful, but for some uses we need a more algebraic definition of the derivative of a function. We will use this definition to calculate the derivatives of several functions and see that the results from the definition agree with our graphical understanding. We will also look at several different interpretations for the derivative, and derive a theorem which will allow us to easily and quickly determine the derivative of any fixed power of $x$
In the last section we found the slope of the tangent line to the graph of the function $f(x)=x^{2}$ at an arbitrary point $(x, f(x))$ by calculating the slope of the secant line through the points $(x, f(x))$ and $(x+h, f(x+h))$,
$\mathrm{m_{sec}}=\frac{\mathrm{f}(\mathrm{x}+\mathrm{h})-\mathrm{f}(\mathrm{x})}{(\mathrm{x}+\mathrm{h})-(\mathrm{x})}$
and then by taking the limit of $m_{sec}$ as h approached 0 (Fig. 1). That approach to calculating slopes of tangent lines is the definition of the derivative of a function.
Definition of the Derivative:
The derivative of a function $\mathrm{f}$ is a new function, $\mathbf{f}$ ' (pronounced "eff prime"),
whose value at $x$ is $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ if the limit exists and is finite.
This is the definition of differential calculus, and you must know it and understand what it says. The rest of this chapter and all of Chapter 3 are built on this definition as is much of what appears in later chapters. It is remarkable that such a simple idea (the slope of a tangent line) and such a simple definition (for the derivative $f '$) will lead to so many important ideas and applications.
Notation: There are three commonly used notations for the derivative of $\mathbf{y}=\mathbf{f}(\mathbf{x})$:
$\mathbf{f}^{\prime}(\mathbf{x})$ emphasizes that the derivative is a function related to $\mathrm{f}$
$D(f)$ emphasizes that we perform an operation on $f$ to get the derivative of $f$
$\frac{d f}{d x} \quad$ emphasizes that the derivative is the limit of $\frac{\Delta f}{\Delta x}=\frac{f(x+h)-f(x)}{h}$.
We will use all three notations so you can get used to working with each of them.
f'$(x)$ represents the slope of the tangent line to the graph of $\mathrm{y}=\mathrm{f}(\mathrm{x})$ at the point $(\mathrm{x}, \mathrm{f}(\mathrm{x}))$ or the instantaneous rate of change of the function $\mathrm{f}$ at the point $(\mathrm{x}, \mathrm{f}(\mathrm{x})$ ).
If, in Fig. 2, we let $x$ be the point $a+h$, then $\mathrm{h}=\mathrm{x}-\mathrm{a}$. As $\mathrm{h} \rightarrow 0$, we see that $\mathrm{x} \rightarrow \mathrm{a}$ and $\lim\limits_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}=\lim\limits_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}$ so
$\mathbf{f}^{\prime}(\mathbf{a})=\lim\limits_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}=\lim\limits_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}$.
We will use whichever of these two forms is more convenient algebraically. | 849 | 2,999 | {"found_math": true, "script_math_tex": 32, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-30 | latest | en | 0.892577 |
https://gssc.esa.int/navipedia/index.php/Mapping_of_Niell | 1,558,323,662,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232255536.6/warc/CC-MAIN-20190520021654-20190520043654-00001.warc.gz | 505,443,543 | 7,882 | If you wish to contribute or participate in the discussions about articles you are invited to join Navipedia as a registered user
# Mapping of Niell
Fundamentals
Title Mapping of Niell
Author(s) J. Sanz Subirana, J.M. Juan Zornoza and M. Hernández-Pajares, Technical University of Catalonia, Spain.
Year of Publication 2011
A mapping function that does not require surface meteorology measurements, but provides comparable accuracy and precision to others that requires such data is presented in [Niell, 1996][1]. This mapping only uses the receiver geographic location and measurement time as inputs.
The Niell's mapping involves two different obliquity factors, for the dry and wet components, which are computed from the equations (1) to (4), where $(E)$ is the elevation of ray and $(H)$ is the receiver height, in kilometres:
Hydrostatic mapping function:
$\begin{array}{l} M_{dry}(E,H)= m(E,a_d,b_d,c_d)+ \Delta m(E,H)\\ \qquad\mbox{with}\\ \Delta m(E,H)= \left[ \frac{1}{sin E} - m(E,a_{ht},b_{ht},c_{ht}) \right ] \cdot H \end{array} \qquad\mbox{(1)}$
Wet Mapping Function:
$M_{wet}(E)= m(E,a_w,b_w,c_w) \qquad\mbox{(2)}$
where, $m(E,a,b,c)$ is the [Marini, 1972][2] mapping normalised to unity at zenith:
$m(E,a,b,c)= \frac{1+\frac{a}{1+\frac{b}{1+c}}}{\sin E+\frac{a}{\sin E+\frac{b}{\sin E+c}}} \qquad\mbox{(3)}$
• The Hydrostatic parameters $a_d$, $b_d$, $c_d$ are time $(t)$ and latitude ($\Phi)$ dependent parameters given by:
$\xi(\phi,t)=\xi_{_{avg}}(\phi)-\xi_{_{amp}}(\phi) \cos \left ( 2\pi \frac{t-T_0}{365.25} \right ) \qquad\mbox{(4)}$
where, $t$ is the time from January $0.0$, in days, and $T_0$ is taken as DoY $28$ (i.e., $T_0=28$). The parameters $\xi_{_{avg}}(\phi_i)$ and $\xi_{_{amp}}(\phi_i)$ are linearly interpolated from table 1 between the nearest $\xi(\Phi_i)$. The $a_{ht}$, $b_{ht}$, $c_{ht}$ are taken directly from the same table 1.
• The wet parameters $a_w$, $b_w$, $c_w$ are latitude dependent and are linearly interpolated from table 2 between the nearest $\xi(\Phi_i)$.
Table 1: Coefficients of the hydrostatic mapping function.
Table 2: Coefficients of the wet mapping function.
## References
1. ^ [Niell, 1996] Niell, A., 1996. Global mapping functions for the atmosphere delay at radio wavelengths. Journal of Geophysical Research. 101, pp. 3227-3246.
2. ^ [Marini, 1972] Marini, J., 1972. Correction of Satellite tracking data for an arbitrary Tropospheric Pro_le. Radio Science. 7(2), pp. 223-231. | 773 | 2,464 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2019-22 | latest | en | 0.738962 |
https://zipfslaw.org/2015/08/15/regression-models-in-french/ | 1,674,779,486,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494852.95/warc/CC-MAIN-20230127001911-20230127031911-00592.warc.gz | 1,136,343,486 | 31,511 | # Regression models in French: Part I
One of the hot topics in linguistics right now is mixed effects models. A mixed effect model is a kind of regression analysis. Regression analysis is a way of building a statistical model of a phenomenon. There are all kinds of things that you might want to build a statistical model of in linguistics, including phonetic relationships, sociolinguistics, syntax, and doubtless many others. I’m going to use this post to put up some links to things that you might find useful in learning about mixed models, and of course we’ll come across some French vocabulary on the way. (A note on the vocabulary in this post: it is mostly not found in dictionaries. I induced it from examples on linguee.fr, an excellent source for finding examples of French technical vocabulary in use.)
The absolute best material for learning about mixed effects models so far is this tutorial by Bodo Winter. If you’re not familiar with simple linear regression (i.e. with fixed effects only), you might want to check out this tutorial of his first. Besides being really clear, Bodo’s tutorial is especially suitable for linguists, because it works through an extended example on F0 (fundamental frequency–roughly, the pitch of your voice) variation in situations of different politeness levels.
Let’s build up to the vocabulary of mixed effects models. First, some basic vocabulary for talking about regression modelling. Bear in mind that regression modelling–well, simple linear regression modelling–is about finding a formula that can predict the value for something on the basis of the value of something else. The figure to the left plots F1 (first formant frequencies–part of what makes a vowel sound like what it sounds like) for female speakers of several language over the F1 for male speakers of the same language. (The data comes from the web site accompanying Keith Johnson’s book Quantitative methods in linguistics.) The line on the plot reflects a formula that will let you predict the F1 of a female speaker if you know the F1 of a male speaker. Not surprisingly, the female frequencies are always higher–one of the determinants of overall patterns of F1 is that all other things being equal, the shorter your vocal tract is, the higher your F1 will be, and all other things being equal, women have shorter vocal tracts than men, on average. What the line says is that you can get pretty close to an accurate prediction of the female F1 if you multiply the male F1 by 1.29. (Yes, we’re glossing over the y intercept.) OK, now on to that basic vocabulary:
• le modèle: model.
• le modèle de régression: regression model.
• la régression linéaire: linear regression.
• la régression logistique: logistic regression.
• la régression linéaire simple: simple linear regression.
That got us through simple linear regression modelling. Recall that in simple linear regression, you’re predicting a value for something on the basis of the value of something else. But, most things don’t have simple one-to-one relationships. Rather, it’s often the case that you need to predict one thing on the basis of multiple other things. For example, suppose that you want to know what affects how long it takes a speaker of a language to respond to the question of whether or not a given sentence is grammatical (i.e., could be said in that language. Colorless green ideas sleep furiously doesn’t mean anything, but you could say it in English. On the other hand, green sleep colorless ideas furiously is something that you couldn’t say in English). You might have to include multiple things in the model–how long the sentence is, how frequent the words in the sentence are, how long the words are, etc. In this case–predicting one thing (response time) from multiple things (sentence length, word frequency, word length)–you need something called multiple linear regression. This brings up more vocabulary:
• la régression multiple: multiple regression.
• la régression linéaire multiple: multiple linear regression.
So far, we know how to talk about linear regression. What both kinds of linear regression have in common is that (a) we’re predicting a value from something else–from one value in the case of simple linear regression, or from multiple values in the case of multiple linear regression–and (b) we can describe the relationship between the value that we’re trying to predict and the value(s) that we’re trying to predict it from on the basis of a (straight) line. Some relationships can’t be described by a straight line, though. A classic example in linguistics is the U-shaped curve in language acquisition by children. This describes a common phenomenon relating age to the percentage of correct productions of some linguistic target–say, irregular plurals, or the past tenses of verbs. Initially, the child has a high percentage of correct productions. Then, the child goes through a stage where the percentage of correct productions drops. (As the figure suggests, this is thought to be because the child has made a transition from “memorizing” the regular and irregular forms to developing a hypothesis about a rule for forming plurals, or past tenses, or whatever.) Finally, the child’s percentage of production of the correct forms climbs again. Now we can’t describe the relationship between what we’re trying to predict (the percentage of correct productions and what we’re trying to predict it from (the child’s age) with a straight line. However, there is another kind of regression that we can use. It is called non-linear regression:
• la régression non linéaire: non-linear regression.
We’ve now talked about three kinds of regression modelling. They all have in common the fact that they are used to predict the value for something from the value(s) for something else. If we’re trying to predict one value from one other value, that’s simple linear regression (la régression linéaire simple). If we’re trying to predict one value from multiple other values, that’s multiple linear regression (la régression linéaire multiple). And, if the relationship between what we’re trying to predict and what we’re trying to predict it from can’t be described by a straight line, then we have non-linear regression (la régression non linéaire). (Before you ask: yes, there is such a thing as non-linear multiple regression, but I don’t know how to say it in French. Heck, I’m not even sure how to say it in English–non-linear multiple regression? Multiple non-linear regression? It’s pretty rare.) There’s one more kind of regression modelling that we need to talk about before we can move on to mixed effects regression modelling: logistic regression.
Logistic regression is used to predict the probability of something from something else. Up ’til now, we’ve been predicting a value; now we’re predicting a probability. What is the probability that a vowel will be unvoiced (whispered)? What is the probability that I will pronounce -ing, versus -in’? These are questions for logistic regression. I’ll leave out the details, but we need to know the vocabulary:
• la régression logistique: logistic regression.
OK, we can talk about a variety of types of regression modelling in French now. But, to talk about mixed effects regression modelling, we also need to be able to talk about effects. This post is already super-long, so let’s save that for next time. In the meantime, here’s a shout-out to Bodo Winter, regression-modelling explainer extraordinaire: https://twitter.com/BodoWinter.
## One thought on “Regression models in French: Part I”
Curative Power of Medical Data
JCDL 2020 Workshop on Biomedical Natural Language Processing
Crimescribe
Criminal Curiosities
BioNLP
Biomedical natural language processing
Mostly Mammoths
but other things that fascinate me, too
Zygoma
Our French Oasis
FAMILY LIFE IN A FRENCH COUNTRY VILLAGE
ACL 2017
PC Chairs Blog
Abby Mullen
A site about history and life
EFL Notes
Random commentary on teaching English as a foreign language
Natural Language Processing
Université Paris-Centrale, Spring 2017
Speak Out in Spanish!
living and loving language
- MIKE STEEDEN -
THE DRIVELLINGS OF TWATTERSLEY FROMAGE
mathbabe
Exploring and venting about quantitative issues | 1,815 | 8,327 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2023-06 | longest | en | 0.878027 |
http://support.sas.com/documentation/cdl/en/statug/68162/HTML/default/statug_quantlife_examples01.htm | 1,537,411,096,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156376.8/warc/CC-MAIN-20180920020606-20180920040606-00379.warc.gz | 219,865,881 | 10,028 | # The QUANTLIFE Procedure
### Example 94.1 Primary Biliary Cirrhosis Study
This example illustrates how to use quantile regression analysis to detect varying covariate effects on survival time. Consider a study of primary biliary cirrhosis, a rare but fatal chronic liver disease, discussed by Fleming and Harrington (1991). Researchers followed 418 patients who had this disease, 161 of whom died during the study.
The data set contains the following variables:
• Time, follow-up time, in years
• Status, event indicator, with value 1 for death time and value 0 for censored time
• Age, age from birth to study registration, in years
• Albumin, serum albumin level, in g/dl
• Bilirubin, serum bilirubin level, in mg/dl
• Edema, edema presence
• Protime, prothrombin time, in seconds
The following statements create the data set PBC, which is used in this example:
data pbc;
input Time Status Age Albumin Bilirubin Edema Protime @@;
label Time="Follow-Up Time in Days";
logAlbumin = log(Albumin);
logBilirubin = log(Bilirubin);
logProtime = log(Protime);
datalines;
400 1 58.7652 2.60 14.5 1.0 12.2 4500 0 56.4463 4.14 1.1 0.0 10.6
1012 1 70.0726 3.48 1.4 0.5 12.0 1925 1 54.7406 2.54 1.8 0.5 10.3
1504 0 38.1054 3.53 3.4 0.0 10.9 2503 1 66.2587 3.98 0.8 0.0 11.0
1832 0 55.5346 4.09 1.0 0.0 9.7 2466 1 53.0568 4.00 0.3 0.0 11.0
2400 1 42.5079 3.08 3.2 0.0 11.0 51 1 70.5599 2.74 12.6 1.0 11.5
3762 1 53.7139 4.16 1.4 0.0 12.0 304 1 59.1376 3.52 3.6 0.0 13.6
... more lines ...
989 0 35.0000 3.23 0.7 0.0 10.8 681 1 67.0000 2.96 1.2 0.0 10.9
1103 0 39.0000 3.83 0.9 0.0 11.2 1055 0 57.0000 3.42 1.6 0.0 9.9
691 0 58.0000 3.75 0.8 0.0 10.4 976 0 53.0000 3.29 0.7 0.0 10.6
;
The next statements fit a linear model for the log of survival time of the PBC patients with the covariates logBilirubin, logProtime, logAlbumin, Age, and Edema:
ods graphics on;
proc quantlife data=pbc log method=na plot=(quantplot survival) seed=1268;
model Time*Status(0)=logBilirubin logProtime logAlbumin Age Edema
/quantile=(.1 .2 .3 .4 .5 .6 .75);
run;
You use the QUANTILE= option to specify a set of quantiles of interest for comparing quantile-specific covariate effects. The METHOD= option specifies the Nelson-Aalen method for estimating the regression parameters.
The QUANTLIFE procedure provides resampling methods for computing confidence limits for the parameters; for more information, see the section Confidence Interval. By default, the repetition number is 200. You can request a different number of repetitions by specifying the NREP= option. You can also use the SEED= option to specify the seed for generating random numbers so that you can later reproduce the results.
Output 94.1.1 displays model information and information about censoring in the data. Out of 418 observations, 257 are censored.
Output 94.1.1: Model Information
The QUANTLIFE Procedure
Model Information
Data Set WORK.PBC
Dependent Variable Log(Time)
Censoring Variable Status
Censoring Value(s) 0
Number of Observations 418
Method Nelson-Aalen
Replications 200
Seed for Random Number Generator 1268
Summary of the Number of Event and Censored
Values
Total Event Censored Percent
Censored
418 161 257 61.48
Output 94.1.2 provides the parameter estimates. Each quantile level has a set of parameter estimates and confidence limits.
Output 94.1.2: Parameter Estimates at Different Quantiles
Parameter Estimates
Quantile Parameter DF Estimate Standard
Error
95% Confidence Limits t Value Pr > |t|
0.1000 Intercept 1 14.8030 4.0967 6.7736 22.8325 3.61 0.0003
logBilirubin 1 -0.4488 0.1485 -0.7398 -0.1578 -3.02 0.0027
logProtime 1 -3.6378 1.4560 -6.4915 -0.7841 -2.50 0.0129
logAlbumin 1 1.9286 0.9756 0.0165 3.8408 1.98 0.0487
Age 1 -0.0244 0.0107 -0.0455 -0.00334 -2.27 0.0237
Edema 1 -1.0712 0.6688 -2.3820 0.2396 -1.60 0.1100
0.2000 Intercept 1 15.1800 2.6664 9.9540 20.4060 5.69 <.0001
logBilirubin 1 -0.6532 0.0886 -0.8268 -0.4796 -7.37 <.0001
logProtime 1 -3.3273 0.9401 -5.1699 -1.4847 -3.54 0.0004
logAlbumin 1 1.6842 0.6888 0.3343 3.0342 2.45 0.0149
Age 1 -0.0291 0.00687 -0.0425 -0.0156 -4.23 <.0001
Edema 1 -0.7265 0.3179 -1.3497 -0.1034 -2.29 0.0228
0.3000 Intercept 1 13.2382 2.5296 8.2804 18.1961 5.23 <.0001
logBilirubin 1 -0.6013 0.0762 -0.7506 -0.4521 -7.90 <.0001
logProtime 1 -2.5816 0.8907 -4.3273 -0.8359 -2.90 0.0039
logAlbumin 1 1.7246 0.7142 0.3248 3.1245 2.41 0.0162
Age 1 -0.0244 0.00716 -0.0385 -0.0104 -3.41 0.0007
Edema 1 -0.8577 0.2763 -1.3992 -0.3163 -3.10 0.0020
0.4000 Intercept 1 13.4716 3.0874 7.4204 19.5228 4.36 <.0001
logBilirubin 1 -0.6047 0.0846 -0.7705 -0.4389 -7.15 <.0001
logProtime 1 -2.1632 1.1726 -4.4615 0.1351 -1.84 0.0658
logAlbumin 1 0.9819 0.7191 -0.4274 2.3912 1.37 0.1728
Age 1 -0.0255 0.00681 -0.0389 -0.0122 -3.74 0.0002
Edema 1 -1.0589 0.3104 -1.6672 -0.4506 -3.41 0.0007
0.5000 Intercept 1 10.9205 2.8047 5.4235 16.4175 3.89 0.0001
logBilirubin 1 -0.5315 0.0904 -0.7087 -0.3543 -5.88 <.0001
logProtime 1 -1.2222 1.2142 -3.6020 1.1577 -1.01 0.3148
logAlbumin 1 1.5700 0.6284 0.3383 2.8016 2.50 0.0129
Age 1 -0.0318 0.00883 -0.0491 -0.0145 -3.60 0.0004
Edema 1 -0.7316 0.3743 -1.4653 0.00202 -1.95 0.0513
0.6000 Intercept 1 11.2381 2.6294 6.0846 16.3917 4.27 <.0001
logBilirubin 1 -0.5701 0.0852 -0.7370 -0.4031 -6.69 <.0001
logProtime 1 -1.3508 1.1402 -3.5856 0.8840 -1.18 0.2368
logAlbumin 1 1.3704 0.5091 0.3726 2.3682 2.69 0.0074
Age 1 -0.0226 0.0109 -0.0440 -0.00111 -2.06 0.0399
Edema 1 -0.5141 0.3088 -1.1193 0.0912 -1.66 0.0968
0.7500 Intercept 1 10.0954 3.1893 3.8445 16.3463 3.17 0.0017
logBilirubin 1 -0.6366 0.1071 -0.8466 -0.4267 -5.94 <.0001
logProtime 1 -0.9670 1.2343 -3.3862 1.4521 -0.78 0.4338
logAlbumin 1 1.8148 0.5883 0.6618 2.9678 3.08 0.0022
Age 1 -0.0203 0.0156 -0.0509 0.0102 -1.30 0.1931
Edema 1 -0.3529 0.3120 -0.9644 0.2586 -1.13 0.2587
For comparison, the following statements use the LIFEREG procedure to fit a Weibull distribution to the data. The LIFEREG procedure fits an accelerated failure time model, which assumes that the independent variables have a multiplicative effect on the event time.
proc lifereg data=pbc;
model Time*Status(0)=logBilirubin logProtime logAlbumin Age Edema;
run;
Output 94.1.3 provides the parameter estimates that are computed by PROC LIFEREG.
Output 94.1.3: Parameter Estimates from PROC LIFEREG
The LIFEREG Procedure
Analysis of Maximum Likelihood Parameter Estimates
Parameter DF Estimate Standard
Error
95% Confidence Limits Chi-Square Pr > ChiSq
Intercept 1 12.2155 1.4539 9.3658 15.0651 70.59 <.0001
logBilirubin 1 -0.5770 0.0556 -0.6861 -0.4680 107.55 <.0001
logProtime 1 -1.7565 0.5248 -2.7850 -0.7280 11.20 0.0008
logAlbumin 1 1.6694 0.4276 0.8313 2.5074 15.24 <.0001
Age 1 -0.0265 0.0053 -0.0368 -0.0162 25.35 <.0001
Edema 1 -0.6303 0.1805 -0.9842 -0.2764 12.19 0.0005
Scale 1 0.6807 0.0430 0.6014 0.7704
Weibull Shape 1 1.4691 0.0928 1.2980 1.6628
The p-value for logProtime is very small. For this same variable, the p-values that result from the quantile regression analysis are 0.3148 for the 0.5th quantile and 0.4338 for the 0.75th quantile, and the p-values are much smaller for the lower quantiles. Apparently, the effect of this covariate depends on which side of the response distribution is being modeled.
The PLOT=QUANTPLOT option in the PROC QUANTLIFE statement requests the quantile process plots in Output 94.1.4 and Output 94.1.5, which plot the estimated regression parameter against the quantile level. You can use these plots to compare quantile-specific covariate effects. If the curve is not constant, it can indicate heterogeneity in the data. The interpretation of the regression coefficients at a given quantile is similar to that of classical regression analysis. That is, the coefficient from a given covariate indicates the effect on log(Time) of a unit change in that covariate, assuming that the other covariates are fixed.
Output 94.1.4: Quantile Processes with 95% Confidence Bands
Output 94.1.5: Quantile Processes with 95% Confidence Bands
In Output 94.1.4, you can see that the effect of logProtime has a negative effect over the lower quantiles, which diminishes in magnitude at the median and upper quantiles. This insight would be missed if you were using the accelerated failure model. | 3,401 | 8,278 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-39 | longest | en | 0.614215 |
https://codelucky.com/python-sum/ | 1,701,612,384,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.23/warc/CC-MAIN-20231203125921-20231203155921-00800.warc.gz | 215,542,595 | 24,244 | # Python sum() Function – Tutorial with Examples
The sum() function in Python is a built-in function that calculates the sum of all the elements in an iterable. It takes an iterable as an argument and returns the sum of its elements. An iterable is any object that can be iterated over, such as lists, tuples, sets, dictionaries, and more.
Table of Contents
## Syntax
```sum(iterable, start=0)
```
### Parameters
• iterable : The iterable to be summed.
• start : The value to start the sum with. It is an optional parameter and its default value is 0.
### Return Value
The sum() function returns the sum of all the elements in the iterable, starting with the value of the ‘start’ parameter.
## Examples
### Example 1: Calculating the sum of elements in a list
```# Sum of elements in a list
numbers = [1, 2, 3, 4, 5]
print(sum(numbers))
```
```15
```
### Example 2: Calculating the sum of elements in a tuple
```# Sum of elements in a tuple
numbers = (1, 2, 3, 4, 5)
print(sum(numbers))
```
```15
```
### Example 3: Using the ‘start’ parameter
```# Using the 'start' parameter
numbers = [1, 2, 3, 4, 5]
print(sum(numbers, 10))
```
```25
```
## Use Cases
• Calculating the total sum of elements in an iterable such as a list, tuple, set, etc.
• Calculating the sum of elements in a sequence such as a range.
• Calculating the sum of values in a dictionary, by using the values() method to return the values as a list.
In conclusion, the sum() function is a simple and efficient way to calculate the sum of elements in an iterable in Python. It is commonly used in a variety of use cases, such as calculating the total sum of elements in a list, tuple, set, or range, and even in dictionaries, by using the values() method to return the values as a list. | 457 | 1,773 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.76996 |
https://community.smartsheet.com/discussion/128218/nested-if-statement-with-index-and-collect-formula-multiple-reference | 1,722,928,444,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640476915.25/warc/CC-MAIN-20240806064139-20240806094139-00829.warc.gz | 148,464,844 | 109,937 | # Nested If Statement with Index and Collect formula (multiple reference)
Options
Hi,
Hoping for some guidance and help in the right direction or alternative solution.
Trying to get Estimated Total in Summary sheet to return the proper calculation of Each Site reference, which are pulled from another master sheet reference.
The Estimated Total is calculated based on Location site reference, Years, Scope type[multi-select fields].
for example:
Summary sheet it is referencing first Year 1, Tropic , Annual should return Estimate : \$1500
Column: Site Reference, Years, Location , Scope Type, Estimate
Year 1, Year 2, Year 3 [Single select fields]
Location A , Location B, Location C [Single select fields]
Scope Type A, Scope Type B, Scope Type C - [Multi-select fields]
Estimate A, Estimate B, Estimate C
Alternatively, (if selected
Years = Year 1
Site Reference: Tropic
Scope Type: Annual, Semi-Annual & 3 Year Test
Estimated Total: \$ 5,800 (getting errors on Summary sheet)
using reference from Master sheet
Master Sheet Reference:
Column: Year, Location , Scope Type, Estimate
Year 1, Year 2, Year 3
Location A , Location B, Location C
Scope Type A, Scope Type B, Scope Type C - [Multi-select fields]
Estimate A, Estimate B, Estimate C
On Main Summary Sheet
=IF([Scope Type]@row = "Annual", INDEX((COLLECT({Estimate Tropic A}, {Year Tropic A}, Years@row, {Location Tropic A}, [Site Reference]@row, {Scope Type Tropic A}, "Annual")), 1)IF([Scope Type]@row = "Semi-Annual", INDEX((COLLECT({Estimate Tropic A}, {Year Tropic A}, Years@row, {Location Tropic A}, [Site Reference]@row, {Scope Type Tropic A}, "Semi-Annual")), 1),IF([Scope Type]@row = "3 Year Test", INDEX((COLLECT({Estimate Tropic A}, {Year Tropic A}, Years@row, {Location Tropic A}, [Site Reference]@row, {Scope Type Tropic A}, "3 Year Test")), 1))
I am using column Estimate Location A, Estimate Location B, Estimate Location C to test that each site reference is calculating properly (like helper columns to check), but if there is a better way to include all three that would be great as well!
is there an easier way to do this? or fix this formula?
I am open to any suggestions,
Thank you!
• ✭✭✭✭
Options
Hi @SamEast
Maybe try the following:
Change your master sheet to have fewer columns
Then, in your summary sheet, you can do one of two things,
1: Change the multi select to separate checkboxes:
Then you can use the following formula:
=IF([Annual Scope]@row = 1, SUM(COLLECT({Test 1_Estimate A}, {Test 1_Year}, Year@row, {Test 1_Scope}, "Annual", {Test 1_Site}, Site@row)), 0) + IF([Semi-Annual Scope]@row = 1, SUM(COLLECT({Test 1_Estimate A}, {Test 1_Year}, Year@row, {Test 1_Scope}, "Semi-Annual", {Test 1_Site}, Site@row)), 0) + IF([3 Year Test Scope]@row = 1, SUM(COLLECT({Test 1_Estimate A}, {Test 1_Year}, Year@row, {Test 1_Scope}, "3 Year Test", {Test 1_Site}, Site@row)), 0)
What the above does, is add the 3 scopes seperately (First bold part for Annual, Italic part for Semi Annual and last bold part for 3 Year scope)
OR
2. Keeping it the way you have it, you can use the same principle as above, but use
=IF(CONTAINS("Annual", Scope@row), SUM(COLLECT({Test 1_Estimate A}, {Test 1_Year}, Year@row, {Test 1_Scope}, "Annual", {Test 1_Site}, Site@row)), 0) + IF(CONTAINS("Semi-Annual", Scope@row), SUM(COLLECT({Test 1_Estimate A}, {Test 1_Year}, Year@row, {Test 1_Scope}, "Semi-Annual", {Test 1_Site}, Site@row)), 0) + IF(CONTAINS("3 Year Test", Scope@row), SUM(COLLECT({Test 1_Estimate A}, {Test 1_Year}, Year@row, {Test 1_Scope}, "3 Year Test", {Test 1_Site}, Site@row)), 0)
I hope this helps
Marcé Holzhauzen
Dare to try
If this helped, help me & the SSC by accepting it and reacting w/💡insightful, ⬆️ Vote Up, and/or ❤️Awesome.
• ✭✭✭✭
Options
Hi @SamEast
What are the rules for the multi select column (`Scope Type A, Scope Type B, Scope Type C - ``[Multi-select fields])`in this formula?
Should al 3 options be selected or is it any of the options?
Marcé Holzhauzen
Dare to try
If this helped, help me & the SSC by accepting it and reacting w/💡insightful, ⬆️ Vote Up, and/or ❤️Awesome.
• edited 07/22/24
Options
thank you for response. The multi-select for Scope Type would have any one of those options or all (should add/sum all the Estimate value(s) associated with these criterias based on YEAR —+— > Location —+-- > Scope type(Single or multiple Scope Type values).
The Estimate amount is dependent on all these three different criterias: Scope type, YEAR , Location , which I have these defined and it is referencing from another smartsheet in an Estimate column.
(details in my original post above for the complete master sheet details on these options of each column Scope type, YEAR and Location …and Estimate values
please let me know if you need more details,
thank you and looking forward for your feedback and anyone who has any suggestions!
• ✭✭✭✭
Options
Hi @SamEast
Maybe try the following:
Change your master sheet to have fewer columns
Then, in your summary sheet, you can do one of two things,
1: Change the multi select to separate checkboxes:
Then you can use the following formula:
=IF([Annual Scope]@row = 1, SUM(COLLECT({Test 1_Estimate A}, {Test 1_Year}, Year@row, {Test 1_Scope}, "Annual", {Test 1_Site}, Site@row)), 0) + IF([Semi-Annual Scope]@row = 1, SUM(COLLECT({Test 1_Estimate A}, {Test 1_Year}, Year@row, {Test 1_Scope}, "Semi-Annual", {Test 1_Site}, Site@row)), 0) + IF([3 Year Test Scope]@row = 1, SUM(COLLECT({Test 1_Estimate A}, {Test 1_Year}, Year@row, {Test 1_Scope}, "3 Year Test", {Test 1_Site}, Site@row)), 0)
What the above does, is add the 3 scopes seperately (First bold part for Annual, Italic part for Semi Annual and last bold part for 3 Year scope)
OR
2. Keeping it the way you have it, you can use the same principle as above, but use
=IF(CONTAINS("Annual", Scope@row), SUM(COLLECT({Test 1_Estimate A}, {Test 1_Year}, Year@row, {Test 1_Scope}, "Annual", {Test 1_Site}, Site@row)), 0) + IF(CONTAINS("Semi-Annual", Scope@row), SUM(COLLECT({Test 1_Estimate A}, {Test 1_Year}, Year@row, {Test 1_Scope}, "Semi-Annual", {Test 1_Site}, Site@row)), 0) + IF(CONTAINS("3 Year Test", Scope@row), SUM(COLLECT({Test 1_Estimate A}, {Test 1_Year}, Year@row, {Test 1_Scope}, "3 Year Test", {Test 1_Site}, Site@row)), 0)
I hope this helps
Marcé Holzhauzen
Dare to try
If this helped, help me & the SSC by accepting it and reacting w/💡insightful, ⬆️ Vote Up, and/or ❤️Awesome.
• edited 07/23/24
Options
Thank you @MarceHolzhauzen ! that works with some modifications in sheets.
Thank you for your help! Much Appreciated!
• ✭✭✭✭
Options
I am so glad I could help! :-)
Marcé Holzhauzen
Dare to try
If this helped, help me & the SSC by accepting it and reacting w/💡insightful, ⬆️ Vote Up, and/or ❤️Awesome. | 2,043 | 6,804 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-33 | latest | en | 0.69551 |
http://www.gamedev.net/topic/646970-projectile-motion/?setlanguage=1&langurlbits=topic/646970-projectile-motion/&langid=1 | 1,477,360,970,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719843.44/warc/CC-MAIN-20161020183839-00052-ip-10-171-6-4.ec2.internal.warc.gz | 457,270,049 | 20,969 | • Create Account
# Projectile Motion
Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.
### #1dribble04 Members - Reputation: 159
Like
0Likes
Like
Posted 23 August 2013 - 12:10 AM
Hey,
Can somebody please tell me how I can simulate a trajectory or projectile motion. I am building a game where a character is moving and as it gains momentum and moves up a slope, I want the character to launch it self.
Any help would be greatly appreciated.
Thank you.
Cheers!!
### #2Álvaro Crossbones+ - Reputation: 19952
Like
2Likes
Like
Posted 24 August 2013 - 02:56 AM
Look up Euler integration'. You model your character/projectile/whatever as a point mass. In each frame, compute the sum of all the forces, compute acceleration (a vector) as force divided by mass, and then do this:
velocity += acceleration * delta_time;
position += velocity * delta_time;`
If you want to do the same thing for rotations, the situation is similar, but make sure you understand this part first.
Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.
PARTNERS | 328 | 1,309 | {"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} | 2.515625 | 3 | CC-MAIN-2016-44 | latest | en | 0.884323 |
http://anomalous-readings.blogspot.com/2015/03/euler-unmasked.html | 1,532,194,787,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592650.53/warc/CC-MAIN-20180721164755-20180721184755-00282.warc.gz | 26,062,598 | 19,645 | ## Saturday, March 21, 2015
We're going from straight philosophy in my last post to straight math in this one. But if you're an ancient Greek thinker type person, math and philosophy are the same thing, anyway.
So about a year and a half ago, I made a post that touched briefly on the relationship between trig functions and exponential functions as a way of justifying my tendency to make things more complex than they need to be. I mentioned there that I didn't have a firm enough mathematical grasp to explain how these two mathy bits are related. Well, the topic of Euler'sidentity came up a little while ago in my writing group, so I decided to do some research and figure out just how it is that trig functions and exponential functions come together.
For those of you that don't click links, Euler's identity says:
This is a pretty remarkable and frankly incredible equation, but it's true. It manages to link probably the three most famous mathematical constants in a very simple way. The identity arises from Euler's formula, which says:
If you replace x with π, then isin(π) = 0 and cos(π) = -1, so with a little rearranging you can get Euler's identity. But this raises the question of why it should be true that exponential functions and trig functions are connected by the imaginary unit.
First, a quick primer for those who need it. In the common parlance, something that is "exponentially" better is "really super" better. This kind of talk tends to aggravate the mathematically aware, however. Really, exponential functions are ones where adding a constant increment to the input multiplies the output by a constant factor.
So if you hear something like, "Kyrgyzstan's GDP has doubled every year for the last ten years," then that's exponential growth. The factor is 2, and the increment is yearly. But this also applies to, say, the interest rate on your savings account, which as we all know is not exactly "really super" better than anything except possibly 0. There, your balance is getting multiplied by something like 1.0025 every year, which is every bit as exponential as Kyrgyzstan's doubling GDP (totally made up).
The point is, however, that exponential functions (with a factor greater than 1) demonstrate constant (monotonic) growth. If you increase the x value, the y value will increase, too.
Trig functions, on the other hand, are the realm of waves, which go up and down and up and down. They are all about rhythmic or periodic behavior. But as their name suggests, the trigonometric functions are actually based on the angles formed by triangles. Trig functions are really expressions of the Pythagorean formula, A2 + B2 = C2. The relationship between this formula and periodic motion is that for some constant value of C, increasing A will decrease B, and vice versa.
So it's hard to see how exponential functions and trig functions could be related. As I hinted up above, the answer is through i.
i, the imaginary unit, is what the square root of negative one is defined to be. Imaginary numbers kind of get a bad rap, partly because of their name. They seem like something mathematicians just made up that couldn't possibly be real. The funny thing is people had the same opinion about negative numbers for a very long time. After all, how can you possibly have -3 apples? On this whole controversy, the great mathematician Carl Friedrich Gauss had this to say:
That this subject [imaginary numbers] has hitherto been surrounded by mysterious obscurity, is to be attributed largely to an ill-adapted notation. If, for instance, +1, -1, √-1 had been called direct, inverse, and lateral units, instead of positive, negative, and imaginary (or even impossible), such an obscurity would have been out of the question.
While his preferred notation might seem somewhat opaque, it does lend itself very well to a geometric interpretation of numbers. If you look at a Cartesian plot, you can think of Gauss's direct, inverse, and lateral numbers this way.
The direct unit (+1) moves you one to the right on the graph. The inverse unit (-1) moves you one to the left. And the lateral unit (√-1) moves you up one. Rather than being on the number line we're used to, imaginary numbers can be thought of as being at right angles to it.
This idea lets you plot numbers that are a combination of "real" and "imaginary." So if you have the complex number 3 + 2i, that's just 3 units to the right and 2 units up.
As you see, plotting numbers this way means you can draw right triangles that are related to those numbers. This is the first way that we can connect imaginary numbers to the trig functions. Getting from imaginary numbers to exponential functions will take a little more work, though.
If i is the square root of -1, we can play around with exponentiation to find an interesting pattern. i2 = (√-1)2, which by definition equals -1. i3 = (√-1)3, or (√-1)* (√-1)2, or i*-1, which just comes out to -i. i4 = i2 * i2, or -1 * -1, which equals 1. Multiply that by i, and you of course have i again. So through exponentiation, we have discovered something of a pattern.
i1 = i
i2 = -1
i3 = -i
i4 = 1
i5 = i
The exponents of i loop back in on themselves. You might even say they exhibit periodic behavior, like the trig functions.
Our next step is probably the toughest bit. Bear with me. So, if you recall from my foray into Fourier, many functions can be expressed as an infinite series of sines and cosines that eventually converge on a desired function. These infinite series turn out to be very useful to mathematicians, because not all patterns can be expressed as "elementary" functions, but only as infinite series of some other type of function. One type of infinite series is the power series, which looks like this:
To get different functions, just plug in different values for the coefficients an. The way you figure out which coefficients correspond to the function you want is basically by assuming your function can fit into some power series and then just playing around for awhile until you find a pattern that fits. Let me demonstrate.
One of the defining features of the exponential function, ex, is that it is its own derivative. This means that its rate of change is equal to its value. So the derivative of ex is also ex, and so on.
One of the first tools you learn in calculus is that the derivative of a power function like x4 is 4x3. You multiply by the exponent, and then lower the exponent by one. If the exponent is already 0, then your derivative is 0. So if you take the derivative of our above model power series, you get:
And if you take the derivative of that, you get:
And if you take the derivative of that, you get:
And one more time, because there's a pattern I want you to see:
Now remember, all of these series are equal to the function ex, because ex is its own derivative. The missing ingredients are the values of an. If we evaluate ex at x=0, we have e0, and anything to the 0th power is equal to 1. In the above series, when x is 0, everything except the leading term is also 0. So we have:
1 = a0 = a1 = 2a2 = 6a3 = 24a4
and so on. So with a little bit of algebra, you can figure out the value of any an. It's just 1 divided by the factor preceding the coefficient. But there's a pattern here. 24 = 4*3*2*1. 6 = 3*2*1. 2 = 2*1. The value of the coefficient is equal to 1 over the index of the coefficient multiplied by each integer lower than it. This is known as a factorial in mathematics and looks like this:
5! = 5*4*3*2*1 = 120
With that information in hand, we know what the power series of the exponential function is:
I've gone through this process once so that you don't think I'm pulling this stuff out of a hat, but you can do the same thing to find the power series of a lot of different functions, including the trig functions. For example, the power series of sin(x) is:
And the power series of cos(x) is:
Weirdly, the sine and cosine power series look kind of similar to the exponential function, but with terms missing and some negative signs thrown in. This curious fact turns out to be very important for connecting exponential and trig functions. Let's remember that the key to that connection is i.
Let's see what happens if we try to find the power series of eix rather than ex. To do that, we just replace all instances of x with ix in our series above. That gets us:
Hey, that means we're finding powers of i. But we already did that up above. That follows a pattern, so we can just fill in from that pattern and get:
Now, just for the heck of it, let's separate our series into terms without i and terms with i. So we have:
Look familiar? That's the power series for cosine plus i times the power series for sine. In other words...
Just as Euler told us.
All of this may seem like some kind of tedious mathematical trick. After all, how do we know that the power series representation of a function behaves identically to the function itself in all instances? The truth is, it doesn't, and that's one of the things you have to be careful of when finding series expansions. It does happen to work in this case, though.
But there are ways in which this proof can help motivate understanding. One way to think of the idea is that the introduction of i into the exponential function breaks the function down into four interacting parts: one increasing in the direction of 1, another increasing in the direction of -1, and two others increasing in the direction of i and -i. Different values of x contribute more to one direction than another, and the whole thing repeats with a period of 2πi.
To see if this picture holds true, let's take another look at the powers of i. We saw that powers of i cycle from i to -1 to -i to 1 and then back to i again. But we were only looking at integer powers of i. What happens if we replace the integer with an unknown variable x? That is, how do we evaluate ix?
A neat tool that can sometimes work in mathematics is to perform some operation on an expression and then also perform the inverse of that operation. Doing so doesn't change the expression, but it does let us look at it in a different light. So how about we take the natural log of ix and then exponentiate the expression. That gets us:
The laws of logarithms mean we can move that x to outside the log, giving us:
We know how to evaluate ex, but it’s not immediately clear how to evaluate ln(i). Here it's useful to remember what ln means. The natural log of some number is the power to which you must raise e in order to get that number. So if you have, say, ln(e2), then our answer is 2, because e to the power of 2 obviously equals e2. So let's look at it this way: e to what power equals i?
Now we bring in Euler's formula again.
eix = i when cos(x) = 0 and isin(x) = i
This is true for x = π/2, because cos(π/2) = 0 and sin(π/2) = 1.
So then ln(i) = iπ/2, which means that ix = eiπx/2 = cos(xπ/2) + isin(xπ/2). With that conversion, we can evaluate i to any power at all, not just integer powers. But to reaffirm that this isn't some trick, let's go ahead and see what evaluating it to integer powers means.
This is the exact same pattern we saw above, but this time through the lens of Euler's formula rather than the logic of manipulating √-1. For non-integer values of x, you get complex numbers that, when treated as vectors on the complex plane, are all a distance of 1 from the origin, creating a circle of radius 1. Through purely algebraic means, this connects back up with the geometrical interpretation of imaginary numbers suggested by Gauss.
Okay, I'm done now. I hope this sheds some light on the interconnectedness of math, which can be demonstrated by taking the rules you're familiar with and applying them to unfamiliar situations. When people speak of the beauty of math, this is it. In the real world, we often find depth and meaning through metaphors that connect disparate ideas. That's what art and literature are all about. Math does the same thing, but with numbers, letters, and funny symbols.
(On the other hand, I may have written this post just to play around with LaTeX.) | 2,827 | 12,147 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2018-30 | latest | en | 0.960129 |
https://documen.tv/question/plz-help-asap-ill-mark-brainliest-what-are-the-leading-coefficient-and-degree-of-the-polynomial-24222836-68/ | 1,627,748,565,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154089.68/warc/CC-MAIN-20210731141123-20210731171123-00542.warc.gz | 221,297,543 | 17,914 | ## PLZ HELP ASAP ILL MARK BRAINLIEST! what are the leading coefficient and degree of the polynomial? 9y+8-y^2+12y^4
Question
PLZ HELP ASAP ILL MARK BRAINLIEST!
what are the leading coefficient and degree of the polynomial?
9y+8-y^2+12y^4
in progress 0
2 weeks 2021-07-19T21:05:06+00:00 2 Answers 0 views 0
The degree of the polynomial is 4.
The leading coefficient of the polynomial is 12.
Step-by-step explanation:
The degree of the polynomial is the highest power in the polynomial.
The highest power in the given polynomial is 4.
The leading coefficient of the polynomial is 12.
2. Polynomial degree : 2 | 176 | 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} | 3.546875 | 4 | CC-MAIN-2021-31 | latest | en | 0.847255 |
https://algorithmist.com/index.php/UVa_10007 | 1,569,101,563,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574665.79/warc/CC-MAIN-20190921211246-20190921233246-00069.warc.gz | 370,862,882 | 7,071 | # UVa 10007
## Summary
The task is to find the count of rooted labeled binary trees on N vertices.
## Explanation
Counts of unlabeled rooted binary trees with N vertices are exactly the famous Catalan numbers, i.e.,
${\displaystyle {\rm {count}}(N)={1 \over N+1}\cdot {2N \choose N}}$.
Once we have an unlabeled rooted binary tree with N vertices, there are exactly N! ways to add the labels. We can do this for each of the trees, thus the final answer is given by the formula
${\displaystyle {\rm {answer}}(N)=N!\cdot {1 \over N+1}\cdot {2N \choose N}={(2N)! \over (N+1)!}}$.
## Implementations
As the example I/O shows, this is intended to be a BigNum problem. You only need to implement integer multiplication, as the answer can be obtained by multiplying the numbers (N+2) to 2N.
As the set of possible inputs is limited, it is possible to precompute the answers in some scripting language that supports big integers (Python, bc) and submit a program that has the answers hard-wired as string constants.
## Input
1
2
10
25
0
## Output
1
4
60949324800
75414671852339208296275849248768000000
## Reference
1. http://mathworld.wolfram.com/CatalanNumber.html | 321 | 1,175 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "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} | 3.09375 | 3 | CC-MAIN-2019-39 | latest | en | 0.834146 |
https://babelnet.org/synset?id=bn%3A02784845n&lang=EN | 1,632,138,124,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057036.89/warc/CC-MAIN-20210920101029-20210920131029-00612.warc.gz | 171,845,544 | 47,423 | bn:02784845n
Noun Concept
Categories: Probability theory, Enumerative combinatorics, Mathematical principles
EN
Inclusion–exclusion principle Principle Of Inclusion and Exclusion Diluted inclusion-exclusion principle Diluted inclusion–exclusion principle Exclusion-inclusion
EN
In combinatorics, a branch of mathematics, the inclusion–exclusion principle is a counting technique which generalizes the familiar method of obtaining the number of elements in the union of two finite sets; symbolically expressed as | A ∪ B | = | A | + | B | − | A ∩ B |, {\displaystyle |A\cup B|=|A|+|B|-|A\cap B|,} where A and B are two finite sets and |S| indicates the cardinality of a set S. Wikipedia
Definitions
Relations
Sources
EN
In combinatorics, a branch of mathematics, the inclusion–exclusion principle is a counting technique which generalizes the familiar method of obtaining the number of elements in the union of two finite sets; symbolically expressed as | A ∪ B | = | A | + | B | − | A ∩ B |, {\displaystyle |A\cup B|=|A|+|B|-|A\cap B|,} where A and B are two finite sets and |S| indicates the cardinality of a set S. Wikipedia
Counting technique in combinatorics Wikidata
IS A
Wikipedia
Wikidata
Wikipedia Redirections
EN
Wikidata Alias | 335 | 1,240 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.739348 |
https://mathoverflow.net/questions/155419/distance-measure-for-noisy-se3-transforms | 1,675,562,433,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500158.5/warc/CC-MAIN-20230205000727-20230205030727-00129.warc.gz | 397,031,679 | 26,292 | # Distance measure for noisy $SE(3)$ transforms
I have a transformation $T \in SE(3)$ parameterized by a mean quaternion $q$ with covariance matrix $\Sigma_q \in R^{4\times4}$ and a mean translation $t \in R^3$ with covariance matrix $\Sigma_t \in R^{3\times3}$.
If I were to transform some point $p_1 \in R^3$ by $T$, I can easily compute its Euclidean distance to some second point $p_2 \in R^3$ as $\lVert T p_1 - p_2\rVert$.
What I would like to do instead is incorporate the covariance matrices of my transformation to compute a distance measure similar in function to a Mahalanobis distance. Is there a closed form way to do this?
# 2D Example
To help clarify, I've plotted the problem in a two dimensional world below, where points are in $R^2$, and transforms are in $SE(2)$. Here, a point $p_1$ can be transformed by $T$ to find a point $T p_1$. However, because we know the covariances of the rotation and translation, we can see that random samples of $T p_1$ would be distributed according to the banana shaped distribution shown in red. What I would like to find then is a distance measure that would will take these covariances into account such that $p_2$ would be significantly "closer" than $p_3$ to $T p_1$.
• Are you trying to define a new distance on the underlying space or a distance between a point and its distribution of transforms ? Mar 1, 2014 at 7:36
Let us write the Euclidean distance of the transformed point to the target as: $$\label{eq1} \tag{1} d(\mathbf{p}_1,\mathbf{p}_2) = {\lVert \mathbf{R}\mathbf{p}_1 + \mathbf{t} - \mathbf{p}_2 \rVert}_2^2$$ where $\mathbf{R}$ and $\mathbf{t}$ are rotation matrix and translation vector, respectively. The Mahalonobis distance gives the distance between two points as: $$$$\label{eq2} \tag{2} d_M(\mathbf{x},\mathbf{y}) = (\mathbf{x}-\mathbf{y})^TC^{-1}(\mathbf{x}-\mathbf{y})$$$$ where $C^{-1}$ is the inverse covariance matrix (or equivalently the conic matrix of the ellipse). We could then plug in $\ref{eq1}$ to $\ref{eq2}$ resulting in :
$$$$\tag{3} d_M(\mathbf{p}_1,\mathbf{p}_2) = (\mathbf{R}\mathbf{p}_1 + \mathbf{t} - \mathbf{p}_2)^TC^{-1}(\mathbf{R}\mathbf{p}_1 + \mathbf{t} - \mathbf{p}_2)$$$$
$\mathbf{R}\mathbf{p}_1 + \mathbf{t}$ can also be viewed as $\mathbf{T}\mathbf{p}_1$, by assembling the augmented matrix $\mathbf{T} = [\mathbf{R}|\mathbf{t}]$. | 736 | 2,349 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2023-06 | latest | en | 0.869563 |
https://discuss.codechef.com/t/atcoder-problem/73855 | 1,618,924,279,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039398307.76/warc/CC-MAIN-20210420122023-20210420152023-00587.warc.gz | 321,289,723 | 4,241 | # Atcoder Problem
The first part of the editorial was clear to me, the second part(forming the dp) wasn’t clear, can someone help me in this?
@galencolin @carre @ssjgz @everule1
Thanks
Sorry, I took a quick look at the problem and editorial just to realize that that’s not enough for me to come up with an answer. I hope tonight (it’s morning on this side of the word right now) I have time to look deeper and see if I can understand it.
You get the bipartite matching right?
The main idea is that the edges nicely give us 2k straight lines. 2 for each value mod k.
Like 1L - 4R - 7L and 1R-4L-7R.
To be clear 1L is node 1 on the left.
We can compute the number of ways to match j nodes in line of length l by dp of length, matches and whether the previous value is used.
Which will work in O(l^2).
I can’t figure the second dp, but you can use NTT.
There will at most be 3 different polynomials. M_ix^i for each line. So you can count the number of lines to use polynomial exponentiation for each type and multiply all to get the final M_i.
The dp is probably some knapsack variant, but I haven’t thought of the details yet. My current idea is O(N^3) | 311 | 1,160 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | latest | en | 0.900666 |
https://www.slideserve.com/zlata/cyclic-properties-of-locally-connected-graphs-with-bounded-vertex-degree | 1,579,337,681,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250592394.9/warc/CC-MAIN-20200118081234-20200118105234-00519.warc.gz | 1,091,404,532 | 13,715 | Cyclic Properties of Locally Connected Graphs with Bounded Vertex Degree
1 / 16
# Cyclic Properties of Locally Connected Graphs with Bounded Vertex Degree - PowerPoint PPT Presentation
## Cyclic Properties of Locally Connected Graphs with Bounded Vertex Degree
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. Cyclic Properties of Locally Connected Graphs with BoundedVertex Degree V. Gordon1, Yu. Orlovich2, C. Potts3 , V.Strusevich4 1 United Institute of Informatics Problems of the NASB, Minsk, Belarus; 2 Institute of Mathematics of the NASB, Minsk, Belarus; 3 University of Southampton, School of Mathematics, Southampton, UK 4University of Greenwich, School of Computer and Mathematical Sciences, London, UK
2. Plan of the talk • Introduction • Locally connected graphs with maximum vertex degree (G) bounded by 4 • Cyclic properties of locally connected graphs with (G) 5 • Complexity of the HAMILTONIAN CYCLE problem for locally connected graphs with (G) 7
3. Introduction Let Gbe a simple undirected graph with the vertex set V(G) and the edge set E(G). • For a vertex u of G, the neighborhoodN(u)of u is the set of all vertices adjacent to u. • A graph is locally connected if for each vertex u, the subgraph induced by N(u) is connected. • The degreedeguof vertex u in Gis the number of edges incident with u or, equivalently, degu =|N(u)|. The minimum and maximumdegrees of verticesin V(G)are denoted by (G)and (G), respectively.
4. Graph G is hamiltonian if Ghas a hamiltonian cycle, i.e. a cycle containing every vertex of G. In 1989, Hendry introduced the following concept. • A cycle C in a graph G is extendable if there exists a cycle C* in Gsuch that V(C) V(C*) and |V(C*)| = |V(C)| + 1. • A connected graph G is fully cyclic extendable if every vertex of G is on a triangle and every nonhamiltonian cycle is extendable. Clearly, any fully cycle extendable graph is hamiltonian.
5. Previous Results Theorem A (Chartrand, Pippert, 1974).LetGbe a connected, locally connectedgraph with (G)4.Then either Gis hamiltonian or isomorphic to K1,1,3 (complete 3-partite graph with two parts of size 1 and one part of size 3). Theorem B (Kikust, 1975).LetGbe a connected, locally connectedgraph with (G) = (G)= 5.Then Gis hamiltonian. Theorem C (Hendry, 1989).LetGbe a connected, locally connectedgraph with (G) 5 and (G) (G) 1.Then Gis fully cyclic extendable.
6. Locally Connected Graphs with (G) 4 Theorem 1.LetGbe a connected, locally connected(not necessary finite)graph with(G)4. The following claims hold. The following theorem explicitly describes connected, locally connected graphs with(G)4. If(G) = (G), then If(G) (G) = 1, then If(G) (G) = 2,then
7. If(G) = (G), then • In Theorem 1, • Cn is the cycle, • Pn is the path, • Kn is the complete graph, • On is the empty graph, • Wn is the wheel on n vertices, • K1,1,q is the complete 3-partite graph with two parts of size 1 and one part of size q,and • Kn e is the graph obtained from Kn by deleting a single edge. If(G) (G) = 1, then If(G) (G) = 2,then For a graph G, is the complement to G and G2 is the square of G. P1, is a one-way infinite path,i.e. a graph with V(P1,)={xk| k N}, E(P1,)={xkxk +1 | k N}. P, is a two-way infinite path,i.e. a graphwith V(P,)={xk| k Z}, E(P,)={xkxk +1 | k Z}.
8. Fig. 1 represents nonstandard graphs H1, H2, H3, and H4 from Theorem 1. H4 H3 H1 H2 Fig. 1. Graphs H1, H2, H3, and H4
9. Cyclic Properties of Locally Connected Graphs with (G) 5 Connected, locally connected graphs with (G)4are completely described (Theorem 1). It is interesting to find the hamiltonian properties of locally connected graphs under (G) 5 enhancing the results of Kikust (Theorem B) and Hendry (Theorem C). First of all, the following question arises naturally: Is it correct that connected, locally connected graphGwith(G) = 5and (G) 3 is hamiltonian? The following theorem answers this question affirmatively.
10. Theorem 2.Let G be a connected, locally connected graph with(G) = 5and (G) 3. Then G is fully cyclic extendable. The stronger inequality (G) 2 cannot be used in Theorem 2 since graph K1,1,4 with = 2 is not hamiltonian. Graph K1,1,4 The following theorem gives another examples of nonhamiltonian graphs with = 2 and together with Theorem 2 enhances the results of Kikust and Hendry on hamiltonicity of locally connected graph with = 5.
11. Theorem 3.Let G be a connected, locally connected graph with (G) = 5 and do not contain graph F from Fig. 2 as induced subgraph. Then eitherGis hamiltonian or G S{G1, G2, G3}. Here graphs G1, G2, and G3are shown in Fig. 2 and S is a class of connected, locally connected graphs H with (H) = 5 and with such four vertices u, v, x, y that deg x = deg y = 2 and uv E(H), ux E(H), uy E(H), vx E(H), vy E(H) but xy E(H). G3 F G1 G2 Fig. 2. Graphs F, G1, G2, and G3
12. y x u v V(H) Class S is the set of all connected, locally connected graphs H with (H) = 5 and with such four vertices u, v, x, y that deg x = deg y = 2 and uv E(H), ux E(H), uy E(H), vx E(H), vy E(H) but xy E(H). It is easy to see that none of the graphs in S is hamiltonian.
13. The extendability of cycles for locally connected k-regular graphs (k 6) are described by the following theorem. Theorem 4.Fork 6, let G be a connected, locally connected k-regular graph and suppose that every edge of G belongs to at least k – 4 triangles. Then Gis fully cyclic extendable. Recall that a graphGis regular of degreek, or simply k-regular, if (G) = (G) = k for some k 1.
14. Complexity of the HAMILTONIAN CYCLE Problem for Locally Connected Graphs with (G) 7 Consider the following well-known decision problem. • HAMILTONIAN CYCLE • Instance: A graph G. • Question: Is Ghamiltonian?
15. HAMILTONIAN CYCLE is NP-complete for general graphs and remains difficult for: • 3-connected cubic (i.e., 3-regular) planar graphs (Garey, Johnson, and Tarjan, 1976); • bipartite planar graphs of maximum degree 3(Akiyama, Nishizeki, and Saito, 1980); • line graphs (Bertossi, 1981); • grid graphs (Itai, Papadimitriou, Szwarcfiter, 1982); • maximal planar graphs (Chvátal, 1985); • chordal bipartite graph (Muller, 1996). • HAMILTONIAN CYCLE is solvable in polynomial time for: • cographs (Corneil, Lerchs, Stewart-Burlingham, 1981); • proper circular arc graphs (Bertossi, 1983); • interval graphs (Keil, 1985); • co-comparability graphs (Deogun, Steiner, 1994).
16. For locally connected graphs with (G) 4 the HAMILTONIAN CYCLE problem is solvable in polynomial time (Theorem 1). Theorem 5.The HAMILTONIAN CYCLE problem is NP-complete for locally connected graphs with(G) 7. Proof is done by the reduction from the HAMILTONIAN CYCLE problem for cubic planar bipartite graphs. Let * be a maximum integer such that the HAMILTONIAN CYCLE problem for locally connected graphs with a restriction * is polynomially solvable. As an immediate consequence of Theorems 1 and 5, we can restrict the range of * to 4 * 6. Conjecture. * = 6. | 2,398 | 7,178 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2020-05 | latest | en | 0.847406 |
https://www.thestudentroom.co.uk/showthread.php?t=2287298 | 1,524,408,060,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945604.91/warc/CC-MAIN-20180422135010-20180422155010-00244.warc.gz | 895,090,087 | 38,030 | x Turn on thread page Beta
You are Here: Home >< Physics
# Mock AS Physics Coursework, Error bars? watch
1. Ok, so basically, I'm doing my mock analysis section later in the week, and we've been told that we need to do error bars for either the stress or strain (Young Modulus experiment), but not both, and that we need to choose the one that generates the most error.
So, if i used a metre stick to measure length and extension, and a micrometer screw gauge to measure diameter, and a top-pan balance to measure the mass of the load, which would be likely to generate the largest percentage uncertainty, stress or strain?
Or should I just work it out when I get to the lesson on Friday? And see which causes the greatest uncertainty by using maximum force and minimum area, and then the converse to find the uncertainty in stress, and then the maximum extension and minimum length, and then the converse to find the uncertainty in strain, and then compare the percentages? Does this method make sense?
2. In all your calculations you take all readings (all have errors) and then substitute them into your formula.
To find the max limit all multipliction terms must be maximised and all division terms must be minimised. This gives you your upper error.
Do the reverse for the minimum error.
The key is you need to find out the errors possible for each type of measuring instrument.
3. (Original post by TeslaCoil)
Ok, so basically, I'm doing my mock analysis section later in the week, and we've been told that we need to do error bars for either the stress or strain (Young Modulus experiment), but not both, and that we need to choose the one that generates the most error.
So, if i used a metre stick to measure length and extension, and a micrometer screw gauge to measure diameter, and a top-pan balance to measure the mass of the load, which would be likely to generate the largest percentage uncertainty, stress or strain?
Or should I just work it out when I get to the lesson on Friday? And see which causes the greatest uncertainty by using maximum force and minimum area, and then the converse to find the uncertainty in stress, and then the maximum extension and minimum length, and then the converse to find the uncertainty in strain, and then compare the percentages? Does this method make sense?
This is for a metal wire, I assume.
You need to have an awareness of where the greatest %age uncertainty lies in individual measurements.
Think:
What sort of value do you expect for the extension of the wire?
If you are measuring this extension with a metre rule, what is the % uncertainty in that? (A meter rule has a scale marked in mm.)
You should find it to be rather large. Certainly larger than the other uncertainties.
I suggest you measure the extension with some sort of vernier scale.
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
This forum is supported by:
Updated: March 14, 2013
Today on TSR
### Loughborough better than Cambridge
Loughborough at number one
### Can I date a girl with no boobs?
Poll
The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.
Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE | 744 | 3,425 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-17 | latest | en | 0.917857 |
http://www.tetsuccesskey.com/2015/09/chapter-1-ray-optics-and-optical.html | 1,519,245,570,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813803.28/warc/CC-MAIN-20180221202619-20180221222619-00257.warc.gz | 561,814,553 | 61,766 | ## Ncert Solutions of Physics Part 2 for Chapter 1 : Ray Optics And Optical Instruments
### Exercise : Solutions of Questions on Page Number : 345
Q1 :
A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?
Size of the candle, h= 2.5 cm
Image size = h'
Object distance, u= - 27 cm
Radius of curvature of the concave mirror, R= - 36 cm
Focal length of the concave mirror,
Image distance = v
The image distance can be obtained using the mirror formula:
Therefore, the screen should be placed 54 cm away from the mirror to obtain a sharp image.
The magnification of the image is given as:
The height of the candle's image is 5 cm. The negative sign indicates that the image is inverted and real.
If the candle is moved closer to the mirror, then the screen will have to be moved away from the mirror in order to obtain the image.
Q2 : A 4.5 cm needle is placed 12 cm away from a convex mirror of focal
Height of the needle, h1= 4.5 cm
Object distance, u = - 12 cm
Focal length of the convex mirror, f= 15 cm
Image distance = v
The value of v can be obtained using the mirror formula:
Hence, the image of the needle is 6.7 cm away from the mirror. Also, it is on the other side of the mirror.
The image size is given by the magnification formula:
Hence, magnification of the image,
The height of the image is 2.5 cm. The positive sign indicates that the image is erect, virtual, and diminished.
Ifthe needle is moved farther from the mirror, the image will also move away from the mirror, and the size of the image will reduce gradually.
Q3 : A tank is filled with water to a height of 12.5 cm. The apparent
Actual depth of the needle in water, h1= 12.5 cm
Apparent depth of the needle in water, h2= 9.4 cm
Refractive index of water = μ
The value of μcan be obtained as follows:
Hence, the refractive index of water is about 1.33.
Water is replaced by a liquid of refractive index,
The actual depth of the needle remains the same, but its apparent depth changes. Let ybe the new apparent depth of the needle. Hence, we can write the relation:
Hence, the new apparent depth of the needle is 7.67 cm. It is less than h2. Therefore, to focus the needle again, the microscope should be moved up.
Distance by which the microscope should be moved up = 9.4 - 7.67
= 1.73 cm
Q4 :
Figures 9.34(a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45º with the normal to a water-glass interface [Fig. 9.34(c)].
As per the given figure, for the glass - air interface:
Angle of incidence, i = 60°
Angle of refraction, r= 35°
The relative refractive index of glass with respect to air is given by Snell's law as:
As per the given figure, for the air - water interface:
Angle of incidence, i= 60°
Angle of refraction, r= 47°
The relative refractive index of water with respect to air is given by Snell's law as:
Using (1) and (2), the relative refractive index of glass with respect to water can be obtained as:
The following figure shows the situation involving the glass - water interface.
Angle of incidence, i= 45°
Angle of refraction = r
From Snell's law, rcan be calculated as:
Hence, the angle of refraction at the water - glass interface is 38.68°.
Q5 : A small bulb is placed at the bottom of a tank containing water to a depth of 80
Actual depth of the bulb in water, d1= 80 cm = 0.8 m
Refractive index of water,
The given situation is shown in the following figure:
Where,
i= Angle of incidence
r= Angle of refraction = 90°
Sincethe bulb is a point source, the emergent light can be considered as a circle of radius,
Using Snell' law, we can write the relation for the refractive index of water as:
Using the given figure, we have the relation:
R = tan 48.75° ×0.8 = 0.91 m
Area of the surface of water = πR2= π(0.91)2 = 2.61 m2
Hence, the area of the surface of water through which the light from the bulb can emerge is approximately 2.61 m2.
Q6 : A prism is made of glass of unknown refractive index. A parallel
Angle of minimum deviation, = 40°
Angle of the prism, A = 60°
Refractive index of water, µ= 1.33
Refractive index of the material of the prism =
The angle of deviation is related to refractive indexas:
Hence,the refractive index of the material of the prism is 1.532.
Since the prism is placed in water, let be the new angle of minimum deviation for the same prism.
The refractive index of glass with respect to water is given by the relation:
Hence, the new minimum angle of deviation is 10.32°.
Q7 : Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20
Refractive index of glass,
Focal length of the double-convex lens,f = 20 cm
Radius of curvature of one face of the lens =R1
Radius of curvature of the other face of the lens = R2
Radius of curvature of the double-convex lens =R
The value of Rcan be calculated as:
Hence, the radius of curvature of the double-convex lens is 22 cm.
Q8 :
A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, and (b) a concave lens of focal length 16 cm?
In the given situation, the object is virtual and the image formed is real.
Object distance, u = +12 cm
(a) Focal length of the convex lens, f= 20 cm
Image distance = v
According to the lens formula, we have the relation:
Hence, the image is formed 7.5 cm away from the lens, toward its right.
(b) Focal length of the concave lens, f= - 16 cm
Image distance = v
According to the lens formula, we have the relation:
Hence, the image is formed 48 cm away from the lens, toward its right.
Q9 :
An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?
Size of the object, h1= 3 cm
Object distance, u= - 14 cm
Focal length of the concave lens, f = - 21 cm
Image distance = v
According to the lens formula, we have the relation:
Hence, the image is formed on the other side of the lens, 8.4 cm away from it. The negative sign shows that the image is erect and virtual.
The magnification of the image is given as:
Hence, the height of the image is 1.8 cm.
If the object is moved further away from the lens, then the virtual image will move toward the focus of the lens, but not beyond it. The size of the image will decrease with the increase in the object distance.
Q10 :
What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.
Focal length of the convex lens, f1= 30 cm
Focal length of the concave lens, f2= - 20 cm
Focal length of the system of lenses = f
The equivalent focal length of a system of two lenses in contact is given as:
Hence, the focal length of the combination of lenses is 60 cm. The negative sign indicates that the system of lenses acts as a diverging lens.
Q11 :
A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?
Focal length of the objective lens, f1= 2.0 cm
Focal length of the eyepiece, f2= 6.25 cm
Distance between the objective lens and the eyepiece, d= 15 cm
(a) Least distance of distinct vision,
Image distance for the eyepiece, v2= - 25 cm
Object distance for the eyepiece = u2
According to the lens formula, we have the relation:
Image distance for the objective lens,
Object distance for the objective lens = u1
According to the lens formula, we have the relation:
Magnitude of the object distance, = 2.5 cm
The magnifying power of a compound microscope is given by the relation:
Hence, the magnifying power of the microscope is 20.
(b) The final image is formed at infinity.
Image distance for the eyepiece,
Object distance for the eyepiece = u2
According to the lens formula, we have the relation:
Image distance for the objective lens,
Object distance for the objective lens = u1
According to the lens formula, we have the relation:
Magnitude of the object distance, = 2.59 cm
The magnifying power of a compound microscope is given by the relation:
Hence, the magnifying power of the microscope is 13.51.
Q12 :
A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope,
Focal length of the objective lens, fo= 8 mm = 0.8 cm
Focal length of the eyepiece, fe= 2.5 cm
Object distance for the objective lens, uo= - 9.0 mm = - 0.9 cm
Least distance of distant vision, d = 25 cm
Image distance for the eyepiece, ve= - d= - 25 cm
Object distance for the eyepiece =
Using the lens formula, we can obtain the value ofas:
We can also obtain the value of the image distance for the objective lens using the lens formula.
The distance between the objective lens and the eyepiece
The magnifying power of the microscope is calculated as:
Hence, the magnifying power of the microscope is 88.
Q13 :
A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?
Focal length of the objective lens, fo= 144 cm
Focal length of the eyepiece, fe= 6.0 cm
The magnifying power of the telescope is given as:
The separation between the objective lens and the eyepiece is calculated as:
Hence, the magnifying power of the telescope is 24 and the separation between the objective lens and the eyepiece is 150 cm.
Q14 :
(a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope?
(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 x 106m, and the radius of lunar orbit is 3.8 x 108m.
Focal length of the objective lens, fo= 15 m = 15 ×102 cm
Focal length of the eyepiece, fe= 1.0 cm
(a) The angular magnification of a telescope is given as:
Hence, the angular magnification of the given refracting telescope is 1500.
(b) Diameter of the moon, d= 3.48 ×106 m
Radius of the lunar orbit, r0= 3.8 ×108 m
Let be the diameter of the image of the moon formed by the objective lens.
The angle subtended by the diameter of the moon is equal to the angle subtended by the image.
Hence, the diameter of the moon's image formed by the objective lens is 13.74 cm
Q15 :
Use the mirror equation to deduce that:
(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
(b) a convex mirror always produces a virtual image independent of the location of the object.
(c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.
(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.
[Note: This exercise helps you deduce algebraically properties of
images that one obtains from explicit ray diagrams.]
(a) For a concave mirror, the focal length (f)is negative.
f < 0
When the object is placed on the left side of the mirror, the object distance (u)is negative.
u< 0
For image distance v, we can write the lens formula as:
The object lies between fand 2f.
Using equation (1), we get:
is negative, i.e., vis negative.
Therefore, the image lies beyond 2f.
(b) For a convex mirror, the focal length (f) is positive.
∴ f > 0
When the object is placed on the left side of the mirror, the object distance (u) is negative.
∴ u < 0
For image distance v, we have the mirror formula:
Thus, the image is formed on the back side of the mirror.
Hence, a convex mirror always produces a virtual image, regardless of the object distance.
(c) For a convex mirror, the focal length (f) is positive.
f> 0
When the object is placed on the left side of the mirror, the object distance (u) is negative,
u< 0
For image distance v, we have the mirror formula:
Hence, the image formed is diminished and is located between the focus (f) and the pole.
(d) For a concave mirror, the focal length (f) is negative.
f< 0
When the object is placed on the left side of the mirror, the object distance (u) is negative.
u< 0
It is placed between the focus (f) and the pole.
For image distance v, we have the mirror formula:
The image is formed on the right side of the mirror. Hence, it is a virtual image.
For u < 0 and v > 0, we can write:
Q16 :
A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab?
Actual depth of the pin, d= 15 cm
Apparent dept of the pin =
Refractive index of glass,
Ratio of actual depth to the apparent depth is equal to the refractive index of glass, i.e.
The distance at which the pin appears to be raised =
For a small angle of incidence, this distance does not depend upon the location of the slab.
Q17 :
(a) Figure 9.35 shows a cross-section of a 'light pipe' made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure.
(b) What is the answer if there is no outer covering of the pipe?
(a) Refractive index of the glass fibre,
Refractive index of the outer covering of the pipe, = 1.44
Angle of incidence = i
Angle of refraction = r
Angle of incidence at the interface = i'
The refractive index (μ) of the inner core - outer core interface is given as:
For the critical angle, total internal reflection (TIR) takes place only when, i.e., i > 59°
Maximum angle of reflection,
Let, be the maximum angle of incidence.
The refractive index at the air - glass interface,
We have the relation for the maximum angles of incidence and reflection as:
Thus, all the rays incident at angles lying in the range 0 < i< 60°will suffer total internal reflection.
(b) If the outer covering of the pipe is not present, then:
Refractive index of the outer pipe,
For the angle of incidence i= 90°, we can write Snell's law at the air - pipe interface as:
.
Q18 :
(a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.
(b) A virtual image, we always say, cannot be caught on a screen.
Yet when we 'see' a virtual image, we are obviously bringing it on to the 'screen' (i.e., the retina) of our eye. Is there a contradiction?
(c) A diver under water, looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman look taller or shorter to the diver than what he actually is?
(d) Does the apparent depth of a tank of water change if viewed obliquely? If so, does the apparent depth increase or decrease?
(e) The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter?
(a) Yes
Plane and convex mirrors can produce real images as well. If the object is virtual, i.e., if the light rays converging at a point behind a plane mirror (or a convex mirror) are reflected to a point on a screen placed in front of the mirror, then a real image will be formed.
(b) No
A virtual image is formed when light rays diverge. The convex lens of the eye causes these divergent rays to converge at the retina. In this case, the virtual image serves as an object for the lens to produce a real image.
(c) The diver is in the water and the fisherman is on land (i.e., in air). Water is a denser medium than air. It is given that the diver is viewing the fisherman. This indicates that the light rays are travelling from a denser medium to a rarer medium. Hence, the refracted rays will move away from the normal. As a result, the fisherman will appear to be taller.
(d) Yes; Decrease
The apparent depth of a tank of water changes when viewed obliquely. This is because light bends on travelling from one medium to another. The apparent depth of the tank when viewed obliquely is less than the near-normal viewing.
(e) Yes
The refractive index of diamond (2.42) is more than that of ordinary glass (1.5). The critical angle for diamond is less than that for glass. A diamond cutter uses a large angle of incidence to ensure that the light entering the diamond is totally reflected from its faces. This is the reason for the sparkling effect of a diamond.
Q19 :
The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?
Distance between the object and the image, d= 3 m
Maximum focal length of the convex lens =
For real images, the maximum focal length is given as:
Hence, for the required purpose, the maximum possible focal length of the convex lens is 0.75 m.
Q20 :
A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.
Distance between the image (screen) and the object, D= 90 cm
Distance between two locations of the convex lens, d = 20 cm
Focal length of the lens = f
Focal length is related to dand D as:
Therefore, the focal length of the convex lens is 21.39 cm.
Q21 :
(a) Determine the 'effective focal length' of the combination of the two lenses in Exercise 9.10, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all?
(b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.
Focal length of the convex lens, f1= 30 cm
Focal length of the concave lens, f2= - 20 cm
Distance between the two lenses, d= 8.0 cm
(a) When the parallel beam of light is incident on the convex lens first:
According to the lens formula, we have:
Where,
= Object distance =
v1= Image distance
The image will act as a virtual object for the concave lens.
Applying lens formula to the concave lens, we have:
Where,
= Object distance
= (30 - d) = 30 - 8 = 22 cm
= Image distance
The parallel incident beam appears to diverge from a point that is from the centre of the combination of the two lenses.
(ii) When the parallel beam of light is incident, from the left, on the concave lens first:
According to the lens formula, we have:
Where,
= Object distance = -
= Image distance
The image will act as a real object for the convex lens.
Applying lens formula to the convex lens, we have:
Where,
= Object distance
= - (20 + d) = - (20 + 8) = - 28 cm
= Image distance
Hence, the parallel incident beam appear to diverge
Q22 :
At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.
The incident, refracted, and emergent rays associated with a glass prism ABC are shown in the given figure.
Angle of prism, A= 60°
Refractive index of the prism, µ= 1.524
= Incident angle
= Refracted angle
= Angle of incidence at the face AC
e= Emergent angle = 90°
According to Snell's law, for face AC, we can have:
It is clear from the figure that angle
According to Snell's law, we have the relation:
Hence, the angle of incidence is 29.75°.
Q23 :
You are given prisms made of crown glass and flint glass with a wide variety of angles. Suggest a combination of prisms which will
(a) deviate a pencil of white light without much dispersion,
(b) disperse (and displace) a pencil of white light without much deviation.
(a)Place the two prisms beside each other. Make sure that their bases are on the opposite sides of the incident white light, with their faces touching each other. When the white light is incident on the first prism, it will get dispersed. When this dispersed light is incident on the second prism, it will recombine and white light will emerge from the combination of the two prisms.
(b)Take the system of the two prisms as suggested in answer (a). Adjust (increase) the angle of the flint-glass-prism so that the deviations due to the combination of the prisms become equal. This combination will disperse the pencil of white light without much deviation.
Q24 :
For a normal eye, the far point is at infinity and the near point of distinct vision is about 25cm in front of the eye. The cornea of the eye provides a converging power of about 40 dioptres, and the least converging power of the eye-lens behind the cornea is about 20 dioptres. From this rough data estimate the range of accommodation (i.e., the range of converging power of the eye-lens) of a normal eye.
Least distance of distinct vision, d= 25 cm
Far point of a normal eye,
Converging power of the cornea,
Least converging power of the eye-lens,
To see the objects at infinity, the eye uses its least converging power.
Power of the eye-lens, P= Pc+ Pe= 40 + 20 = 60 D
Power of the eye-lens is given as:
To focus an object at the near point, object distance (u)= - d= - 25 cm
Focal length of the eye-lens = Distance between the cornea and the retina
= Image distance
Hence, image distance,
According to the lens formula, we can write:
Where,
= Focal length
Power of the eye-lens = 64 - 40 = 24 D
Hence, the range of accommodation of the eye-lens is from 20 D to 24 D.
Q25 :
Does short-sightedness (myopia) or long-sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of accommodation? If not, what might cause these defects of vision?
A myopic or hypermetropic person can also possess the normal ability of accommodation of the eye-lens. Myopia occurs when the eye-balls get elongated from front to back. Hypermetropia occurs when the eye-balls get shortened. When the eye-lens loses its ability of accommodation, the defect is called presbyopia.
Q26 :
A myopic person has been using spectacles of power -1.0 dioptre for distant vision. During old age he also needs to use separate reading glass of power + 2.0 dioptres. Explain what may have happened.
The power of the spectacles used by the myopic person, P= - 1.0 D
Focal length of the spectacles,
Hence, the far point of the person is 100 cm. He might have a normal near point of 25 cm. When he uses the spectacles, the objects placed at infinity produce virtual images at 100 cm. He uses the ability of accommodation of the eye-lens to see the objects placed between 100 cm and 25 cm.
During old age, the person uses reading glasses of power,
The ability of accommodation is lost in old age. This defect is called presbyopia. As a result, he is unable to see clearly the objects placed at 25 cm.
Q27 :
A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to? How is such a defect of vision corrected?
In the given case, the person is able to see vertical lines more distinctly than horizontal lines. This means that the refracting system (cornea and eye-lens) of the eye is not working in the same way in different planes. This defect is called astigmatism. The person's eye has enough curvature in the vertical plane. However, the curvature in the horizontal plane is insufficient. Hence, sharp images of the vertical lines are formed on the retina, but horizontal lines appear blurred. This defect can be corrected by using cylindrical lenses.
Q28 :
A man with normal near point (25 cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5 cm.
(a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass?
(b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope?
(a) Focal length of the magnifying glass, f = 5 cm
Least distance of distance vision, d = 25 cm
Closest object distance = u
Image distance, v= - d= - 25 cm
According to the lens formula, we have:
Hence, the closest distance at which the person can read the book is 4.167 cm.
For the object at the farthest distant (u'), the image distance
According to the lens formula, we have:
Hence, the farthest distance at which the person can read the book is
5 cm.
(b) Maximum angular magnification is given by the relation:
Minimum angular magnification is given by the relation:
Q29 :
A card sheet divided into squares each of size 1 mm2is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye.
(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?
(b) What is the angular magnification (magnifying power) of the lens?
(c) Is the magnification in (a) equal to the magnifying power in (b)?
Explain.
(a) Area of each square, A= 1 mm2
Object distance, u= - 9 cm
Focal length of a converging lens, f = 10 cm
For image distance v, the lens formula can be written as:
Magnification,
Area of each square in the virtual image = (10)2A
= 102 ×1 = 100 mm2
= 1 cm2
(b) Magnifying power of the lens
(c) The magnification in (a)is not the same as the magnifying power in (b).
The magnification magnitude is and the magnifying power is.
The two quantities will be equal when the image is formed at the near point (25 cm).
Q30 :
(a) At what distance should the lens be held from the figure in
Exercise 9.29 in order to view the squares distinctly with the maximum possible magnifying power?
(b) What is the magnification in this case?
(c) Is the magnification equal to the magnifying power in this case?
Explain.
(a) The maximum possible magnification is obtained when the image is formed at the near point (d= 25 cm).
Image distance, v= - d= - 25 cm
Focal length, f= 10 cm
Object distance = u
According to the lens formula, we have:
Hence, to view the squares distinctly, the lens should be kept 7.14 cm away from them.
(b) Magnification =
(c) Magnifying power =
Since the image is formed at the near point (25 cm), the magnifying power is equal to the magnitude of magnification.
Q31 :
What should be the distance between the object in Exercise 9.30 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm2. Would you be able to see the squares distinctly with your eyes very close to the magnifier?
[Note: Exercises 9.29 to 9.31 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.]
Area of the virtual image of each square, A= 6.25 mm2
Area of each square, A0= 1 mm2
Hence, the linear magnification of the object can be calculated as:
Focal length of the magnifying glass, f= 10 cm
According to the lens formula, we have the relation:
The virtual image is formed at a distance of 15 cm, which is less than the near point (i.e., 25 cm) of a normal eye. Hence, it cannot be seen by the eyes distinctly.
Q32 :
(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?
(b) In viewing through a magnifying glass, one usually positions one's eyes very close to the lens. Does angular magnification change if the eye is moved back?
(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?
(d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?
(e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?
(a)Though the image size is bigger than the object, the angular size of the image is equal to the angular size of the object. A magnifying glass helps one see the objects placed closer than the least distance of distinct vision (i.e., 25 cm). A closer object causes a larger angular size. A magnifying glass provides angular magnification. Without magnification, the object cannot be placed closer to the eye. With magnification, the object can be placed much closer to the eye.
(b) Yes, the angular magnification changes. When the distance between the eye and a magnifying glass is increased, the angular magnification decreases a little. This is because the angle subtended at the eye is slightly less than the angle subtended at the lens. Image distance does not have any effect on angular magnification.
(c) The focal length of a convex lens cannot be decreased by a greater amount. This is because making lenses having very small focal lengths is not easy. Spherical and chromatic aberrations are produced by a convex lens having a very small focal length.
(d) The angular magnification produced by the eyepiece of a compound microscope is
Where,
fe= Focal length of the eyepiece
It can be inferred that if feis small, then angular magnification of the eyepiece will be large.
The angular magnification of the objective lens of a compound microscope is given as
Where,
= Object distance for the objective lens
= Focal length of the objective
The magnification is large when >. In the case of a microscope, the object is kept close to the objective lens. Hence, the object distance is very little. Since is small, will be even smaller. Therefore, and are both small in the given condition.
(e)When we place our eyes too close to the eyepiece of a compound microscope, we are unable to collect much refracted light. As a result, the field of view decreases substantially. Hence, the clarity of the image gets blurred.
The best position of the eye for viewing through a compound microscope is at the eye-ring attached to the eyepiece. The precise location of the eye depends on the separation between the objective lens and the eyepiece.
Q33 :
An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?
Focal length of the objective lens,= 1.25 cm
Focal length of the eyepiece, fe= 5 cm
Least distance of distinct vision, d= 25 cm
Angular magnification of the compound microscope = 30X
Total magnifying power of the compound microscope, m= 30
The angular magnification of the eyepiece is given by the relation:
The angular magnification of the objective lens (mo) is related to meas:
= m
Applying the lens formula for the objective lens:
The object should be placed 1.5 cm away from the objective lens to obtain the desired magnification.
Applying the lens formula for the eyepiece:
Where,
= Image distance for the eyepiece = - d= - 25 cm
= Object distance for the eyepiece
Separation between the objective lens and the eyepiece
Therefore, the separation between the objective lens and the eyepiece should be 11.67 cm.
Q34 :
An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?
Focal length of the objective lens,= 1.25 cm
Focal length of the eyepiece, fe= 5 cm
Least distance of distinct vision, d= 25 cm
Angular magnification of the compound microscope = 30X
Total magnifying power of the compound microscope, m= 30
The angular magnification of the eyepiece is given by the relation:
The angular magnification of the objective lens (mo) is related to meas:
= m
Applying the lens formula for the objective lens:
The object should be placed 1.5 cm away from the objective lens to obtain the desired magnification.
Applying the lens formula for the eyepiece:
Where,
= Image distance for the eyepiece = - d= - 25 cm
= Object distance for the eyepiece
Separation between the objective lens and the eyepiece
Therefore, the separation between the objective lens and the eyepiece should be 11.67 cm.
Q35 :
A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when
(a) the telescope is in normal adjustment (i.e., when the final image
is at infinity)?
(b) the final image is formed at the least distance of distinct vision
(25 cm)?
Focal length of the objective lens,= 140 cm
Focal length of the eyepiece, fe = 5 cm
Least distance of distinct vision, d = 25 cm
(a) When the telescope is in normal adjustment, its magnifying power is given as:
(b) When the final image is formed atd,the magnifying power of the telescope is given as:
Q36 :
(a) For the telescope described in Exercise 9.34 (a), what is the separation between the objective lens and the eyepiece?
(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?
(c) What is the height of the final image of the tower if it is formed at 25 cm?
Focal length of the objective lens, fo= 140 cm
Focal length of the eyepiece, fe= 5 cm
(a) In normal adjustment, the separation between the objective lens and the eyepiece
(b) Height of the tower, h1= 100 m
Distance of the tower (object) from the telescope, u= 3 km = 3000 m
The angle subtended by the tower at the telescope is given as:
The angle subtended by the image produced by the objective lens is given as:
Where,
h2= Height of the image of the tower formed by the objective lens
Therefore, the objective lens forms a 4.7 cm tall image of the tower.
(c) Image is formed at a distance, d= 25 cm
The magnification of the eyepiece is given by the relation:
Height of the final image
Hence, the height of the final image of the tower is 28.2 cm.
Q37 :
A Cassegrain telescope uses two mirrors as shown in Fig. 9.33. Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be?
The following figure shows a Cassegrain telescope consisting of a concave mirror and a convex mirror.
Distance between the objective mirror and the secondary mirror, d= 20 mm
Radius of curvature of the objective mirror, R1= 220 mm
Hence, focal length of the objective mirror,
Radius of curvature of the secondary mirror, R1 = 140 mm
Hence, focal length of the secondary mirror,
The image of an object placed at infinity, formed by the objective mirror, will act as a virtual object for the secondary mirror.
Hence, the virtual object distance for the secondary mirror,
Applying the mirror formula for the secondary mirror, we can calculate image distance (v)as:
Hence, the final image will be formed 315 mm away from the secondary mirror.
Q38 :
Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. 9.36. A current in the coil produces a deflection of 3.5°of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?
Angle of deflection, θ = 3.5°
Distance of the screen from the mirror, D= 1.5 m
The reflected rays get deflected by an amount twice the angle of deflection i.e., 2θ= 7.0°
The displacement (d) of the reflected spot of light on the screen is given as:
Hence, the displacement of the reflected spot of light is 18.4 cm.
Q39 :
Figure 9.37 shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid?
Focal length of the convex lens, f1= 30 cm
The liquid acts as a mirror. Focal length of the liquid = f2
Focal length of the system (convex lens + liquid), f = 45 cm
For a pair of optical systems placed in contact, the equivalent focal length is given as:
Let the refractive index of the lens be and the radius of curvature of one surface be R. Hence, the radius of curvature of the other surface is - R.
R can be obtained using the relation:
Let be the refractive index of the liquid.
Radius of curvature of the liquid on the side of the plane mirror =
Radius of curvature of the liquid on the side of the lens, R= - 30 cm
The value of can be calculated using the relation:
Hence, the refractive index of the liquid is 1.33.
<< Back to NCERT/CBSE Notes | 9,771 | 38,256 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2018-09 | latest | en | 0.870533 |
https://us.metamath.org/mpeuni/clatlem.html | 1,713,949,842,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819089.82/warc/CC-MAIN-20240424080812-20240424110812-00730.warc.gz | 535,563,148 | 5,401 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > clatlem Structured version Visualization version GIF version
Theorem clatlem 17723
Description: Lemma for properties of a complete lattice. (Contributed by NM, 14-Sep-2011.)
Hypotheses
Ref Expression
clatlem.b 𝐵 = (Base‘𝐾)
clatlem.u 𝑈 = (lub‘𝐾)
clatlem.g 𝐺 = (glb‘𝐾)
Assertion
Ref Expression
clatlem ((𝐾 ∈ CLat ∧ 𝑆𝐵) → ((𝑈𝑆) ∈ 𝐵 ∧ (𝐺𝑆) ∈ 𝐵))
Proof of Theorem clatlem
StepHypRef Expression
1 clatlem.b . . 3 𝐵 = (Base‘𝐾)
2 clatlem.u . . 3 𝑈 = (lub‘𝐾)
3 simpl 485 . . 3 ((𝐾 ∈ CLat ∧ 𝑆𝐵) → 𝐾 ∈ CLat)
41fvexi 6686 . . . . . . 7 𝐵 ∈ V
54elpw2 5250 . . . . . 6 (𝑆 ∈ 𝒫 𝐵𝑆𝐵)
65biimpri 230 . . . . 5 (𝑆𝐵𝑆 ∈ 𝒫 𝐵)
76adantl 484 . . . 4 ((𝐾 ∈ CLat ∧ 𝑆𝐵) → 𝑆 ∈ 𝒫 𝐵)
8 clatlem.g . . . . . . . 8 𝐺 = (glb‘𝐾)
91, 2, 8isclat 17721 . . . . . . 7 (𝐾 ∈ CLat ↔ (𝐾 ∈ Poset ∧ (dom 𝑈 = 𝒫 𝐵 ∧ dom 𝐺 = 𝒫 𝐵)))
109biimpi 218 . . . . . 6 (𝐾 ∈ CLat → (𝐾 ∈ Poset ∧ (dom 𝑈 = 𝒫 𝐵 ∧ dom 𝐺 = 𝒫 𝐵)))
1110adantr 483 . . . . 5 ((𝐾 ∈ CLat ∧ 𝑆𝐵) → (𝐾 ∈ Poset ∧ (dom 𝑈 = 𝒫 𝐵 ∧ dom 𝐺 = 𝒫 𝐵)))
1211simprld 770 . . . 4 ((𝐾 ∈ CLat ∧ 𝑆𝐵) → dom 𝑈 = 𝒫 𝐵)
137, 12eleqtrrd 2918 . . 3 ((𝐾 ∈ CLat ∧ 𝑆𝐵) → 𝑆 ∈ dom 𝑈)
141, 2, 3, 13lubcl 17597 . 2 ((𝐾 ∈ CLat ∧ 𝑆𝐵) → (𝑈𝑆) ∈ 𝐵)
1511simprrd 772 . . . 4 ((𝐾 ∈ CLat ∧ 𝑆𝐵) → dom 𝐺 = 𝒫 𝐵)
167, 15eleqtrrd 2918 . . 3 ((𝐾 ∈ CLat ∧ 𝑆𝐵) → 𝑆 ∈ dom 𝐺)
171, 8, 3, 16glbcl 17610 . 2 ((𝐾 ∈ CLat ∧ 𝑆𝐵) → (𝐺𝑆) ∈ 𝐵)
1814, 17jca 514 1 ((𝐾 ∈ CLat ∧ 𝑆𝐵) → ((𝑈𝑆) ∈ 𝐵 ∧ (𝐺𝑆) ∈ 𝐵))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 398 = wceq 1537 ∈ wcel 2114 ⊆ wss 3938 𝒫 cpw 4541 dom cdm 5557 ‘cfv 6357 Basecbs 16485 Posetcpo 17552 lubclub 17554 glbcglb 17555 CLatccla 17719 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1796 ax-4 1810 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2116 ax-9 2124 ax-10 2145 ax-11 2161 ax-12 2177 ax-ext 2795 ax-rep 5192 ax-sep 5205 ax-nul 5212 ax-pow 5268 ax-pr 5332 This theorem depends on definitions: df-bi 209 df-an 399 df-or 844 df-3an 1085 df-tru 1540 df-ex 1781 df-nf 1785 df-sb 2070 df-mo 2622 df-eu 2654 df-clab 2802 df-cleq 2816 df-clel 2895 df-nfc 2965 df-ne 3019 df-ral 3145 df-rex 3146 df-reu 3147 df-rab 3149 df-v 3498 df-sbc 3775 df-csb 3886 df-dif 3941 df-un 3943 df-in 3945 df-ss 3954 df-nul 4294 df-if 4470 df-pw 4543 df-sn 4570 df-pr 4572 df-op 4576 df-uni 4841 df-iun 4923 df-br 5069 df-opab 5131 df-mpt 5149 df-id 5462 df-xp 5563 df-rel 5564 df-cnv 5565 df-co 5566 df-dm 5567 df-rn 5568 df-res 5569 df-ima 5570 df-iota 6316 df-fun 6359 df-fn 6360 df-f 6361 df-f1 6362 df-fo 6363 df-f1o 6364 df-fv 6365 df-riota 7116 df-lub 17586 df-glb 17587 df-clat 17720 This theorem is referenced by: clatlubcl 17724 clatglbcl 17726
Copyright terms: Public domain W3C validator | 1,674 | 2,836 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-18 | latest | en | 0.119309 |
http://mathematica.stackexchange.com/questions/tagged/fitting?page=3&sort=newest&pagesize=15 | 1,467,400,068,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783403823.74/warc/CC-MAIN-20160624155003-00062-ip-10-164-35-72.ec2.internal.warc.gz | 204,807,624 | 24,392 | Tagged Questions
Questions on the use of Mathematica to construct models for approximating empirical data.
837 views
Exponential fitting - isn't actually a BUG there?
There are many questions on this site about wrong exponential fitting in Mathematica but no one considers this well-known problem as a potential bug. Usually people suggest well-known workarounds ...
327 views
Findfit does not find the best fit
I am working on a project that needs curve fitting. The experimental data should be fitted with a model that calculated with an integration equation. At first, I enter my experimental data ...
95 views
How to fit parameters, not functions?
Given a set of parameters ...
70 views
How to realize Global Fitting with shared parameters, same as OriginPro9.0? [closed]
I have an experimental data from Michelson interference and I can fit the envelope in OriginPro9.0 using Global Fitting with shared parameters. The figure by OriginPro9.0 is like this: However, I ...
75 views
Paramter Scans and $\chi^2$ Test in Mathematica
Is there any in-built function or a recommended package that enables one to find, say, 10 Parameters, which are used as boundary conditions in NDSolve, that minimize $\chi^2$? ...
496 views
Fitting a curve to an image object: not a superellipse, not an egg curve.. what else?
I'm facing the problem of quantifying the morphology of objects with such a shape (nannoconids... they are fossils). One of the final goals is to produce a solid of revolution and compute volume/...
94 views
Getting image data values with coordinates in circular region
I have a number of gray level images (600, with size: 2400x2400 pixel, small instance see above). In this image there is a circular region of interest. The aim is to fit a function to the gray value ...
75 views
Fitting data in Log-Log scale
I want to fit my data from a txt file to a curve in log-log scale. What I tried to do was this: ...
99 views
Fitting data with unique variable per measurement
Imagine we are in the following situation, we have a nonlinear system of functions $f$ with coördinates $x_i$, and unknown parameters $\delta_i$'s and $t$. Let this system be defined as, \begin{...
34 views
How to calculate a required definition only once for each iteration of NonlinearModelFit
This is my first post here. I've been using Mathematica for about 4 years now to analyze data, but am still very new and have no background in formal programming. This is the first question I couldn't ...
89 views
How to improve the nonlinear fit to a data set [closed]
Some data to begin with ...
57 views
Multivariate FindFormula? [closed]
I am using Mathematica Online and want to do FindFormula with two independent variables, but I cannot figure out if it's possible and if so what the syntax is. Here's my dataset: ...
85 views
Find the linear combination of two functions to fit to a third
I have been trying to solve a simple enough problem, but I can't seem to find the proper method. The problem is given as, Find 0 < x < 1 for which, h(t) $\approx$ (1-x) f(t)+x g(t), where is ...
81 views
Fitting with constraints including the independent variable
I would like to write the following statement: data2fit = NonlinearModelFit[data2, {a Sqrt[b x + c] + d, b x + c > 0}, {a, b, c, d}, {x}] However, it returns ...
188 views
How to fit data to a convolution equation
How to fit data to a convolution equation: ...
227 views
Multiple regression curve fitting
I've got some data with 2 independent variables and 1 dependent variable I'm trying to fit a curve to. I'm really kind of a dummy with Mathematica except for the most basic things, so I looked ...
77 views
113 views
Why is TimeSeriesForecast linear?
When I use FinancialData with TimeSeriesForecast, the resulting graph is always linear and thus not very accurate. ...
92 views
FindFit and log norm function
I'm trying to fit a function to data using FindFit, but I can't find a way to replace norm function RMSE with a RMSE in log space: ...
92 views
Fitting real and imaginary parts of a function
I have an analytic form of a function G[w_] = -I*z0*w/(1 + I*z0*w*a) I know this function can be obtained (up to some overall normalization) from solving a ...
217 views
Fitting does not work
I have a set of points, which is basically one minus a Lorentz curve. And the NonlinearModelFit ...
96 views
Fitting a set of functions using same basis functions
I am trying to fit the functions in the plot below, but I need them to be fitted using the same basis functions. I want to interpolate between the functions by interpolating between their coefficients,...
80 views
Boundary parameter in Beta Distribution
I use EasyFit to find the probability distribution of my data. It turns out the data follows BetaDistribution with 0.54022,0.70601 shape parameters and 1.33,6.83 continuous boundary parameters I am ...
52 views
Displacement around the x axis, how to change a function location
While plotting a function I want to change the location of x. I am doing a regression with the logistic function ...
80 views
Data fitted by two linear fits [duplicate]
can anybody help me please and tell me how to fit my set of points by two linear fits? It is always exactly two of them and are always linear and should cover all the data. The second line should ...
71 views
problems of Lorentizan fit with Nonlinear fit model [closed]
I tried to do a Lorentzian fit for a Spectroscopy data.I found that I have to actually give the precise value for this fit.[i do not like to call this as fit ,if that is the case]? Am I missing ...
87 views
What´s wrong with my FindFit
I updated the data which is a table I deleted the first coordinate as it was the quote from the document. ...
78 views
Invert and fit implicitly defined curve
I need to fit an implicitly defined curve. I thought I could get some data out of Solve, and then using FindFit. Therefore, I ...
95 views
Fitting in Mathematica: Which function? [closed]
in other programs, there is a possibility to take a data set and to find the function which fits best. Is there also such a possibility in Mathematica? I only could find ...
96 views
NonlinearModelFit gives imaginary numbers [duplicate]
I am having trouble using NonlinearModelFit to fit to data generated by the equation I am trying to fit to. ...
95 views
Is there a built-in function implementing the minimum and maximum gradient?
Suppose I have a set of data. Mathematica has fitting functions to determine the line and corresponding equation to the best fit given any arbitrary set of data. But in experimental techniques, ...
87 views
Fitting data using ParametricNDSolveValue and NonlinearModelFit [closed]
I'm trying to evaluate the kinetics of the chemical reaction. For that I need to fit my experimental data to the kinetic equation. These are my steps in Mathematica. ...
71 views
Detecting and measuring an ellipse in a glaucoma photo? [duplicate]
I've tried the solutions given in How to find a circular object in the image glaucoma? to detect the ellipse, but the detection results are not quite what I wanted. In glaucoma we have a cup to disk ...
1k views
Detecting an ellipse in a glaucoma photo
For glaucoma diagnosis it is common to determine a "cup to disk ratio" which compares the diameter of the optic disk (VDD) and optic cup (VCD). The optical disk is visible as a circular red feature (...
156 views
Fitting to the system of equations with no analytical solution
I'm trying to fit linear combination of 2 functions to my experimental data set (it represents voltage of 2 photodiodes - Voc1 and Voc2 - connected in series under illumination - S) ...
167 views
How to find analytical expresion for data from image
I need to find an analytical expression for the data collected from an image. I found a relevant topic on how to extract the data in here. But there are some differences in my question. I have this ...
48 views
Incorporating sophisticated constraints on the approximating function into NMinimize
Following the approach as given in the answer of Belisarius here I just wonder how one could incorporate some more sophisticated requirements on the approximating function. Suppose the approximating ...
143 views
Estimated Variance in LinearModelFit
Suppose that lmModel is a linear model which is fit using LinearModelFit. How come ...
190 views
FindFit and NMinimize to fit a parametric model (minimize the distance of two curves)
I am trying to find a fit to the cumulative distribution of a set of points using FindFit or NMinimize. In particular, I would like to find the parameters of the cdf of the Beta Distribution that ...
406 views
Maxwell fit in 2 dimensions
The histogram below shows the 2D-velocities v of a micrometer sized particle in a plasma measured from image to image (60 Hz) over 33,33 sec. I would like to fit ...
59 views
FindFit with a sophisticated function (2), with corrected question and code
I am trying to find a fit to the distribution function (empiricial data) in terms of a function which is itself an integral of a product of two simplier functions. In particular, I observe T(x) (this ...
32 views
EstimatedBackground function on Mathematica 9
Is there a way to replicate the EstimatedBackground function on Mathematica 9?
108 views
FindFit algorithm?
Does anyone know how to program a mathematica algorithm that does the same thing that FindFit does? Is there documentation of it somewhere? I assume it is a least ... | 2,161 | 9,526 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 2.671875 | 3 | CC-MAIN-2016-26 | latest | en | 0.906666 |
https://calculator.academy/maximum-angular-velocity-calculator/ | 1,695,289,881,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233505362.29/warc/CC-MAIN-20230921073711-20230921103711-00568.warc.gz | 169,697,778 | 56,845 | Enter the total angular change in position and the time into the calculator to determine maximum angular velocity.
## Maximum Angular Velocity Formula
The following equation is used to calculate the Maximum Angular Velocity.
MAV = Δ θ / Δ t
• Where MAV is the maximum angular velocity (rad/s)
• Δ θ is the change in angular position (radians)
• Δ t is the change in time (seconds)
To calculate maximum angular velocity, divide the change in angular position by the change in time.
## What is a Maximum Angular Velocity?
Definition:
A maximum angular velocity describes the maximum rate of change of angular position with respect to time.
## How to Calculate Maximum Angular Velocity?
Example Problem:
The following example outlines the steps and information needed to calculate Maximum Angular Velocity.
First, determine the change in angular position. In this example, the change in angular position is found to be 100 radians.
Next, determine the change in time. For this problem, the change in time is found to be 20 seconds.
Finally, calculate the Maximum Angular Velocity using the formula above:
MAV = Δ θ / Δ t
MAV = 100 / 20 | 239 | 1,150 | {"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} | 3.75 | 4 | CC-MAIN-2023-40 | latest | en | 0.844935 |
http://www.justanswer.co.uk/computer/9airk-i-m-using-excel-need-help-following.html | 1,490,314,299,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218187227.84/warc/CC-MAIN-20170322212947-00356-ip-10-233-31-227.ec2.internal.warc.gz | 603,353,065 | 25,497 | ### How JustAnswer Works:
• Ask an Expert
• Get a Professional Answer
• 100% Satisfaction Guarantee
## Ask Jess M. Your Own Question
Jess M., Computer Support Specialist
Category: Computer
Satisfied Customers: 6395
Experience: Computer Hardware & Software Specialist - 10yrs Exp, Web Designer & Developer
49766785
Type Your Computer Question Here...
Jess M. is online now
# I'm using excel and need help with the following in an
### Resolved Question:
Hi, I'm using excel and need help with the following in an inventory list.
1) If the quantity on board is added, deduct the value from quantity on order
2) When quantity on board is deducted add this to quantity in stock
3) The quantity on board (unit price per item) is increased, add to stock value - this will be a currency value based on items x unit price + to stock value.
Any help is much appreciated.
Thanks
Nadine
Submitted: 1 year ago.
Category: Computer
Expert: Jess M. replied 1 year ago.
Hi, welcome and thank you for your question. My name is ***** ***** I am glad to assist you today.I am very sorry about your issues. Can you you please send me a sample or dummy Excel file with few sample data that we can work on? You can attach the sample file using the paper clip icon in the reply box below.Please let me know so that I can help you further.Best regards,Jess
Customer: replied 1 year ago.
Attachment: 2015-09-06_122147_dummyexcelfile_sep15.xlsx
Hi Jess
File attached for info.
Regards
Nadine
Expert: Jess M. replied 1 year ago.
Hi Nadine,Thank you for the sample file. Please give me a moment to check your requirements.
Expert: Jess M. replied 1 year ago.
Hi Nadine, I want to clarify things first. So your only data entry point in the sample file is H3. You entered 5 in H3 and you want 5 to be deducted from G3 and 5 to be added to F3.Then Available Stock is computed in J3 by adding F3 and H3 and Excel will compute K3 by multiplying J3 and the landed cost.Is this the entire flow of your calculations?Please let me know so that I can help you further.Jess
Customer: replied 1 year ago.
Hi Jess
Thanks for getting back to me. I'm in the UAE hence the delay in getting to you. My response to your queries are given below.
1. That's right and also to be added to J3.
2. Correct.
3. At this moment, yes. We will input from stock take what goes into F3.
Regards
Nadine
Expert: Jess M. replied 1 year ago.
Thank you for writing back with that information. I am from another time zone so I just started my day (Monday). Please give me a moment to work on your requirements.Jess
Customer: replied 1 year ago.
Hi Jess
Wondered how you were getting on with the queries.
Regards
Nadine
Expert: Jess M. replied 1 year ago.
Nadine,Thank you for writing back. I have completed the file requirements but I believe there is something missing. Please tell me the following:In F3 (Quantity in Stock), you want to add the value you enter in H3 (Quantity on Board). But is this G3 a blank cell originally? Or does it contain a value that is a result of a formula or calculation?Also, in like manner, in G3, you want to subtract the value in H3, but is this G3 cell originally blank? Or holds a value that is a result of some calculation?I am asking these because I believe you need a dynamic data that automatically updates when you enter a value in H3.
Customer: replied 1 year ago.
Hi Jess
1.In F3 (Quantity in Stock), you want to add the value you enter in H3 (Quantity on Board).
But is this G3 a blank cell originally? Or does it contain a value that is a result of a
formula or calculation?
A: My mistake. We want to add the Quantity on Board (H3) to Available Stock (J3).
Quantity on Order (G3) is manually input at the point of raising a purchase order.
2.Also, in like manner, in G3, you want to subtract the value in H3, but is this G3
cell originally blank? Or holds a value that is a result of some calculation?
A: We need to add Quantity on Order (G3) to Quantity on Board (H3).
We would like anything that is deducted from Available Stock (J3) to be added to deducted from
Quantity in Stock (F3).
Thanks again,
Nadine
Expert: Jess M. replied 1 year ago.
Nadine,Thank you for writing back with that information. I am almost complete with your inventory file but let me clarify things first. I better do it by definition. Also, I am asking these questions because I want to exactly know what COLUMNS require you to enter values manually (data entry cells) and which columns are CALCULATED or results of formulas.Col J (Available Stock) = the sum of Quantity on Stock and Quantity on BoardCol H (Quantity on Board) = data entry cellsCol G (Quantity on Order) = data entry cellsCol F (Quantity in Stock) = Quantity on Board + Available StockIs that correct?
Customer: replied 1 year ago.
Hi Jess
This is what we've worked out from this side.
a) Quantity on board (H3) + Quantity in stock (F3) = Available stock (J3) (includes stock at sea)
b) Once the goods have landed, they will be deducted from Available Stock (J3) and need to be added to Quantity in Stock (F3).
c) We think we are missing a step. Anything that is in Pending Orders (I3) should be deducted from Available Stock (J3).
Hopefully that should clear things up.
Thanks again,
Nadine
Expert: Jess M. replied 1 year ago.
Nadine,Thank you for the clarification and I believe I am seeing it clearly this time. Let me tell you what I understood about your process to confirm, so that I can re-create your inventory sheet based on your data and requirements. Yes, I mentioned "re-create" because we might need to add some "helper columns" to hold calculation data in order to derive your needed information.So Available Stock is the sum of Quantity On Board and Quantity In Stock. You mentioned "when goods have landed", are you referring to the Quantity on Board? Are these the goods that "land"? If yes, then, when the "quantity on board" lands, it will be deducted from Available Stock and then ADDED to Quantity in Stock. Let us take an example.Product A:Quantity On Board = 25Quantity in Stock = 50Therefore, Available Stock = 75Now, when this 25 items that are "on board" lands, this is what happens:Quantity On Board = 0Quantity in Stock = 75Available Stock = 50Is that correct?Also, please tell me more about "Pending Orders" and "Quantity of Orders". Do you manually enter the value for Pending Orders? Or are these referring to some values in your columns like "Quantity of Orders"? Keep me posted.Jess
Customer: replied 1 year ago.
Hi Jess
When goods have landed, yes, this is quantity on board which has been added into the available stock minus pending orders. Once landed it becomes quantity in stock. Point 2 is correct. Yes, it will be as described.
In the example, no. This is because quantity on board (25), quantity in stock (50), however, this may be on two containers. Therefore, may possibly only (15) pieces are landed. So, quantity in stock becomes (65) quantity on board becomes (10) but available stock remains (75).
Hope this clarifies things.
Thanks
Nadine
Expert: Jess M. replied 1 year ago.
Nadine, thank you for the clarification. So it means that quantity on board IS NOT always equal to landed quantity. Please give me some time to rebuild your inventory sheet. Thanks. Jess
Customer: replied 1 year ago.
Hi Jess
I've just received a 'rate the service' email but haven't received the rebuilt inventory sheet as mentioned above. Can you confirm if I'm missing an answer.
Thanks and regards
Nadine
Expert: Jess M. replied 1 year ago.
Hi Nadine, Please disregard this email yet -- please allow me to finish your worksheet first so that you can rate me positively. I will be sending you the final file in a moment.Jess
Expert: Jess M. replied 1 year ago.
Hi Nadine,Thank you for patiently waiting. I have completed the recreation of your Inventory Sheet with few additions which I will explain here.First, I have created 2 sheets for the 2 options. In either options, I added "helper" columns to keep track of the values that you want to monitor. I am referring to the Quantity in Stock (QS) and the Quantity on Board (QB) values. I named the "helper" columns as "Actual QS", "Actual QB" and "Quantity Landed".Here is how it works and why you need these helper columns:For instance, for Product A your QS is 50 and QB is 25. Available Stock (AS) is 75 assuming there are no pending orders to subtract to.Now, if 15 of the QB landed, the original or raw QB must be retained and the updated (current, actual or running are appropriate terms) QB will then become 10. Also, QS will be retained and the new or updated QS will be 65. While the AS will remain 75 since there are no orders.If we do not use helper functions for there, there is NO WAY that we can modify QS and QB since these values are manually entered -- they are not results of formulas.Now regarding the 2 worksheets, I suggest Option 2. In this option, before any items land, QS and Actual QS are the same. The same with QB and Actual QB. Once you enter the Quantity Landed, that is when the calculation begins. In option 1 though, you need to enter 0 in all cells to show the calculation results, otherwise, they are blank.Please see the attached final file. I hope that helped.Please remember to rate my service positively (3-5 stars/faces) if this helped. Tips are always highly appreciated!If you need further assistance, please do not rate me negatively with 1 or 2 faces. Instead, please reply to me so that I can help you further.Thank you!Best regards,Jess
Customer: replied 1 year ago.
Hi Jess
Many thanks for this. I need to look over it but it's the weekend here at the moment. Please give me a bit of time before I rate it. I'm sure it's absolutely fine. Thanks again and I'll get back to you as soon as (hopefully tomorrow) with the rating.
Best regards
Nadine
Expert: Jess M. replied 1 year ago.
Hi Nadine,Thank you for writing back and I am glad you got the modified file I sent you. Yes, please take your time and check the file and see if it fits your requirements. If you need further clarification with that file I gave you, please let me know.Best regards,Jess
Customer: replied 1 year ago.
Hi Jess
Completely aware that I haven't rated your service as yet. It's the weekend here and I haven't had a chance to look through what you sent in details. I hope to get back to you tomorrow, or at the very latest Monday.
Thanks for your patience.
Regards
Nadine
Expert: Jess M. replied 1 year ago.
Hi Nadine, Thank you for writing back. Please take your time. If you received an email inviting you to rate me, that is system generated. No rush, you can rate me when you have tested the file I gave you. Thank you.Best regards, Jess
Jess M., Computer Support Specialist
Category: Computer
Satisfied Customers: 6395
Experience: Computer Hardware & Software Specialist - 10yrs Exp, Web Designer & Developer
Jess M. and 6 other Computer Specialists are ready to help you
### What Customers are Saying:
• My Expert answered my question promptly and he resolved the issue totally. This is a great service. I am so glad I found it I will definitely use the service again if needed. One Happy Customer
< Previous | Next >
• My Expert answered my question promptly and he resolved the issue totally. This is a great service. I am so glad I found it I will definitely use the service again if needed. One Happy Customer
• Wonderful service, prompt, efficient, and accurate. Couldn't have asked for more. I cannot thank you enough for your help. Mary C.
• This expert is wonderful. They truly know what they are talking about, and they actually care about you. They really helped put my nerves at ease. Thank you so much!!!! Alex
• Thank you for all your help. It is nice to know that this service is here for people like myself, who need answers fast and are not sure who to consult. GP
• I couldn't be more satisfied! This is the site I will always come to when I need a second opinion. Justin
• Just let me say that this encounter has been entirely professional and most helpful. I liked that I could ask additional questions and get answered in a very short turn around. Esther
• Wonderful service, prompt, efficient, and accurate. Couldn't have asked for more. I cannot thank you enough for your help. Mary C.
• ### Kamil Anwar
#### Satisfied Customers:
156
8+ Years of Experience. / CCNA (S), CCNA (W), CCNA (RS), MCTS, MBCs.
< Previous | Next >
• http://ww2.justanswer.com/uploads/JA/JACUSTOMERf8udkdxk/2013-8-3_15150_323738101505074393259301621172992o.64x64.jpg Kamil Anwar's Avatar
### Kamil Anwar
#### Satisfied Customers:
156
8+ Years of Experience. / CCNA (S), CCNA (W), CCNA (RS), MCTS, MBCs.
• http://ww2.justanswer.com/uploads/WI/Windowspcfix/2013-8-19_153826_ja12.64x64.jpg James K.'s Avatar
### James K.
#### Satisfied Customers:
93
Technical Director of IT Company
• http://ww2.justanswer.com/uploads/JO/JonTech/2011-10-25_111835_A2c.64x64.jpg Jon-Tech's Avatar
### Jon-Tech
#### Satisfied Customers:
61
20 yrs: Computers, Networking
• http://ww2.justanswer.com/uploads/JA/jamieratliff/2012-2-8_194551_jamie.64x64.jpg jamieratliff's Avatar
### jamieratliff
#### Satisfied Customers:
21
Jamie has worked in the Information Technology field for over a decade.
• http://ww2.justanswer.com/uploads/JA/JACUSTOMERleg4q7o8/2012-3-6_203036_meja.64x64.png TheDoctor's Avatar
### TheDoctor
#### Satisfied Customers:
16
Experienced Software Engineer
• http://ww2.justanswer.com/uploads/FI/FisherEngineering/2012-1-18_194422_Fisher375Headshotcompressed239K.64x64.jpg Bill Fisher's Avatar
### Bill Fisher
#### Satisfied Customers:
1
Computer system design, networks, integration, general support.
• http://ww2.justanswer.com/uploads/ZE/zeyank/2014-12-9_221254_zeyank.64x64.png Ryan H.'s Avatar
### Ryan H.
#### Satisfied Customers:
1739
A+ Certified Technician - 10 Years experience working with all types of computer systems. | 3,446 | 13,956 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-13 | longest | en | 0.863334 |
https://brainmass.com/business/accounting/cost-accounting-342849 | 1,627,537,343,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153816.3/warc/CC-MAIN-20210729043158-20210729073158-00609.warc.gz | 157,714,195 | 75,093 | Explore BrainMass
# Cost accounting
Not what you're looking for? Search our solutions OR ask your own Custom question.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
Can you help me get started on this assignment?
Acquatic Manufacturing uses a job order cost system in each of its three manufacturing departments. Manufacturing overhead is applied to jobs on the basis of direct labor cost in Department A, direct labor hours in Department B, and machine hours in Department C.
In establishing the predetermined overhead rates for 2005 the following estimates were made for the year.
Instructions:
(a) Compute the predetermined overhead rate for each department.
(b) Compute the total manufacturing costs assigned to jobs in January in each department.
(c) Compute the under- or overapplied overhead for each department at January 31.
Manufacturing overhead (A) 92000 (B)86000 (C)64000
Direct labor cost (A) 48000, (B) 35000 ,(C) 50400
Direct Labor Hours, (A)4000,(B)3500,(C)4200
Machine Hours (A)8000,(B)10500,(C)12600
#### Solution Preview
Acquatic Manufacturing uses a job order cost system in each of its three manufacturing departments. Manufacturing overhead is applied to jobs on the basis of direct labor cost in Department A, direct labor hours in Department B, and machine hours in Department C.
In establishing the predetermined overhead rates for 2005 the following estimates were made for the year.
Department
A B C | 334 | 1,503 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.939594 |
https://www.physicsforums.com/threads/point-p-is-a-point-of-inflection.117679/ | 1,545,018,198,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828056.99/warc/CC-MAIN-20181217020710-20181217042710-00177.warc.gz | 980,691,759 | 12,647 | # Point P is a point of inflection?
1. Apr 16, 2006
### Hootenanny
Staff Emeritus
I just want to check something I'm not sure on. If at point P; $f'(x) = 0$ and $f''(x) = 0$ and $f'''(x) \neq 0$ can we definatly say that point P is a point of inflection?
Regards,
~Hoot
2. Apr 16, 2006
### Hootenanny
Staff Emeritus
"bump"
3. Apr 16, 2006
### Zurtex
That depends how you define "point of inflection", some times that's just how it is defined.
Easy way to test, if you consider x5 to be a point of inflection at x=0 then your test fails.
However if you don't consider it to be a point of inflection then your test is fine.
Last edited: Apr 16, 2006
4. Apr 16, 2006
### Hootenanny
Staff Emeritus | 239 | 708 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-51 | latest | en | 0.895953 |
https://math.stackexchange.com/questions/3074801/counting-with-principle-of-inclusion-exclusion | 1,563,743,139,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527204.71/warc/CC-MAIN-20190721205413-20190721231413-00046.warc.gz | 471,417,037 | 36,647 | # Counting with Principle of inclusion-exclusion
Let $$n$$ be a positive integer. Find the number of permutations of $$(1, 2,...,n)$$ such that no number remains in its original place.
Solution: To do this, first we are going to count the number of permutations where at least one number remains in its place, according to the inclusion-exclusion principle, we must first add the permutations where a given number is fixed, then subtract the permutations where 2 given numbers are fixed and so on.
To find a permutation that fixes $$k$$ given elements, we only have to arrange the rest, which can be done in $$(n − k)!$$ ways. However, if we do this for every choice of $$k$$ elements, we are counting $${n \choose k}(n − k)! = \frac{n!}{k!}$$ permutations. Since in total there are $$n!$$ permutations we get as our result:$$n!-\left(\frac{n!}{1!}+\frac{n!}{2!}-\frac{n!}{3!}+...+(-1)^{n+1}\frac{n!}{n!}\right)$$ I'm a bit confused about a point of the solution, when we fix $$k$$ elements and rearrange the other $$n-k$$, some of the remaining elements will stay fixed in their position right? so why does this work?
• This works because you are counting the number of permutations with at least $k$ elements in there original position. – Daniel Mathias Jan 15 at 19:09
• Yes, that is the point of the adding and subtracting (inclusion-exclusion): to fix the over-counting that results because "some of the remaining elements will stay fixed in their position," as you noted. – angryavian Jan 15 at 19:09
Your confusion seems to be with the inclusion-exclusion principle in itself: when we say that $$|A\cup B|=|A|+|B|-|A\cap B|$$ it is precisely because $$A$$ and $$B$$ have some elements in common, those of $$A\cap B$$, and we are counting them twice, once in $$|A|$$ and again in $$|B|$$. In your problem it is the same, you look first at those permutations which have at least one fixed element, so you continue with the inclusion-exclusion principle until you take into account all the common elements''.
• Okay so, when I count the number of permutations that fix at least one element I have ${n \choose 1}(n-1)!$ , so what happens is we fix only the first element and permute the others, then we fix only the second and permute and so on... once I've done this, if I take $n! - {n \choose 1}(n-1)!$ I would remove all the permutations s.t. at least one element is fixed, but, when I do so, some of the cases that occured during that $n$ permutations will be overcounted because when we permute the $n-1$ remaining elements we'll end up with some cases that were previously counted, that's it right? – Spasoje Durovic Jan 15 at 19:24
• @SpasojeDurovic That's more or less it! I'll try to give a more precise explanation (since you are using "at least one element fixed" in the wrong place). When you remove from all permutations the first $n!/1!$ term, you are removing those which have one fixed element and $n-1$ arbitrary. But if they are arbitrary, some repetitions will occur, so you are not removing all permutations s.t. at least one element is fixed, but only a lower bound of that. Then you realize that you have overcounted permutations with at least $2$ fixed elements, and you bound that quantity by below by $n!/2!$... – Jose Brox Jan 15 at 19:47
• @SpasojeDurovic ...but then you have overcounted permutations with at least 3 fixed elements when you tried to fix the problem with 2-fixed permutations from 1-fixed ones; so you approximate the error by $n!/3!$ and keep going. – Jose Brox Jan 15 at 19:52
• @SpasojeDurovic In the same vein, when you have 3 sets, $|A\cup B\cup C|\leq|A|+|B|+|C|$ but we have counted $|A\cap B|,|A\cap C|,|B\cap C|$ twice each, so we remove them once, but then we have removed $|A\cap B\cap C|$ thrice and we included it thrice in our first approximation, so we have no $|A\cap B\cap C|$ contribution and we have to include it, to get $$|A\cup B\cup C|=|A|+|B|+|C|-(|A\cap B|+|A\cap C|+|B\cap C|)+|A\cap B\cap C|$$ – Jose Brox Jan 15 at 19:53 | 1,121 | 4,001 | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 16, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2019-30 | latest | en | 0.900895 |
https://www.cfd-online.com/Forums/fluent/40670-negative-volume-2d.html | 1,503,065,434,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104636.62/warc/CC-MAIN-20170818121545-20170818141545-00246.warc.gz | 885,431,260 | 15,672 | # negative volume 2d
Register Blogs Members List Search Today's Posts Mark Forums Read
April 20, 2006, 11:43 negative volume 2d #1 AJG Guest Posts: n/a Sponsored Links Hi, I'm meshing a square domain with a hole in the middle. Is 2d and very simple, but when I check the grid in fluent it tells me that I have a negative volume! what can I do? Thanks
April 21, 2006, 03:29 Re: negative volume 2d #2 gernot Guest Posts: n/a Is it axysymetric? Because then you must not have a grid with negativ y-cordinates.
April 21, 2006, 04:33 Re: negative volume 2d #3 Claud Guest Posts: n/a Hi, the best thing is to work only in positive directions for x- and y-axis. In case you want to rotate your square, then your rotation axis has to be the x-axis! Claud
April 21, 2006, 04:42 Re: negative volume 2d #4 philipov Guest Posts: n/a I had a similar problem. Increase number of nodes and try different meshing scheme.
April 21, 2006, 06:52 Re: negative volume 2d #5 Rajesh Guest Posts: n/a This is more common when you have highly curved regions in your domain. In this case you have a circle where the grid may generate the negative volume on it. This can be elimintaed by dividing circle into more no. of segments and then mesh it. You need to try with diffrent grid spacings also. regards Rajesh
Thread Tools Display Modes Linear Mode
Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are On Pingbacks are On Refbacks are On Forum Rules
Similar Threads Thread Thread Starter Forum Replies Last Post Ganesh FLUENT 13 January 22, 2014 05:11 AlmostSurelyRob OpenFOAM 3 June 24, 2011 13:06 gaottino OpenFOAM Native Meshers: blockMesh 7 July 19, 2010 14:11 paean OpenFOAM Running, Solving & CFD 0 November 14, 2008 22:14 Rasmus Gjesing (Gjesing) OpenFOAM Native Meshers: blockMesh 10 April 2, 2007 14:00 | 556 | 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} | 2.578125 | 3 | CC-MAIN-2017-34 | longest | en | 0.899145 |
https://oneconvert.com/unit-converters/power-converter/btu-it-minute-to-joule-hour | 1,701,480,366,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100309.57/warc/CC-MAIN-20231202010506-20231202040506-00320.warc.gz | 518,520,495 | 46,887 | # Convert Btu (IT)/minute to joule/hour
## How to Convert Btu (IT)/minute to joule/hour
1 Btu,min
=
63303 J,h
1 J,h
=
0 Btu,min
Example: convert 15 Btu,min to J,h:
15 Btu,min
=
15
x
63303 J,h
=
94955e+1 J,h
## Btu (IT)/minute to joule/hour Conversion Table
Btu (IT)/minute (Btu,min)
joule/hour (J,h)
0.01 Btu,min633.0335116 J,h
0.1 Btu,min6330.335116 J,h
1 Btu,min63303.35116 J,h
2 Btu,min126606.7023 J,h
3 Btu,min189910.0535 J,h
5 Btu,min316516.7558 J,h
10 Btu,min633033.5116 J,h
20 Btu,min1266067.023 J,h
50 Btu,min3165167.558 J,h
100 Btu,min6330335.116 J,h
1000 Btu,min63303351.16 J,h
## Popular Unit Conversions Power
The most used and popular units of power conversions are presented for quick and free access.
Watt to EW
OneConvert - online calculator and file converter. Provides free unit conversion, convert PDF files, and other documents to help you solve problems of all types.
Share us
2023 OneConvert | 326 | 923 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-50 | longest | en | 0.563626 |
http://betterlesson.com/lesson/resource/2815040/homework-sample-1-jpg | 1,487,524,373,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170186.50/warc/CC-MAIN-20170219104610-00189-ip-10-171-10-108.ec2.internal.warc.gz | 26,916,830 | 22,537 | ## homework sample 1.JPG - Section 5: Class Discussion
homework sample 1.JPG
# The Punk Prank Payback
Unit 10: Exploring Area & Perimeter
Lesson 2 of 16
## Big Idea: Students need to use prior knowledge that opposite sides of rectangles are equal.
Print Lesson
2 teachers like this lesson
Standards:
59 minutes
### Erica Burnison
##### Similar Lessons
###### Density of Gases
8th Grade Science » Heat Transfer and Interactions of Matter
Big Idea: This investigation uses a simple method that allows for capturing the gas released from a chemical reaction to help students determine the density of the released gas.
Favorites(54)
Resources(14)
Brookline, MA
Environment: Urban
###### Perimeter of Irregular Rectilinear Shapes
Big Idea: Students use the properties of rectangles to discover missing dimensions.
Favorites(7)
Resources(19)
New Orleans, LA
Environment: Urban
###### Area of Irregular Figures - How do you break up a figure?
7th Grade Math » Geometric Measurement
Big Idea: Expert students will lead the lesson as students apply prior knowledge of area to composite figures.
Favorites(23)
Resources(17)
Elon, NC
Environment: Suburban | 260 | 1,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-09 | latest | en | 0.771195 |
https://kopiandproperty.com/2020/04/12/lets-be-millionaire-just-by-working/ | 1,624,548,672,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488556133.92/warc/CC-MAIN-20210624141035-20210624171035-00231.warc.gz | 322,120,940 | 54,569 | # Let’s all become a millionaire just by working 9-5 shall we?
What are the chances of everyone becoming a millionaire by just working 9-5 everyday for the next 30 years? In one word? Possible. In two words? Yes, possible. Let’s not look at the T20 households in Malaysia. Let’s just look at the M40 households instead. (click here to know what’s a T20 and M40 and also another category known as B40) These households earn a median income of RM6,275 per month.
How about saving just 20% of that salary and we assume we will have no further increments until we retire 30 years later? Simple calculation as follows: We round up the salary to RM6,300 per month.
RM6,300 x 22.5% = RM1,417 x 12 (months) = RM17,000 x 30 years = RM510,300 x 2 (conservative estimate for interest rate returns on savings) = RM1,020,600. We have just become a millionaire just by savings alone. By the way, this does not yet include our EPF savings which would have given us another RM1,000,000 based on 11% savings from us and 12% savings from our employer which is a total of 23% per month. See, life’s not that bad, IF we at least save enough of the sum.
I know, some will say that by the time one’s salary reaches RM6,300 the person may have already worked more than 10 years. Noted but like I say, the above calculation is also very conservative as it assumes that when your salary reaches RM6,300 it will then stop increasing… It also assumes that the returns on your investment is pretty low yeah. In fact lower than the fixed deposit rates too. So, give and take lah, don’t be too calculative.
We could also supplement our savings by spending less on food. Here’s an earlier article:Eating cheaper is still possible, depends on us actually Oh yeah, the above also assumes that we are someone who continue to improve, work hard and well keep climbing up the corporate career ladder. In case you like to use EPF’s calculator to understand what would the savings be like, please do try out their friendly calculator here.
Remember yeah, the above are all about savings alone. If we intend to do much better, then investment must come into play. Here’s an earlier article: Are we earning, saving, investing and preserving? It does not even start with property investment. We must remember to protect what we have through insurance too. That’s an investment too to ensure we do not suddenly lose a lot of our money when we have to be hospitalised.
Happy working and happy understanding that it’s possible to become a millionaire or multi-millionaire (if we include EPF savings too) by working 9-5 for the next 30 years.
Next suggested article: Moderating property price increase. Thus just 2.5%. Still a wow for me
This site uses Akismet to reduce spam. Learn how your comment data is processed.
#### kopiandproperty.com
kopiandproperty.com is everything about property related writings and news. Enjoy reading with a latte.
### How much savings is enough in future? RM3 million?
#### LIKE us for property news update, FREE.
An article a day, keeps updated all the way.
Join 1,509 other subscribers
## Property investment news everyday?
An article a day, keeps updated all the way. Subscribe for free! | 742 | 3,197 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.968263 |
http://k8schoollessons.com/2-digit-from-3-digit-subtraction-with-borrowing-method/ | 1,510,999,858,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804724.3/warc/CC-MAIN-20171118094746-20171118114746-00335.warc.gz | 159,270,047 | 12,692 | Online Homeschooling Website
“When the power of love overcomes the love of power, the world will know peace” – Jimi Hendrix
# 2-digit from 3-digit Subtraction with borrowing method
Subtraction with regrouping worksheets 5: Practise to subtract a 2 digit number from a 3 digit number. Use the empty boxes to type the correct workings and answers.
## Subtraction with regrouping worksheets 5:
Subtracting 2-digit from 3-digit with regrouping online math practice worksheet. Learn 2 digit subtraction from 3 digit numbers with regrouping method. online math practice worksheets to learn how to take away a 2 digit number from a 3 digit number with the help of workings.
Subtraction with regrouping worksheets 5: This is a 3rd grade math worksheet for online math practice on subtracting a 2 digit number from a 3 digit number with borrowing or regrouping. Use these online math practice worksheets to be fast and smart in subtracting 2 digit numbers from 3 digit numbers. These subtraction with regrouping or subtraction with borrowing online math practice worksheets are ideal to test your elementary child’s understanding of the fundamentals of simple subtraction.
Regrouping in subtraction or borrowing in subtraction: Let’s now practise regrouping which is also called ‘borrowing’ in subtraction. Try our great collection of online math practice worksheets to learn regrouping in subtraction or borrowing in subtraction! Make it easier for your child to understand the concept of subtraction with regrouping or subtraction with borrowing. Let your child be quick in subtraction with regrouping or subtraction with borrowing.
Click here to try more subtraction with regrouping or subtraction with borrowing online practice math worksheets.
[{ "question": [{ "hundreds": "3", "tens": "5", "ones": "2" }, { "hundreds": " ", "tens": "7", "ones": "3" } ] }, { "question": [{ "hundreds": "5", "tens": "8", "ones": "2" }, { "hundreds": " ", "tens": "9", "ones": "3" } ] }, { "question": [{ "hundreds": "6", "tens": "7", "ones": "0" }, { "hundreds": " ", "tens": "9", "ones": "1" } ] }, { "question": [{ "hundreds": "5", "tens": "5", "ones": "2" }, { "hundreds": " ", "tens": "7", "ones": "7" } ] }, { "question": [{ "hundreds": "4", "tens": "6", "ones": "7" }, { "hundreds": " ", "tens": "7", "ones": "8" } ] }, { "question": [{ "hundreds": "9", "tens": "7", "ones": "3" }, { "hundreds": " ", "tens": "8", "ones": "4" } ] }, { "question": [{ "hundreds": "8", "tens": "2", "ones": "5" }, { "hundreds": " ", "tens": "3", "ones": "6" } ] }, { "question": [{ "hundreds": "7", "tens": "6", "ones": "5" }, { "hundreds": " ", "tens": "8", "ones": "6" } ] }, { "question": [{ "hundreds": "4", "tens": "6", "ones": "1" }, { "hundreds": " ", "tens": "7", "ones": "4" } ] }, { "question": [{ "hundreds": "9", "tens": "6", "ones": "5" }, { "hundreds": " ", "tens": "7", "ones": "6" } ] }, { "question": [{ "hundreds": "5", "tens": "1", "ones": "3" }, { "hundreds": " ", "tens": "5", "ones": "5" } ] }, { "question": [{ "hundreds": "9", "tens": "2", "ones": "1" }, { "hundreds": " ", "tens": "3", "ones": "7" } ] } ] | 948 | 3,106 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2017-47 | latest | en | 0.782337 |
https://3dm3.com/tutorials/zbrush/making-of-amazon-warrior/ | 1,675,338,002,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500017.27/warc/CC-MAIN-20230202101933-20230202131933-00754.warc.gz | 92,711,907 | 25,033 | # Making of Amazon Warrior
Fabio Bautista shows us how to create a female amazon warrior
### CONCEPT
This project started as an entry for a Sculpting Challenge. The brief of the challenge was to create a female amazon warrior and I thought about doing something with the proportions of comic book characters, a tall young woman with slender figure but at the same time looking like a strong warrior.
So first task was to gather references and Google’s image search was great for this. Amazon warrior and native women images were my primary research goal. This gave me some ideas for the design of the character.
### STEP 1: CREATING THE BASE MESH
I usually start working on a base geometry in Max, but this time I decided to use a different technique. ZBrush gives you different methods to create a base mesh and one of these ways to do it is using ZSpheres. With ZSpheres you can create easily structures that can be converted into a basic form. It uses the Adaptive Skin feature that creates a surface of polygons around these ZSpheres structures. Well, I started selecting the ZSphere tool and drawing on the canvas. Then I went to the Edit mode and added more ZSpheres building up the structure of the body. Once it was completed I used Make Adaptive Skin to convert the figure to an editable mesh. Although the distribution of this geometry was not the best, it was a good starting point (Fig. 01).
I used the same method for the hair, but it was created as an independent Subtool. An important thing to note here is that by creating separate objects, the memory resources are used in a better way. This give you the possibility to increase the subdivision of each object much more than using the whole model. Thus, each subtool can have a different subdivision level and you can work with a greater level of detail on objects that need it.
### STEP 2: FIRST STAGE OF SCULPTING
After the base mesh is finished, I started searching the overall proportions using the Move brush. As the idea was to achieve a slender figure, first I extended a little the neck and trunk and adjusted the limbs according to them. The key to a well-structured body is to pay special attention on the proportions and shapes. Here for example, I used as reference the joints of the arms and the waist to shaping out the legs and the overall body.
After getting the overall shape, I increased the level of subdivision and began to adjust the overall details of the body. For this task I used mostly Move and Standard. Later I added a subdivision more to refine each area of the body. Here, I used the Standard, Clay and Smooth brushes. Then I divided the model one more time to get more definition on the face and chest (Fig. 02).
When I was happy with the shape I exported a version of this early model to Max to optimize and retopologize. There I adjusted the distribution of geometry and created the base mesh for the accessories starting from primitive objects. Finally I exported the model with clothes and accessories to ZBrush (Fig. 03).
Back in ZBrush and once established the overall details, I started working on each specific area of the body. At this point, I focused especially on the face. I wanted to give it some thin features, but at the same time it proved security and strength. Then I added two subdivisions to the model and began to get some more definition to the face refining the eyelids, ears, nose and lips. To carry out this task I used mainly the Clay, Inflat, Pinch and Standard brushes (Fig. 04).
The same process was used to give shape to the other parts of the body, this time I focused on areas like the torso, hands, knees and feet, but without adding too much detail or skin imperfections, for a smooth and soft look (Fig. 05).
After this, I went up one level of division more and continued refining the face, giving some more details (like in the hair, eyebrows and lips) and then working in the same way with the rest of the body (Fig. 06).
Clothing was the next step. I divided it to five levels of subdivision and added some small details to clothing and the other accessories (Fig. 07).
As for the hair I paid special attention first to define locks of hair to add some fluency and dynamism. For this I used the SnakeHook brush to create the tips and give shape. Then I used a customized brush created from the Standard brush, to block in and define the layers of hair. To complete this process, I added details on the hair using alphas and Lazy Mouse mode (Fig. 08).
At the end of the sculpting process, the model was at level six of subdivisiуn and already had all the detail that I wanted. It was ready for the last step. (Fig. 09)
### STEP 4: POSING THE MODEL
The final step was to give a pose to the model. As it was consisted of several elements, it was a little bit difficult to adjust one by one. So I used the Transpose Master plugin. It is found in the Zplugin menu and you only need to click on the TPoseMesh button to create a low resolution mesh conformed of all subtools included in the original high-res model (Fig. 10).
With this low poly mesh, the task of posing the whole model was pretty simple. At this point I used the Transpose Tool to reshape and adjust the model This tool is activated by pressing any of deformation modes (Move, Rotate or Scale). The process consisted of making a selection, shading the part which I did not want to deform and creating one Action Line to deform the model. So I used the Move and Rotate modes to adjust each part of the body (Fig. 11).
Once I got the final pose of the model I went back to Transpose Master and used the TPose>SubT button to transfer the changes to the original model (Fig. 12).
I also tested some other more dynamic poses for the model by using this method (Fig. 13).
Finally this was the result after some little tweaking in Photoshop (Fig 14) | 1,278 | 5,862 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-06 | latest | en | 0.950827 |
https://math.stackexchange.com/questions/2747943/does-rx-cong-sx-imply-r-cong-s?noredirect=1 | 1,558,826,934,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258453.85/warc/CC-MAIN-20190525224929-20190526010929-00059.warc.gz | 552,809,953 | 33,578 | # Does $R[[x]] \cong S[[x]]$ imply $R\cong S$
Let $R,S$ be commutative unitary rings. Is it true that
$$R[[x]] \cong S[[x]] \quad \Rightarrow \quad R\cong S.$$
Here by $R[[x]], S[[x]]$ I mean the ring of formal power series and the isomorphisms as isomorphisms of rings.
In fact, in this question Does $R[x] \cong S[x]$ imply $R \cong S$? the answer for the polynomial ring is negative and I became curious about the formal power series case.
• not an answer, but here is a paper from around the same time as the cited one. The result was "author does not know a counterexample." – Andres Mejia Apr 21 '18 at 22:49
This is not true.
An explicit counterexample is given on p. 154 of the paper On power-invariance, by Eloise A. Hamann, Pacific J. Math., Volume 61, Number 1 (1975), 153-159. The example was given by Andy Magid at the Commutative Algebra Conference in June of 1974, based on Hochster's counterexample for polynomial rings.
More precisely, Hochster's counterexample provides (using hairy ball theorem) a projective non-free module $$P$$ over the commutative ring $$A = \Bbb R[x,y,z]/(x^2+y^2+z^2-1)$$ such that $$A^3 \cong A \oplus P$$ as $$A$$-modules.
Let $$R$$ be the complete symmetric algebra of the $$A$$-modules $$P$$, i.e. the completion of the symmetric algebra $$S_A(P)$$ with respect to the ideal generated by $$P$$. Let $$S := A[[X,Y]]$$. Then, one can show that $$R[[T]] \cong A[[X,Y,Z]] \cong S[[T]]$$ as rings (not necessarily as topological rings), but $$R \not \cong A[[X,Y]] = S.$$
However, theorem 4 in Hamann's paper states that if $$R$$ has finitely many maximal ideals, then $$R[\![x]\!] \cong S[\![x]\!] \implies R \cong S$$, for every commutative ring $$S$$.
Moreover, theorem 1.4 in Joong Ho Kim's 1974 paper Power invariant rings shows that the same property holds if $$R$$ has a nilpotent Jacobson radical (i.e. the intersection $$J$$ of all maximal ideals of $$R$$ satisfies $$J^n=0$$ for some $$n>0$$). For instance, this holds if $$R$$ is artinian (e.g. a field) or semiprimitive (e.g. $$\Bbb Z$$).
• Wow, that is quite a complete answer. Thank you very much. – Severin Schraven May 11 '18 at 19:15 | 671 | 2,153 | {"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": 22, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-22 | latest | en | 0.870746 |
https://www.ams.org/journals/proc/1995-123-05/S0002-9939-1995-1233969-4/home.html | 1,669,492,917,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446708046.99/warc/CC-MAIN-20221126180719-20221126210719-00686.warc.gz | 688,908,821 | 15,617 | # Proceedings of the American Mathematical Society
Published by the American Mathematical Society, the Proceedings of the American Mathematical Society (PROC) is devoted to research articles of the highest quality in all areas of pure and applied mathematics.
ISSN 1088-6826 (online) ISSN 0002-9939 (print)
The 2020 MCQ for Proceedings of the American Mathematical Society is 0.85.
What is MCQ? The Mathematical Citation Quotient (MCQ) measures journal impact by looking at citations over a five-year period. Subscribers to MathSciNet may click through for more detailed information.
## On a sequence transformation with integral coefficients for Euler’s constantHTML articles powered by AMS MathViewer
by C. Elsner
Proc. Amer. Math. Soc. 123 (1995), 1537-1541 Request permission
## Abstract:
Let $\gamma$ denote Euler’s constant, and let ${s_n} = \left ( {1 + \frac {1}{2} + \cdots + \frac {1}{{n - 1}}} \right ) - \log n\quad (n \geq 2).$ We prove by Ser’s formula for the remainder $\gamma - {s_n}$ that for all integers $n \geq 1$ and $\tau \geq 2$ there are integers ${\mu _{n,0,}}{\mu _{n,1}}, \ldots ,{\mu _{n,n}}$ such that ${\mu _{n,0}}{s_\tau } + {\mu _{n,1}}{s_{\tau + 1}} + \cdots + {\mu _{n,n}}{s_{\tau + n}} = \gamma + {O_\tau }({(n(n + 1)(n + 2) \bullet \cdots \bullet (n + \tau ))^{ - 1}}),$ where the constant in ${O_\tau }$ depends only on $\tau$. The coefficients ${\mu _{n,k}}$ are explicitly given and are bounded by ${2^{3n + \tau - 1}}$.
References
Similar Articles
• Retrieve articles in Proceedings of the American Mathematical Society with MSC: 11Y60, 40A05, 65B05
• Retrieve articles in all journals with MSC: 11Y60, 40A05, 65B05 | 504 | 1,664 | {"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} | 2.84375 | 3 | CC-MAIN-2022-49 | longest | en | 0.74472 |
http://www.ck12.org/algebra/Multiplication-of-Polynomials-by-Binomials/?difficulty=at+grade&by=community | 1,449,034,371,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448399326483.8/warc/CC-MAIN-20151124210846-00080-ip-10-71-132-137.ec2.internal.warc.gz | 362,173,685 | 16,121 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Multiplication of Polynomials by Binomials
## Multiply vertically to keep terms organized
Levels are CK-12's student achievement levels.
Basic Students matched to this level have a partial mastery of prerequisite knowledge and skills fundamental for proficient work.
At Grade (Proficient) Students matched to this level have demonstrated competency over challenging subject matter, including subject matter knowledge, application of such knowledge to real-world situations, and analytical skills appropriate to subject matter.
Advanced Students matched to this level are ready for material that requires superior performance and mastery.
## Multiplication of Polynomials by Binomials
Learn how to distribute a binomial to a polynomial.
0
## Multiplication of Polynomials by Binomials A.APR.1
by Christine Corbley //at grade
Multiply two binomials or a polynomial by a binomial.
0
## Multiplication of Polynomials by Binomials
by Sandra Bowling //at grade
Multiply two binomials or a polynomial by a binomial.
0
## Multiplication of Polynomials by Binomials
by Carolyn Carroll //at grade
Multiply two binomials or a polynomial by a binomial.
0
## Multiplication of Polynomials by Binomials
by EPISD Algebra 1 Team //at grade
0 | 312 | 1,380 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-48 | longest | en | 0.853816 |
https://www.thestudentroom.co.uk/showthread.php?t=2949983 | 1,624,552,056,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488556133.92/warc/CC-MAIN-20210624141035-20210624171035-00042.warc.gz | 941,195,254 | 34,713 | Grade Boundaries Maths Higher Edexcel
Watch
Announcements
Thread starter 6 years ago
#1
Does anyone know the grade boundaries for GCSE MATHS HIGHER ..... How many marks do I need to get a C grade ?
Exams on the 5th november 2014
Would appreciate your help thank you .
0
6 years ago
#2
u - >9%
d- 10%
c -30%
b - 50%
a- 70%
a*- 85%
0
6 years ago
#3
(Original post by Yasmeen11)
Does anyone know the grade boundaries for GCSE MATHS HIGHER ..... How many marks do I need to get a C grade ?
Exams on the 5th november 2014
Would appreciate your help thank you .
The grade boundaries are different for every exam and are worked out after the papers are marked and analysed.
Those shown above are a decent estimation of roughly what they are likely to be.
Posted from TSR Mobile
0
6 years ago
#4
Won't be able to know until much later on in the year but you can have a look at previous grade boundaries to see where the rough boundary is. The most recent grade boundary were:
A*: 164 - 200,
A: 133 - 163,
B: 95 - 132,
C: 57 - 94,
D: 28 - 56,
E: 13 - 27,
U: 0 - 12
All the grade boundaries can be found on the Edexcel website.
0
6 years ago
#5
Im doing this exam as well and am in yr 12 really stressing out what do you think the topics in the first paper could be??
0
Thread starter 6 years ago
#6
(Original post by rohan taneja)
Im doing this exam as well and am in yr 12 really stressing out what do you think the topics in the first paper could be??
Topics like :
- Factor and Primes
-Indices
-Ratio
-standard form
-Factorising
-Expanding brackets
-Translation,reflection,rotation,Enlargement
These topic do come up on the non cal higher tier as well as other subject that would be easy to solve without using calculator ....
Hope this helps and good luck with the exam
1
6 years ago
#7
Thanks Really hoping for a B do you think the boundaries will be high or low?
(Original post by Yasmeen11)
Topics like :
- Factor and Primes
-Indices
-Ratio
-standard form
-Factorising
-Expanding brackets
-Translation,reflection,rotation,Enlargement
These topic do come up on the non cal higher tier as well as other subject that would be easy to solve without using calculator ....
Hope this helps and good luck with the exam
0
Thread starter 6 years ago
#8
(Original post by rohan taneja)
Thanks Really hoping for a B do you think the boundaries will be high or low?
Grade boundaries change every year so I don't really know, better chance of passing if it is low ....
Do you know any websites that do past papers for higher tier ?
0
6 years ago
#9
yeah mathsgenie.com
x
0
6 years ago
#10
You shouldn't be aiming for a C. You should aim for an A*. If you get a C for GCSE Higher math, you will get a big shock come A levels. I got an A* for IGCSE Higher level math and I an having a bit of a struggle with Scottish Higher level math(AS Level equivalent).
0
5 years ago
#11
(Original post by gwagon)
You shouldn't be aiming for a C. You should aim for an A*. If you get a C for GCSE Higher math, you will get a big shock come A levels. I got an A* for IGCSE Higher level math and I an having a bit of a struggle with Scottish Higher level math(AS Level equivalent).
Maths is hard and I do not think that you are in the position of telling people what they should be aiming for. A 'C' is still a pass and that is all that counts. You do not need to pick maths for A levels it is not always compulsory. Everyone knows that an IGCSE has lower grade boundaries compared to edexcel. Quite frankly, nobody cares what you achieved in maths so stop blowing your own horn!
1
5 years ago
#12
(Original post by 11jojo11)
Maths is hard and I do not think that you are in the position of telling people what they should be aiming for. A 'C' is still a pass and that is all that counts.
Wow...you have very low expectations of yourself, don't you?
0
3 years ago
#13
I wanna know what is our passing grade boundaries are edexcel igcse??
0
3 years ago
#14
Grade boundaries for the recent exams will be published at the same time as results
Previous boundaries can be found on the Edexcel web site
0
3 years ago
#15
What topics are going to come up on the new gcse exam ?
0
3 years ago
#16
(Original post by Hhsilva)
What topics are going to come up on the new gcse exam ?
Everything you have learned in maths lessons will appear on the exams.
You'll find a list of topics, revision videos and practice questions on the www.mathsgenie.co.uk web site
0
X
new posts
Back
to top
Latest
My Feed
Oops, nobody has postedin the last few hours.
Why not re-start the conversation?
see more
See more of what you like onThe Student Room
You can personalise what you see on TSR. Tell us a little about yourself to get started.
Poll
Join the discussion
Do you think receiving Teacher Assessed Grades will impact your future?
I'm worried it will negatively impact me getting into university/college (207)
43.86%
I'm worried that I'm not academically prepared for the next stage in my educational journey (54)
11.44%
I'm worried it will impact my future career (36)
7.63%
I'm worried that my grades will be seen as 'lesser' because I didn't take exams (100)
21.19%
I don't think that receiving these grades will impact my future (48)
10.17%
I think that receiving these grades will affect me in another way (let us know in the discussion!) (27)
5.72% | 1,394 | 5,322 | {"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} | 2.9375 | 3 | CC-MAIN-2021-25 | latest | en | 0.920812 |
https://it.mathworks.com/matlabcentral/profile/authors/5758333?detail=all&page=2 | 1,716,059,580,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057494.65/warc/CC-MAIN-20240518183301-20240518213301-00399.warc.gz | 289,908,232 | 5,773 | Risposto
how can i find a fuction which define input and output of a system?
Use: p = polyfit(x,y,n) % n is degree of polynomial, use a value that best fit your data To check the fitting: y_fi...
oltre 9 anni fa | 2
| accettato
Risposto
How to concatenate several matrices into one matrix?
Perhaps, you mistyped. It is C = [A; B] for different number of rows, but similar number of columns. Anyway, how your matrices a...
oltre 9 anni fa | 3
| accettato
Risposto
Multi level sorting and filtering
[~, ~, D] = xlsread('Multilevel_sorting.xlsx'); A = D(3:end,1:2); % your data in table 1 G = {'Group1','Group2','Gro...
oltre 9 anni fa | 1
Risposto
Acceleration vs Time data into FFT
Define the number of DFT points N, and then perform the fft. N = pow2(nextpow2(length(signal))); y = fft(signal,N); ...
oltre 9 anni fa | 1
Risposto
Recode strings in Cell array
indx = 0; for k = 1:size(codes,1) if strcmp(codes(k,1),'Home') && strcmp(codes(k,2),'5') indx = indx+1; ...
oltre 9 anni fa | 1
| accettato
Risposto
Solving system of nonlinear euations
syms x y [x,y] = vpasolve(x^2 + y^3 == 1, 2*x + 3*y == 4, x,y)
oltre 9 anni fa | 1
| accettato
Risposto
manually draw shapes (like using a pencil to draw on figure) on opened image
Try: roipoly
oltre 9 anni fa | 1
| accettato
Risposto
How Can I Assign variable Name?
x = 1:5; eval([datestr(date,'mmmm_dd_yyyy') '= x']); x is stored in a new variable named based on today's date. If you r...
oltre 9 anni fa | 1
Risposto
Compound Poisson Distribution Model
I am not sure if I understand your question correctly. I guess once you get P, then you would like to choose its elements random...
oltre 9 anni fa | 1
| accettato
Risposto
How is this an exponential curve?
Convert your difference equation into differential equation: y(i+1) = y(i)-y(i)*h; y(i+1) - y(i) = -y(i)*h; (y(i+1) -...
oltre 9 anni fa | 1
| accettato
Risposto
Stuck with electronic circuit simulation..
R = 1; L = 1; C = 1; % replace by your values w = 50; % replace by frequency Z = R+1j*w*L+1./(1j*w*C); % impedence of th...
oltre 9 anni fa | 2
| accettato
Risposto
Generating row combination from a set of data
T = rand(37,25); fil = 1:37; b = combntns(fil,3); for k = 1:size(b,1) out(:,:,k) = T(b(k,:),:); end
oltre 9 anni fa | 1
Risposto
Combination of the value of different vectors into one matrix ?
out = combvec(A,B,C)'
oltre 9 anni fa | 2
Risposto
Generate an equation from a 3d surface
You can play a trick here. Use your code first: Eq = @(x,y) x.*y.*(y>=0 & y<3) + 2*x.*y.*(y>=3 & y<5) + 3*x.*y.*(y>=5 & y...
oltre 9 anni fa | 0
Risposto
Generate an equation from a 3d surface
In Matlab, you can compact the three equations into one as written below. It is not a theoretical modeling anyway, but simply a ...
oltre 9 anni fa | 1
Risposto
Summation equations in matlab
You can try with this: SOAM = sum(CP(1:n-3))/sum(abs(P(2:n)-P(1:n-1))) The 2nd sum is confusing because of the indexin...
oltre 9 anni fa | 2
| accettato
Risposto
Regarding importing excel in matlab
xlsread import excel data with cell type. Convert them to double before plotting. In this case u can use cell2mat function: ...
oltre 9 anni fa | 1
| accettato
Risposto
What is the problem in this code?
Your n has a length of 9, as n=0:8, so i will have length of 9 as well,and hence, you cannot write i(1:81). There are many ways ...
oltre 9 anni fa | 1
| accettato
Risposto
binary matrix that has different orders of 0 and 1
If the ordering of rows is not a concern, then use: n = 3; A = dec2bin(0:2^n-1); B = double(A)-48 Note: 48 is the ...
oltre 9 anni fa | 2
Risposto
create 3-dimension matrix
If you use your current code, then call your function create1() (from command window, or in another m-file): c = create1();...
oltre 9 anni fa | 1
| accettato
Risposto
problem in Hanning window with FFT
Here are few things: Your plot shows the amplitude of Fourier transform, not the original signal. So, the amplitude may not be ...
oltre 9 anni fa | 1
| accettato
Risposto
How can I break line in comment for HTML publication?
%% Test section %% % After the break!
oltre 9 anni fa | 3
Risposto
how can i plot using variables from a loop?
You can do this without using a for loop: er=2.2;h=0.1588; freq={11 2 3 4 5 6 7 8 9 10}; i=1:1:length(freq); ...
oltre 9 anni fa | 1
Risposto
Producing an error vector with a loop
Where is your LogLike_Error_Vector? Is it LogLike_DT_Data? If so, you can try using a for loop: indx = 0; for MinLeaf = ...
oltre 9 anni fa | 2
| accettato
Risposto
I want to plot the auto correlation function for 15 numbers without using the inbuilt function. My series is P=[ 47,64,23,71,38,64,55,41,59,48,71,35,57,40,58 ] and the plot should be autocorrelation function vs lags.
Replace acf=covark/covar0 ; by *acf(k)=covark/covar0;* Also no need to use acf after that line. After the end of for loop,...
oltre 9 anni fa | 3
| accettato
Risposto
How to separate signals with diffrent frequencies
Add the following lines of code at the bottom of your code (after y=x+randn(size(x)).*sigma;): ffty = fft(y); ffty = abs...
oltre 9 anni fa | 5
| accettato
Risposto
Why am I getting imaginary values with ode45 ??
Your I has both negative and positive values, which are in the range of 10^(-7), but Is is in 10^(-12). So, I/Is gives high nega...
oltre 9 anni fa | 1
Risposto
Summation of a series without for loop
Yes you can do this! Use vectorization technique. Try to understand the line F = ... in the following code. If you get any hard ...
oltre 9 anni fa | 3
| accettato
Risposto
Rotation of 3d image
I am not sure what your aim is. You can try the following code: for az = -37.5:5:322.5 view(az,30) pause(0.1)...
oltre 9 anni fa | 1
| accettato
Risposto
how can i solve this one below?!! PS: the most important to me is to know how to take the values from the user and put them into a matrix and please i want the simplest way for it
N = input('N='); for i = 1:N for j = 1:3 A(i,j) = input(['row ',num2str(i),' and column ',num2str(j),' elemen...
oltre 9 anni fa | 1 | 1,998 | 6,059 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-22 | latest | en | 0.586823 |
https://www.physicsforums.com/threads/magnetic-field-and-its-energy.187757/ | 1,524,526,322,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946256.50/warc/CC-MAIN-20180423223408-20180424003408-00455.warc.gz | 879,160,667 | 14,792 | # Homework Help: Magnetic field and its energy?
1. Sep 28, 2007
### pivoxa15
1. The problem statement, all variables and given/known data
If a magnetic field has length 1m and strength 1T. What is its energy contained in it?
3. The attempt at a solution
I have no idea as I only know how to calculate energy from a volume basis.
2. Sep 29, 2007
### chaoseverlasting
$$F=\frac{-dU}{dx}$$, If you solve for U, you get your general expression and plug in the values there.
3. Sep 29, 2007
### Gokul43201
Staff Emeritus
This question looks nonsensical. Did you write it down exactly as it was given to you? Where is this from?
4. Sep 29, 2007
### pivoxa15
The question was about the magnetic prominence which is a magnetic loop. I just simplified the data but its about a magnetic loop and finding how much energy it can contain. | 230 | 838 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2018-17 | longest | en | 0.936883 |
https://www.thenational.academy/pupils/lessons/using-bar-models-to-solve-word-problems-part-2-6cupat/video | 1,716,995,436,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059246.67/warc/CC-MAIN-20240529134803-20240529164803-00277.warc.gz | 896,614,181 | 22,748 | # Lesson video
In progress...
Hi everyone.
Ms. Hill here.
Today we are doing a lesson using bar models to solve word problems. It's going to be a really fun lesson today, so make sure you've got those mathematical hats on, so put on your hats, tighten your ties, and tell the computer "Now I'm a mathematician".
Let's get started.
Here is our lesson agenda for today.
First off we're going to our star words.
We're then going to be looking at a big picture, before having a think and doing some exploring, before moving on to independent task and the quiz.
Before we get started, please make sure you have a pencil, a piece of paper, and a ruler with you.
If you need to pause this video, then please do and get all the bits that you need.
Let's have a look at our star words.
Well done.
So we have our strategy, total, pounds, pence, part, whole, bar model.
Can you show me pounds? Can you show me pence? Can you show your bar model? Great job everyone.
So here we have our big picture.
This is a river in Cambridge, and as you can see there are some students on the punts, and they're pushing people around the canals.
At every university, there is a student union.
And in the student union you can buy some items. So we have rulers which are 42 pence.
We have 27p pencils, rubbers, which are 40 pence, pens are 35 pence, and paperclips that are 8 pence.
This student is from King's College.
He bought a rubber and a ruler.
But how much did he spend? This is an opportunity for us to use our bar models in order to find the total amount of money spent.
Here we have our two parts and we need to add them together to create our whole.
And I'm going to use some of our number knowledge.
If I know that four add four is equal to eight, then I know that 40 add 40 will be equal to 80.
And then we have two ones.
80 add two is equal to 82.
But remember to use your units of measurement, which in this case is pence.
So our student spent 82 pence.
So it's your turn to go shopping now.
Here we have the same items that he might have bought.
The ruler, the pencil, the rubber, the pen, and the paperclip.
And what you need to do is you're going to do some shopping.
Let me show you what I would do.
So I've got my pen and I've got a whiteboard instead of a piece of paper.
I'm going to pick the rubber at 40 p.
I'm also going to pick, hmm what do I need? I need a pencil, which is 27 p.
So here you can see I have got one part, which is 40 pence, and I have a second part, which is 27 pence, and I need to work out the total.
So I'm going to add 40 pence and 27 pence.
And I'm going to use my number knowledge.
If I know that 4 add 2 is equal to 6, then I know that 40 add 20 is equal to 60.
And 60 add 7 pence is equal to 67 pence.
So I spent 67 pence in the shop.
How much are you going to spend? Pause this video before you do some shopping.
Great job.
Let's have a look at some word problems together.
On a trip to Cambridge, Andrew had a 20 pound note to spend in the gift shop.
He bought a T-shirt for seven pounds and a map for eight pounds.
But how much change did he get? So what we need to do is to follow our steps to success.
Step two, identify key information.
Well, my key information is that he has 20 pounds to spend, he's bought a T-shirt worth seven pounds, and a map worth eight pounds.
So I know that the equation I need to do is seven pounds plus eight pounds.
And I've even drawn my bar model.
So we have seven pounds plus eight pounds.
There's a much nicer one there.
And when you add the two together, you get the total of 15 pounds.
However, I haven't finished, because he spent 15 pounds, and I need to work out the change.
So 20 take away 15, we need to find the difference.
The difference is five pounds.
You have three questions.
So first you need to read the question carefully, identify and underline the key information, write the equation and create your bar model, and then you're going to answer it.
Remember to pause this video while completing the task.
Great job everybody.
Let's have a look at these questions together.
Question number one.
He bought a T-shirt for seven pounds, and a key ring for four pounds.
How much change did he get? Well we know our two parts in seven and four, and I know that seven add four is 11 pounds.
But we're not finished there.
I need to remember to work out the change.
20 take away 11 pounds is equal to nine pounds.
Question number two.
She bought a mug for five pounds, a bookmark for three pounds, and a key ring for four pounds.
How much change did she get? So I know I need to add three parts together.
Five add four is equal to nine, add three is equal to 12.
But have I finished there? No I haven't, have I? I still need to work out the change.
And I can use a number bond here.
Because if I know that two add eight is equal to 20, Ooh! No it's not.
Two add eight is equal to ten, then I know 12 add eight is equal to 20.
So 20 take away eight is equal to 12.
After she bought her two items she had three pound change.
One item she bought was a model punt for 11 pounds.
How much did the other item cost? Now this is an interesting one because its worded a bit differently.
So we know she had 20 pounds to spend.
And we know she got three pounds change.
So in total, she spent.
Well 20 take away three is equal to 17.
One item she bought was 11 pounds.
How much did the other item cost? So if the total was 17, and one of the parts was 11, then the other part must be six.
And we can check it, 20 take away 17 is equal to three. | 1,390 | 5,592 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-22 | latest | en | 0.972754 |
https://quizlet.com/explanations/questions/based-on-a-sample-of-50-us-citizens-the-american-film-institute-found-that-a-typical-american-spent-78-hours-watching-movies-last-year-the-s-ae8f84c2-033607df-c4e7-40d0-b3c7-90eae863dbcd | 1,675,629,738,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500288.69/warc/CC-MAIN-20230205193202-20230205223202-00179.warc.gz | 497,944,262 | 44,629 | Question
The American Film Institute discovered that the average American spent 78 hours viewing movies last year based on a study of 50 US people. This sample's standard deviation was nine hours. With the help of this, develop a 95% confidence interval for the population mean number of hours spent watching movies last year.
Solution
Verified
Answered 9 months ago
Answered 9 months ago
Step 1
1 of 4
The confidence interval for the population mean when the population distribution is normal and the population standard deviation is unknown is given by
$\overline{x}\pm t\cdot\dfrac{s}{\sqrt{n}},$
where $\overline{x}$ is the sample mean which is a point estimate of the population mean, the value of $t$ is found in the Student's $t$ Distribution table for the given confidence level and degrees of freedom, $s$ is the sample standard deviation, and $n$ is the number of observations in the sample. | 204 | 908 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 6, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2023-06 | latest | en | 0.899741 |
http://www.sceneadvisor.com/North-Carolina/maximum-error-formula.html | 1,575,769,206,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540503656.42/warc/CC-MAIN-20191207233943-20191208021943-00481.warc.gz | 229,682,848 | 7,563 | Address 811 Washington Square Mall, Washington, NC 27889 (252) 940-4898
# maximum error formula Chocowinity, North Carolina
Be careful to not assume this is a large error. On the surface it looks large, however if we compute the actual volume for we get . So, in comparison the You can click on any equation to get a larger view of the equation. Please do not email asking for the solutions/answers as you won't get them from me. Algebra/Trig Review Common Math Errors Complex Number Primer How To Study Math Close the Menu Current Location : Calculus I (Notes) / Applications of Derivatives / Differentials Calculus I [Notes] [Practice
What can I do to fix this? So, because I can't help everyone who contacts me for help I don't answer any of the emails asking for help. Put Internet Explorer 10 in Compatibility Mode Look to the right side of the address bar at the top of the Internet Explorer window. This will present you with another menu in which you can select the specific page you wish to download pdfs for.
Given a function we call dy and dx differentials and the relationship between them is given by, Note that if we are just given then the differentials are Here are the solutions. Not much to do here other than take a derivative and don’t forget to add on the second differential to the derivative. (a) (b) From Download Page All pdfs available for download can be found on the Download Page. If you are a mobile device (especially a phone) then the equations will appear very small.
Example 1 Compute the differential for each of the following. (a) (b) (c) Solution Before working any of these we should first discuss just what we’re being Show Answer Yes. Put Internet Explorer 11 in Compatibility Mode Look to the right side edge of the Internet Explorer window. How do I download pdf versions of the pages?
I am hoping they update the program in the future to address this. Here's why. Long Answer : No. These often do not suffer from the same problems.
I've found a typo in the material. Let’s compute a couple of differentials. Some of the equations are too small for me to see! My first priority is always to help the students who have paid to be in one of my classes here at Lamar University (that is my job after all!).
Show Answer Short Answer : No. If you want a printable version of a single problem solution all you need to do is click on the "[Solution]" link next to the problem to get the solution to I am attempting to find a way around this but it is a function of the program that I use to convert the source documents to web pages and so I'm You will be presented with a variety of links for pdf files associated with the page you are on.
Also, when I first started this site I did try to help as many as I could and quickly found that for a small group of people I was becoming a Click on this to open the Tools menu. It's kind of hard to find the potential typo if all you write is "The 2 in problem 1 should be a 3" (and yes I've gotten handful of typo reports Those are intended for use by instructors to assign for homework problems if they want to.
Terms of Use - Terms of Use for the site. I would love to be able to help everyone but the reality is that I just don't have the time. Calculus I - Complete book download links Notes File Size : 3.98 MB Last Updated : Friday August 12, 2016 Practice Problems File Size : 563 KB Last Updated : Thursday From Content Page If you are on a particular content page hover/click on the "Downloads" menu item.
FAQ - A few frequently asked questions. Privacy Statement - Privacy statement for the site. Click on this and you have put the browser in Compatibility View for my site and the equations should display properly. Long Answer with Explanation : I'm not trying to be a jerk with the previous two answers but the answer really is "No".
Close the Menu The equations overlap the text! Download Page - This will take you to a page where you can download a pdf version of the content on the site. It is especially true for some exponents and occasionally a "double prime" 2nd derivative notation will look like a "single prime". We can use the fact that in the following way.
Clicking on the larger equation will make it go away. My Students - This is for students who are actually taking a class from me at Lamar University. Unfortunately there were a small number of those as well that were VERY demanding of my time and generally did not understand that I was not going to be available 24 Solution First, recall the equation for the volume of a sphere. Now, if we start with and use then should give us maximum error.
Show Answer Answer/solutions to the assignment problems do not exist. Show Answer If you have found a typo or mistake on a page them please contact me and let me know of the typo/mistake. You can access the Site Map Page from the Misc Links Menu or from the link at the bottom of every page. Also most classes have assignment problems for instructors to assign for homework (answers/solutions to the assignment problems are not given or available on the site).
Linear Approximations Previous Section Next Section Newton's Method Derivatives Previous Chapter Next Chapter Integrals Calculus I (Notes) / Applications of Derivatives / Differentials [Notes] [Practice Problems] [Assignment Problems] Another option for many of the "small" equation issues (mobile or otherwise) is to download the pdf versions of the pages. | 1,197 | 5,515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-51 | latest | en | 0.894677 |
https://www.jiskha.com/display.cgi?id=1310094748 | 1,502,897,819,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102307.32/warc/CC-MAIN-20170816144701-20170816164701-00680.warc.gz | 916,301,679 | 3,619 | # CAL
posted by .
consider the area between the graphs x+3y=1 and x+9=y^2. this area can be computed in two different ways using integrals. Please integral both ways both respect to x and to y. ANd please show your work and thank you
## Similar Questions
1. ### calculus showed work
find the area of the rgion bounded by the graphs of y=x^3-2x and g(x)=-x i drew the graph and half of the graph is above the xaxis and the other half is below the axis. so the integrals i came up with are two because i broke them up …
2. ### MATH 2B Calculus
Consider the area between the graphs x+4y=14 and x+7=y^2. This area can be computed in two different ways using integrals. First of all it can be computed as a sum of two integrals They ask to use two integrals so i put f(x) from -7 …
3. ### Cal
consider the area between the graphs x+3y=1 and x+9=y^2. this area can be computed in two different ways using integrals. First of all it can be computed as a sum of two integrals where a=,b=,c= and f(x0= and g(x)=. Alterntaively this …
4. ### Cal
consider the area between the graphs x+3y=1 and x+9=y^2. this area can be computed in two different ways using integrals. First of all it can be computed as a sum of two integrals where a=,b=,c= and f(x0= and g(x)=. Alterntaively this …
5. ### Calculus
Consider the area between the graphs x+1y=12 and x+8=y2 . This area can be computed in two different ways using integrals First of all it can be computed as a sum of two integrals Interval a to b f(x)= i put 12-x but it's wrong Interval …
6. ### math, calculus 2
Consider the area between the graphs x+y=16 and x+4= (y^2). This area can be computed in two different ways using integrals. First of all it can be computed as a sum of two integrals integrate from a to b of f(x)dx + integrate from …
7. ### Calculus Area between curves
Consider the area between the graphs x+6y=8 and x+8=y2. This area can be computed in two different ways using integrals First of all it can be computed as a sum of two integrals where a= , b=, c= and f(x)= g(x)= I found a, but not … | 582 | 2,065 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2017-34 | latest | en | 0.909443 |
https://curatedsql.com/2015/12/11/goodness-of-fit/ | 1,529,748,800,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864957.2/warc/CC-MAIN-20180623093631-20180623113631-00057.warc.gz | 588,458,719 | 11,475 | # Goodness Of Fit
2015-12-11
Steve Bolton is one of my favorite long-form analytics bloggers, and his ongoing goodness of fit series is a testament as to why.
Goodness-of-fit tests are also sometimes applicable to regression models, which I introduced in posts like A Rickety Stairway to SQL Server Data Mining, Algorithm 2: Linear Regression and A Rickety Stairway to SQL Server Data Mining, Algorithm 4: Logistic Regression. I won’t rehash the explanations here for the sake of brevity; suffice it to say that regressions can be differentiated from probability distributions by looking at them as line charts which point towards the predicted values of one or more variables, whereas distributions are more often represented as histograms representing the full range of a variable’s actual or potential values. I will deal with methods more applicable to regression later in this series, but in this article I’ll explain some simple methods for implementing the more difficult concept of a probability distribution.
As I found out the hard way, the difficult part with implementing these visual aids is not in representing the data in Reporting Services, but in calculating the deceptively short formulas in T-SQL. For P-P Plots, we need to compare two cumulative distribution functions (CDFs). That may be a mouthful, but one that is not particularly difficult to swallow once we understand how to calculate probability distribution functions. PDFs[2] are easily depicted in histograms, where we can plot the probability of the occurrence of each particular value in a distribution from left to right to derive such familiar shapes as the bell curve. Since probabilities in stochastic theory always start at 0 and sum to 1, we can plot them a different way, by summing them in succession for each associated value until we reach that ceiling. Q-Q Plots are a tad more difficult because they involve comparing the inverse of the CDFs, using what is alternately known as quantile or percent point functions[3], but not terribly so. Apparently the raison d’etre for these operations is to distill distributions like the Gaussian down to the uniform distribution, i.e. a flat line in which all outcomes are equally likely, for easier comparison.[4]
The most well-known extension of these somewhat forgotten stats is the Jarque-Bera Test, which only dates back to the 1970s despite being one of earliest examples of normality testing. All of these measures have fallen out of favor with statisticians to some extent, for reasons that will be apparent shortly, but one of the side effects of this is that it is a little more difficult to find variations on them that are more suited to the unique needs of the SQL Server community. One of the strengths of data mining on database servers like SQL Server is that you typically have such an enormous number of records to draw from that you can actually perform calculations on the full population, or a proportion close to it. In ordinary statistics, however, you’re often limited to making inferences based on small samples of just a few dozen or a few hundred rows, out of a much larger population that is often of unknown size; the results can still be logically valid, but often only if other preconditions are met on the data (including normality tests, which are often not performed). For that reason, I usually prefer to leverage SQL Server’s fast set-based retrieval methods to quickly calculate statistics on full populations whenever possible, especially when there are simpler versions of the mathematical formulas available for the full dataset.
Steve doesn’t post very frequently, but if you have a few hours on a lazy Friday, check him out.
## Comparing Keras In Python Versus R
2018-06-20
Dmitry Kisler performs image classification using Keras in both Python and R: From the plots above, one can see that: the accuracy of your model doesn’t depend on the language you use to build and train it (the plot shows only train accuracy, but the model doesn’t have high variance and the bias accuracy is […]
## Auto-Encoders And KernelML
2018-06-20
Rohan Kotwani gives us an example where KernelML might be better than TensorFlow or PyTorch: So what’s the point of using KernelML? 1. The parameters in each layer can be non-linear 2. Each parameter can be sampled from a different random distribution 3. The parameters can be transformed to meet certain constraints 4. Network combinations […]
### 1 Comment
• Steve Bolton on 2015-12-11
Hey Kevin, thanks! This made my day – it’s always good to know there are people out there getting some benefit from my misadventures LOL. 🙂 I’ve gotta finish this stuff on Goodness-of-Fit testing first, but you might like my next two series even more – I’m going to tackle Implementing Fuzzy Sets in SQL Server and Information Measurement with SL Server, two completely neglected topics that are nonetheless insanely useful in DIY data mining. Keep the comments coming – I’m trying to teach myself these technologies through the school of hard knocks, so it’s good to know that I’m making some headway 🙂 Thanks again — Steve | 1,061 | 5,130 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2018-26 | longest | en | 0.942798 |
http://sagemath.org/doc/reference/combinat/sage/combinat/composition_signed.html | 1,394,416,367,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394010547566/warc/CC-MAIN-20140305090907-00018-ip-10-183-142-35.ec2.internal.warc.gz | 158,401,396 | 4,126 | # Signed Compositions¶
class sage.combinat.composition_signed.SignedCompositions(n)
The class of signed compositions of $$n$$.
EXAMPLES:
sage: SC3 = SignedCompositions(3); SC3
Signed compositions of 3
sage: SC3.cardinality()
18
sage: len(SC3.list())
18
sage: SC3.first()
[1, 1, 1]
sage: SC3.last()
[-3]
sage: SC3.random_element()
[1, -1, 1]
sage: SC3.list()
[[1, 1, 1],
[1, 1, -1],
[1, -1, 1],
[1, -1, -1],
[-1, 1, 1],
[-1, 1, -1],
[-1, -1, 1],
[-1, -1, -1],
[1, 2],
[1, -2],
[-1, 2],
[-1, -2],
[2, 1],
[2, -1],
[-2, 1],
[-2, -1],
[3],
[-3]]
TESTS:
sage: SC = SignedCompositions(3)
sage: TestSuite(SC).run()
cardinality()
Return the number of elements in self.
The number of signed compositions of $$n$$ is equal to
$\sum_{i=1}^{n+1} \binom{n-1}{i-1} 2^i$
TESTS:
sage: SC4 = SignedCompositions(4)
sage: SC4.cardinality() == len(SC4.list())
True
sage: SignedCompositions(3).cardinality()
18
Combinations
#### Next topic
Integer compositions | 369 | 956 | {"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": 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} | 2.78125 | 3 | CC-MAIN-2014-10 | longest | en | 0.568512 |
http://www.canarina.com/desaneamientoambiental.htm | 1,542,663,539,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746112.65/warc/CC-MAIN-20181119212731-20181119234731-00452.warc.gz | 394,503,246 | 5,318 | CANARINA:
DISPER:
Algorithmen II · DISPER software
## Flüsse von Impuls und Nachhaltigkeit:
(Briggs, 1975, p. 63):
Fb=gvsds2(DT/4Ts) (7)
DT = Ts - Ta, Ts
Fm=gvs2ds2(Ta/4Ts) (8)
Fb<55,
(DT)c=0.0297 Ts(vs/ds2)1/3 (9)
Fb > 55,
(DT)c=0.00575 Ts(vs2/ds)1/3 (10)
Fb < 55:
xf=49Fb5/8 (11)
Fb > 55:
xf=119Fb2/5 (12)
l(Briggs, 1971, p. 1031), Fb < 55:
he=hs+(21.425 Fb3/4/us) (13)
Fb > 55:
he=hs+(38.71 Fb3/5/us) (14)
(Briggs, 1969, p. 59):
he=hs+3ds(vs/us) (15)
(Briggs, 1971, p. 1031):
s=g[(dT/dz)/Ta] (16)
Briggs (1975, p. 96):
(DT)c=0.019582 Ts vs s1/2 (17)
Briggs, (1975), p. 96:
xf=2.0715 us s-1/2 (18)
(Briggs, 1975, p. 96):
he=hs+2.6 [Fb/(uss)]1/3 (19)
Briggs, (1969), p. 59:
he=hs+1.5[Fm/(uss1/2)]1/3 (20)
Briggs (1972), p. 1030:
he=hs+1.60 [(Fb x2)1/3/us] (21)
(Bowers, et al, 1979)
a) instabile:
he=hs+[3Fmx/(betj2us2)]1/3 (22)
xmax=4ds(vs+3us)/(vsus) para Fb=0 (23)
xmax=49 Fb5/8 para 0 < Fb < 55 m2s3 (24)
xmax=119 Fb2/5 para Fb > 55 m2s3 (25)
b) stabile:
he=hs+(3Fm)1/3{sin[x s1/2/us]}1/3[betj2uss1/2]-1/3 (26)
xmax=0.5 pi us/s1/2 (27)
betj=(1/3)+(us/vs) (28)
castellano: italiano:
français: português:
deutsch: english: | 658 | 1,353 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2018-47 | latest | en | 0.258907 |
https://www.vedantu.com/question-answer/if-fleft-x-right-log-eleft-dfrac1-x1-+-x-class-11-maths-cbse-5fb341698d8dbc5aa191250c | 1,726,428,815,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651647.78/warc/CC-MAIN-20240915184230-20240915214230-00170.warc.gz | 979,006,478 | 28,141 | Courses
Courses for Kids
Free study material
Offline Centres
More
Store
If $f\left( x \right) = {\log _e}\left( {\dfrac{{1 - x}}{{1 + x}}} \right)$,$\left| x \right| < 1$, then$f\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$is equal to:A.$2f\left( x \right)$B.$2f\left( {{x^2}} \right)$C.${\left( {f\left( x \right)} \right)^2}$D.${\left( {f\left( x \right)} \right)^3}$
Last updated date: 15th Sep 2024
Total views: 420.6k
Views today: 5.20k
Verified
420.6k+ views
Hint: Here we will find the value of $f\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$ by substituting the value of $x$ as $\dfrac{{2x}}{{1 + {x^2}}}$ in the function $f\left( x \right)$. Then we will simplify the equation to get the answer in terms of the main function that is $f\left( x \right)$.
Given function is $f\left( x \right) = {\log _e}\left( {\dfrac{{1 - x}}{{1 + x}}} \right)$……………….$\left( 1 \right)$
We will find the value of $f\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$. Therefore, substituting the value of $x$ as $\dfrac{{2x}}{{1 + {x^2}}}$ in the equation$\left( 1 \right)$, we get
$\Rightarrow f\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) = {\log _e}\left( {\dfrac{{1 - \dfrac{{2x}}{{1 + {x^2}}}}}{{1 + \dfrac{{2x}}{{1 + {x^2}}}}}} \right)$
Now, we will solve and simplify the above equation. So, by taking $1 + {x^2}$ common in both the numerator and denominator, we get
$\Rightarrow f\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) = {\log _e}\left( {\dfrac{{\dfrac{{1 + {x^2} - 2x}}{{1 + {x^2}}}}}{{\dfrac{{1 + {x^2} + 2x}}{{1 + {x^2}}}}}} \right) = {\log _e}\left( {\dfrac{{1 + {x^2} - 2x}}{{1 + {x^2} + 2x}}} \right)$
Now, we can clearly see that the numerator and the denominator is the perfect square of $1 - x$ and $1 + x$ respectively. So, we get
$\Rightarrow f\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) = {\log _e}\left( {\dfrac{{{{\left( {1 - x} \right)}^2}}}{{{{\left( {1 + x} \right)}^2}}}} \right) = {\log _e}{\left( {\dfrac{{1 - x}}{{1 + x}}} \right)^2}$
Now as we know this the property of the logarithmic function that $\log {a^b} = b\log a$.
Applying the property of logarithmic function, we get
$\Rightarrow f\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) = 2{\log _e}\left( {\dfrac{{1 - x}}{{1 + x}}} \right)$
Now from the equation $\left( 1 \right)$ we know that ${\log _e}\left( {\dfrac{{1 - x}}{{1 + x}}} \right)$ is equal to $f\left( x \right)$. Therefore, we can write
$\Rightarrow f\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) = 2f\left( x \right)$
Hence, $f\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$ is equal to $2f\left( x \right)$.
So, option A is the correct option.
Note: Here, it is important to rewrite the function whose value is to be found out in such a way that the function changes in terms of the given value. So, that we can easily substitute the values and find the answer. Also to solve this question we need to keep in mind the basic logarithmic properties. Few properties of the logarithmic function is:
(1)$\log a + \log b = \log ab$
(2)$\log a - \log b = \log \dfrac{a}{b}$
(3)$\log {a^b} = b\log a$ | 1,139 | 3,019 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-38 | latest | en | 0.58888 |
https://discourse.julialang.org/t/check-the-sum-of-a-certain-array-index/100552 | 1,702,157,015,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100972.58/warc/CC-MAIN-20231209202131-20231209232131-00426.warc.gz | 233,276,079 | 10,092 | # Check the sum of a certain array index
Hello everyone,
I would like to check the sum of certain indexes in the array. For example, I have the array:
``````a = [0, 0, 0, 1, 0, 1, 0, 1]
b = [1, 2, 3, 5]
``````
I will check the sum of a[1], a[2], a[3], and a[5]. If the sum is = 0 then a is correct. I am looking forward to any help and replies.
The easiest way is probably
``````sum(a[b])
``````
if you want to avoid the extra copy, you could do
``````sum(@view a[b])
``````
2 Likes
Less easy, but benchmarked faster than `@view` method:
``````sum(Base.Fix1(getindex,a),b)
``````
2 Likes
Thank you very much for the replies, @SteffenPL and @Dan . I have tried both of your solutions, and they worked very well. Once again, thank you very much.
P.S.:
I wanted to mark both of the replies as solutions, but I can’t. I am really sorry of it @SteffenPL
2 Likes
@divon Don’t worry, you should mark the most useful answer as the solution (and it’s not a competition )
@Dan Given that @divon is maybe new to Julia, I would maybe recommend
``````x = zero(eltype(a)) # or x = 0 (for integers) or x = 1.0 (floats)
for i in b
x += a[i]
end
``````
which is (usually) equally fast and much easier to read (IMO) than the `Base.Fix1` solution.
5 Likes
but the “good” vector a could also be like this` [-1,2,-1,99,0,3,3,3]`?
Thank you very much for your understanding, @SteffenPL. As you said, I am new to Julia. And it is easy to understand your recommendation.
I don’t know about the “good” vector, @rocco_sprmnt21. I wanted to check if the sum of some positions in the array was equal to 0.
I used the term “good” because I didn’t remember the one you used: “correct”.
The meaning of my question is if, given that the example of a=[…] that you proposed has only ‘0’ and ‘1’, the sum must be ‘0’ or all the values must be ‘0’ which is a tighter condition and easier to test, possibly.
Okay, I get what you mean, @rocco_sprmnt21. Actually, in my simulation, I will have:
``````a = [1, 0, 0, 1, 0, 1, 0, 1]
b = [1, 2, 3, 5]
z = [0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 1; 0 0 0 0 0 0 1 0; 0 0 0 0 0 0 1 1; ...; 1 1 1 1 1 1 1 1]
``````
I will do an XOR between a and z, then check that the sum of the results of the XOR is equal to 0. If it is, then I will store which z is the culprit (I hope that you understand).
I would try
``````sum(a[i] for i in b)
``````
(That’s a response to the original question, not the latest. I don’t understand the latest question, unfortunately.)
2 Likes
Thank you very much for your reply, @DNF. I have tried it in my simulation, and it returns the same result as other solutions. Also, your solution gives me a new perspective on the for loop.
No, I’m sorry, but I don’t understand what you’re doing and how you transform the data.
If you give a complete example, maybe it’s easier for me to understand
In any case if a contains only 1 and 0 (or even boolean or even only positive numbers ) you could also use a different solution than sum
``````julia> @btime sum(a[i] for i in b)
171.777 ns (1 allocation: 16 bytes)
1
julia> @btime all(i->a[i]==0,b)
120.948 ns (0 allocations: 0 bytes)
false
julia> @btime findfirst(i->a[i]!=0,b) != nothing
119.912 ns (0 allocations: 0 bytes)
true
``````
EDIT: ignore the code below, I forgot that you had to index one array with the values of the other. The best notation is the anonymous function one because you want to a method that takes a function and avoids creating an intermediary array.
This can be more succinctly written as `all(iszero, b)`.
This can be written as `!isnothing(findfirst(!iszero, b))` or `!isnothing(findfirst(>(0), b))` (if only positive numbers are allowed).
You can also replace any `iszero` by `==(0)` or `!iszero` by `!=(0)` but I find those much less legible.
b contains the indexes not the values to check.
In any case it is useful to know the existence of these functions.
1 Like
I am sorry for not being detailed, I will try to explain it to you @rocco_sprmnt21.
In this task, the array is:
``````a = [1, 0, 0, 1, 0, 1, 0, 1]
``````
Then I need to XOR array a with each of the arrays below:
``````z = [0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 1; 0 0 0 0 0 0 1 0; 0 0 0 0 0 0 1 1;
0 0 0 0 0 1 0 0; 0 0 0 0 0 1 0 1; 0 0 0 0 0 1 1 0; 0 0 0 0 0 1 1 1;
0 0 0 0 1 0 0 0; 0 0 0 0 1 0 0 1; 0 0 0 0 1 0 1 0; 0 0 0 0 1 0 1 1;
0 0 0 0 1 1 0 0; 0 0 0 0 1 1 0 1; 0 0 0 0 1 1 1 0; 0 0 0 0 1 1 1 1;
0 0 0 1 0 0 0 0; 0 0 0 1 0 0 0 1; 0 0 0 1 0 0 1 0; 0 0 0 1 0 0 1 1;
0 0 0 1 0 1 0 0; 0 0 0 1 0 1 0 1; 0 0 0 1 0 1 1 0; 0 0 0 1 0 1 1 1;
0 0 0 1 1 0 0 0; 0 0 0 1 1 0 0 1; 0 0 0 1 1 0 1 0; 0 0 0 1 1 0 1 1;
0 0 0 1 1 1 0 0; 0 0 0 1 1 1 0 1; 0 0 0 1 1 1 1 0; 0 0 0 1 1 1 1 1;
0 0 1 0 0 0 0 0; 0 0 1 0 0 0 0 1; 0 0 1 0 0 0 1 0; 0 0 1 0 0 0 1 1;
0 0 1 0 0 1 0 0; 0 0 1 0 0 1 0 1; 0 0 1 0 0 1 1 0; 0 0 1 0 0 1 1 1;
0 0 1 0 1 0 0 0; 0 0 1 0 1 0 0 1; 0 0 1 0 1 0 1 0; 0 0 1 0 1 0 1 1;
0 0 1 0 1 1 0 0; 0 0 1 0 1 1 0 1; 0 0 1 0 1 1 1 0; 0 0 1 0 1 1 1 1;
0 0 1 1 0 0 0 0; 0 0 1 1 0 0 0 1; 0 0 1 1 0 0 1 0; 0 0 1 1 0 0 1 1;
0 0 1 1 0 1 0 0; 0 0 1 1 0 1 0 1; 0 0 1 1 0 1 1 0; 0 0 1 1 0 1 1 1;
0 0 1 1 1 0 0 0; 0 0 1 1 1 0 0 1; 0 0 1 1 1 0 1 0; 0 0 1 1 1 0 1 1;
0 0 1 1 1 1 0 0; 0 0 1 1 1 1 0 1; 0 0 1 1 1 1 1 0; 0 0 1 1 1 1 1 1;
0 1 0 0 0 0 0 0; 0 1 0 0 0 0 0 1; 0 1 0 0 0 0 1 0; 0 1 0 0 0 0 1 1;
0 1 0 0 0 1 0 0; 0 1 0 0 0 1 0 1; 0 1 0 0 0 1 1 0; 0 1 0 0 0 1 1 1;
0 1 0 0 1 0 0 0; 0 1 0 0 1 0 0 1; 0 1 0 0 1 0 1 0; 0 1 0 0 1 0 1 1;
0 1 0 0 1 1 0 0; 0 1 0 0 1 1 0 1; 0 1 0 0 1 1 1 0; 0 1 0 0 1 1 1 1;
0 1 0 1 0 0 0 0; 0 1 0 1 0 0 0 1; 0 1 0 1 0 0 1 0; 0 1 0 1 0 0 1 1;
0 1 0 1 0 1 0 0; 0 1 0 1 0 1 0 1; 0 1 0 1 0 1 1 0; 0 1 0 1 0 1 1 1;
0 1 0 1 1 0 0 0; 0 1 0 1 1 0 0 1; 0 1 0 1 1 0 1 0; 0 1 0 1 1 0 1 1;
0 1 0 1 1 1 0 0; 0 1 0 1 1 1 0 1; 0 1 0 1 1 1 1 0; 0 1 0 1 1 1 1 1;
0 1 1 0 0 0 0 0; 0 1 1 0 0 0 0 1; 0 1 1 0 0 0 1 0; 0 1 1 0 0 0 1 1;
0 1 1 0 0 1 0 0; 0 1 1 0 0 1 0 1; 0 1 1 0 0 1 1 0; 0 1 1 0 0 1 1 1;
0 1 1 0 1 0 0 0; 0 1 1 0 1 0 0 1; 0 1 1 0 1 0 1 0; 0 1 1 0 1 0 1 1;
0 1 1 0 1 1 0 0; 0 1 1 0 1 1 0 1; 0 1 1 0 1 1 1 0; 0 1 1 0 1 1 1 1;
0 1 1 1 0 0 0 0; 0 1 1 1 0 0 0 1; 0 1 1 1 0 0 1 0; 0 1 1 1 0 0 1 1;
0 1 1 1 0 1 0 0; 0 1 1 1 0 1 0 1; 0 1 1 1 0 1 1 0; 0 1 1 1 0 1 1 1;
0 1 1 1 1 0 0 0; 0 1 1 1 1 0 0 1; 0 1 1 1 1 0 1 0; 0 1 1 1 1 0 1 1;
0 1 1 1 1 1 0 0; 0 1 1 1 1 1 0 1; 0 1 1 1 1 1 1 0; 0 1 1 1 1 1 1 1;
1 0 0 0 0 0 0 0; 1 0 0 0 0 0 0 1; 1 0 0 0 0 0 1 0; 1 0 0 0 0 0 1 1;
1 0 0 0 0 1 0 0; 1 0 0 0 0 1 0 1; 1 0 0 0 0 1 1 0; 1 0 0 0 0 1 1 1;
1 0 0 0 1 0 0 0; 1 0 0 0 1 0 0 1; 1 0 0 0 1 0 1 0; 1 0 0 0 1 0 1 1;
1 0 0 0 1 1 0 0; 1 0 0 0 1 1 0 1; 1 0 0 0 1 1 1 0; 1 0 0 0 1 1 1 1;
1 0 0 1 0 0 0 0; 1 0 0 1 0 0 0 1; 1 0 0 1 0 0 1 0; 1 0 0 1 0 0 1 1;
1 0 0 1 0 1 0 0; 1 0 0 1 0 1 0 1; 1 0 0 1 0 1 1 0; 1 0 0 1 0 1 1 1;
1 0 0 1 1 0 0 0; 1 0 0 1 1 0 0 1; 1 0 0 1 1 0 1 0; 1 0 0 1 1 0 1 1;
1 0 0 1 1 1 0 0; 1 0 0 1 1 1 0 1; 1 0 0 1 1 1 1 0; 1 0 0 1 1 1 1 1;
1 0 1 0 0 0 0 0; 1 0 1 0 0 0 0 1; 1 0 1 0 0 0 1 0; 1 0 1 0 0 0 1 1;
1 0 1 0 0 1 0 0; 1 0 1 0 0 1 0 1; 1 0 1 0 0 1 1 0; 1 0 1 0 0 1 1 1;
1 0 1 0 1 0 0 0; 1 0 1 0 1 0 0 1; 1 0 1 0 1 0 1 0; 1 0 1 0 1 0 1 1;
1 0 1 0 1 1 0 0; 1 0 1 0 1 1 0 1; 1 0 1 0 1 1 1 0; 1 0 1 0 1 1 1 1;
1 0 1 1 0 0 0 0; 1 0 1 1 0 0 0 1; 1 0 1 1 0 0 1 0; 1 0 1 1 0 0 1 1;
1 0 1 1 0 1 0 0; 1 0 1 1 0 1 0 1; 1 0 1 1 0 1 1 0; 1 0 1 1 0 1 1 1;
1 0 1 1 1 0 0 0; 1 0 1 1 1 0 0 1; 1 0 1 1 1 0 1 0; 1 0 1 1 1 0 1 1;
1 0 1 1 1 1 0 0; 1 0 1 1 1 1 0 1; 1 0 1 1 1 1 1 0; 1 0 1 1 1 1 1 1;
1 1 0 0 0 0 0 0; 1 1 0 0 0 0 0 1; 1 1 0 0 0 0 1 0; 1 1 0 0 0 0 1 1;
1 1 0 0 0 1 0 0; 1 1 0 0 0 1 0 1; 1 1 0 0 0 1 1 0; 1 1 0 0 0 1 1 1;
1 1 0 0 1 0 0 0; 1 1 0 0 1 0 0 1; 1 1 0 0 1 0 1 0; 1 1 0 0 1 0 1 1;
1 1 0 0 1 1 0 0; 1 1 0 0 1 1 0 1; 1 1 0 0 1 1 1 0; 1 1 0 0 1 1 1 1;
1 1 0 1 0 0 0 0; 1 1 0 1 0 0 0 1; 1 1 0 1 0 0 1 0; 1 1 0 1 0 0 1 1;
1 1 0 1 0 1 0 0; 1 1 0 1 0 1 0 1; 1 1 0 1 0 1 1 0; 1 1 0 1 0 1 1 1;
1 1 0 1 1 0 0 0; 1 1 0 1 1 0 0 1; 1 1 0 1 1 0 1 0; 1 1 0 1 1 0 1 1;
1 1 0 1 1 1 0 0; 1 1 0 1 1 1 0 1; 1 1 0 1 1 1 1 0; 1 1 0 1 1 1 1 1;
1 1 1 0 0 0 0 0; 1 1 1 0 0 0 0 1; 1 1 1 0 0 0 1 0; 1 1 1 0 0 0 1 1;
1 1 1 0 0 1 0 0; 1 1 1 0 0 1 0 1; 1 1 1 0 0 1 1 0; 1 1 1 0 0 1 1 1;
1 1 1 0 1 0 0 0; 1 1 1 0 1 0 0 1; 1 1 1 0 1 0 1 0; 1 1 1 0 1 0 1 1;
1 1 1 0 1 1 0 0; 1 1 1 0 1 1 0 1; 1 1 1 0 1 1 1 0; 1 1 1 0 1 1 1 1;
1 1 1 1 0 0 0 0; 1 1 1 1 0 0 0 1; 1 1 1 1 0 0 1 0; 1 1 1 1 0 0 1 1;
1 1 1 1 0 1 0 0; 1 1 1 1 0 1 0 1; 1 1 1 1 0 1 1 0; 1 1 1 1 0 1 1 1;
1 1 1 1 1 0 0 0; 1 1 1 1 1 0 0 1; 1 1 1 1 1 0 1 0; 1 1 1 1 1 0 1 1;
1 1 1 1 1 1 0 0; 1 1 1 1 1 1 0 1; 1 1 1 1 1 1 1 0; 1 1 1 1 1 1 1 1;
``````
The XOR operation is using:
``````c = zeros(8)
for i in 1 : 2^8
for j in 1 : 8
c[j] = xor(a[j], z[i, j]
end
end
``````
Using one of the solutions above, I will check the sum of array c
``````res = []
b = [1, 2, 3, 5]
sum_c = sum(Base.Fix1(getindex,a), b)
if sum_c == 0
res = c
end
``````
And use the array that has sum = 0 as the result.
``````z = [1, 0, 0, 0, 0, 0, 0, 0]
c = [0, 0, 0, 1, 0, 1, 0, 1]
``````
The operation is actually done in the while loop, and when sum_c = 0, the loop will stop and give the result.
(I hope you understand now because it is really hard to create array z)
If this is the actual code, it could be greatly simplified using the following function. You don’t really need to create and store the array `z` before hand. Create the binary representation vector of each byte one by one until your criterion is satisfied. This will be simple and fast because you will return as early as possible.
``````function test_xor_sum(a, b)
for i in 0:255
all(a[j] == z[j] for j in b) && return z
end
end
a = [1, 0, 0, 1, 0, 1, 0, 1]
b = [1, 2, 3, 5]
@btime test_xor_sum(\$a, \$b) # 363.462 ns (2 allocations: 256 bytes)
``````
1 Like
Thank you very much for the solution, @Seif_Shebl. I have tried it and it worked well because, in my previous method (where I needed to generate array z), Julia returned “Out of Memory Error” when I tried to generate an array with length 2^32.
This could even be made faster if you like `Iterators`. Immutable containers like Tuples allocate zero memory, that’s why it’s faster.
``````function test_xor_sum(a, b)
for z in Iterators.product(ntuple(_->0:1,8)...)
all(a[j] == z[j] for j in b) && return z
end
end
a = (1, 0, 0, 1, 0, 1, 0, 1)
b = (1, 2, 3, 5)
@btime test_xor_sum(\$a, \$b) # 140.418 ns (0 allocations: 0 bytes)
``````
Another way, is to forego all the iterations and consider what the solution should be:
``````julia> a = (1, 0, 0, 1, 0, 1, 0, 1)
(1, 0, 0, 1, 0, 1, 0, 1)
julia> b = (1, 2, 3, 5)
(1, 2, 3, 5)
julia> function test_xor_sum2(a, b)
ntuple(i -> i ∈ b ? a[i] : 0, length(a))
end
test_xor_sum2 (generic function with 1 method)
julia> test_xor_sum2(a,b)
(1, 0, 0, 0, 0, 0, 0, 0)
julia> @btime test_xor_sum2(\$a,\$b);
19.748 ns (0 allocations: 0 bytes)
``````
This is at least a solution, there can be an issue of which solution is returned for which iteration order.
I still don’t understand the context.
The `xor` of `a` with `z` reproduces all `z` in random order (that is, the bit octets of the numbers from ‘0’ to ‘255’).
Associating `b` to the number 232 = 2^7+2^6+2^5+2^3 a number which certainly has the bits at ‘0’ in the positions (1,2,3,5) from left to rigth is 255-232=23. | 6,561 | 11,166 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2023-50 | latest | en | 0.952008 |
https://www.reddit.com/r/statisticalconsulting/ | 1,534,409,315,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210559.6/warc/CC-MAIN-20180816074040-20180816094040-00594.warc.gz | 985,571,128 | 46,406 | Press J to jump to the feed. Press question mark to learn the rest of the keyboard shortcuts
View
1
Posted by11 days ago
We conducted an experiment with 5 treatments, and attached are photos of the statistical analysis of our results via Duncan’s Multiple Range Test. We need some help in interpreting the data so we can come up with a conclusion of whether accepting our hypothesis or not.
What we cannot entirely understand are what does the ranking at probability level .01 and .05 mean and what does it really mean two treatments share the same letter.
In addition, our project is producing a natural fungicide and the treatments have different concentrations of it. T4 is made with purely of the fungicide alone and T5 is the commercially available fungicide (positive control). T1, T2, and T3 are concentrations of the fungicide mixed with different amounts of distilled water.
1
Posted by5 months ago
In Probability Of Success, a statistical consultant use a variety of statistical science tools to solve various practical business problems. An online Statistics Tutor also helps project managers and architects to get financial support from the board for various projects by using statistical facts capable of proving their point.
1
Posted by2 years ago
Archived
I'm currently an undergraduate psychology student here in the Philippines and for our thesis we made an intervention program for a public high school here. We're having trouble finding the correct statistical treatment for our results.
I'm not sure if I'm in the correct subreddit but I thought I would try my luck. Here are the details of our study
We have 1 independent variable (the intervention program), and 2 dependent variables (happiness and psychological well-being).
We gathered our respondents through purposive sampling, the sample is not normally distributed, we used a pretest and post-test design, and it was a within groups design with one group.
Do you guys know what statistical treatment we should use that would measure the effect of our IV to our two DV's simultaneously?
Any help would be appreciated!
Community Details
24
Subscribers
4
Online
Discussion, articles, and other links of interest for the practicing applied statistician.
Moderators | 442 | 2,257 | {"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} | 2.921875 | 3 | CC-MAIN-2018-34 | latest | en | 0.947275 |
https://www.algomentor.in/2021/03/recursion-in-java.html | 1,701,977,727,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100686.78/warc/CC-MAIN-20231207185656-20231207215656-00435.warc.gz | 723,903,070 | 58,638 | ### Ticker
20/recent/ticker-posts
# Recursion in java
## Recursion
Recursion is the process of repeating items in a self-similar way. If a function has a function call to itself then it is said to be a recursive function.
### How recursion can be used effectively?
Recursion can be useful if sub-problems of a problem are repetitive and can be useful if the problem has many branches and it's hard for an iterative approach.
### Stack overflow error in recursion
Every recursive approach has a base case and recursive steps with backtracking to it. If a function doesn't reach the base case stack overflow problem may arise.
### Direct and Indirect recursion
Direct recursion is approaching when the same function calls itself. This is like one step recursion the function makes a call within its own body.
Ex:
class App {
public static void main(String[] args) {
// calling indirectRecursion1()
}
static int directRecursion(int n) {
// function calls indirectRecursion1
// indirectRecursion1(n);
// some code
}
}
Indirect recursion is approach when the function calls another function recursively.
class App {
public static void main(String[] args) {
// calling indirectRecursion1()
}
static int indirectRecursion1(int n) {
// function calls indirectRecursion2
// indirectRecursion1(n);
// some code
}
static int indirectRecursion2(int n) {
// function calls indirectRecursion1
// indirectRecursion1(n);
// some code
}
}
### What is tail recursion?
A recursive function is tail recursive when recursive call is the last thing executed by the function.
class App {
public static void main(String[] args) {
tailRecursion(2);
}
static void tailRecursion(int n) {
if (n < 0) {
return;
}
System.out.println(n);
tailRecursion(n - 1);
}
}
### Non-tail recursion
A recursive function is tail recursive when recursive call is executed by the function at the beginning itself.
class App {
public static void main(String[] args) {
nonTailRecursion(2);
}
static void nonTailRecursion(int n) {
if (n < 0) {
return;
}
nonTailRecursion(n - 1);
System.out.println(n);
}
}
### How memory is allocated to recursion?
When any function is called from the main function, the memory is allocated to it on the stack. A recursive function calls itself, the memory for a called function is allocated on top of memory allocated to calling function and different copy of local variables is created for each function call. When the base case is reached, the function returns its value to the function by whom it is called and memory is de-allocated and the process continues.
There are a lot of recursion examples but this example is considered as a simple one with just printing the variable given a number of times.
Ex:
class App {
public static void main(String[] args) {
Recursion(2);
//The value 2 is passed to recursive function
}
static void Recursion(int n) {
//In every recursion call it checks if n is less than zero and acts as a baser condition.
if (n < 0) {
return;
}
// prints the current value of n
System.out.println(n);
//n is further decremented by one
// It again calls the recursive function with the decremented value.
Recursion(n - 1);
}
}
I hope you guys liked this article and here are some articles you may like: | 822 | 3,243 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-50 | latest | en | 0.55613 |
https://askfilo.com/math-question-answers/the-length-of-a-tangent-from-a-point-mathrma-at-a-distance-5-mathrm~cm-from-the | 1,695,873,120,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510358.68/warc/CC-MAIN-20230928031105-20230928061105-00379.warc.gz | 138,228,214 | 21,232 | World's only instant tutoring platform
Question
Easy
Solving time: 2 mins
# The length of a tangent from a point at a distance from the centre of the circle is 4 . The radius of the circle is
A
B
C
D
## Text solution
Let AT be the tangent drawn from point A to a circle with center
and and . Since tangent at a point is
perpendicular to the radius through the point of contact
from right-angled
radius of the circle .
Was this solution helpful?
111
Share
Report
Stuck on the question or explanation?
Connect with our Mathematics tutors online and get step by step solution of this question.
231 students are taking LIVE classes
Question Text The length of a tangent from a point at a distance from the centre of the circle is 4 . The radius of the circle is Topic Circles Subject Mathematics Class Class 9 Answer Type Text solution:1 Upvotes 111 | 197 | 859 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-40 | latest | en | 0.852049 |
https://www.tec-science.com/category/thermodynamics/heat/ | 1,611,605,437,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703644033.96/warc/CC-MAIN-20210125185643-20210125215643-00638.warc.gz | 989,467,032 | 29,061 | # Heat
### Calorimeter to determine the specific heat capacities of liquids
0
Calorimetry deals with the measurement of heat energy.These measurements are based on temperature changes, which are used to determine the amount of heat involved. In calorimeters, for example, the specific heat...
0
### Specific heat capacity of water
0
The specific heat capacity is not a material constant for a substance, but depends on the temperature and above all on the state of matter. The dependence of the specific heat...
### Specific heat capacity of selected substances
0
In this article, learn more about the specific heat capacity of different materials and how it affects the change in temperature over time during a heat transfer. Definition of the specific heat...
### Important remarks on the specific heat capacity
0
Definition of the specific heat capacity The specific heat capacity (c) describes the relationship between a transfer of heat (Q) and the associated temperature change (Delta T) of a substance of mass...
### Specific heat capacity (derivation and definition)
0
The specific heat capacity indicates how much heat must be absorbed by a substance of mass 1 kg in order to increase its temperature by 1 K (1 °C). Learn more...
### Gulf Stream & global ocean conveyor belt
0
The Gulf Stream is an ocean current in the Atlantic Ocean which, as part of the earth's global conveyor belt, has a decisive influence on the climate in Northern and Western...
### Difference between thermal conductivity, diffusivity, transmittance, resistance and heat transfer coefficient
0
Learn more in this article about the differences and importance of thermal conductivity, thermal diffusivity, heat transfer coefficient, thermal transmittance and thermal resistance, etc. Thermal conductivity Thermal conductivity (lambda) describes the heat transfer...
### Why are radiators usually located under a window?
0
Central heating systems use the principle of thermal convection. The water is heated by a central heater and then transferred to the radiators by means of pumps. It is thus a...
0
### Thermal and concentration boundary layer
0
In addition to the hydrodynamic boundary layer, the thermal boundary layer and the concentration boundary layer also have a decisive influence on the entire heat and mass transport in a flow....
0
0
### Calculation of the Nusselt numbers for forced flows over plates and in pipes
0
In this article you will find formulas for calculating the local and average Nusselt numbers for forced flows over plates and in pipes with circular cross sections. This article provides answers to...
0
0
0
### Temperature profiles and heat flows through different geometries
0
In this article we discuss temperature curves and heat flows through a plane wall, through a cylindrical pipe and through a hollow sphere. This article provides answers to the following questions, among...
0
### Laser-Flash method for determining thermal conductivity (LFA)
0
With the Laser-Flash method (Laser Flash Analyser, LFA), the thermal conductivity is determined by the temperature rise in a test sample that is heated by a short laser pulse from one...
### Transient-Hot-Wire method method for determining thermal conductivity (THW)
0
With the Transient-Hot-Wire method (THW), the thermal conductivity is determined by the change in temperature over time at a certain distance from a heating wire. Learn more about the experimental determination...
0
### Guarded-Hot-Plate method for determining thermal conductivity (GHP)
0
With the Guarded-Hot-Plate method (GHP) the thermal conductivity is determined by the electrical power output of a hot plate with guided heat conduction. Learn more about this measurement method in this...
0
### Thermal conductivity of gases
0
The thermal conductivity of ideal gases is not dependent on pressure for gases that are not too strongly diluted. This is no longer the case for gases with low pressure. Learn...
0
### Thermal transmittance (U-value)
0
The thermal transmittance (U-value or U-factor) describes the heat transfer through a solid object, which is located between two fluids (gas or liquid) with different temperatures. Learn more about it in...
### How does a thermos work? Design of a vacuum flask!
0
The reason why hot tea or coffee stays warm for so long in a thermos is the special design of the flask. This design limits the different mechanisms of heat transfer...
0
0
### Heat transfer by thermal conduction
0
Heat transfer by thermal conduction means that heat is conducted through a material. Heat energy is transferred from molecule to molecule at the atomic level. More information on this topic can...
### Rate of heat flow: Definition and direction
0
The rate of heat flow refers to the heat energy transferred per unit of time ("heat output"). The drive for the heat flow is a temperature difference. Learn more about the...
0
### Heat and thermodynamic equilibrium
0
In thermodynamics, heat is the transport of energy due to a temperature difference. Heat in this respect is never "contained" in an object! Learn more about the thermodynamic concept of heat...
### Heat transfer (heat transport)
0
Heat transfer is the transport of thermal energy from a warmer object to a cooler object. A distinction is made between thermal convection, thermal conduction and thermal radiation. You can read...
### Why does metal feel colder than wood (human thermal response)
0
Find out in this article why metal feels colder than wood of the same temperature, while at higher temperatures the metal suddenly feels warmer than wood. The property of an object to... | 1,141 | 5,675 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-04 | longest | en | 0.874274 |
https://java4school.com/isc-computer-science-sample-paper-1 | 1,600,961,569,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400219221.53/warc/CC-MAIN-20200924132241-20200924162241-00686.warc.gz | 454,655,766 | 32,090 | Answer all questions in Part I (compulsory) and seven questions from Part-II, choosing two questions from Section-A, two from Section-B and two from Section-C.
All working, including rough work, should be done on the same sheet as the rest of the answer.
The intended marks for questions or parts of questions are given in brackets [ ].
———————————————————————————————————————–
PART I
Question1:
1. Convert the following sum of product expression into its corresponding POS [1]
F(A,B,C)=A’.B’.C’+A’.B’.C+A’.B.C+A.B’.C
1. Show that AB+(AC)’+AB’C(AB+C) is always 1. [1]
2. Simplify expression AB’CD’+AB’CD+ABCD’+ABCD [1]
3. Convert the following into canonical SOP form (X’Y)’.(X’Z’)’ [1]
4. Find the complement of,F(A,B,C,D)=(A+((B+C).(B’+D’))) [1]
Question 2:
1. Draw the logic gate diagram and truth table for XOR gate. [2]
2. Name the Universal gates ,Explain with truth table And Diagram [2]
3. Prove that (x.y)’=x’+y’ using truth table [2]
4. State the two idempotent laws of Boolean algebra. Verify any one. [2]
5. Draw the logic gate diagram for the following function using NOR gates only. [2]
F(A,B,C)=A’B+AB’+C
Question 3 [5]
The following function comb() and combi() are a part of some class. Give the output when the combi() is called.
Show the dry run/working.
void combi()
{
for(int i=0; i<5; i++)
{
for(int j=0;j<i+1; j++)
System.out.print(“ “+comb(i,j));
System.out.println();
}
}
long comb(int n, int k)
{
long c=1;
for(int i=1;i<=k;i++,n–)
c=c*n/i;
return c;
}
PART II
Attempt any two questions from Section-A, two from Section-B and two from Section-C
Section A
Question 4
(a) Given the Boolean function: [5]
F ( P, Q, R, S ) = ∑ ( 0, 1, 3, 4, 5, 6, 7, 9, 10, 11, 13, 15 )
Use Karnaugh’s map to reduce the function F, using the SOP form. Draw a logic gate diagram for the reduced SOP form. You may use gates with more than two inputs. Assume that the variable and their complements are available as inputs.
(b) Given G(A,B,C,D)= (A+B+C+D) (A+B+C+D’) (A+B+C’+D’) (A+B+C’+D) (A+B’+C+D’) (A+B’+C’+D’) (A’+B+C+D) (A’+B+C’+D)
1. i) Reduce the above expression by using 4 variable Karnaugh map, showing the various groups (i.e.octal, quads and pairs). [4]
2. ii) Draw the logic gate diagram for the reduced expression. Assume that the variables and their complements are available as inputs. [1]
Question 5.
• What is Full adder? Write the SOP expressions for sum and carry. Simplify the expressions algebraically using Boolean algebra. [5]
• Draw the truth table and logic circuit diagram for a Hexadecimal to Binary encoder. [5]
Question 6.
• Write the equivalent Boolean expression for the propositions F=X→Y and F=X↔Y. [2]
• Reduce the following Boolean expressions to the simplest form: [3]
A.[B+C.(A.B+A.C)’]
(c) Draw the truth table and logic circuit diagram for a 3×8 decoder. What is the difference between a
multiplexer and a decoder? [5]
Section B
Each program should be written in such a way that it clearly depicts the logic of the problem. This can be achieved by using mnemonic names and comments in the program. (Flowcharts and algorithms are not required)
The programs must be written in Java.
Question 7. [10]
In a class VowelWord to accept sentence and calculate the frequency of words that begin with a vowel. The words in the input string are separated by a single blank space and terminated by a full stop. The description of the class is given below:
Class name VowelWord
Data members
str to store the sentence
freq store the frequency of the words beginning with a vowel
Member functions
VowelWord() constructor to initialize data members to legal initial value.
void readstr() to accept a sentence
void freqvowel() counts the frequency of the words that begin with a vowel
void display() to display the original string and the frequency of the words
that begin with a vowel.
Specify the class VowelWord giving details of the constructors and all the functions. Also write the main function
Question 8. [10]
A class RecFact defines a recursive function to find the factorial of a number. The details of the class are given below: Class name : RecFact Data members/instance variables n : stores the number whose factorial is required r : stores an integer Member functions RecFact( ) : default constructor void readnum( ) : to enter values for ‘n’ and ‘r’ int factorial(int) : returns the factorial of the number using the recursive technique. void factseries() : to calculate and display the value of n!+(r-2)! ( n − r)!
Specify the class RecFact giving the details of the constructor and member functions void readnum( ), int factorial(int) and void factseries( ). Also define the main function to create an object and call methods accordingly to enable the task.
Question 9. [10]
A perfect square is an integer which is the square of another integer. For example, 4, 9,16 are perfect squares. Design a Class Perfect with the following description:
Class name Perfect Data members/instance variables N stores an integer number Member functions: Perfect( ) default constructor Perfect(int) parameterized constructor to assign a value to ‘n’. void perfect_sq() to display the first 5 perfect squares larger than ‘n’ (if n = 15, the next 3 perfect squares are 16, 25, 36) void sum_of() to display all combinations of consecutive integers whose sum is equal to n. ( the number n = 15 can be expressed as 1 2 3 4 5 4 5 6 7 8
Specify the class Perfect giving details of the constructors, void perfect_sq( ) and void sum_of(). Also define the main function to create an object and call methods accordingly to enable the task.
Section C
Each program should be written in such a way that it clearly depicts the logic of the problem step wise.This can also be achieved by using comments in the program and mnemonic names or pseudo codes for algorithms. The program must be written in Java and the algorithms must be written in general/standard form, wherever required/ specified. (Flowcharts are not required.)
Question 10 [5]
A class OctaDeci has been defined to convert an Octal number into its equivalent Decimal number. Some of the members of the class are given below:
Data members :
dec : integer to store decimal number.
N : integer octal to be converted to its decimal form.
Member functions :
void getOctal() : to accept Octal integer N.
void recursiveDec(int): to find the decimal equivalent of the octal number stored in N and store
in dec using Recursive Technique.
void putdata() : to display the octal number ‘N’ and its decimal equivalent.
Specify the class OctaDeci giving details of constructor and functions. You do not need to write the main function.
Question 11. [5]
A class Revstr defines a recursive function to reverse a string and check whether it is a palindrome. The details of the class are given below:
Class name : Revstr
Data members Str : stores the string.
Revst : stores the reverse of the string.
Member functions void getStr() : to accept the string.
void recReverse(int) : to reverse the string using recursive technique.
void check() : to display the original string, its reverse and whether the string is a palindrome or not.
Specify the class Revstr giving details of the functions void getStr(), void recReverse(int) and void check(). The main function need not be written.
Question 12. [5]
Draw a truth table with a three input combination which outputs 1 if there are odd number of 0’s. Also derive an SOP expression using K map. | 2,139 | 9,748 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2020-40 | longest | en | 0.631476 |
https://webdva.github.io/a-return-to-binary-search-trees/ | 1,603,936,547,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107902683.56/warc/CC-MAIN-20201029010437-20201029040437-00133.warc.gz | 585,889,773 | 4,874 | # A return to binary search trees
Having made progress in the once obscure data structure, I’ve decided to share my experience with binary search trees this week.
I developed an algorithm that finds, for a binary search tree, a specified node’s most adjacent neighbor. I believe that it could have implications in efficiently finding cheap alternatives to missing resources in some large scale infrastructure.
This is what the algorithm looks like in simple or general terms:
1. Obtain a list of all the binary search tree’s nodes in ascending order.
2. Maintain a list of all the distances between each node and the specified node.
3. Determine the most adjacent node as the one with the smallest distance between it and the specified node.
I decided to use Python instead of the half-quirky Javascript. So, I found a nice Python module called `binarytree` to help me in this binary search tree endeavor.
To install that sole dependency, I simply used the Python package installer named pip in a terminal.
``````pip install binarytree
``````
Then, I was able to use the library to aid me in my efforts to create the algorithm with just this as the sole import of the Python file
``````from binarytree import bst
``````
The next step involved designing the function that encapsulates the algorithm. I decided to name the function `findNearestNeighborNode` and make it take two parameters, `rootNode` and `specifiedNodeValue`.
`rootNode` must be defined as a binary search tree created with the `binarytree` module. And `specifiedNodeValue` would be an integer value that specifies which node to find an adjacent neighbor for.
The function would then return an integer value that points to the target node in question, i.e., the node that is most adjacent to the specified node.
``````def findNearestNeighborNode(rootNode, specifiedNodeValue):
``````
The algorithm requires a list of all the binary search trees node’s in ascending order. I noticed that the inorder method for traversing binary search trees would yield such an effect. [1][2] Thanks to the `binarytree` module, all that is needed to be done by the developer is to provide the root node of a `binarytree` created binary search tree. Inorder traversal could then be performed with a single, simple statement, like
``````allNodes = rootNode.inorder
``````
Now, as the algorithm needs to maintain a list of the differences between each node and the specified node, a list must be created in the function. A difference represents a node’s distance from the specified node, by the way.
``````differences = []
``````
Having created an array that can maintain the list of differences, the algorithm then computes each node’s distance from the specified node and adds the result of each determination to the array of differences.
``````for node in allNodes:
differences.append(abs(specifiedNodeValue - node.value))
``````
Finally, the algorithm finds the node whose distance with the specified node is the smallest and then the function returns that node’s location.
``````lowest_difference = min(differences)
for index, difference in enumerate(differences):
if difference == lowest_difference:
return allNodes[index].value
``````
To test to see if the algorithm works and to see its effects, I created a random binary search tree to test the algorithm on.
``````test_bst = bst(height = 3, is_perfect = False)
``````
The `binarytree` module can print a very nice graph of a binary search tree like this for us, with the statement `print test_bst`.
`````` ____6__
/ \
__3__ 10___
/ \ / \
1 5 8 _14
/ \ / /
0 2 4 12
``````
Using `findNearestNeighborNode(test_bst, 7)` on the above binary search tree, the algorithm would find the closest node to a non-existant 7 node to be the 6 node.
Here’s the full source code of the algorithm, complete with very informative comments:
``````from binarytree import bst
# Function to find the nearest neighbor node of a specified node in a binary search tree. Returns the target node's value in question.
# rootNode is a binary search tree created from the bst module and specifiedNodeValue is an integer value of a possible node that could exist.
def findNearestNeighborNode(rootNode, specifiedNodeValue):
# Use the in order method of binary search tree traversing to get a list of all nodes in ascending order.
allNodes = rootNode.inorder
# For each node, find the arithmetic difference between it and the specified node, then place the result into an array for finding the lowest difference.
# The node that has the lowest difference will be the node that will be considered the nearest neighbor node.
differences = [] # The array that will contain all the differences.
for node in allNodes:
differences.append(abs(specifiedNodeValue - node.value))
# Find the first lowest difference and then deem the first node with that difference as the node that is the nearest neighbor to the specfied node.
lowest_difference = min(differences)
for index, difference in enumerate(differences): # Finding the node with the same index as the index of the lowest difference.
if difference == lowest_difference: # Note that the allNodes array and differences array map or correspond to each other.
return allNodes[index].value
``````
# Stumbling upon the answer to inverting a binary search tree
Noticing the effect that the inorder traversal had, I think I finally found the solution for reversing a binary search tree by considering inorder traversal.
But I don’t know how want to implement a traversal method myself nor implement a binary search tree data structure (complete with accessor functions) from scratch right now.
# Flaming chicken in hand
I’m still learning how to use my Effectual Transformation skill. And I forgot about having knowledge in binary search trees. I must not squander these great fortunes that I have as they can create opportunities that I can exploit. I will strive to leverage these effects in my endeavors.
March 16, 2018 | 1,305 | 6,049 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2020-45 | latest | en | 0.879134 |
http://docs.julia.tokyo/ja/latest/stdlib/collections.html | 1,505,982,744,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687711.44/warc/CC-MAIN-20170921082205-20170921102205-00561.warc.gz | 93,039,944 | 22,039 | # Collections and Data Structures¶
## Iteration¶
Sequential iteration is implemented by the methods start(), done(), and next(). The general for loop:
for i = I # or "for i in I"
# body
end
is translated into:
state = start(I)
while !done(I, state)
(i, state) = next(I, state)
# body
end
The state object may be anything, and should be chosen appropriately for each iterable type. See the manual section on the iteration interface for more details about defining a custom iterable type.
start(iter) → state
Get initial iteration state for an iterable object
done(iter, state) → Bool
Test whether we are done iterating
next(iter, state) → item, state
For a given iterable object and iteration state, return the current item and the next iteration state
zip(iters...)
For a set of iterable objects, returns an iterable of tuples, where the ith tuple contains the ith component of each input iterable.
Note that zip() is its own inverse: collect(zip(zip(a...)...)) == collect(a).
enumerate(iter)
An iterator that yields (i, x) where i is an index starting at 1, and x is the ith value from the given iterator. It’s useful when you need not only the values x over which you are iterating, but also the index i of the iterations.
julia> a = ["a", "b", "c"];
julia> for (index, value) in enumerate(a)
println("$index$value")
end
1 a
2 b
3 c
rest(iter, state)
An iterator that yields the same elements as iter, but starting at the given state.
countfrom(start=1, step=1)
An iterator that counts forever, starting at start and incrementing by step.
take(iter, n)
An iterator that generates at most the first n elements of iter.
drop(iter, n)
An iterator that generates all but the first n elements of iter.
cycle(iter)
An iterator that cycles through iter forever.
repeated(x[, n::Int])
An iterator that generates the value x forever. If n is specified, generates x that many times (equivalent to take(repeated(x), n)).
Fully implemented by:
• Range
• UnitRange
• NDRange
• Tuple
• Number
• AbstractArray
• IntSet
• ObjectIdDict
• Dict
• WeakKeyDict
• EachLine
• AbstractString
• Set
• Task
## General Collections¶
isempty(collection) → Bool
Determine whether a collection is empty (has no elements).
julia> isempty([])
true
julia> isempty([1 2 3])
false
empty!(collection) → collection
Remove all elements from a collection.
length(collection) → Integer
For ordered, indexable collections, the maximum index i for which getindex(collection, i) is valid. For unordered collections, the number of elements.
endof(collection) → Integer
Returns the last index of the collection.
julia> endof([1,2,4])
3
Fully implemented by:
• Range
• UnitRange
• Tuple
• Number
• AbstractArray
• IntSet
• Dict
• WeakKeyDict
• AbstractString
• Set
## Iterable Collections¶
in(item, collection) → Bool
∈(item, collection) → Bool
∋(collection, item) → Bool
∉(item, collection) → Bool
∌(collection, item) → Bool
Determine whether an item is in the given collection, in the sense that it is == to one of the values generated by iterating over the collection. Some collections need a slightly different definition; for example Sets check whether the item isequal() to one of the elements. Dicts look for (key,value) pairs, and the key is compared using isequal(). To test for the presence of a key in a dictionary, use haskey() or k in keys(dict).
eltype(type)
Determine the type of the elements generated by iterating a collection of the given type. For associative collection types, this will be a Pair{KeyType,ValType}. The definition eltype(x) = eltype(typeof(x)) is provided for convenience so that instances can be passed instead of types. However the form that accepts a type argument should be defined for new types.
indexin(a, b)
Returns a vector containing the highest index in b for each value in a that is a member of b . The output vector contains 0 wherever a is not a member of b.
findin(a, b)
Returns the indices of elements in collection a that appear in collection b
unique(itr[, dim])
Returns an array containing only the unique elements of the iterable itr, in the order that the first of each set of equivalent elements originally appears. If dim is specified, returns unique regions of the array itr along dim.
reduce(op, v0, itr)
Reduce the given collection ìtr with the given binary operator op. v0 must be a neutral element for op that will be returned for empty collections. It is unspecified whether v0 is used for non-empty collections.
Reductions for certain commonly-used operators have special implementations which should be used instead: maximum(itr), minimum(itr), sum(itr), prod(itr), any(itr), all(itr).
The associativity of the reduction is implementation dependent. This means that you can’t use non-associative operations like - because it is undefined whether reduce(-,[1,2,3]) should be evaluated as (1-2)-3 or 1-(2-3). Use foldl or foldr instead for guaranteed left or right associativity.
Some operations accumulate error, and parallelism will also be easier if the reduction can be executed in groups. Future versions of Julia might change the algorithm. Note that the elements are not reordered if you use an ordered collection.
reduce(op, itr)
Like reduce(op, v0, itr). This cannot be used with empty collections, except for some special cases (e.g. when op is one of +, *, max, min, &, |) when Julia can determine the neutral element of op.
foldl(op, v0, itr)
Like reduce(), but with guaranteed left associativity. v0 will be used exactly once.
foldl(op, itr)
Like foldl(op, v0, itr), but using the first element of itr as v0. In general, this cannot be used with empty collections (see reduce(op, itr)).
foldr(op, v0, itr)
Like reduce(), but with guaranteed right associativity. v0 will be used exactly once.
foldr(op, itr)
Like foldr(op, v0, itr), but using the last element of itr as v0. In general, this cannot be used with empty collections (see reduce(op, itr)).
maximum(itr)
Returns the largest element in a collection.
maximum(A, dims)
Compute the maximum value of an array over the given dimensions.
maximum!(r, A)
Compute the maximum value of A over the singleton dimensions of r, and write results to r.
minimum(itr)
Returns the smallest element in a collection.
minimum(A, dims)
Compute the minimum value of an array over the given dimensions.
minimum!(r, A)
Compute the minimum value of A over the singleton dimensions of r, and write results to r.
extrema(itr)
Compute both the minimum and maximum element in a single pass, and return them as a 2-tuple.
indmax(itr) → Integer
Returns the index of the maximum element in a collection.
indmin(itr) → Integer
Returns the index of the minimum element in a collection.
findmax(itr) -> (x, index)
Returns the maximum element and its index.
findmax(A, dims) -> (maxval, index)
For an array input, returns the value and index of the maximum over the given dimensions.
findmin(itr) -> (x, index)
Returns the minimum element and its index.
findmin(A, dims) -> (minval, index)
For an array input, returns the value and index of the minimum over the given dimensions.
findmax!(rval, rind, A, [init=true]) -> (maxval, index)
Find the maximum of A and the corresponding linear index along singleton dimensions of rval and rind, and store the results in rval and rind.
findmin!(rval, rind, A, [init=true]) -> (minval, index)
Find the minimum of A and the corresponding linear index along singleton dimensions of rval and rind, and store the results in rval and rind.
maxabs(itr)
Compute the maximum absolute value of a collection of values.
maxabs(A, dims)
Compute the maximum absolute values over given dimensions.
maxabs!(r, A)
Compute the maximum absolute values over the singleton dimensions of r, and write values to r.
minabs(itr)
Compute the minimum absolute value of a collection of values.
minabs(A, dims)
Compute the minimum absolute values over given dimensions.
minabs!(r, A)
Compute the minimum absolute values over the singleton dimensions of r, and write values to r.
sum(itr)
Returns the sum of all elements in a collection.
sum(A, dims)
Sum elements of an array over the given dimensions.
sum!(r, A)
Sum elements of A over the singleton dimensions of r, and write results to r.
sum(f, itr)
Sum the results of calling function f on each element of itr.
sumabs(itr)
Sum absolute values of all elements in a collection. This is equivalent to sum(abs(itr)) but faster.
sumabs(A, dims)
Sum absolute values of elements of an array over the given dimensions.
sumabs!(r, A)
Sum absolute values of elements of A over the singleton dimensions of r, and write results to r.
sumabs2(itr)
Sum squared absolute values of all elements in a collection. This is equivalent to sum(abs2(itr)) but faster.
sumabs2(A, dims)
Sum squared absolute values of elements of an array over the given dimensions.
sumabs2!(r, A)
Sum squared absolute values of elements of A over the singleton dimensions of r, and write results to r.
prod(itr)
Returns the product of all elements of a collection.
prod(A, dims)
Multiply elements of an array over the given dimensions.
prod!(r, A)
Multiply elements of A over the singleton dimensions of r, and write results to r.
any(itr) → Bool
Test whether any elements of a boolean collection are true.
any(A, dims)
Test whether any values along the given dimensions of an array are true.
any!(r, A)
Test whether any values in A along the singleton dimensions of r are true, and write results to r.
all(itr) → Bool
Test whether all elements of a boolean collection are true.
all(A, dims)
Test whether all values along the given dimensions of an array are true.
all!(r, A)
Test whether all values in A along the singleton dimensions of r are true, and write results to r.
count(p, itr) → Integer
Count the number of elements in itr for which predicate p returns true.
any(p, itr) → Bool
Determine whether predicate p returns true for any elements of itr.
all(p, itr) → Bool
Determine whether predicate p returns true for all elements of itr.
julia> all(i->(4<=i<=6), [4,5,6])
true
map(f, c...) → collection
Transform collection c by applying f to each element. For multiple collection arguments, apply f elementwise.
julia> map((x) -> x * 2, [1, 2, 3])
3-element Array{Int64,1}:
2
4
6
julia> map(+, [1, 2, 3], [10, 20, 30])
3-element Array{Int64,1}:
11
22
33
map!(function, collection)
In-place version of map().
map!(function, destination, collection...)
Like map(), but stores the result in destination rather than a new collection. destination must be at least as large as the first collection.
mapreduce(f, op, v0, itr)
Apply function f to each element in itr, and then reduce the result using the binary function op. v0 must be a neutral element for op that will be returned for empty collections. It is unspecified whether v0 is used for non-empty collections.
mapreduce() is functionally equivalent to calling reduce(op, v0, map(f, itr)), but will in general execute faster since no intermediate collection needs to be created. See documentation for reduce() and map().
julia> mapreduce(x->x^2, +, [1:3;]) # == 1 + 4 + 9
14
The associativity of the reduction is implementation-dependent. Additionally, some implementations may reuse the return value of f for elements that appear multiple times in itr. Use mapfoldl() or mapfoldr() instead for guaranteed left or right associativity and invocation of f for every value.
mapreduce(f, op, itr)
Like mapreduce(f, op, v0, itr). In general, this cannot be used with empty collections (see reduce(op, itr)).
mapfoldl(f, op, v0, itr)
Like mapreduce(), but with guaranteed left associativity. v0 will be used exactly once.
mapfoldl(f, op, itr)
Like mapfoldl(f, op, v0, itr), but using the first element of itr as v0. In general, this cannot be used with empty collections (see reduce(op, itr)).
mapfoldr(f, op, v0, itr)
Like mapreduce(), but with guaranteed right associativity. v0 will be used exactly once.
mapfoldr(f, op, itr)
Like mapfoldr(f, op, v0, itr), but using the first element of itr as v0. In general, this cannot be used with empty collections (see reduce(op, itr)).
first(coll)
Get the first element of an iterable collection. Returns the start point of a Range even if it is empty.
last(coll)
Get the last element of an ordered collection, if it can be computed in O(1) time. This is accomplished by calling endof() to get the last index. Returns the end point of a Range even if it is empty.
step(r)
Get the step size of a Range object.
collect(collection)
Return an array of all items in a collection. For associative collections, returns Pair{KeyType, ValType}.
collect(element_type, collection)
Return an array of type Array{element_type,1} of all items in a collection.
issubset(a, b)
⊆(a, b) → Bool
⊈(a, b) → Bool
⊊(a, b) → Bool
Determine whether every element of a is also in b, using in().
filter(function, collection)
Return a copy of collection, removing elements for which function is false. For associative collections, the function is passed two arguments (key and value).
filter!(function, collection)
Update collection, removing elements for which function is false. For associative collections, the function is passed two arguments (key and value).
## Indexable Collections¶
getindex(collection, key...)
Retrieve the value(s) stored at the given key or index within a collection. The syntax a[i,j,...] is converted by the compiler to getindex(a, i, j, ...).
setindex!(collection, value, key...)
Store the given value at the given key or index within a collection. The syntax a[i,j,...] = x is converted by the compiler to (setindex!(a, x, i, j, ...); x).
Fully implemented by:
• Array
• BitArray
• AbstractArray
• SubArray
• ObjectIdDict
• Dict
• WeakKeyDict
• AbstractString
Partially implemented by:
• Range
• UnitRange
• Tuple
## Associative Collections¶
Dict is the standard associative collection. Its implementation uses hash() as the hashing function for the key, and isequal() to determine equality. Define these two functions for custom types to override how they are stored in a hash table.
ObjectIdDict is a special hash table where the keys are always object identities.
WeakKeyDict is a hash table implementation where the keys are weak references to objects, and thus may be garbage collected even when referenced in a hash table.
Dicts can be created by passing pair objects constructed with =>() to a Dict constructor: Dict("A"=>1, "B"=>2). This call will attempt to infer type information from the keys and values (i.e. this example creates a Dict{ASCIIString, Int64}). To explicitly specify types use the syntax Dict{KeyType,ValueType}(...). For example, Dict{ASCIIString,Int32}("A"=>1, "B"=>2).
As with Arrays, Dicts may be created with comprehensions. For example, [i => f(i) for i = 1:10].
Given a dictionary D, the syntax D[x] returns the value of key x (if it exists) or throws an error, and D[x] = y stores the key-value pair x => y in D (replacing any existing value for the key x). Multiple arguments to D[...] are converted to tuples; for example, the syntax D[x,y] is equivalent to D[(x,y)], i.e. it refers to the value keyed by the tuple (x,y).
Dict([itr])
Dict{K,V}() constructs a hash table with keys of type K and values of type V.
Given a single iterable argument, constructs a Dict whose key-value pairs are taken from 2-tuples (key,value) generated by the argument.
julia> Dict([("A", 1), ("B", 2)])
Dict{ASCIIString,Int64} with 2 entries:
"B" => 2
"A" => 1
Alternatively, a sequence of pair arguments may be passed.
julia> Dict("A"=>1, "B"=>2)
Dict{ASCIIString,Int64} with 2 entries:
"B" => 2
"A" => 1
haskey(collection, key) → Bool
Determine whether a collection has a mapping for a given key.
get(collection, key, default)
Return the value stored for the given key, or the given default value if no mapping for the key is present.
get(f::Function, collection, key)
Return the value stored for the given key, or if no mapping for the key is present, return f(). Use get!() to also store the default value in the dictionary.
This is intended to be called using do block syntax:
get(dict, key) do
# default value calculated here
time()
end
get!(collection, key, default)
Return the value stored for the given key, or if no mapping for the key is present, store key => default, and return default.
get!(f::Function, collection, key)
Return the value stored for the given key, or if no mapping for the key is present, store key => f(), and return f().
This is intended to be called using do block syntax:
get!(dict, key) do
# default value calculated here
time()
end
getkey(collection, key, default)
Return the key matching argument key if one exists in collection, otherwise return default.
delete!(collection, key)
Delete the mapping for the given key in a collection, and return the collection.
pop!(collection, key[, default])
Delete and return the mapping for key if it exists in collection, otherwise return default, or throw an error if default is not specified.
keys(collection)
Return an iterator over all keys in a collection. collect(keys(d)) returns an array of keys.
values(collection)
Return an iterator over all values in a collection. collect(values(d)) returns an array of values.
merge(collection, others...)
Construct a merged collection from the given collections. If necessary, the types of the resulting collection will be promoted to accommodate the types of the merged collections. If the same key is present in another collection, the value for that key will be the value it has in the last collection listed.
julia> a = Dict("foo" => 0.0, "bar" => 42.0)
Dict{ASCIIString,Float64} with 2 entries:
"bar" => 42.0
"foo" => 0.0
julia> b = Dict(utf8("baz") => 17, utf8("bar") => 4711)
Dict{UTF8String,Int64} with 2 entries:
"bar" => 4711
"baz" => 17
julia> merge(a, b)
Dict{UTF8String,Float64} with 3 entries:
"bar" => 4711.0
"baz" => 17.0
"foo" => 0.0
julia> merge(b, a)
Dict{UTF8String,Float64} with 3 entries:
"bar" => 42.0
"baz" => 17.0
"foo" => 0.0
merge!(collection, others...)
Update collection with pairs from the other collections
sizehint!(s, n)
Suggest that collection s reserve capacity for at least n elements. This can improve performance.
keytype(collection)
For associative collection types, this will be the type of the Key, This is not defined for non-associative collections
valtype(collection)
For associative collection types, this will be the type of the Value, This is not defined for non-associative collections
Fully implemented by:
• ObjectIdDict
• Dict
• WeakKeyDict
Partially implemented by:
• IntSet
• Set
• EnvHash
• Array
• BitArray
## Set-Like Collections¶
Set([itr])
Construct a Set of the values generated by the given iterable object, or an empty set. Should be used instead of IntSet for sparse integer sets, or for sets of arbitrary objects.
IntSet([itr])
Construct a sorted set of positive Ints generated by the given iterable object, or an empty set. Implemented as a bit string, and therefore designed for dense integer sets. Only Ints greater than 0 can be stored. If the set will be sparse (for example holding a few very large integers), use Set instead.
union(s1, s2...)
∪(s1, s2...)
Construct the union of two or more sets. Maintains order with arrays.
union!(s, iterable)
Union each element of iterable into set s in-place.
intersect(s1, s2...)
∩(s1, s2)
Construct the intersection of two or more sets. Maintains order and multiplicity of the first argument for arrays and ranges.
setdiff(s1, s2)
Construct the set of elements in s1 but not s2. Maintains order with arrays. Note that both arguments must be collections, and both will be iterated over. In particular, setdiff(set,element) where element is a potential member of set, will not work in general.
setdiff!(s, iterable)
Remove each element of iterable from set s in-place.
symdiff(s1, s2...)
Construct the symmetric difference of elements in the passed in sets or arrays. Maintains order with arrays.
symdiff!(s, n)
The set s is destructively modified to toggle the inclusion of integer n.
symdiff!(s, itr)
For each element in itr, destructively toggle its inclusion in set s.
symdiff!(s1, s2)
Construct the symmetric difference of sets s1 and s2, storing the result in s1.
complement(s)
Returns the set-complement of IntSet s.
complement!(s)
Mutates IntSet s into its set-complement.
intersect!(s1, s2)
Intersects sets s1 and s2 and overwrites the set s1 with the result. If needed, s1 will be expanded to the size of s2.
issubset(A, S) → Bool
⊆(A, S) → Bool
Return true if A is a subset of or equal to S.
Fully implemented by:
• IntSet
• Set
Partially implemented by:
• Array
## Dequeues¶
push!(collection, items...) → collection
Insert one or more items at the end of collection.
julia> push!([1, 2, 3], 4, 5, 6)
6-element Array{Int64,1}:
1
2
3
4
5
6
Use append!() to add all the elements of another collection to collection. The result of the preceding example is equivalent to append!([1, 2, 3], [4, 5, 6]).
pop!(collection) → item
Remove the last item in collection and return it.
julia> A=[1, 2, 3, 4, 5, 6]
6-element Array{Int64,1}:
1
2
3
4
5
6
julia> pop!(A)
6
julia> A
5-element Array{Int64,1}:
1
2
3
4
5
unshift!(collection, items...) → collection
Insert one or more items at the beginning of collection.
julia> unshift!([1, 2, 3, 4], 5, 6)
6-element Array{Int64,1}:
5
6
1
2
3
4
shift!(collection) → item
Remove the first item from collection.
julia> A = [1, 2, 3, 4, 5, 6]
6-element Array{Int64,1}:
1
2
3
4
5
6
julia> shift!(A)
1
julia> A
5-element Array{Int64,1}:
2
3
4
5
6
insert!(collection, index, item)
Insert an item into collection at the given index. index is the index of item in the resulting collection.
julia> insert!([6, 5, 4, 2, 1], 4, 3)
6-element Array{Int64,1}:
6
5
4
3
2
1
deleteat!(collection, index)
Remove the item at the given index and return the modified collection. Subsequent items are shifted to fill the resulting gap.
julia> deleteat!([6, 5, 4, 3, 2, 1], 2)
5-element Array{Int64,1}:
6
4
3
2
1
deleteat!(collection, itr)
Remove the items at the indices given by itr, and return the modified collection. Subsequent items are shifted to fill the resulting gap. itr must be sorted and unique.
julia> deleteat!([6, 5, 4, 3, 2, 1], 1:2:5)
3-element Array{Int64,1}:
5
3
1
julia> deleteat!([6, 5, 4, 3, 2, 1], (2, 2))
ERROR: ArgumentError: indices must be unique and sorted
in deleteat! at array.jl:546
splice!(collection, index[, replacement]) → item
Remove the item at the given index, and return the removed item. Subsequent items are shifted down to fill the resulting gap. If specified, replacement values from an ordered collection will be spliced in place of the removed item.
julia> A = [6, 5, 4, 3, 2, 1]; splice!(A, 5)
2
julia> A
5-element Array{Int64,1}:
6
5
4
3
1
julia> splice!(A, 5, -1)
1
julia> A
5-element Array{Int64,1}:
6
5
4
3
-1
julia> splice!(A, 1, [-1, -2, -3])
6
julia> A
7-element Array{Int64,1}:
-1
-2
-3
5
4
3
-1
To insert replacement before an index n without removing any items, use splice!(collection, n:n-1, replacement).
splice!(collection, range[, replacement]) → items
Remove items in the specified index range, and return a collection containing the removed items. Subsequent items are shifted down to fill the resulting gap. If specified, replacement values from an ordered collection will be spliced in place of the removed items.
To insert replacement before an index n without removing any items, use splice!(collection, n:n-1, replacement).
julia> splice!(A, 4:3, 2)
0-element Array{Int64,1}
julia> A
8-element Array{Int64,1}:
-1
-2
-3
2
5
4
3
-1
resize!(collection, n) → collection
Resize collection to contain n elements. If n is smaller than the current collection length, the first n elements will be retained. If n is larger, the new elements are not guaranteed to be initialized.
julia> resize!([6, 5, 4, 3, 2, 1], 3)
3-element Array{Int64,1}:
6
5
4
julia> resize!([6, 5, 4, 3, 2, 1], 8)
8-element Array{Int64,1}:
6
5
4
3
2
1
0
0
append!(collection, collection2) → collection.
Add the elements of collection2 to the end of collection.
julia> append!([1],[2,3])
3-element Array{Int64,1}:
1
2
3
julia> append!([1, 2, 3], [4, 5, 6])
6-element Array{Int64,1}:
1
2
3
4
5
6
Use push!() to add individual items to collection which are not already themselves in another collection. The result is of the preceding example is equivalent to push!([1, 2, 3], 4, 5, 6).
prepend!(collection, items) → collection
Insert the elements of items to the beginning of collection.
julia> prepend!([3],[1,2])
3-element Array{Int64,1}:
1
2
3
Fully implemented by:
• Vector (a.k.a. 1-dimensional Array)
• BitVector (a.k.a. 1-dimensional BitArray)
## PriorityQueue¶
The PriorityQueue type is available from the Collections module. It provides a basic priority queue implementation allowing for arbitrary key and priority types. Multiple identical keys are not permitted, but the priority of existing keys can be changed efficiently.
PriorityQueue(K, V[, ord])
Construct a new PriorityQueue, with keys of type K and values/priorites of type V. If an order is not given, the priority queue is min-ordered using the default comparison for V.
enqueue!(pq, k, v)
Insert the a key k into a priority queue pq with priority v.
dequeue!(pq)
Remove and return the lowest priority key from a priority queue.
peek(pq)
Return the lowest priority key from a priority queue without removing that key from the queue.
PriorityQueue also behaves similarly to a Dict in that keys can be inserted and priorities accessed or changed using indexing notation.
julia> # Julia code
pq = Collections.PriorityQueue();
julia> # Insert keys with associated priorities
pq["a"] = 10; pq["b"] = 5; pq["c"] = 15; pq
Base.Collections.PriorityQueue{Any,Any,Base.Order.ForwardOrdering} with 3 entries:
"c" => 15
"b" => 5
"a" => 10
julia> # Change the priority of an existing key
pq["a"] = 0; pq
Base.Collections.PriorityQueue{Any,Any,Base.Order.ForwardOrdering} with 3 entries:
"c" => 15
"b" => 5
"a" => 0
## Heap Functions¶
Along with the PriorityQueue type, the Collections module provides lower level functions for performing binary heap operations on arrays. Each function takes an optional ordering argument. If not given, default ordering is used, so that elements popped from the heap are given in ascending order.
heapify(v[, ord])
Return a new vector in binary heap order, optionally using the given ordering.
heapify!(v[, ord])
In-place heapify().
isheap(v[, ord])
Return true iff an array is heap-ordered according to the given order.
heappush!(v, x[, ord])
Given a binary heap-ordered array, push a new element x, preserving the heap property. For efficiency, this function does not check that the array is indeed heap-ordered.
heappop!(v[, ord])
Given a binary heap-ordered array, remove and return the lowest ordered element. For efficiency, this function does not check that the array is indeed heap-ordered. | 7,013 | 27,156 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 2.71875 | 3 | CC-MAIN-2017-39 | longest | en | 0.745557 |
https://nrich.maths.org/6071/inde | 1,638,386,992,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360881.12/warc/CC-MAIN-20211201173718-20211201203718-00607.warc.gz | 488,594,459 | 4,957 | ### Five Coins
Ben has five coins in his pocket. How much money might he have?
### The Puzzling Sweet Shop
There were chews for 2p, mini eggs for 3p, Chocko bars for 5p and lollypops for 7p in the sweet shop. What could each of the children buy with their money?
### Little Man
The Man is much smaller than us. Can you use the picture of him next to a mug to estimate his height and how much tea he drinks?
# Stop the Clock
##### Age 5 to 7Challenge Level
This is a game for two players. You can use the interactivity below, or you could print off a page of blank clock faces in Word or as a pdf.
Set the time on the clock to 6 o'clock to start the game.
Decide who will go first (player 1) and who will go second (player 2).
Take it in turns to choose to move the hands of the clock on by $\frac{1}{2}$ hour or by 1 hour. For example, player 1 could choose $\frac{1}{2}$ hour, so the clock hands move to 6.30, then player 2 might choose 1 hour, moving the clock hands to 7.30... etc.
The winner is the player who moves the hands exactly onto 12 o'clock.
Can you work out a winning strategy so that you can always beat your opponent? | 311 | 1,143 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2021-49 | latest | en | 0.960531 |
https://algebra-class-ecourse.com/question/will-give-brainliest-what-can-you-say-about-the-end-behavior-of-the-function-f-log2-4-8-16-a-one-11535184-36/ | 1,638,223,540,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358842.4/warc/CC-MAIN-20211129194957-20211129224957-00346.warc.gz | 162,797,505 | 12,122 | ## WILL GIVE BRAINLIEST what can you say about the end behavior of the function f(x)=log2(4x+8)+16? A.one end increases
Question
WILL GIVE BRAINLIEST
what can you say about the end behavior of the function f(x)=log2(4x+8)+16?
A.one end increases and One end decreases
B.one end decreases and one end approaches a constant
C.both ends increase
D.both ends decrease
NEXT QUESTION WILL BE 25 POINTS
0 | 115 | 403 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-49 | latest | en | 0.705096 |
https://moda-tamshi.com/qa/question-what-does-110-feel-like.html | 1,603,527,923,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107882103.34/warc/CC-MAIN-20201024080855-20201024110855-00357.warc.gz | 434,773,514 | 9,804 | # Question: What Does 110 Feel Like?
## What is real feel?
The RealFeel Temperature is an index that describes what the temperature really feels like.
It is a unique composite of the effects of temperature, wind, humidity, sunshine intensity, cloudiness, precipitation and elevation on the human body–everything that affects how warm or cold a person feels..
## What is the difference between heat index and feels like?
“It’s not the heat, it’s the humidity”. That’s a partly valid phrase you may have heard in the summer, but it’s actually both. The heat index, also known as the apparent temperature, is what the temperature feels like to the human body when relative humidity is combined with the air temperature.
## How do you calculate the real feel temperature?
The AccuWeather RealFeel Temperature takes into account the effects of multiple parameters, including air temperature, wind speed, solar intensity, humidity, precipitation intensity/type, elevation and atmospheric pressure.
## How much hotter does the sun make it feel?
Rather, being in direct sunlight and solar radiation makes the air feel 10 to 15 degrees warmer than it actually is, said Jim Lushine, a retired weather service meteorologist. “So, conversely, it would feel that much cooler in the shade,” he said.
## How do you know if yeast is still active?
There’s an Easy Way to Check Proof your yeast to find out if it’s still active by adding 1 teaspoon of sugar and 2 1/4 teaspoons of yeast (one envelope) to 1/4 cup of warm water. Then, wait 10 minutes. If the mixture bubbles and develops a yeasty aroma, the yeast is still good. Want to store yeast longer?
## How can it feel hotter than it is?
When a human being perspires, the water in his or her sweat evaporates. This results in the cooling of the body as heat is carried away from it. When humidity is high, the rate of evaporation and cooling is reduced, resulting in it feeling hotter than it actually is.
## Did Kuwait record 63 degrees?
Last Saturday, Kuwait reportedly recorded the highest temperature on Earth at 63°C under sunlight (and 52.2°C in the shadows). … One temperature claim found invalid was of 58ºC, said to be recorded at El Azizia, Libya, on September 13, 1922.
## Can a hot bath raise body temperature?
Soaking in a warm bath will raise your body temperature, and exiting will more rapidly cool it down, thus instigating the production of melatonin, and better preparing you for sleep.
## Which yeast is best?
Which SAF yeast to use, Red or Gold? SAF Red is your best choice for all-around baking, from sandwich loaves to crusty no-knead bread to freeze-and-bake dinner rolls. SAF Gold is formulated for one specific type of dough: sweet dough. Think Portuguese Sweet Bread, Hawaiian Buns, Panettone, Raisin Challah, and the like.
## Is 110 degrees Fahrenheit hot?
Depending on the source, you will find differing opinions on what temperature range constitutes lukewarm water. Some references say it is between 100 and 110 F (36.5 to 40.5 C). … Run the water over your wrist and if it feels warmer than your body temperature, but not hot, that should be just about right.
## Is room temperature water better than cold water?
Room temperature water maintains hydration. By drinking room temperature water throughout the day, you’ll feel less thirsty compared to drinking it cold. The downside to this is not drinking enough water. To stay cool, your body will sweat and lose the minimal water you’re intaking.
## What temperature does it feel like with humidity?
That relative humidity, combined with the temperature, then determines the heat index or an estimate of how hot the air actually feels. So 85 degrees at 10% humidity, the temperature feels closer to 79 degrees, but at 90% humidity, it will feel closer to 100 degrees outside.
## How can you tell if water is 110 degrees?
To proof, add your yeast to your warm water. The water should be between 100 and 110 degrees. If you don’t have a thermometer, use your wrist to test the water temperature. If it feels very warm on your wrist, it’s perfect for the yeast.
## Why we Cannot see Sun night?
The Sun is our nearest star. … From Earth, the Sun looks like it moves across the sky in the daytime and appears to disappear at night. This is because the Earth is spinning towards the east. The Earth spins about its axis, an imaginary line that runs through the middle of the Earth between the North and South poles.
## Why cold showers are bad for you?
The cons of cold showers: It could actually make you even colder and increase the amount of time it will take for your body to warm back up. They may not be a good idea if you’re sick, either. Initially, the cold temperature might be too hard on your immune system, so it’s best to ease into the cooler temperatures.
## What temperature is a hot bath?
What Is the Best Water Temperature for Your Bath or Shower? While some advocate the benefits of bathing in very hot or very cold water, the magic number is 112 degrees.
## What do I do if my yeast isn’t foaming?
Once you see the foam, you’re ready to use your yeast in any recipe it calls for. If you DON’T see foam and you’ve been patient (given it 15 minutes or so), try again with another packet. If you made your water hot, try reducing that heat a bit and give it another try.
## Can I use old yeast?
Remember that yeast, like a lot of other baking products, usually has a best before date and not a use by date or expiration date. Because of this distinction, you may safely use yeast for your baking needs for a time after the best before date has lapsed. | 1,236 | 5,618 | {"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} | 2.515625 | 3 | CC-MAIN-2020-45 | latest | en | 0.929343 |
https://www.pbiusergroup.com/discussion/show-value-from-selected-date-range?ReturnUrl=%2Fcommunities%2Fcommunity-home%2Fdigestviewer%3FListKey%3D92819279-4c76-4531-a009-69222395538c | 1,670,018,763,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710916.70/warc/CC-MAIN-20221202215443-20221203005443-00024.warc.gz | 988,697,448 | 298,080 | # Power BI Exchange
View Only
## Show value from selected date range
• #### 1. Show value from selected date range
Bronze Contributor
Posted Nov 01, 2022 07:10 AM
Hi.
I am trying to create a report showing the development in purchase prices over a given period. In inclosed report, I have the purchase prices in one table and a date table.
When I select a period (based on date table) from March 12 2021 to March 30 2022 I would like to see all puchase prices in this period. However, if I have a purchase price valid from January 1 2021 to April 30 2021 this price wont show in the report as the start date for the price is before the start date in the date table.
How can I solve this? The price is valid on March 12...
Pls. see enclosed file - I have used article "Test 4" in above description.
Helen
------------------------------
Helen Wrensted
------------------------------
Attachment(s)
Historic_Gross_Price.xlsx 11 KB 1 version
CostPrices.pbix 25.17 MB 1 version
• #### 2. RE: Show value from selected date range
Posted 30 days ago
Hi Helen,
Here's what you should do:
a. Create another date table and connect to valid from column ---> this is just have your date always at the beginning of the month (you can use valid from date directly as well)
b. Create inactive relationship between main table and existing date table (connection: Valid to and Date)
c. Create a formula to incorporate the "valid to" values to your existing results (I removed the unnecessary formula btw).
PricePerStartOfMonth =
Var MaxDate = CALCULATE(MAX('Date'[Date]),ALLSELECTED('Date'[Date]))
Var MinDate = CALCULATE(Min('Date'[Date]),ALLSELECTED('Date'[Date]))
VAR ValidTo = CALCULATE (
MAX ( 'Historic_Gross_Price'[Price] ),
USERELATIONSHIP(Historic_Gross_Price[Valid to],'Date'[Date]),
MinDate>=Historic_Gross_Price[Valid from] && MinDate <=Historic_Gross_Price[Valid to])
RETURN
CALCULATE(MAX ( 'Historic_Gross_Price'[Price] ),'Date'[First day] = "Yes") +
ValidTo
------------------------------
Senior Finance Analyst
------------------------------
Attachment(s)
CostPrices (1).pbix 76 KB 1 version
• #### 3. RE: Show value from selected date range
Bronze Contributor
Posted 30 days ago
Hi.
It works great! However, I have a problem with the prices that are effective from dates that are not the first day of the month. In enclosed file, I have added a simple matrix where we would like to see all cost prices in selected period. For article Test 4 it is fine, but if you look at article Test 3 (where some of the prices are effective from mid month), prices are missing.
How can this be soved?
------------------------------
Helen Wrensted
------------------------------
Attachment(s)
CostPrices (1).pbix 76 KB 1 version
• #### 4. RE: Show value from selected date range Best Answer
Posted 30 days ago
Hi Helen, In that case, just remove the filter context in the formula.
PricePerStartOfMonth =
Var MaxDate = CALCULATE(MAX('Date'[Date]),ALLSELECTED('Date'[Date]))
Var MinDate = CALCULATE(Min('Date'[Date]),ALLSELECTED('Date'[Date]))
VAR ValidTo = CALCULATE (
MAX ( 'Historic_Gross_Price'[Price] ),
USERELATIONSHIP(Historic_Gross_Price[Valid to],'Date'[Date]),
MinDate>=Historic_Gross_Price[Valid from] && MinDate <=Historic_Gross_Price[Valid to])
RETURN
CALCULATE(MAX ( 'Historic_Gross_Price'[Price] )) +
ValidTo
------------------------------
Senior Finance Analyst
------------------------------
• #### 5. RE: Show value from selected date range
Bronze Contributor
Posted 26 days ago
It works perfectly - thanks a lot!
------------------------------
Helen Wrensted
------------------------------
• #### 6. RE: Show value from selected date range
Bronze Contributor
Posted 27 days ago
From the title of the measure, "Price per start of month", I assume that you want the price to be displayed not on the actual valid from date, but at the first month start date on or after the valid from date. In the attached file I have set up a solution based on that assumption.
------------------------------
Tomas
------------------------------
Attachment(s)
CostPrices 2.pbix 60 KB 1 version
• #### 7. RE: Show value from selected date range
Bronze Contributor
Posted 26 days ago
Hi Tomas.
Thanks a lot for the solution, which works great for the first month start date.
------------------------------
Helen Wrensted
------------------------------
• #### 8. RE: Show value from selected date range
Posted 27 days ago
Edited by Kaz Shakir 26 days ago
@Helen Wrensted,
I think that you should get the result that you want if you simply delete the relationship between your Date table and the Historic_Gross_Price table. Then the measures you have created will give you the results that you are expecting. So, your data model will simply look like this:
And after you delete that relationship, your resulting matrix will look like this:
Another approach would be to create a new historic gross price table, where the prices for each "Material" are shown for each date. This new table will be a cross join between that dates of the date table and different material. Here is the formula I used to create that table:
``````Daily_Historic_Gross_Price =
CROSSJOIN(
Calendar( Date(2017, 1, 1), Date(2022,3,30)),
VALUES(Historic_Gross_Price[Material])
)``````
Next, I would create calculated columns in this new Daily_Historic_Gross_Price table that show the price that's effective on that given date for that given material, and another to show the currency that is effective during that time as well. Here are the formulas for those columns:
``````Effective_Price =
VAR currentDate = 'Daily_Historic_Gross_Price'[Date]
VAR currentMaterial = 'Daily_Historic_Gross_Price'[Material]
VAR filteredData =
FILTER(
Historic_Gross_Price,
Historic_Gross_Price[Material] = currentMaterial && Historic_Gross_Price[Valid from] <= currentDate && Historic_Gross_Price[Valid to] >= currentDate
)
VAR relatedPrice =
MAXX(
filteredData,
Historic_Gross_Price[Price]
)
RETURN
relatedPrice``````
``````Effective_Currency =
VAR currentDate = 'Daily_Historic_Gross_Price'[Date]
VAR currentMaterial = 'Daily_Historic_Gross_Price'[Material]
VAR filteredData =
FILTER(
Historic_Gross_Price,
Historic_Gross_Price[Material] = currentMaterial && Historic_Gross_Price[Valid from] <= currentDate && Historic_Gross_Price[Valid to] >= currentDate
)
VAR relatedCurrency =
MAXX(
filteredData,
Historic_Gross_Price[Currency]
)
RETURN
relatedCurrency``````
Then, in your data model, you would connect the Date table to this new Daily_Historic_Gross_Price table:
Finally, in your table visualization, you would use the columns from the Date table and the Daily_Historic_Gross_Price table, so the result would look like this:
I am attaching my solution here for your review.
Please let us know if this gets you the results you were looking for.
Kaz.
------------------------------
Kaz Shakir
Sr. Program Manager, Asset Planning
TN
------------------------------
Attachment(s)
CostPrices_r3.pbix 25.19 MB 1 version
• #### 9. RE: Show value from selected date range
Bronze Contributor
Posted 26 days ago
Hi Kaz.
Thanks a lot for your solution - it works great!
------------------------------
Helen Wrensted | 1,735 | 7,242 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.769015 |
https://eduzip.com/ask/question/find-the-angles-x-and-y160in-the-figure-aoob-520115 | 1,642,599,240,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301341.12/warc/CC-MAIN-20220119125003-20220119155003-00205.warc.gz | 291,490,620 | 8,890 | Mathematics
Find the angles x and y in the figure. $AO=OB$
SOLUTION
Since,
$AO=OB\Rightarrow \angle ABO=\angle x..........(1)$
Now, $\angle AOB=92^o$ as it is vertically oppsite angle to $92^o$
Also, According to Sum of Opposite angle Rule
$x+92^o=y...................(2)$
Also, $\angle ABO+y=180^o\Rightarrow \angle ABO=180^o-y...................(3)$
From (1),(2) and (3)
$x=180^o-y\Rightarrow x=180^o-(x+92^o)\Rightarrow 2x=88^o\Rightarrow x=44^o$
$y=180^o-x=180^o-44^o=136^0$
You're just one step away
Subjective Medium Published on 09th 09, 2020
Questions 120418
Subjects 10
Chapters 88
Enrolled Students 86
Realted Questions
Q1 Subjective Easy
Find the measure of an angle which is 36$^\circ$ more than its complement.
Asked in: Mathematics - Lines and Angles
1 Verified Answer | Published on 09th 09, 2020
Q2 Subjective Medium
In the given figure, $AB\parallel CD$ and $CD\parallel EF$. If $\angle BEF=64^{o}$. Find the value of $x,y,z$
Asked in: Mathematics - Straight Lines
1 Verified Answer | Published on 17th 08, 2020
Q3 Single Correct Medium
Mark the correct alternative of the following.
Two complementary angles are in the ratio $2:3$. The measure of the larger angle is?
• A. $60^o$
• B. $66^o$
• C. $48^o$
• D. $54^o$
Asked in: Mathematics - Lines and Angles
1 Verified Answer | Published on 09th 09, 2020
Q4 Single Correct Medium
If figure, $AB\parallel CD$. Determine $\angle a$
• A. $53^o$
• B. $73^o$
• C. $83^o$
• D. $93^o$
Asked in: Mathematics - Lines and Angles
1 Verified Answer | Published on 23rd 09, 2020
Q5 Subjective Medium
What is an Obtuse Angle ?
Asked in: Mathematics - Lines and Angles
1 Verified Answer | Published on 09th 09, 2020 | 555 | 1,692 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2022-05 | latest | en | 0.690462 |
http://forum.allaboutcircuits.com/showthread.php?t=28288 | 1,406,703,638,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510268734.38/warc/CC-MAIN-20140728011748-00159-ip-10-146-231-18.ec2.internal.warc.gz | 107,591,154 | 12,690 | Register Blogs FAQ Members List Today's Posts Search Today's Posts Mark Forums Read
Homework Help Stuck on a textbook question or coursework? Cramming for a test and need help understanding something? Post your questions and attempts here and let others help.
#1
10-02-2009, 11:26 PM
hasie Junior Member Join Date: Sep 2009 Location: South Africa Posts: 35
Hi Guys
Im new to this site, and im really looking forward to explore further. By the topics I have read sofar, this forum really seems informative.
Dont really know your full rules yet, but I would really appreciate it if anybody could give me help on my problem.
I did run a search on the forum, but did not get my answer that i am wondering about.
I am currently busy trying to build a RC Phase Shift Oscillator.
I see there is alot of info available on the net, but not any info sofar that I specifically found that I am looking for.
Most sites explain the works of the phase shift oscialltor with op amps.
My instructions are:
* Must be designed with BC547 transistor
* Frequency of oscillation must be 1KHz
* Must be as close as possible to a perfect sine wave
* Signal Amplitude must be atleast 2v(p2p)
* Must be operated with a 9v supply
I am not too farmiliar with these oscillators, so this week searched for as much info as I could.
I found a circuit at http://www.talkingelectronics.com.au...TrCcts.html#37 that somebody told me works perfectly.
I then used the formula to get the required resistance values and cap values for the required frequency.( F = 1/((2PI)(R)(C)(sqrt6))
an accepable value to the frequency to caps that I have with me, was 4.7nF capacitors, and 13.8K resistors(12k and 1k8 in series).
The other 2 resistors I left as is at 1M and 3k3.
On my simulation program it runs smoothly and just short of 1KHz, allthough when I built the circuit it gives a saw tooth(triangular wave) waveform.
I have no clue as to how to continue. Can anybody maybe shed abit of light that maybe know, or maybe have a better circuit for me?
ANy help much apprecated
Many Thanks
#2
10-03-2009, 12:46 AM
Audioguru Banned Join Date: Dec 2007 Location: Ontario, Canada Posts: 9,411
The simple single transistor phase-shift-oscillator should give a clipped sine-wave output. The output is clipped because the circuit has nothing to prevent the transistor from going into saturation. The transistor does not have much negative feedback so it produces distortion.
Check the pins of your transistor and the wiring of your circuit to see why it produces a sawtooth/triangle waveform.
#3
10-03-2009, 02:14 AM
Bill_Marsden Super Moderator Join Date: Mar 2008 Location: Dallas, TX (GMT-5 w/ DST) Posts: 19,022 Blog Entries: 5
A simple light bulb can be used to create an AGC (automatic gain control) effect. Why not post your simulation?
__________________
..
"Good enough is enemy of the best." An old engineering saying, Author unknown.
General info:
If you have a question, please start a thread/topic. I do not provide gratis assistance via PM nor E-mail, as that would violate the intent of this Board, which is sharing knowledge ... and deprives you of other knowledgeable input. Thanks for the verbage Wookie.
#4
10-03-2009, 08:50 AM
hasie Junior Member Join Date: Sep 2009 Location: South Africa Posts: 35
My simulation on proteus?
How do I post my simultion? WIth a screenshot?
Many thanks guys, will recheck my circuit again quickly allthough I am almost sure its right..
How do I post simulation? Sorry im new to this
Many thanks
#5
10-03-2009, 09:49 AM
hasie Junior Member Join Date: Sep 2009 Location: South Africa Posts: 35
And is this circuit relatively stable and accurate? Or is there a better circuit I can use to get a stable 1KHz sine wave?
Many thx
#6
10-03-2009, 09:51 AM
Bill_Marsden Super Moderator Join Date: Mar 2008 Location: Dallas, TX (GMT-5 w/ DST) Posts: 19,022 Blog Entries: 5
An RC oscillator is never going to be as accurate as a crystal oscillator, but it is accurate enough. I was asking for the schematic locally with the adjusted values. It might give us some insight as to what's happening.
Looking at your schematic, try adding a variable resistor to the emitter. This is better negitive feedback than the base resistor going to the collector. It will control the gain much better.
This transistors biasing needs a lot of work. This is the fundimental reason the waveform is so off.
__________________
..
"Good enough is enemy of the best." An old engineering saying, Author unknown.
General info:
If you have a question, please start a thread/topic. I do not provide gratis assistance via PM nor E-mail, as that would violate the intent of this Board, which is sharing knowledge ... and deprives you of other knowledgeable input. Thanks for the verbage Wookie.
Last edited by Bill_Marsden; 10-03-2009 at 09:57 AM.
#7
10-03-2009, 10:15 AM
hasie Junior Member Join Date: Sep 2009 Location: South Africa Posts: 35
A RC oscillator was askes so I just hope to get a good design.
Here is my circuit at this moment
Ill play around with a variable resistor on the emitter.
many thanks for the help sofar
Any other suggestions?
Thanks
#8
10-03-2009, 10:20 AM
hasie Junior Member Join Date: Sep 2009 Location: South Africa Posts: 35
When adding a 100ohm var resistor to the emitter the amplitude on the oscilloscope gets smaller.
And then when I remove the 1M resistor going from the base to the collector it does not oscillate at all?
Im going to rebuild this circuit again quickly on breadboard. and test it again. any input appreciated.
Many Thanks
Last edited by hasie; 10-03-2009 at 10:42 AM.
#9
10-03-2009, 11:02 AM
hasie Junior Member Join Date: Sep 2009 Location: South Africa Posts: 35
Rebuilt the whole circuit on different breadboard, different components. Still doesnt work.
See also the thing that doesnt make sense to me, is that there are 3 caps in the RC network but only 2 resistors. is the f = 1/((2PI)(R)(C)sqrt6) still correct?
I can use this circuit. http://www.electronics-tutorials.ws/...scillator.html
But I have no idea what to make R4, R5, R6 and C4?
Would this circuit be better to use?
Any ideas appreciated,
Last edited by hasie; 10-03-2009 at 11:15 AM.
#10
10-03-2009, 12:13 PM
Bill_Marsden Super Moderator Join Date: Mar 2008 Location: Dallas, TX (GMT-5 w/ DST) Posts: 19,022 Blog Entries: 5
I like the second circuit much better, it addresses my points about the transistor bias. Thing is, a 3 leg RC oscillator should have matching RC legs, the first design only had 2, the third they were hoping the transistor would be close enough to the correct impedance.
The second circuit actually puts a 3rd RC leg in there. Try using similar bias resistors, and see it if works. You can adjust the gain of this circuit by putting a variable resistor in series with the capacitor, which will allow you to adjust the gain of the transistor amp. While I don't know what the resistor needs to be it will be quite low in value.
Here's Wikipedia's take on it.
http://en.wikipedia.org/wiki/RC_Phase_shift_Oscillator
__________________
..
"Good enough is enemy of the best." An old engineering saying, Author unknown.
General info:
If you have a question, please start a thread/topic. I do not provide gratis assistance via PM nor E-mail, as that would violate the intent of this Board, which is sharing knowledge ... and deprives you of other knowledgeable input. Thanks for the verbage Wookie.
Tags oscillator, phase, shift
Related Site Pages Section Title Worksheet Opamp oscillator circuits Worksheet Positive feedback opamp circuits Worksheet Thyristor application circuits Worksheet Oscillator circuits Worksheet Series and parallel AC circuits Textbook Oscillator circuits -- INCOMPLETE : Practical Analog Semiconductor Circuits
Similar Threads Thread Thread Starter Forum Replies Last Post The Electrician General Electronics Chat 4 09-13-2009 01:58 PM kris_maher Embedded Systems and Microcontrollers 5 08-23-2009 04:58 PM leoed General Electronics Chat 3 02-27-2009 09:07 AM fancy102 The Projects Forum 41 03-22-2008 12:43 AM mksa Homework Help 2 05-09-2006 05:41 AM
Thread Tools Display Modes Linear Mode
Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Forum Rules
Forum Jump User Control Panel Private Messages Subscriptions Who's Online Search Forums Forums Home Electronics Forums General Electronics Chat The Projects Forum Homework Help Electronics Resources Software, Microcomputing, and Communications Forums Programmer's Corner Embedded Systems and Microcontrollers Computing and Networks Radio and Communications Circuits and Projects The Completed Projects Collection Abstract Forums Math Physics General Science All About Circuits Commmunity Forums Off-Topic The Flea Market Feedback and Suggestions
All times are GMT. The time now is 07:00 AM. | 2,216 | 9,046 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-23 | latest | en | 0.948404 |
https://brighterly.com/blog/halloween-math-activities/ | 1,675,067,722,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499804.60/warc/CC-MAIN-20230130070411-20230130100411-00194.warc.gz | 171,293,797 | 17,662 | # 7 Halloween Math Activities for Kids to Engage in During Holidays
Halloween is the first major holiday of the academic year and a fantastic experience for teachers and students. Creative math teachers capitalize on students’ high spirits and enthusiasm to introduce fun learning exercises during the holiday. If you are a parent or teacher interested in ensuring your kids don’t stop learning even while having spooky fun, consider the following seven fun Halloween math activities.
### Halloween Trivia Game
If you need exciting Halloween math activities 5th grade students will enjoy, the Halloween Trivia game is an excellent pick. Rather than reviewing math concepts, the game gets kids working together to answer Halloween-themed questions. When a group answers a question, they check it with the response of the other group and bet on the correct answer. Children can earn points for their answers and the ones they wagered. Many people refer to this activity as Guesses and Wagers.
1:1 Math Lessons
Want to raise a genius?
Start learning Math with Brighterly
### Coordinate Graphing
This Halloween math activity is mainly for 8th graders. Because there’s much graphing in the 8th grade, this game offers something kids can do after their benchmark test. Kids don’t know what they are drawing until they trace the coordinates. Coordinate Graphing typically takes a while to complete, but the suspense makes it worth the wait.
Depending on the graphing activity you want to use and bearing in mind that some require more work than others, you can differentiate the guidelines of these activities. It is critical to give kids a heads-up on locating the coordinate point beforehand. And you can let the kids color the graph to produce a masterpiece.
### Counting Hands
The Counting Hands Dice activity is one of the great Halloween math activities 3rd grade kids may enjoy. This exercise requires kids to create scary manipulatives with a bag of mixed beans, gloves, dice, a spoon, and rubber bands. This activity gives kids much-needed visuals while counting.
Fill the gloves with beans, but not too much so that the glove fingers would be tough to fold. Tie the wrist end with rubber bands and place the gloves on a board or black surface with plus and minus signs between them. Then fold specific fingers on the glove while the kids add or subtract the ones still standing, just like they would while counting on their hands.
### Halloween guessing game
Halloween guessing game is one of the several free Halloween math activities you can practice with your kids. It involves kids guessing the weight of a candy bag. You need a scale and pencil to play this game.
Kids play this game by guessing the weight of a candy bag after a trick-or-treat stash. After guessing, you should weigh the bag to determine its actual weight. You can decide to let the kids write down the weight or determine who has the heaviest bag.
If more than one child plays the game, competition may arise. In that case, rather than weighing candy bags one after another, you can put the candies in a bowl and then ask the participants to guess the total weight of the sweets and weigh them afterwards. That way, everyone wins.
### Pumpkin farm Halloween math game
This exciting Halloween math game is reminiscent of the old pumpkin farm. It is one of the many Halloween math activities 2nd grade kids will enjoy, and you can download and print the game pages and instructions online. You need a pencil or marker, a visual barrier or file folders, and scissors.
Cut the pumpkin games pages from the printout. Help the kids set up a visual block with the file folder or any other opaque barrier to ensure opponents can’t see their boards. After each kid gets a game board and a few pumpkins to hide, they should try to guess where each other’s pumpkin is growing. Correct answer to where a fat pumpkin grows gives 5 points, while a thin pumpkin is worth 2 points.
### Candy Craze Comparisons
The candy craze comparison game is one of the Halloween math activities 4th grade students should learn. The game is perfect for the whole class or small groups and combines skills and luck. Each kid should get a mat with seven spaces made with tape or glue. Then, they must take turns drawing cards with the numbers 1-9 to achieve the highest sum.
Each time a kid chooses a card, they will place it on their mat and can’t move it anymore. After the kids have filled their carpet holes, they can record and compare the numbers. To ensure that this Halloween activity has the ideal impact on kids, the participants should read the numbers aloud as they arrange them from the lowest to the highest.
### Trick or Treat
Tricky or treat is one of the most famous Halloween math activities for preschoolers and kids of all ages. Most people don’t play it with the intention of learning math, but you can up the ante by creating and writing True or False math statements on a card. Then place the card in a Halloween-themed bag.
Afterward, pull out a True or False statement from the bag throughout the day. Present these questions for the kids to answer, and if they answer correctly, they can take a treat from the basket. This activity will ensure the kids actively participate in the Halloween festivities while rewarding their math skills.
## Conclusion
It is evident that there are countless activities to keep your kids learning math irrespective of season. Suffice to say that learning math should be enjoyable for students, rather than anxiety-inducing and tear-dropping. Brighterly has loads of math activities for kindergarten students and kids of all ages which guarantee to rekindle their interest in math.
Book 1 to 1 Math Lesson
• Specify your child’s math level
• Get practice worksheets for self-paced learning
• Your teacher sets up a personalized math learning plan for your child | 1,185 | 5,891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2023-06 | longest | en | 0.927331 |
https://www.dewetron.com/2022/02/mechanical-motor-analysis/ | 1,726,794,717,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652073.91/warc/CC-MAIN-20240919230146-20240920020146-00334.warc.gz | 673,807,650 | 36,705 | # Dewetron blog news
the measurable difference
## Mechanical Motor Analysis
### What is mechanical motor analysis?
Apart from electrical parameters, mechanical variables play a major role in practice. When thinking of an electric motor, for example, quantities such as current and voltage first come to mind. However, other characteristics are also important for the operation of electric motors: How fast does the motor work? What noise does it generate? Do vibrations occur? These are all mechanical variables that one tries to observe by means of mechanical motor analysis.
Mechanical motor analysis obviously finds application in the automotive industry. However, mechanical properties of engines are also important in other industries, for example when operating an emergency power generator. In many cases, mechanical characteristics are even more important than electrical ones, such as when flying a propeller-driven airplane.
### What variables play a role?
What do you look for when buying a car? For many, the horsepower or even speed of the car is a crucial factor. To determine these quantities, the torque and rotational speed are identified in the mechanical motor analysis. Depending on these two properties, one can calculate the power and efficiency of an engine.
In mechanical motor analysis, it is also very important to determine the optimum operating point. An engine can often have power losses at very high or low speeds. In everyday use of a car, however, these losses should be as low as possible, because otherwise the gas station bill may be high. Mechanical motor analysis can now be used in the development of engines to minimize power loss in typical ranges.
Another important part of mechanical motor analysis is the occurrence of vibrations and noise. As already explained in the blog post on NVH, minimizing these variables improves the driving experience, and can therefore reduce the number of returns on already sold models. But vibration also creates some hazard potential. If the vibration of an engine is too strong, this leads to premature aging of components. If a vibration also encounters a resonant frequency, the vibration can be amplified to such an extent that components may even break down while the vehicle is in motion. Engineers have therefore to ensure that this never happens.
### DEWETRON’s solution for mechanical motor analysis
DEWETRON is a manufacturer of high-precision and modular measuring and testing equipment. Apart from this, we develop the OXYGEN software, which is ideally suited for mechanical motor analysis.
OXYGEN offers a wide range of features. For example, you can directly combine speed and torque to form a performance indicator. OXYGEN also makes it easy for you to evaluate sound pressure level and vibration. You can choose between logarithmic and linear scales, take the frequency-weighted sound pressure level into account and even display the data in a matrix map.
The matrix map is also the perfect tool for creating a power characteristics map of a motor. This is necessary to find the optimum operating point of an engine. The following figure shows how such a performance map may look like.
Matrix map for analyzing efficiency (color relates to efficiency)
OXYGEN also supports a variety of basic analysis tools. For example, you can create FFT or CPB spectra. To bring you even closer to this topic, we have already produced a whole video tutorial on mechanical motor analysis, which you can find here.
If we have caught your interest, you can find all information about OXYGEN and our measurement modules on our DEWETRON website. Video tutorials, whitepapers and much more can be found there as well. To stay up to date, you can also follow us on Twitter or LinkedIn. | 718 | 3,773 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.935998 |
https://www.meritnation.com/ask-answer/question/half-the-perimeter-of-a-rectangular-garden-whose-length-is/pair-of-linear-equations-in-two-variables/6294057 | 1,653,215,167,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545326.51/warc/CC-MAIN-20220522094818-20220522124818-00320.warc.gz | 1,036,005,934 | 8,383 | # Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Let the width of the garden be x and length be y.
According to the question,
yx = 4 (1)
y + x = 36 (2)
yx = 4
y = x + 4
x 0 8 12 y 4 12 16
y + x = 36
x 0 36 16 y 36 0 20
Hence, the graphic representation is as follows.
From the figure, it can be observed that these lines are intersecting each other at only point i.e., (16, 20). Therefore, the length and width of the given garden is 20 m and 16 m respectively.
• -2
What are you looking for? | 184 | 588 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2022-21 | latest | en | 0.903631 |
https://proofwiki.org/wiki/Paracompact_Space_is_Countably_Paracompact | 1,695,986,671,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510501.83/warc/CC-MAIN-20230929090526-20230929120526-00875.warc.gz | 518,368,562 | 10,680 | # Paracompact Space is Countably Paracompact
## Theorem
Let $T = \struct {S, \tau}$ be a paracompact space.
Then $T$ is a countably paracompact space.
## Proof
From the definition, $T$ is paracompact space if and only if every open cover of $T$ has an open refinement which is locally finite.
This also applies to all countable open covers.
So every countable open cover of $T$ has an open refinement which is locally finite.
This is precisely the definition for a countably paracompact space.
$\blacksquare$ | 138 | 517 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-40 | latest | en | 0.911767 |
http://www.jiskha.com/display.cgi?id=1284580618 | 1,498,363,033,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320395.62/warc/CC-MAIN-20170625032210-20170625052210-00389.warc.gz | 578,280,144 | 3,658 | # Math.. sorry for posting so many, I just don't get
posted by .
Let W represent the width of a rectangle, the length is 7cm more then the width
A. 7cm more than four times the width
How do I write that into an algebraic expression?
• Math.. sorry for posting so many, I just don't get -
4W + 7 | 82 | 299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-26 | latest | en | 0.917804 |
http://fungrim.org/entry/8a316c/ | 1,582,623,173,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146064.76/warc/CC-MAIN-20200225080028-20200225110028-00482.warc.gz | 60,272,460 | 4,739 | # Fungrim entry: 8a316c
$\theta_{3}^{4}\!\left(0, \tau\right) = 1 + 8 \sum_{n=0}^{\infty} \frac{2 n {q}^{2 n}}{1 + {q}^{2 n}} + 8 \sum_{n=0}^{\infty} \frac{\left(2 n + 1\right) {q}^{2 n + 1}}{1 - {q}^{2 n + 1}}\; \text{ where } q = {e}^{\pi i \tau}$
Assumptions:$\tau \in \mathbb{H}$
TeX:
\theta_{3}^{4}\!\left(0, \tau\right) = 1 + 8 \sum_{n=0}^{\infty} \frac{2 n {q}^{2 n}}{1 + {q}^{2 n}} + 8 \sum_{n=0}^{\infty} \frac{\left(2 n + 1\right) {q}^{2 n + 1}}{1 - {q}^{2 n + 1}}\; \text{ where } q = {e}^{\pi i \tau}
\tau \in \mathbb{H}
Definitions:
Fungrim symbol Notation Short description
Pow${a}^{b}$ Power
JacobiTheta$\theta_{j}\!\left(z , \tau\right)$ Jacobi theta function
Sum$\sum_{n} f(n)$ Sum
Infinity$\infty$ Positive infinity
Exp${e}^{z}$ Exponential function
Pi$\pi$ The constant pi (3.14...)
ConstI$i$ Imaginary unit
HH$\mathbb{H}$ Upper complex half-plane
Source code for this entry:
Entry(ID("8a316c"),
Formula(Equal(Pow(JacobiTheta(3, 0, tau), 4), Where(Add(Add(1, Mul(8, Sum(Div(Mul(Mul(2, n), Pow(q, Mul(2, n))), Add(1, Pow(q, Mul(2, n)))), For(n, 0, Infinity)))), Mul(8, Sum(Div(Mul(Add(Mul(2, n), 1), Pow(q, Add(Mul(2, n), 1))), Sub(1, Pow(q, Add(Mul(2, n), 1)))), For(n, 0, Infinity)))), Equal(q, Exp(Mul(Mul(Pi, ConstI), tau)))))),
Variables(tau),
Assumptions(Element(tau, HH)))
## Topics using this entry
Copyright (C) Fredrik Johansson and contributors. Fungrim is provided under the MIT license. The source code is on GitHub.
2020-01-31 18:09:28.494564 UTC | 626 | 1,483 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 10, "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} | 2.71875 | 3 | CC-MAIN-2020-10 | latest | en | 0.346381 |
https://duolifeusa.com/index.php/2021/08/06/the-secret-to-getting-the-best-numbers-to-play-keno-using-numbers-1-to-eighty-6/ | 1,701,623,919,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.42/warc/CC-MAIN-20231203161435-20231203191435-00017.warc.gz | 270,070,500 | 37,450 | Keno is a game of chance. The principle objective of the game is to match as many numbers as potential out of the 20 numbers that will be drawn. Mathematical analysis show that the odds of getting 1 number alone is one in 86,446 or roughly 1.1%. Getting all 20 numbers is close to unimaginable statistically. How then can one decide the perfect numbers to play Keno using numbers 1 to 80? Statistically, his could also be unattainable, but listed below are just a few ideas that can help improve the odds of profitable in Keno.
Overview Earlier Successful Numbers
In order to see if there’s a pattern, one can evaluation previous Keno results. This, however, may be very tedious, because this means getting tons of Keno outcomes and trying to investigate if all these numbers make a pattern. Evaluation can involve just the day gone by’s results, a week’s price of results, or a month’s value of results. Those that are really into the game may even get a full yr’s data for analysis. This may also help a player decide if certain numbers always make it to the lower, or if there are some numbers that don’t make it in any respect, since you may also win in Keno if not one of the numbers picked matched the drawn numbers. | 273 | 1,224 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2023-50 | longest | en | 0.949855 |
http://www.askiitians.com/forums/Mechanics/10/3977/kinematics.htm | 1,516,268,903,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887224.19/warc/CC-MAIN-20180118091548-20180118111548-00123.warc.gz | 402,387,276 | 27,405 | Click to Chat
1800-2000-838
+91-120-4616500
CART 0
• 0
MY CART (5)
Use Coupon: CART20 and get 20% off on all online Study Material
ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.
There are no items in this cart.
Continue Shopping
` Two boats A and B move away from a bauy anchored at the middle of a river along the mutually perpendicular straight lines,the boat A along the river and the boat B across the river.Having moved off equal distance from the bauy the boats returned.Find the ratio of times of boats Ta/Tb if the velocity of both the boats with respect to water is n=1.2 times greater than the stream velocity.`
8 years ago
8 Points
``` Considering for the boat B to have it's velocity in such a direction so that it travels straight across the river on the perpendicular bisector of river's axis(path of boat A).Then for the same distance travelled D
Value of Ta=(D/V)*[1/2.2+1/.2] For Downstream velocity of boat A=vel. of A+vel. of stream ;vel. of stream=V
while Upstream vel. of A=vel. of A-vel. of stream ;vel. of A (&B)=1.2*vel. of stream
Tb=(D/V)*[2/0.663] Magnitude of vel. of boat B is 1.2V of whose component in the direction of flow of river
=stream vel.=V
or,(1.2V)Cosθ=V (θ=angle made by 1.2V with flow of river but in such a way that it is
opposite to stream vel. V)
From above we get (1.2V)Sinθ which is the across component to cross distance D.
Ta/Tb=[1/2.2+1/.2]/[2/0.663] =1.81
```
8 years ago
Think You Can Provide A Better Answer ?
## Other Related Questions on Mechanics
View all Questions »
• Complete Physics Course - Class 12
• OFFERED PRICE: Rs. 2,756
• View Details
• Complete Physics Course - Class 11
• OFFERED PRICE: Rs. 2,968
• View Details | 520 | 1,742 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2018-05 | latest | en | 0.854469 |
http://mathhelpforum.com/algebra/106025-graphing-absolute-value-inequalities.html | 1,529,793,000,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865250.0/warc/CC-MAIN-20180623210406-20180623230406-00622.warc.gz | 214,489,964 | 11,658 | # Thread: Graphing absolute value inequalities
1. ## Graphing absolute value inequalities
I've reached a standstill on my math homework, can someone explain how to do this? I'll post one of my questions and if possible can you show me the graph and how you would reach that point? I'd appreciate it. Also can you explain what each part of the equation means?
$\displaystyle f(x)=3-|x-2|$
What would it change if that $\displaystyle 3$ was a fraction such as $\displaystyle 1/2$?
What about if there was a number after it being added? Subtracted?
2. See this. Can you graph $\displaystyle 3-(x-2)$ and $\displaystyle 3+(x-2)$? When you want to plot functions with absolute values you should consider different cases. You should see that if $\displaystyle x>2$ then you can leave the absolute value and if $\displaystyle x<2$ then $\displaystyle |x-2|=-(x-2)$. Maybe you can take a look at this as well.
To your questions: there is no basic difference in the graphing method if you change 3 or equivalently add/subtract numbers to/from it. You can try plotting in WolframAlpha.
3. Originally Posted by james_bond
See this. Can you graph $\displaystyle 3-(x-2)$ and $\displaystyle 3+(x-2)$? When you want to plot functions with absolute values you should consider different cases. You should see that if $\displaystyle x>2$ then you can leave the absolute value and if $\displaystyle x<2$ then $\displaystyle |x-2|=-(x-2)$. Maybe you can take a look at this as well.
To your questions: there no basic difference in the graphing method if you change 3 or equivalently add/subtract numbers to/from it. You can try plotting in WolframAlpha.
Wolfram should be excessively helpful for the rest of this year but let me take a second to understand, you just make an input output system and using the absolute value it should eventually make a V shape... hard to describe but... I think I get it. Hm.
4. Originally Posted by jscalalamoboy
I've reached a standstill on my math homework, can someone explain how to do this? I'll post one of my questions and if possible can you show me the graph and how you would reach that point? I'd appreciate it. Also can you explain what each part of the equation means?
$\displaystyle f(x)=3-|x-2|$
What would it change if that $\displaystyle 3$ was a fraction such as $\displaystyle 1/2$?
What about if there was a number after it being added? Subtracted?
It should be easier to plot the graph of -|x| (reflect |x| in the x axis blue line) and then move it two units to the right on the x axis (green line).
Finally move it 3 up on the y axis (red line)
5. Originally Posted by e^(i*pi)
It should be easier to plot the graph of -|x| (reflect |x| in the x axis blue line) and then move it two units to the right on the x axis (green line).
Finally move it 3 up on the y axis (red line)
That's how the teacher explained it. I'll remember that, thanks. Haha, you're using Linux! What kind? I got UBUNTU and Grub wiped itself off my PC, so I have to use my Mac for now until I can find my Livedisk... Anyways, is there any possible way to get KM Plot on a Mac and how much does KM Plot cost?
6. Originally Posted by jscalalamoboy
That's how the teacher explained it. I'll remember that, thanks. Haha, you're using Linux! What kind? I got UBUNTU and Grub wiped itself off my PC, so I have to use my Mac for now until I can find my Livedisk... Anyways, is there any possible way to get KM Plot on a Mac and how much does KM Plot cost?
I use Arch+KDEmod
You may be able to get KmPlot as part of the KDE suite for mac but it's likely to end up being bloated and not really worth it. May be better to find the live disk first . Also KmPlot is free as in both speech and beer
7. Originally Posted by e^(i*pi)
I use Arch+KDEmod
You may be able to get KmPlot as part of the KDE suite for mac but it's likely to end up being bloated and not really worth it. May be better to find the live disk first . Also KmPlot is free as in both speech and beer
Yeah every time I boot up I get a GRUB error that says it can't find the disk.
8. Originally Posted by jscalalamoboy
Yeah every time I boot up I get a GRUB error that says it can't find the disk.
Chances are you removed the partition grub part 2 was residing on. Nonetheless to avoid spamming this thread it would be better to continue via PM or VM
9. Originally Posted by e^(i*pi)
Chances are you removed the partition grub part 2 was residing on. Nonetheless to avoid spamming this thread it would be better to continue via PM or VM
Okay thanks for your help anyway! I finished that part of my homework now. Test on Monday.
"Is the ordered pair $\displaystyle (2,-1)$ a solution for the inequality $\displaystyle 5x-3y$<$\displaystyle -10$?"
I just put $\displaystyle 2$ in for $\displaystyle x$ and $\displaystyle -1$ in for $\displaystyle y$, correct?
10. Originally Posted by jscalalamoboy
Okay thanks for your help anyway! I finished that part of my homework now. Test on Monday.
"Is the ordered pair $\displaystyle (2,-1)$ a solution for the inequality $\displaystyle 5x-3y$<$\displaystyle -10$?"
I just put $\displaystyle 2$ in for $\displaystyle x$ and $\displaystyle -1$ in for $\displaystyle y$, correct?
Yes you do.
I make it that the ordered pair is not a solution | 1,356 | 5,268 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-26 | latest | en | 0.898434 |
https://bookofproofs.github.io/branches/geometry/euclidean-geometry/elements-euclid/book--1-plane-geometry/triangles-of-equal-area-i.html | 1,702,285,320,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103810.88/warc/CC-MAIN-20231211080606-20231211110606-00222.warc.gz | 165,976,903 | 3,307 | # Proposition: 1.37: Triangles of Equal Area I
### (Proposition 37 from Book 1 of Euclid's “Elements”)
Triangles which are on the same base and between the same parallels are equal to one another. * Let $ABC$ and $DBC$ be triangles on the same base $BC$, and between the same parallels $AD$ and $BC$. * I say that triangle $ABC$ is equal to triangle $DBC$.
### Modern Formulation
Triangles ($$\triangle{ABC}$$, $$\triangle{DBC}$$) on the same base ($$\overline{BC}$$) and standing between the same parallels ($$\overline{AD}$$, $$\overline{BC}$$) are equal in area.
Proofs: 1
Proofs: 1 2
Thank you to the contributors under CC BY-SA 4.0!
Github:
non-Github:
@Calahan
@Casey
@Fitzpatrick
### References
#### Adapted from CC BY-SA 3.0 Sources:
1. Callahan, Daniel: "Euclid’s 'Elements' Redux" 2014 | 244 | 807 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-50 | latest | en | 0.76378 |
http://www.uopetutors.com/tags/-fin-575-contract-paper | 1,620,450,552,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988837.67/warc/CC-MAIN-20210508031423-20210508061423-00268.warc.gz | 89,414,734 | 54,197 | # FIN 575 Final Exam Questions and Answers free Calculator FIN 575 Project Budget & Finance
Students, we are informing that a huge number of study resources you can take for your final exam 2016-2017. The FIN 575 Final Exam study guides are available and the University of Phoenix has been recently released a notification for various specialized offers on all courses. Aspirants should join Uop E Tutors as this offer get expired. Here we are provided platform for final exam 2017. Study material and test papers etc. are as given at tutorial. For more details, www.uopetutors.com Go to the site and check the fabulous courses in a range. Great experience are waiting for you.
## Start studying Financing Final Exam Review. Learn from study guides, test series, and more with study tools.
### 1.During the project initiation, a project charter is created. The project charter should include which of the following?
• Project managers expenses
• Analysis of budget
• Selection of the senior project manager
• Projects high-level deliverables
### 2.A project's budget should be based on a company’s
• strategy and financial goals
• profitability
• financial goals and equity
• debt load and equity
Download FIN 575 Final Exam complete solutions with FIN 575 Final Exam Answers
### 3.Earned value management is a technique used to integrate projects
• resources
• scope, schedule, and resources
• schedule, costs, and benefits
• costs and profits
### 4.Bill’s Billiards has total assets of \$8 million and a total asset turnover of 2.9 times. If the return on assets is 11%, what is Bill's profit margin?
• 11%
• 4.10%
• 2.50%
• 3.79%
Complete FIN 575 Final Exam Answers right here FIN 575 Final Exam Calculator
### 5.What are the acceptance criteria for NPV?
• If the NPV is less that \$0, accept the project.
• If the NPV is greater than \$0, accept the project.
• If the IRR is equal to 0%, reject the project.
• If the NPV is equal to the discounted payback, accept the project.
### 6.The risk response plan answers what question?
• What can be done if risk occurs? What is the backup plan?
• What are project costs?
• There is no need to plan for risk seldom occurs in a project.
• How risk is to be managed
### 7.For the most recent year, Cal’s Cats had sales of \$380,000, cost of goods sold of \$93,000, depreciation expense of \$47,000, and additions to retained earnings of \$61,420. The firm had \$52,000 in interest expense, and 34% tax rate. What were the times interest earned ratio?
• 2.2
• 5.8
• 4.61
• 2.8
FIN 575 Final Exam on here just a click away FIN 575 contract paper
• 14%
• 12%
• 51%
• 23.40%
### 9. What are the components of project planning that need monitoring?
• Resource procurement and quality
• Project cost and risk
• Project cost, risk, resource procurement and quality
• Quality and control
Want to see the complete FIN 575 Final Exam..?? Click FIN 575 project budget and finance
### 10.During project planning, the project team creates a work breakdown structure that details work tasks that must be completed. The work breakdown structure should include a
• schedule of when every task will start and be completed
• schedule of project staff meetings
• set of management tasks
• budget analysis
• 1
• 0.54
• 5.4
• 1.8
### 12.Why is the communication plan a crucial factor in project success?
• Ensures the timely generation, collection, storage, and disposition of project information
• Facilitates upper management communication with the workers
• Reduces rumors in the organization
• Communicates the economic value of the project to management
### 13.A company’s assets are financed with
• debt
• equity
• equity or debt
• equity and debt
Click on FIN 575 Final Exam and download FIN 575 Final Exam Answers free
### 14.Part of financial planning for projects involves the understanding of the inflows and outflows of cash that will be created by the project. What tool can be used to track these cash flows?
• A NPV flow sheet
• Profitability work sheet.
• Project cash flow worksheet
• Cash flow table
• 1.89
• 1.13
• 1.21
• 2.1
### 16.What ratio measures a firm’s degree of indebtedness?
• Debt ratio
• Quick ratio
• Fixed coverage ratio
• Times interest earned ratio
FIN 575 Final Exam related help? Click to download FIN 575 Final Exam University of phoenix
### 17.Which one of these terms is a type of debt financing?
• Stock repurchases plans
• Collateral
• Bearer bonds
### 18.The sum of the percentage of equity and debt multiplied by their respective cost is called
• weighted average cost of capital
• capital asset pricing model
• market value added
• economic value added.
FIN 575 Final Exam download now uop FIN 575 Final Exam
### 19.Profitability ratios all have what same figure in the numerator?
• Book value per
• Net income
• Price per share
• Total assets
• 1.27
• 0.41
• 0.82
• 1.82
### 21.An investment in a project should be undertaken only if the expected return is greater than the
• NPV
• WACC
• payback method
• economic value added
Search FIN 575 Final Exam Answers click here FIN 575 Final Exam
• 25 days
• 32 days
• 28 days
• 14 days
### 23.You are considering a project with an initial cash outlay of \$160,000 and expected free cash flows of \$40,000 at the end of each year for 6 years. The required rate of return for this project is 10%. What is the project’s payback period?
• 4 years
• 4.5 years
• 6 years
• 5 years
Buy FIN 575 Final Exam Tutorial Guide Step by Step
### 25.What is the primary weakness commonly associated with the use of the payback method to evaluate a proposed investment?
• This approach fails to take into account the time factor in the time value of money.
• The payback method uses the discounted cash flow process.
• The payback method is able to recognize cash flows that occur after the payback period.
• The payback method is not appropriate for evaluating small projects.
### 26.Fijisawa, Inc. is considering a major expansion of its product line and has estimated the following free cash flows associated with such an expansion. The initial outlay associated with the expansion would be \$1,950,000, and the project would generate free cash flows of \$450,000 per year for 6 years. The appropriate required rate of return is 9%. Calculate the net present value and the internal rate of return.
• NPV=\$66,098, IRR=10.5
• NPV=\$72,097, IRR=9.5
• NPV=\$68,663, IRR=10.2
• NPV=\$69,368, IRR=10
Get FIN 575 Final Exam Questions and Answers free
### 27.Cost normally falls into the domain of managerial accounting and has 4 essential proposes. Select the answer that is an essential function of cost.
• Used to calculate earned value cost
• Used to calculate executive stock options
• Used to calculate inventory costs
• Used for planning future activities or budgets
### 28.Select the answer that is an example of a cost classification?
• Credit cost
• Fixed cost
• Retail cost
• Inventory cost
Find the FIN 575 Final Exam right here FIN 575 Final Exam 2014
### 29.What are the four secondary processes in project control?
• Schedule control, change control, risk control, and quality assurance control
• Value control, Inventory control, schedule control and quality control
• Organizational control, cost control, inventory control, and risk control
• Stakeholder control, organization control, risk control, and change control
### 30.Stokes, Inc. has net working capital of \$7,900, current liabilities of \$5,220, and inventory of \$2,000. What is the current ratio?
• 2.1
• 0.77
• 1.89
• 1.51
Whole FIN 575 Final Exam solutions right here FIN 575 final exam
FIN 575 Final Exam & FIN/575 project budget and finance
FIN 575 Final Exam / fin/575 project budget and finance : Good news for the candidates who want to do preparing of FIN 575 Final Exam. Good news that UOP will going to release soon fin/575 project budget and finance. Eligible and interested candidates who want to do great marks in University of Phoenix can buy online FIN 575 Final Exam Questions and Answers free. Many Lakhs Candidates are eagerly waiting for the fin 575 final exam tutorial. Now wait is over and FIN 575 Final Exam Questions With Answers Tutorial has come out.
Some of Uop E Tutor’s assignments and final exam for the University of Phoenix are given below:-
FIN 575 Final Exam
LAW 421 Final Exam
LAW 575 Final Exam
LDR 300 Final Exam
BUS 475 Capstone Final Examination Part 1
FIN 571 Connect Problems
STR 581 Capstone Final Exam part 1 | 2,089 | 8,505 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2021-21 | latest | en | 0.877194 |
http://davidstutz.de/understanding-deep-image-representations-by-inverting-them-mahendran-and-vedaldi/ | 1,513,134,775,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948521188.19/warc/CC-MAIN-20171213030444-20171213050444-00073.warc.gz | 70,013,440 | 9,332 | # DAVIDSTUTZ
Check out the latest superpixel benchmark — Superpixel Benchmark (2016) — and let me know your opinion! @david_stutz
27thSEPTEMBER2017
Aravindh Mahendran, Andrea Vedaldi. Understanding deep image representations by inverting them. CVPR, 2015.
Mahendran and Vedaldi propose a visualization technique allowing to visualize higher level features within deep representations. Essentially, the idea is to compute a reconstruction (based on an adequate image prior) which most closely results in the given representation. The approach is applied to AlexNet [] as well as convolutional neural networks mimicking DSIFT [][] and HoG [].
The underlying optimization problem takes the form
$x^\ast = \arg\min_{x\in \mathbb{R}^{H \times W \times C}} l(\Phi(x), \Phi_0) + \lambda \mathcal{R}(x)$
where $\Phi(x)$ refers to the representation obtained on image $x$, and $\Phi(x_0) = \Phi_0$ is the representation about to be visualized. The loss is the Euclidean distance and as regularization, Mahendran and Vedaldi use a combination of the $\alpha$-norm
$\mathcal{R}_\alpha (x) = \|x\|_\alpha^\alpha$
and total variation (in its discrete form):
$\mathcal{R}_{V^\beta}(x) = \sum_{i,j}\left((x_{i,j + 1} - x_{i,j})^2 + (x_{i + 1,j} - x_{i,j})^2\right)^{\frac{\beta}{2}}$
regarding the balance of these three terms, some caveats need to be considered. First, the Euclidean distance is normalized by $\|\Phi_0\|_2^2$. Furthermore, $\Phi(x)$ is replaced by a scaled version $\Phi(\sigma x)$ in order to address the first convolutional layers being not completely insensitive to scaling. $\sigma$ is set to the average Euclidean norm of the images. The final objective takes the form
$\|\Phi(\sigma x) - \Phi_0\|_2^2/\|\Phi_0\|_2^2 + \lambda_\alpha \mathcal{R}_alpha(x) + \lambda_{V^\beta}\mathcal{R}_{V^\beta}(x)$.
With appropriate weighting parameters as detailed in the paper. The objective is minimized using gradient descent with momentum.
For experiments, they consider AlexNet [] as well as convolutional neural networks reconstructingDSIFT [17,20] and HoG [4]. In particular, they detail how DSIFT and HoG can be expressed as convolutional neural networks by converting the individual operations to commonly used layers. Details can be found in the paper.
Qualitative results in Figure 1 show the reconstruction of an input image for representations obtained from the different layers in AlexNet. Note that for each layer, the weighting parameters are chosen separately.
• [] N. Dalal and B. Triggs. Histograms of oriented gradients for human detection. In CVPR, 2005.
• [] D. G. Lowe. Object recognition from local scale-invariant features. In ICCV, 1999.
• [] E. Nowak, F. Jurie, and B. Triggs. Sampling strategies for bag-of-features image classification. In ECCV, 2006.
• [] A. Krizhevsky, I. Sutskever, and G. E. Hinton. Imagenet classification with deep convolutional neural networks. In NIPS, 2012.
What is your opinion on the summarized work? Or do you know related work that is of interest? Let me know your thoughts in the comments below or get in touch with me: | 812 | 3,090 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2017-51 | longest | en | 0.851326 |
http://gmatclub.com/forum/last-mgmat-test-52703.html#p373117 | 1,474,889,214,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738660746.32/warc/CC-MAIN-20160924173740-00262-ip-10-143-35-109.ec2.internal.warc.gz | 110,094,314 | 46,072 | Find all School-related info fast with the new School-Specific MBA Forum
It is currently 26 Sep 2016, 04:26
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Last MGMAT test 690
Author Message
TAGS:
### Hide Tags
CEO
Joined: 29 Mar 2007
Posts: 2583
Followers: 19
Kudos [?]: 387 [0], given: 0
### Show Tags
22 Sep 2007, 16:14
Well I took my last MGMAT test today and scored a 690. First I wanna be honest to say that yes I did occasionally use the pause button in the quant section. Not in verbal cept when dog decided to poop on the floor
=(
The more I experience MGMAT quant, I really just have to say it is not representative of at least GMATPrep quant.
The math in MGMAT is not only harder, but takes so much more time to solve.
My first GMATPrep I was amazed at how easy the Quant was. Although I messed up in the last 8 questions, this had nothing to do w/ my abilities. Just timing. My second GMATPrep, proved this.
I dunno just venting, although 690 is nothing to complain about, just wish I had more GMATPreps =(.
Senior Manager
Joined: 28 Jun 2007
Posts: 462
Followers: 3
Kudos [?]: 6 [0], given: 0
### Show Tags
23 Sep 2007, 08:35
I actually differ. The first few MGMAT CATs were insane. But the last two MGMAT ones were pretty simple.
After the first test, I gave the quants section 90 minutes and I still struggled to finish on time. For the 5th one , I gave myself 80 minutes and finished with 10 minutes to go. For the sixth one, I gave myself 75 minutes and I still finished with 6-7 minutes left.
I also ended up with a score of 51 on both cases. Now, the questions must have turned easy for me to see a positive change in both time and score especially because I haven't done any prep for quants. (I have exactly 24 hrs to go for the test though).
I think their last few CATs started getting more and more representative of the real GMAT.
In Verbal... I started with a 34(19 wrong)... finished with a 42(11 wrong). This sounds more credible than my math improvement.
Manager
Joined: 07 Mar 2007
Posts: 200
Followers: 1
Kudos [?]: 25 [0], given: 0
### Show Tags
23 Sep 2007, 13:38
Good Score GMATBLACKBELT....you'll do just great on the real exam!
Similar topics Replies Last post
Similar
Topics:
1 610 to 690 with MGMAT class 8 04 Jul 2011, 19:52
Practice Tests Similar to MGMAT 2 11 May 2008, 09:05
MGMAT Test 1: Not pleased 7 19 Apr 2008, 11:46
Kaplan paper test! 690 5 29 Mar 2008, 04:42
MGMAT Tests 1 18 Mar 2008, 14:43
Display posts from previous: Sort by | 845 | 3,019 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2016-40 | latest | en | 0.953676 |
http://mathhelpforum.com/statistics/217782-best-way-tell-difference-t-z-hypothesis-testing.html | 1,480,915,578,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541525.50/warc/CC-MAIN-20161202170901-00368-ip-10-31-129-80.ec2.internal.warc.gz | 177,192,904 | 12,712 | # Thread: Best Way To tell The Difference from T&Z for Hypothesis Testing?
1. ## Best Way To tell The Difference from T&Z for Hypothesis Testing?
This confuses me quite a bit. What are certain words I need to look for to know. I know the rule with if the STD is known or not... Sometimes it's given to me in S and the other in the STD symbol. But can I get some help on a way to really tell the difference, please?
2. ## Re: Best Way To tell The Difference from T&Z for Hypothesis Testing?
Usually S^2 is the sample variance (variance estimated from a sample) and SD refers to a population standard deviation.
You will need to read each source carefully to know exactly what is what though in general.
3. ## Re: Best Way To tell The Difference from T&Z for Hypothesis Testing?
Originally Posted by chiro
Usually S^2 is the sample variance (variance estimated from a sample) and SD refers to a population standard deviation.
You will need to read each source carefully to know exactly what is what though in general.
But let me give you a better example of both.
A psychologist obtains a random sample of 20 mothers in the first trimester of their pregnancy. the mothers are asked to play Mozart in the house at least 30 minutes each day until they give birth. After 5 years, the child is administred an IQ test. We know that IQ's are normally distribute with a mean of 100 and STD of 15. If the IQ's of the 20 children in the study result in a sample mean of 104.2 is there evidence that the children have higher IQ's? Use the (alpha symbol) =.05 level of significance.
In 2001, the mean contract interest rate for a conventional 30 year first loan for the purchase of a single home was 6.3%, according to the US federal Housing Board. A real estate agent believes that interest rates lower today and obains a random sample of 41 recent 30 year convetional loans. The mean interest rate was found to be 6.05% with a STD of 1.75 percent. Is this enough evidence at the the alpha=.05 level of significance?
First one you will use, Z, the second one T. So I am wondering how do you really tell the difference?
Another example, I got wrong on a previous exam because I took the wrong one.
Stacie Statistic is doing her BSP on the time that -full time local student spend on HW perweek. Her group has a random sample of 40 full time students with a mean 32 hours and STD of 8 hours.
I picked Z for this but it was suppose to be T.
4. ## Re: Best Way To tell The Difference from T&Z for Hypothesis Testing?
If you are talking about underlying distributions then you just need to know what kind of test statistic you are using.
The T distribution typically uses the sample mean and sample variance while a lot of results for using Z are based on the Central Limit Theorem and the Asymptotic results like the Wald Statistic and others.
If you know what kind of information you are using and the nature of the test statistic, you will know whether its a t-distribution or a Normal distribution.
5. ## Re: Best Way To tell The Difference from T&Z for Hypothesis Testing?
Originally Posted by chiro
If you are talking about underlying distributions then you just need to know what kind of test statistic you are using.
The T distribution typically uses the sample mean and sample variance while a lot of results for using Z are based on the Central Limit Theorem and the Asymptotic results like the Wald Statistic and others.
If you know what kind of information you are using and the nature of the test statistic, you will know whether its a t-distribution or a Normal distribution.
Thank you for replying btw. I have 1 more question.
Another example(If you do not mind), I got wrong on a previous exam because I took the wrong one.
Stacie Statistic is doing her BSP on the time that -full time local student spend on HW perweek. Her group has a random sample of 40 full time students with a mean 32 hours and STD of 8 hours.
I picked Z for this but it was suppose to be T. I am wondering why was this suppose to be T instead of Z?
Each chapter it tells me which one I am going to be using but when it comes to the exam both the questions sounds the same and I get them mixed up.
6. ## Re: Best Way To tell The Difference from T&Z for Hypothesis Testing?
The key word here is "sample". Your mean and standard deviation are sample estimates, not population estimates.
7. ## Re: Best Way To tell The Difference from T&Z for Hypothesis Testing?
Yeah, I am really sorry. Wen I look at some of the other questions, they are essentially written the same but yet one is Z the other T(like the ones above). Yeah, I am just trying to go back and re-reading the difference from population and sample.
8. ## Re: Best Way To tell The Difference from T&Z for Hypothesis Testing?
Originally Posted by chiro
The key word here is "sample". Your mean and standard deviation are sample estimates, not population estimates.
Just one thing(kind of going senile here looking at my book and computer for the longest) what do you mean by sample estimates and not population estimates?
9. ## Re: Best Way To tell The Difference from T&Z for Hypothesis Testing?
Sample means that you have a sample drawn from a population. Typically populations are infinite (or really large) and samples are just a small representation of tha ist population.
If you collect data, then it is typically a sample. The only time when it is considered a population would be if it was say a national census or something similar. Otherwise its a sample.
Sample estimates are point estimates of population parameters using the sample. This is the case if the estimator is unbiased. If its biased, you correct it to make it unbiased.
The sample variance estimates the population variance and the sample mean estimates the population mean. Estimators are random variables so they have a distribution and variance. | 1,319 | 5,890 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2016-50 | longest | en | 0.953183 |
https://discuss.codecademy.com/t/help-meee-6-just-weight-and-see/71585 | 1,539,620,498,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583509326.21/warc/CC-MAIN-20181015142752-20181015164252-00017.warc.gz | 686,276,513 | 5,850 | # HELP MEEE 6. Just Weight and See
#1
The exercise that I'm stuck on:
6. Just Weight and See
Errors:
Oops, try again. There seems to be something wrong with your code, see the console window for the error message!
``````lloyd = {
"name": "Lloyd",
"homework": [90.0, 97.0, 75.0, 92.0],
"quizzes": [88.0, 40.0, 94.0],
"tests": [75.0, 90.0]
}
alice = {
"name": "Alice",
"homework": [100.0, 92.0, 98.0, 100.0],
"quizzes": [82.0, 83.0, 91.0],
"tests": [89.0, 97.0]
}
tyler = {
"name": "Tyler",
"homework": [0.0, 87.0, 75.0, 22.0],
"quizzes": [0.0, 75.0, 78.0],
"tests": [100.0, 100.0]
}
students = [lloyd, alice, tyler]
def average(numbers):
total = sum(numbers)
total = float(total)
total =total / len(numbers)
def get_average(student):
homework = average(student["homework"])
quizzes = average(student["quizzes"])
tests = average(students["tests"]
weighted = ("homework"*.1)+("quizzes"*.3)+("tests"*.6)
return weighted``````
#2
look at the line above:
``tests = average(students["tests"]``
in particular, look at the parentheses. That should fix your current issue
#3
``````lloyd = {
"name": "Lloyd",
"homework": [90.0, 97.0, 75.0, 92.0],
"quizzes": [88.0, 40.0, 94.0],
"tests": [75.0, 90.0]
}
alice = {
"name": "Alice",
"homework": [100.0, 92.0, 98.0, 100.0],
"quizzes": [82.0, 83.0, 91.0],
"tests": [89.0, 97.0]
}
tyler = {
"name": "Tyler",
"homework": [0.0, 87.0, 75.0, 22.0],
"quizzes": [0.0, 75.0, 78.0],
"tests": [100.0, 100.0]
}
students = [lloyd, alice, tyler]
def average(numbers):
total = sum(numbers)
total = float(total)
total =total / len(numbers)
def get_average(student):
homework = average(students["homework"])
quizzes = average(students["quizzes"])
tests = average(students["tests"])
weighted = ("homework"*.1)+("quizzes"*.3)+("tests"*.6)
return weighted``````
What??????? I'm confused +_+
#4
Try:
weighted = .1 * homework + .3 * quizzes + .6 * tests
#5
@siik, why? What difference does it make?
@th3b33, here:
`````` homework = average(students["homework"])
quizzes = average(students["quizzes"])
tests = average(students["tests"])``````
students is a list which contains all students. You need to use `student` instead, this is the function parameter and will contain a single student (which is a dictionary), then you can use the keys to get the values
#6
Where?? I don't get it 0-0
#7
#8
did you get it?
#9
I dont know where to put the code
#10
#11
i showed you the lines where you wrongly used students? There are in my reply:
#12
... So I'm supposed to put student names?
#13
you need to use `student` (the function parameter) instead of `students`, see my explanations in that reply
#14
I dont know what the function parameter is...
#15
and im using student now and i dont know what im supposed to do =(
#16
#17
you have one more mistake
see
weighted = ("homework".1)+("quizzes".3)+("tests"*.6)
`homework` ,`quizzes`, `tests` what are these?variable or string?
string need `" "` variable don't
#18
Here is my updated code
``````lloyd = {
"name": "Lloyd",
"homework": [90.0, 97.0, 75.0, 92.0],
"quizzes": [88.0, 40.0, 94.0],
"tests": [75.0, 90.0]
}
alice = {
"name": "Alice",
"homework": [100.0, 92.0, 98.0, 100.0],
"quizzes": [82.0, 83.0, 91.0],
"tests": [89.0, 97.0]
}
tyler = {
"name": "Tyler",
"homework": [0.0, 87.0, 75.0, 22.0],
"quizzes": [0.0, 75.0, 78.0],
"tests": [100.0, 100.0]
}
students = [lloyd, alice, tyler]
def average(numbers):
total = sum(numbers)
total = float(total)
total =total / len(numbers)
def get_average(student):
homework = average(student["homework"])
quizzes = average(student["quizzes"])
tests = average(student["tests"])
weighted = ("homework"*.1)+("quizzes"*.3)+("tests"*.6)
return weighted``````
#19
Thanks dude
I just made the changes and it works, here is my updated code
``````lloyd = {
"name": "Lloyd",
"homework": [90.0, 97.0, 75.0, 92.0],
"quizzes": [88.0, 40.0, 94.0],
"tests": [75.0, 90.0]
}
alice = {
"name": "Alice",
"homework": [100.0, 92.0, 98.0, 100.0],
"quizzes": [82.0, 83.0, 91.0],
"tests": [89.0, 97.0]
}
tyler = {
"name": "Tyler",
"homework": [0.0, 87.0, 75.0, 22.0],
"quizzes": [0.0, 75.0, 78.0],
"tests": [100.0, 100.0]
}
students = [lloyd, alice, tyler]
def average(numbers):
total = sum(numbers)
total = float(total)
total =total / len(numbers) | 1,564 | 4,282 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2018-43 | latest | en | 0.723534 |
https://essaysgate.com/2022/04/26/question-1-you-have-been-contacted-by-a-businesswoman-who-wants-you-to-evaluate-her-companys-leadership-training-program-she-wants-to-know-whether-there-are-pretest-influences-on-leadership/ | 1,669,936,442,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710870.69/warc/CC-MAIN-20221201221914-20221202011914-00176.warc.gz | 278,613,020 | 19,190 | Question 1 You have been contacted by a businesswoman who wants you to evaluate her company’s leadership training program. She wants to know whether there are pretest influences on leadership assessme
Question 1
1. You have been contacted by a businesswoman who wants you to evaluate her company’s leadership training program. She wants to know whether there are pretest influences on leadership assessment. She wants a strong design. What design will you suggest for her?
Solomon four group design
Posttest only control or comparison group design
Pretest-posttest design
Posttest-only design with nonequivalent groups
0.5 points
Question 2
1. Non-experimental qualitative designs have which of the following characteristics?
An active independent variable
Sample a broad range of participants
Relies heavily on verbal rather than numerical data
Uses mostly inferential statistical techniques
0.5 points
Question 3
1. A researcher wants to understand the sub-culture of the KKK in the Deep South. He joins the Klan and studies their behavior as an inside member of their society. This approach would be considered
Phenomenological
Ethnographic
Narrative
Quasi-experimental
0.5 points
Question 4
1. A primary data collection method for case studies is
Interviews
Paper and pencil tests
Behavioral ratings
Physiological measures
0.5 points
Question 5
1. We can be most confident that two variables are causally related when
The independent variable (IV) precedes the dependent variable (DV)
The independent variable (IV) and the dependent variable (DV) are associated
There are no extraneous variables that can explain the relationship between IV and DV
All of the above are present
0.5 points
Question 6
1. The randomized experimental approach generally eliminates bias in assigning participants.
True
False
0.5 points
Question 7
1. When matching is used, the characteristic that is matched must NOT be related to the dependent variable.
True
False
0.5 points
Question 8
1. Purposive sampling is an example of probability sampling.
True
False
0.5 points
Question 9
1. With a stratified random probability sample, the sample needs to be weighted according to the population proportions to most adequately describe the population.
True
False
0.5 points
Question 10
1. A convenience sample always selects participants who are representative of the accessible population.
True
False
Calculate the price of your order
550 words
We'll send you the first draft for approval by September 11, 2018 at 10:52 AM
Total price:
\$26
The price is based on these factors:
Number of pages
Urgency
Basic features
• Free title page and bibliography
• Unlimited revisions
• Plagiarism-free guarantee
• Money-back guarantee
On-demand options
• Writer’s samples
• Part-by-part delivery
• Overnight delivery
• Copies of used sources
Paper format
• 275 words per page
• 12 pt Arial/Times New Roman
• Double line spacing
• Any citation style (APA, MLA, Chicago/Turabian, Harvard)
Our guarantees
Delivering a high-quality product at a reasonable price is not enough anymore.
That’s why we have developed 5 beneficial guarantees that will make your experience with our service enjoyable, easy, and safe.
Money-back guarantee
You have to be 100% sure of the quality of your product to give a money-back guarantee. This describes us perfectly. Make sure that this guarantee is totally transparent.
Zero-plagiarism guarantee
Each paper is composed from scratch, according to your instructions. It is then checked by our plagiarism-detection software. There is no gap where plagiarism could squeeze in.
Free-revision policy
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. | 828 | 3,808 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-49 | longest | en | 0.915101 |
https://www.analystforum.com/t/quiz-impact-of-non-consideration-of-pension-assets-liab-in-wacc/51593 | 1,601,048,739,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400227524.63/warc/CC-MAIN-20200925150904-20200925180904-00355.warc.gz | 721,246,118 | 4,528 | # QUIZ - Impact of non-consideration of pension assets/liab in WACC
I have not seen this in Schweser. but there is something about PE multipliers and Impact of non-consideration of pension assets/liab in WACC. If you do not consider pension assets in WACC- You a) Overestimate PE multiplier? b) underestimate PE multiplier? c) there is no link of PE multiplier and (non)/consideration of pension liabilities? Which is the answer ? Also explain why ?
B) Underestimate Price??? My explanation: by not considering the pension assets, you overestimate the WACC, because of which you underestimate price. Lower price/Earnings as opposed to higher price by considering pension assets. That would be my half-baked answer.
B) Underestimate Price
you need to know if pension assets are riskier than firm assets
^ True, but at least in Schweser I think the guidance is that WACC is usually overestimated if you don’t include pension assets. So, I would assume WACC is overestimated unless we see something that contradicts that. I think B should be the right answer here.
Underestimate PE is the right choice. Right now I do not have the CFAI texts. I will type the relevant 3 sentences relating to this from CFAI text, tonight. you guys are doing great!
even without WACC, you can get to this the formulaic way, By not considering pension assets. Sales turnover ratio is higher --> RoE is higher (Dupont model) --> Higher Earnings (EPS) --> Lower P/E
The CFAI readings say that you are underestimating the P/E multiples of these companies, because the WACC goes into the denominator of the DCF calculation: so if WACC is overstated, then PE is understated. Also they say “generally” the WACC is overstated without accounting for the pensions. Meaning, I guess, it could be understated (maybe if the plan’s 100% allocated to a specific stock?? so, adding the pensions increases the overall net firm beta…) | 436 | 1,903 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.941018 |
https://www.12000.org/my_notes/CAS_integration_tests/reports/rubi_4_16_1_graded/test_cases/1_Algebraic_functions/1.2_Trinomial_products/1.2.1_Quadratic/1.2.1.2-d+e_x-%5Em-a+b_x+c_x%5E2-%5Ep/rese954.htm | 1,695,483,065,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506481.17/warc/CC-MAIN-20230923130827-20230923160827-00776.warc.gz | 709,967,075 | 4,540 | ### 3.954 $$\int (d+e x)^m \sqrt{d^2-e^2 x^2} \, dx$$
Optimal. Leaf size=67 $\frac{(d-e x) \sqrt{d^2-e^2 x^2} (d+e x)^{m+1} \, _2F_1\left (1,m+3;m+\frac{5}{2};\frac{d+e x}{2 d}\right )}{d e (2 m+3)}$
[Out]
((d - e*x)*(d + e*x)^(1 + m)*Sqrt[d^2 - e^2*x^2]*Hypergeometric2F1[1, 3 + m, 5/2 + m, (d + e*x)/(2*d)])/(d*e*(
3 + 2*m))
________________________________________________________________________________________
Rubi [A] time = 0.0469089, antiderivative size = 83, normalized size of antiderivative = 1.24, number of steps used = 3, number of rules used = 3, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.125, Rules used = {680, 678, 69} $-\frac{2^{m+\frac{3}{2}} \left (d^2-e^2 x^2\right )^{3/2} (d+e x)^m \left (\frac{e x}{d}+1\right )^{-m-\frac{3}{2}} \, _2F_1\left (\frac{3}{2},-m-\frac{1}{2};\frac{5}{2};\frac{d-e x}{2 d}\right )}{3 d e}$
Antiderivative was successfully verified.
[In]
Int[(d + e*x)^m*Sqrt[d^2 - e^2*x^2],x]
[Out]
-(2^(3/2 + m)*(d + e*x)^m*(1 + (e*x)/d)^(-3/2 - m)*(d^2 - e^2*x^2)^(3/2)*Hypergeometric2F1[3/2, -1/2 - m, 5/2,
(d - e*x)/(2*d)])/(3*d*e)
Rule 680
Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(d^IntPart[m]*(d + e*x)^FracPart[m]
)/(1 + (e*x)/d)^FracPart[m], Int[(1 + (e*x)/d)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && EqQ[c*d
^2 + a*e^2, 0] && !IntegerQ[p] && !(IntegerQ[m] || GtQ[d, 0])
Rule 678
Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(d^(m - 1)*(a + c*x^2)^(p + 1))/((1
+ (e*x)/d)^(p + 1)*(a/d + (c*x)/e)^(p + 1)), Int[(1 + (e*x)/d)^(m + p)*(a/d + (c*x)/e)^p, x], x] /; FreeQ[{a,
c, d, e, m}, x] && EqQ[c*d^2 + a*e^2, 0] && !IntegerQ[p] && (IntegerQ[m] || GtQ[d, 0]) && !(IGtQ[m, 0] && (
IntegerQ[3*p] || IntegerQ[4*p]))
Rule 69
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))
Rubi steps
\begin{align*} \int (d+e x)^m \sqrt{d^2-e^2 x^2} \, dx &=\left ((d+e x)^m \left (1+\frac{e x}{d}\right )^{-m}\right ) \int \left (1+\frac{e x}{d}\right )^m \sqrt{d^2-e^2 x^2} \, dx\\ &=\frac{\left ((d+e x)^m \left (1+\frac{e x}{d}\right )^{-\frac{3}{2}-m} \left (d^2-e^2 x^2\right )^{3/2}\right ) \int \left (1+\frac{e x}{d}\right )^{\frac{1}{2}+m} \sqrt{d^2-d e x} \, dx}{\left (d^2-d e x\right )^{3/2}}\\ &=-\frac{2^{\frac{3}{2}+m} (d+e x)^m \left (1+\frac{e x}{d}\right )^{-\frac{3}{2}-m} \left (d^2-e^2 x^2\right )^{3/2} \, _2F_1\left (\frac{3}{2},-\frac{1}{2}-m;\frac{5}{2};\frac{d-e x}{2 d}\right )}{3 d e}\\ \end{align*}
Mathematica [A] time = 0.0639009, size = 86, normalized size = 1.28 $-\frac{2^{m+\frac{3}{2}} (d-e x) \sqrt{d^2-e^2 x^2} (d+e x)^m \left (\frac{e x}{d}+1\right )^{-m-\frac{1}{2}} \, _2F_1\left (\frac{3}{2},-m-\frac{1}{2};\frac{5}{2};\frac{d-e x}{2 d}\right )}{3 e}$
Antiderivative was successfully verified.
[In]
Integrate[(d + e*x)^m*Sqrt[d^2 - e^2*x^2],x]
[Out]
-(2^(3/2 + m)*(d - e*x)*(d + e*x)^m*(1 + (e*x)/d)^(-1/2 - m)*Sqrt[d^2 - e^2*x^2]*Hypergeometric2F1[3/2, -1/2 -
m, 5/2, (d - e*x)/(2*d)])/(3*e)
________________________________________________________________________________________
Maple [F] time = 0.481, size = 0, normalized size = 0. \begin{align*} \int \left ( ex+d \right ) ^{m}\sqrt{-{e}^{2}{x}^{2}+{d}^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((e*x+d)^m*(-e^2*x^2+d^2)^(1/2),x)
[Out]
int((e*x+d)^m*(-e^2*x^2+d^2)^(1/2),x)
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{-e^{2} x^{2} + d^{2}}{\left (e x + d\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x+d)^m*(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")
[Out]
integrate(sqrt(-e^2*x^2 + d^2)*(e*x + d)^m, x)
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{-e^{2} x^{2} + d^{2}}{\left (e x + d\right )}^{m}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x+d)^m*(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")
[Out]
integral(sqrt(-e^2*x^2 + d^2)*(e*x + d)^m, x)
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )^{m}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x+d)**m*(-e**2*x**2+d**2)**(1/2),x)
[Out]
Integral(sqrt(-(-d + e*x)*(d + e*x))*(d + e*x)**m, x)
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{-e^{2} x^{2} + d^{2}}{\left (e x + d\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x+d)^m*(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")
[Out]
integrate(sqrt(-e^2*x^2 + d^2)*(e*x + d)^m, x) | 2,612 | 5,534 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2023-40 | latest | en | 0.095822 |
http://nrich.maths.org/public/leg.php?code=1&cl=3&cldcmpid=5027 | 1,432,844,401,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207929561.98/warc/CC-MAIN-20150521113209-00277-ip-10-180-206-219.ec2.internal.warc.gz | 171,075,302 | 10,027 | Search by Topic
Resources tagged with Number - generally similar to Weekly Problem 17 - 2007:
Filter by: Content type:
Stage:
Challenge level:
There are 46 results
Broad Topics > Numbers and the Number System > Number - generally
Is it Magic or Is it Maths?
Stage: 3 Challenge Level:
Here are three 'tricks' to amaze your friends. But the really clever trick is explaining to them why these 'tricks' are maths not magic. Like all good magicians, you should practice by trying. . . .
Happy Numbers
Stage: 3 Challenge Level:
Take any whole number between 1 and 999, add the squares of the digits to get a new number. Make some conjectures about what happens in general.
Converging Means
Stage: 3 Challenge Level:
Take any two positive numbers. Calculate the arithmetic and geometric means. Repeat the calculations to generate a sequence of arithmetic means and geometric means. Make a note of what happens to the. . . .
Quick Times
Stage: 3 Challenge Level:
32 x 38 = 30 x 40 + 2 x 8; 34 x 36 = 30 x 40 + 4 x 6; 56 x 54 = 50 x 60 + 6 x 4; 73 x 77 = 70 x 80 + 3 x 7 Verify and generalise if possible.
Chocolate Maths
Stage: 3 Challenge Level:
Pick the number of times a week that you eat chocolate. This number must be more than one but less than ten. Multiply this number by 2. Add 5 (for Sunday). Multiply by 50... Can you explain why it. . . .
Differs
Stage: 3 Challenge Level:
Choose any 4 whole numbers and take the difference between consecutive numbers, ending with the difference between the first and the last numbers. What happens when you repeat this process over and. . . .
Stage: 3 Challenge Level:
Replace the letters with numbers to make the addition work out correctly. R E A D + T H I S = P A G E
Eleven
Stage: 3 Challenge Level:
Replace each letter with a digit to make this addition correct.
Think of Two Numbers
Stage: 3 Challenge Level:
Think of two whole numbers under 10. Take one of them and add 1. Multiply by 5. Add 1 again. Double your answer. Subract 1. Add your second number. Add 2. Double again. Subtract 8. Halve this number. . . .
Cross-country Race
Stage: 3 Challenge Level:
Eight children enter the autumn cross-country race at school. How many possible ways could they come in at first, second and third places?
Card Trick 2
Stage: 3 Challenge Level:
Can you explain how this card trick works?
Inequalities
Stage: 3 Challenge Level:
A bag contains 12 marbles. There are more red than green but green and blue together exceed the reds. The total of yellow and green marbles is more than the total of red and blue. How many of. . . .
Reverse to Order
Stage: 3 Challenge Level:
Take any two digit number, for example 58. What do you have to do to reverse the order of the digits? Can you find a rule for reversing the order of digits for any two digit number?
Calendar Capers
Stage: 3 Challenge Level:
Choose any three by three square of dates on a calendar page. Circle any number on the top row, put a line through the other numbers that are in the same row and column as your circled number. Repeat. . . .
N000ughty Thoughts
Stage: 4 Challenge Level:
Factorial one hundred (written 100!) has 24 noughts when written in full and that 1000! has 249 noughts? Convince yourself that the above is true. Perhaps your methodology will help you find the. . . .
Squaresearch
Stage: 4 Challenge Level:
Consider numbers of the form un = 1! + 2! + 3! +...+n!. How many such numbers are perfect squares?
Novemberish
Stage: 4 Challenge Level:
a) A four digit number (in base 10) aabb is a perfect square. Discuss ways of systematically finding this number. (b) Prove that 11^{10}-1 is divisible by 100.
Stage: 3 Challenge Level:
Visitors to Earth from the distant planet of Zub-Zorna were amazed when they found out that when the digits in this multiplication were reversed, the answer was the same! Find a way to explain. . . .
An Investigation Based on Score
Stage: 3
Class 2YP from Madras College was inspired by the problem in NRICH to work out in how many ways the number 1999 could be expressed as the sum of 3 odd numbers, and this is their solution.
Back to the Planet of Vuvv
Stage: 3 Challenge Level:
There are two forms of counting on Vuvv - Zios count in base 3 and Zepts count in base 7. One day four of these creatures, two Zios and two Zepts, sat on the summit of a hill to count the legs of. . . .
Card Trick 1
Stage: 3 Challenge Level:
Can you explain how this card trick works?
Calendars
Stage: 1, 2 and 3
Calendars were one of the earliest calculating devices developed by civilizations. Find out about the Mayan calendar in this article.
Ways of Summing Odd Numbers
Stage: 3
Sanjay Joshi, age 17, The Perse Boys School, Cambridge followed up the Madrass College class 2YP article with more thoughts on the problem of the number of ways of expressing an integer as the sum. . . .
Mathematical Symbols
Stage: 1, 2 and 3
A brief article written for pupils about mathematical symbols.
Find the Fake
Stage: 4 Challenge Level:
There are 12 identical looking coins, one of which is a fake. The counterfeit coin is of a different weight to the rest. What is the minimum number of weighings needed to locate the fake coin?
Latin Numbers
Stage: 4 Challenge Level:
Let N be a six digit number with distinct digits. Find the number N given that the numbers N, 2N, 3N, 4N, 5N, 6N, when written underneath each other, form a latin square (that is each row and each. . . .
Two Much
Stage: 3 Challenge Level:
Explain why the arithmetic sequence 1, 14, 27, 40, ... contains many terms of the form 222...2 where only the digit 2 appears.
Stage: 2, 3, 4 and 5
This article for the young and old talks about the origins of our number system and the important role zero has to play in it.
Water Lilies
Stage: 3 Challenge Level:
There are some water lilies in a lake. The area that they cover doubles in size every day. After 17 days the whole lake is covered. How long did it take them to cover half the lake?
Multiplication Magic
Stage: 4 Challenge Level:
Given any 3 digit number you can use the given digits and name another number which is divisible by 37 (e.g. given 628 you say 628371 is divisible by 37 because you know that 6+3 = 2+7 = 8+1 = 9). . . .
The Secret World of Codes and Code Breaking
Stage: 2, 3 and 4
When you think of spies and secret agents, you probably wouldn’t think of mathematics. Some of the most famous code breakers in history have been mathematicians.
Pairs
Stage: 3 Challenge Level:
Ann thought of 5 numbers and told Bob all the sums that could be made by adding the numbers in pairs. The list of sums is 6, 7, 8, 8, 9, 9, 10,10, 11, 12. Help Bob to find out which numbers Ann was. . . .
Mod 3
Stage: 4 Challenge Level:
Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.
More Mods
Stage: 4 Challenge Level:
What is the units digit for the number 123^(456) ?
Largest Product
Stage: 3 and 4 Challenge Level:
Which set of numbers that add to 10 have the largest product?
The Patent Solution
Stage: 3 Challenge Level:
A combination mechanism for a safe comprises thirty-two tumblers numbered from one to thirty-two in such a way that the numbers in each wheel total 132... Could you open the safe?
Factorial
Stage: 4 Challenge Level:
How many zeros are there at the end of the number which is the product of first hundred positive integers?
Mathematical Swimmer
Stage: 3 Challenge Level:
Twice a week I go swimming and swim the same number of lengths of the pool each time. As I swim, I count the lengths I've done so far, and make it into a fraction of the whole number of lengths I. . . .
Do Unto Caesar
Stage: 3 Challenge Level:
At the beginning of the night three poker players; Alan, Bernie and Craig had money in the ratios 7 : 6 : 5. At the end of the night the ratio was 6 : 5 : 4. One of them won \$1 200. What were the. . . .
Stage: 2, 3 and 4 Challenge Level:
Three dice are placed in a row. Find a way to turn each one so that the three numbers on top of the dice total the same as the three numbers on the front of the dice. Can you find all the ways to do. . . .
Big and Small Numbers in Biology
Stage: 4 Challenge Level:
Work with numbers big and small to estimate and calulate various quantities in biological contexts.
Funny Factorisation
Stage: 3 Challenge Level:
Some 4 digit numbers can be written as the product of a 3 digit number and a 2 digit number using the digits 1 to 9 each once and only once. The number 4396 can be written as just such a product. Can. . . .
Euler's Officers
Stage: 4 Challenge Level:
How many different solutions can you find to this problem? Arrange 25 officers, each having one of five different ranks a, b, c, d and e, and belonging to one of five different regiments p, q, r, s. . . . | 2,264 | 8,840 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-22 | longest | en | 0.879738 |
https://www.indiabix.com/aptitude/average/006001 | 1,713,381,871,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817171.53/warc/CC-MAIN-20240417173445-20240417203445-00371.warc.gz | 741,265,096 | 8,645 | # Aptitude - Average
Exercise : Average - General Questions
1.
In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 overs to reach the target of 282 runs?
6.25
6.5
6.75
7
Answer: Option
Explanation:
Required run rate = 282 - (3.2 x 10) = 250 = 6.25 40 40
Video Explanation: https://youtu.be/GhK9d8tcqvA
2.
A family consists of two grandparents, two parents and three grandchildren. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. What is the average age of the family?
28 4 years 7
31 5 years 7
32 1 years 7
None of these
Answer: Option
Explanation:
Required average
= 67 x 2 + 35 x 2 + 6 x 3 2 + 2 + 3
= 134 + 70 + 18 7
= 222 7
= 31 5 years. 7
Video Explanation: https://youtu.be/OXLnoItd0MA
3.
A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500?
Rs. 4991
Rs. 5991
Rs. 6001
Rs. 6991
Answer: Option
Explanation:
Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009.
Required sale = Rs. [ (6500 x 6) - 34009 ]
= Rs. (39000 - 34009)
= Rs. 4991.
Video Explanation: https://youtu.be/tP4hszReksU
4.
The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero?
0
1
10
19
Answer: Option
Explanation:
Average of 20 numbers = 0.
Sum of 20 numbers (0 x 20) = 0.
It is quite possible that 19 of these numbers may be positive and if their sum is a then 20th number is (-a).
5.
The average weight of 8 person's increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person?
76 kg
76.5 kg
85 kg
Data inadequate
None of these
Answer: Option
Explanation:
Total weight increased = (8 x 2.5) kg = 20 kg.
Weight of new person = (65 + 20) kg = 85 kg.
Video Explanation: https://youtu.be/ceg2jvHsiJU | 680 | 2,003 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.873696 |
https://socratic.org/questions/how-do-you-find-the-integral-ln-x-x | 1,607,170,938,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141747774.97/warc/CC-MAIN-20201205104937-20201205134937-00281.warc.gz | 486,933,320 | 5,816 | # How do you find the integral ln x / x?
Sep 12, 2015
$\int \ln \frac{x}{x} = {\left(\ln x\right)}^{2} / 2 + C$
#### Explanation:
int ln x / x = ?
This is a classic application of u -substitution.
Let $u = \ln x$. Then $\mathrm{du} = \frac{1}{x} \mathrm{dx}$.
Now we have
$\int \ln \frac{x}{x} \mathrm{dx} = \int u \mathrm{du}$
We can evaluate this easily by using the power rule:
$\int u \mathrm{du} = {u}^{2} / 2 + C$
Now, substituting back $u$, we find
${u}^{2} / 2 + C = {\left(\ln x\right)}^{2} / 2 + C$
And there is our solution.
$\int \ln \frac{x}{x} = {\left(\ln x\right)}^{2} / 2 + C$ | 233 | 607 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2020-50 | longest | en | 0.558633 |
https://www.cpalms.org/PreviewStandard/Preview/15529 | 1,725,784,991,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650976.41/warc/CC-MAIN-20240908083737-20240908113737-00675.warc.gz | 688,893,343 | 19,236 | # MA.912.NSO.1.1
Extend previous understanding of the Laws of Exponents to include rational exponents. Apply the Laws of Exponents to evaluate numerical expressions and generate equivalent numerical expressions involving rational exponents.
### Clarifications
Clarification 1: Instruction includes the use of technology when appropriate.
Clarification 2: Refer to the K-12 Formulas (Appendix E) for the Laws of Exponents.
Clarification 3: Instruction includes converting between expressions involving rational exponents and expressions involving radicals.
Clarification 4:Within the Mathematics for Data and Financial Literacy course, it is not the expectation to generate equivalent numerical expressions.
General Information
Subject Area: Mathematics (B.E.S.T.)
Strand: Number Sense and Operations
Status: State Board Approved
## Benchmark Instructional Guide
• Base
• Exponent
• Expression
### Vertical Alignment
Previous Benchmarks
Next Benchmarks
### Purpose and Instructional Strategies
In grade 8, students generated equivalent numerical expressions and evaluated expressions using the Laws of Exponents with integer exponents. In Algebra I, students work with rational-number exponents. In later courses, students extend the Laws of Exponents to properties of logarithms.
• Instruction includes using the terms Laws of Exponents and properties of exponents interchangeably.
• Instruction includes student discovery of the patterns and the connection to mathematical operations and the inverse relationship between powers and radicals (MTR.5.1)
• Problem types include having a fraction, integer or whole number as an exponent.
• Students should make the connection of the root being equivalent to unit fraction exponent (MTR.4.1).
• For example, $\sqrt[3]{8}$ = $\sqrt[3]{2³}$ is equivalent to the equation $\sqrt[3]{8}$ = (2³)$\frac{\text{1}}{\text{3}}$ which is equivalent to the equation $\sqrt[3]{8}$ = 2$\frac{\text{3}}{\text{1}}$.$\frac{\text{1}}{\text{3}}$ which is equivalent to the equation $\sqrt[3]{8}$ = 21 which is equivalent to the equation $\sqrt[3]{8}$ = 2.
• When evaluating, students should be encouraged to approach from different entry points and discuss how they are different but equivalent strategies (MTR.2.1).
• For example, if evaluating (-27)$\frac{\text{2}}{\text{3}}$ students can either take the cube root of -27 first or raise -27 to the second power first.
### Common Misconceptions or Errors
• Students may not understand the difference between an expression and an equation.
• Students may try to perform operations on bases as well as exponents.
• Students may multiply the base by the exponent instead of understanding that the exponent is the number of times the base occurs as a factor.
• Students may not truly understand exponents that are zero or negative.
### Strategies to Support Tiered Instruction
• Teacher provides a review of the relationship between the base and the exponent by modeling an example of operations using a base and exponent.
• For example, determine the numerical value of 63
• 63 which is equivalent to 6 6 6 which is equivalent to 216.
• Teacher provides exploration of the rules of exponents through patterns. A strategy for developing meaning for integer exponents by making use of patterns is shown below:
• Teacher provides exploration of the rules of rational exponents through patterns. A strategy for developing meaning for rational exponents by making use of patterns is shown below:
• Part A. Think about when solving an equation with a radical. What is the inverse operation of a square root? Of a cube root?
• Part B. Given the expression $\sqrt[3]{27}$, express 27 as a prime number with natural-number exponent.
• Part C. How can we use the information from Part A and B to convert $\sqrt[3]{27}$ to exponential form?
• Part A. Evaluate 64$\frac{\text{1}}{\text{3}}$ by first writing 64 as a power of 2 and using the properties of exponents.
• Part B. Evaluate 64$\frac{\text{1}}{\text{3}}$ using a calculator.
• Part C. Explain your process in both Part A and Part B. Define powers with fractional exponents in your own words.
• Part A. Given $f$($x$) = 32$x$, evaluate $f$(0), $f$(0.2), $f$(0.4), $f$(0.8) and $f$(1) without the use of a calculator.
• Part B. Graph the function $f$ in the domain 0 ≤ $x$ ≤ 1.
• Part C. Between which two values in Part A would $f$(0.5) be? Which one would it be closer to on the graph and why?
### Instructional Items
Instructional Item 1
• Evaluate the numerical expression (64)$\frac{\text{4}}{\text{3}}$
Instructional Item 2
• Rewrite 80.5 2$\frac{\text{2}}{\text{5}}$ as a single power of 2.
Instructional Item 3
• Choose all of the expressions that are equivalent to 7$\frac{\text{5}}{\text{12}}$
• a. (49$\frac{\text{1}}{\text{3}}$)(7$\frac{\text{1}}{\text{4}}$)
• b. (7$\frac{\text{2}}{\text{3}}$)(7$\frac{\text{1}}{\text{4}}$
• c. 7(7$\frac{\text{1}}{\text{4}}$)
• d.
• e.
Instructional Item 4
• Evaluate the numerical expression (−$\frac{\text{729}}{\text{64}}$)$\frac{\text{2}}{\text{3}}$.
?
?
?
*The strategies, tasks and items included in the B1G-M are examples and should not be considered comprehensive.
## Related Courses
This benchmark is part of these courses.
1200310: Algebra 1 (Specifically in versions: 2014 - 2015, 2015 - 2022, 2022 and beyond (current))
1200320: Algebra 1 Honors (Specifically in versions: 2014 - 2015, 2015 - 2022, 2022 and beyond (current))
1200380: Algebra 1-B (Specifically in versions: 2014 - 2015, 2015 - 2022, 2022 and beyond (current))
1200400: Foundational Skills in Mathematics 9-12 (Specifically in versions: 2014 - 2015, 2015 - 2022, 2022 and beyond (current))
7912090: Access Algebra 1B (Specifically in versions: 2014 - 2015, 2015 - 2018, 2018 - 2019, 2019 - 2022, 2022 and beyond (current))
1200315: Algebra 1 for Credit Recovery (Specifically in versions: 2014 - 2015, 2015 - 2022, 2022 and beyond (current))
1200385: Algebra 1-B for Credit Recovery (Specifically in versions: 2014 - 2015, 2015 - 2022, 2022 and beyond (current))
7912075: Access Algebra 1 (Specifically in versions: 2014 - 2015, 2015 - 2018, 2018 - 2019, 2019 - 2022, 2022 and beyond (current))
1200388: Mathematics for Data and Financial Literacy Honors (Specifically in versions: 2022 and beyond (current))
1200384: Mathematics for Data and Financial Literacy (Specifically in versions: 2022 and beyond (current))
7912120: Access Mathematics for Data and Financial Literacy (Specifically in versions: 2022 - 2023, 2023 and beyond (current))
1200710: Mathematics for College Algebra (Specifically in versions: 2022 and beyond (current))
## Related Access Points
Alternate version of this benchmark for students with significant cognitive disabilities.
MA.912.NSO.1.AP.1: Evaluate numerical expressions involving rational exponents.
## Related Resources
Vetted resources educators can use to teach the concepts and skills in this benchmark.
## Formative Assessments
Roots and Exponents:
Students are asked to rewrite the square root of five in exponential form and justify their choice of exponent.
Type: Formative Assessment
Rational Exponents and Roots:
Students asked to show that two forms of an expression (exponential and radical) are equivalent.
Type: Formative Assessment
## Lesson Plan
This lesson unit is intended to help you assess how well students are able to:
• Use the properties of exponents, including rational exponents, and manipulate algebraic statements involving radicals.
• Discriminate between equations and identities.
There is also an opportunity to consider the role of the imaginary number , but this is optional.
Type: Lesson Plan
## MFAS Formative Assessments
Rational Exponents and Roots:
Students asked to show that two forms of an expression (exponential and radical) are equivalent.
Roots and Exponents:
Students are asked to rewrite the square root of five in exponential form and justify their choice of exponent.
## Student Resources
Vetted resources students can use to learn the concepts and skills in this benchmark.
## Parent Resources
Vetted resources caregivers can use to help students learn the concepts and skills in this benchmark. | 2,085 | 8,189 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 35, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2024-38 | latest | en | 0.79148 |
http://angband.oook.cz/forum/printthread.php?s=a897a1f407b772b8046653120894b943&t=9070&pp=10&page=2 | 1,597,259,939,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738913.60/warc/CC-MAIN-20200812171125-20200812201125-00229.warc.gz | 6,581,214 | 3,859 | Angband Forums (http://angband.oook.cz/forum/index.php)
- Vanilla (http://angband.oook.cz/forum/forumdisplay.php?f=3)
- - how does stealth work? (http://angband.oook.cz/forum/showthread.php?t=9070)
Pete Mack January 24, 2020 05:46
I am kinda surprised the game doesn't use ounces. At every other opportunity, the original coding used shift operators as much as possible.
DavidMedley January 24, 2020 05:53
How does distance affect stealth?
Question: How does distance affect stealth?
Quick Answer: The expected number of turns until a monster wakes is proportional to your distance from it. If you are twice as far, it takes twice as many turns to wake up.
We know from the monster recall that there's a max distance at which a creature can hear you. This is calculated essentially the same as walking distance i.e., down hallways, around corners, through open/closed/broken doors, over passable rubble, but not through walls or impassable rubble. Additionally, a monster can see you, which ignores all light levels but otherwise works as you would expect (20 squares max, no visual obstructions).
If a monster can "notice [hear] you from 60 feet" you are safe from being heard at 6 squares away or more. In addition, every 3 full points of stealth you have reduces this by 1 square. All monsters can also notice [see] you from 210 feet, so you are safe from being seen at 21 squares away or if LOS is blocked.
If you are within this detection range, distance does not affect your chance to disturb the monster (stealth does). But distance can greatly affect how much you wake the monster per turn. The amount a monster wakes (loses sleepiness) is inversely proportional to the distance: 100/distance (round down). If a monster reaches 0 sleepiness, it wakes.
I'll address just how many sleepiness points a monster has and what your chance to disturb a monster is elsewhere.
Quick word on smell: If you look up a monster's stats (in monster.txt or elsewhere), smell is listed near hearing for most non-humans. This is not used for waking a monster at all. It is used to help an awake monster decide where to go to find you. Stealth has no effect.
-----
Some relevant code snippets
In the code below, local_noise is the distance noise has to travel from the player to the monster in game squares. Shows distance is inversely proportional to sleep reduction (closer wakes more).
Code:
```int sleep_reduction = 1; int local_noise = c->noise.grids[mon->grid.y][mon->grid.x]; /* Test - wake up faster in hearing distance of the player * Note no dependence on stealth for now */ if ((local_noise > 0) && (local_noise < 50)) { sleep_reduction = (100 / local_noise); }```
The code below shows that monsters cannot hear you at exactly their "notice from" distance, and that stealth can reduce that distance.
Code:
```static bool monster_can_hear(struct chunk *c, struct monster *mon) { int base_hearing = mon->race->hearing - player->state.skills[SKILL_STEALTH] / 3; if (c->noise.grids[mon->grid.y][mon->grid.x] == 0) { return false; } return base_hearing > c->noise.grids[mon->grid.y][mon->grid.x]; }```
Monsters also cannot smell you at their max smell distance
Code:
```static bool monster_can_smell(struct chunk *c, struct monster *mon) { if (c->scent.grids[mon->grid.y][mon->grid.x] == 0) { return false; } return mon->race->smell > c->scent.grids[mon->grid.y][mon->grid.x]; }```
DavidMedley January 24, 2020 05:54
Quote:
Originally Posted by Pete Mack (Post 142638) I do notice that perfect stealth is not actually possible, as there is a 1/1024 chance of disturbance even when stealth is 30.
I'm pretty sure it's actually 2/1024
Pete Mack January 24, 2020 13:38
randint starts at 1.
DavidMedley January 24, 2020 23:17
It may have changed
Code:
```int player_noise = 1 << (30 - stealth); int notice = randint0(1024); ... if ((notice * notice * notice) <= player_noise) {reduce sleep}```
notice can be 0 or 1, so notice^3 can be, too. player_noise can't get less than 1. If they're both 1, sleep is reduced.
All times are GMT +1. The time now is 20:18. | 1,060 | 4,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2020-34 | latest | en | 0.933532 |
https://hcondes.com/qa/question-what-are-the-two-conditions-for-omitted-variable-bias.html | 1,628,018,520,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154471.78/warc/CC-MAIN-20210803191307-20210803221307-00568.warc.gz | 302,659,881 | 8,225 | # Question: What Are The Two Conditions For Omitted Variable Bias?
## What does omitted variable bias mean?
(Learn how and when to remove this template message) In statistics, omitted-variable bias (OVB) occurs when a statistical model leaves out one or more relevant variables.
The bias results in the model attributing the effect of the missing variables to those that were included..
## What is bias in linear regression?
1. In Linear regression analysis, bias refer to the error that is introduced by approximating a real-life problem, which may be complicated, by a much simpler model.
## What is dummy variable trap in machine learning?
The Dummy variable trap is a scenario where there are attributes which are highly correlated (Multicollinear) and one variable predicts the value of others. When we use one hot encoding for handling the categorical data, then one dummy variable (attribute) can be predicted with the help of other dummy variables.
## What does omitted mean?
transitive verb. 1 : to leave out or leave unmentioned omits one important detail You can omit the salt from the recipe. 2 : to leave undone : fail —The patient omitted taking his medication.
## What is an omitted variable in economics?
The term omitted variable refers to any variable not included as an independent variable in the regression that might influence the dependent variable.
## What is the direction of bias?
The direction of bias is towards the null if fewer cases are considered to be exposed or if fewer exposed are considered to have the health outcome. The direction of bias is away from the null if more cases are considered to be exposed or if more exposed are considered to have the health outcome.
## How do you identify omitted variable bias?
How to Detect Omitted Variable Bias and Identify Confounding Variables. You saw one method of detecting omitted variable bias in this post. If you include different combinations of independent variables in the model, and you see the coefficients changing, you’re watching omitted variable bias in action!
## When there is an omitted variable in the regression that is a determinant of the dependent variable then?
Question: 6. When There Are Omitted Variables In The Regression, Which Are Determinants Of The Dependent Variable, Then (a) This Has No Effect On The Estimator Of Your Included Variable Because The Other Variable Is Not Included.
## What is Overfitting and Underfitting?
Overfitting: Good performance on the training data, poor generliazation to other data. Underfitting: Poor performance on the training data and poor generalization to other data.
## What is upward bias?
[¦əp·wərd ′bī·əs] (statistics) The overestimation or overstatement by a statistical measure of the event it is attempting to describe.
## What does bias mean in econometrics?
In statistics, the bias (or bias function) of an estimator is the difference between this estimator’s expected value and the true value of the parameter being estimated. An estimator or decision rule with zero bias is called unbiased.
## What does R Squared mean?
coefficient of determinationR-squared is a statistical measure of how close the data are to the fitted regression line. It is also known as the coefficient of determination, or the coefficient of multiple determination for multiple regression. … 100% indicates that the model explains all the variability of the response data around its mean.
## What causes omitted variable bias?
Intuitively, omitted variable bias occurs when the independent variable (the X) that we have included in our model picks up the effect of some other variable that we have omitted from the model. The reason for the bias is that we are attributing effects to X that should be attributed to the omitted variable.
## Is OLS unbiased?
OLS estimators are BLUE (i.e. they are linear, unbiased and have the least variance among the class of all linear and unbiased estimators). … So, whenever you are planning to use a linear regression model using OLS, always check for the OLS assumptions.
## What are the consequences of having an omitted variable?
An omitted variable leads to biased and inconsistent coefficient estimate. And as we all know, biased and inconsistent estimates are not reliable.
## What is an irrelevant variable?
Definition. A variable is irrelevant if its true coefficient is zero. Effects. The coefficient estimate is unbiased, but is an unbiased estimate of zero. The factor highlighted in blue is greater than one and unnecessary.
## What causes OLS estimators to be biased?
The only circumstance that will cause the OLS point estimates to be biased is b, omission of a relevant variable. Heteroskedasticity biases the standard errors, but not the point estimates.
## How do you find bias in linear regression?
Bias and variance for various regularization valuesBias is computed as the distance from the average prediction and true value — true value minus mean(predictions)Variance is the average deviation from the average prediction — mean(prediction minus mean(predictions)) | 1,004 | 5,090 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.916093 |
https://leanprover-community.github.io/mathlib_docs/combinatorics/simple_graph/partition.html | 1,709,031,402,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474674.35/warc/CC-MAIN-20240227085429-20240227115429-00473.warc.gz | 350,866,806 | 43,526 | # mathlib3documentation
combinatorics.simple_graph.partition
# Graph partitions #
THIS FILE IS SYNCHRONIZED WITH MATHLIB4. Any changes to this file require a corresponding PR to mathlib4.
This module provides an interface for dealing with partitions on simple graphs. A partition of a graph G, with vertices V, is a set P of disjoint nonempty subsets of V such that:
• The union of the subsets in P is V.
• Each element of P is an independent set. (Each subset contains no pair of adjacent vertices.)
Graph partitions are graph colorings that do not name their colors. They are adjoint in the following sense. Given a graph coloring, there is an associated partition from the set of color classes, and given a partition, there is an associated graph coloring from using the partition's subsets as colors. Going from graph colorings to partitions and back makes a coloring "canonical": all colors are given a canonical name and unused colors are removed. Going from partitions to graph colorings and back is the identity.
## Main definitions #
• simple_graph.partition is a structure to represent a partition of a simple graph
• simple_graph.partition.parts_card_le is whether a given partition is an n-partition. (a partition with at most n parts).
• simple_graph.partitionable n is whether a given graph is n-partite
• simple_graph.partition.to_coloring creates colorings from partitions
• simple_graph.coloring.to_partition creates partitions from colorings
## Main statements #
• simple_graph.partitionable_iff_colorable is that n-partitionability and n-colorability are equivalent.
structure simple_graph.partition {V : Type u} (G : simple_graph V) :
A partition of a simple graph G is a structure constituted by
• parts: a set of subsets of the vertices V of G
• is_partition: a proof that parts is a proper partition of V
• independent: a proof that each element of parts doesn't have a pair of adjacent vertices
Instances for simple_graph.partition
def simple_graph.partition.parts_card_le {V : Type u} {G : simple_graph V} (P : G.partition) (n : ) :
Prop
Whether a partition P has at most n parts. A graph with a partition satisfying this predicate called n-partite. (See simple_graph.partitionable.)
Equations
def simple_graph.partitionable {V : Type u} (G : simple_graph V) (n : ) :
Prop
Whether a graph is n-partite, which is whether its vertex set can be partitioned in at most n independent sets.
Equations
def simple_graph.partition.part_of_vertex {V : Type u} {G : simple_graph V} (P : G.partition) (v : V) :
set V
The part in the partition that v belongs to
Equations
theorem simple_graph.partition.mem_part_of_vertex {V : Type u} {G : simple_graph V} (P : G.partition) (v : V) :
v
theorem simple_graph.partition.part_of_vertex_ne_of_adj {V : Type u} {G : simple_graph V} (P : G.partition) {v w : V} (h : G.adj v w) :
Create a coloring using the parts themselves as the colors. Each vertex is colored by the part it's contained in.
Equations
Like simple_graph.partition.to_coloring but uses set V as the coloring type.
Equations
def simple_graph.coloring.to_partition {V : Type u} {G : simple_graph V} {α : Type v} (C : G.coloring α) :
Creates a partition from a coloring.
Equations
@[simp]
theorem simple_graph.coloring.to_partition_parts {V : Type u} {G : simple_graph V} {α : Type v} (C : G.coloring α) :
@[protected, instance]
The partition where every vertex is in its own part.
Equations | 815 | 3,440 | {"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} | 3.09375 | 3 | CC-MAIN-2024-10 | latest | en | 0.85388 |
https://medium.com/berkeleyischool/how-to-create-a-visualization-showing-events-on-time-series-data-in-excel-96abbc1475e0 | 1,561,025,865,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999200.89/warc/CC-MAIN-20190620085246-20190620111246-00367.warc.gz | 517,334,223 | 34,595 | # How to create a visualization showing events on time series data in Excel?
What’s in this article: A step-by-step guide I have laid out, for creating a time-series chart with events markers that change dynamically. While some of us use Python libraries, the vast majority of the world still uses Excel for good reasons, and hence the guide (or at least this first version) is for Excel. At the end of this tutorial, you will be able to produce something like this:
Showing change over time is a very common data visualization need for many analysts and researchers. Usually this is done by showing a line graph with X-axis representing time and Y-axis for the quantity of interest. Sometimes, we need to show particular events in addition to the time-series data, in order to highlight certain points. There are many ways to do this in Excel:
1. The simplest but least flexible method is to just draw some lines and text boxes onto the line graph (not recommended).
2. Another method is to use Y-error bars for some data points, which is a decent method but gets cumbersome as soon as your start making changes to data.
3. The method shown here relies on creating two series on different axes — one for actual quantity of interest and another for event markers. Once done, minimal changes are required for updates.
The completed version of this tutorial in Excel format is available here if you wish to practice on your own. So let’s get started.
#### Step 1: Get the data into shape
For this example, I used publicly available health insurance coverage data of California (available here) and identified some key events from the timeline given here. While I had to use some quick pivot table tricks to get the data into the required shape, those steps depend on your data source and are not discussed. Basically the data should have the following columns:
• Time series data for X-axis: In the figure below, this is the Year column but can be any time-scale (Yes there are two year columns, I will explain shortly).
• Data for quantity of interest: Here, this is the Coverage column, representing percent of population that has health insurance in the corresponding year.
• Events: This column contains names of event to be shown, against a specific year (Event column in figure below).
• Event marker length: This column contains the length of the event marker bar/line (Event_line column in figure below). In this example, I am setting this column equal to the Y-axis data point for the same event, and only for the data points where there is an event. But once you understand this, you can try many interesting variations. Important: where there is no event to be shown, both Event and Event_line are blank.
Why are there two Year columns? Because I wanted to explain a fairly common issue with how Excel treats dates and numbers differently. The first Year column (Column B) in the example above is simple a list of numbers I pasted from the source data. Excel does not recognize it as a date. The second column (Column C) is the actual Year column being used for the graph; it contains a formatted date that Excel recognizes on a date-time scale. To get the second column values from the first column, we need a simple formula. E.g. the formula in cell C2 is: = DATE (B2, 1, 1) which basically takes the number 1995 in cell B2 and converts into the date Jan 1st, 1995. The cell C2 is then formatted using a custom date format of Excel (yyyy) to show only the year part (see below).
The date formatting steps are dependent on the scale you are using, but you can adapt the same method depending on whether you are using a month, day, or even hour, minute or second scale.
Why do this instead of using the numbers for X-axis? We could skip this, but then it will not work whenever the date/time sequence is uneven. With date-time recognized in Excel, it can deal with missing dates nicely.
#### Step 2: Create a line chart
Select the two columns containing the time-series data and the quantity of interest (Columns C and D in figure below), and press Alt + F1. This is the quickest way to create a default chart using the selection.
Next, change the chart type to Line and select the Line with Markers option (see below).
Now you should have a simple line chart with markers showing the change over time.
#### Step 3: Create event markers
Now we need to add the event markers as another series to the chart. Remember that we need to set the values of event markers (Event_line column) same as the value of the Y-variable at the same year. I have done this simply by referring to the Y-values for the event years. E.g. cell F4 has the formula =D4. To add this Event_line column to the chart, right-click the chart and click “Select Data…”. In the dialogue box, click on the “Add” button as shown below. Then in the next dialogue box, for the “Series values” select all the values of the Event_line column (Column F) including the empty ones. Click OK and OK to finish adding series.
At this point, the event markers are added as another set of dots on top of the original Y-values and are hardly visible. But we want bars dropping from these points on the line graph down to the X-axis. For that, we need to convert this second series of data points to a cluster bar type. To do so, click on Change Chart Type option for the chart, select “Combo” from the left column in the dialogue box, and then make sure that the original Y-values series (Coverage in this example) is set to “Line with Markers” and the event markers series (named Event_line here) is set to “Clustered Columns” and Secondary Axis is also selected for the latter.
Now we are beginning to see light at the end of the tunnel. The bars however are now on a secondary axis. In order for the bars to exactly touch the dot markers on the line chart, both primary and secondary Y-axes need to be set exactly same. To do this, double click on the left Y-axis to open up the properties box and then set the Minimum and Maximum values of the axes as desired. In the figure below, I’ve set these to 0.75 (or 75%) and 0.90 (or 90%). Then repeat the same process for the right Y-axis.
Now, select the right sided Y-axis and press the Delete key to remove it. At this point, your chart should look something like this:
#### Step 4: Beautify (simplify)
Technically, the above chart is showing what we want to show, but now we shall apply some design principles to clean it up and improve the appearance significantly. This is what separates an amateurish attempt from a professional looking chart. Let’s do the following to make it awesome:
• Improve readability of line graph: Double click anywhere on the line to open its properties and change the color of line to very light gray and that of dot markers to a darker gray (but never full black). Also change the weight of the line to 1 pt. If you feel like it, increase the marker size to 6 pt.
• Make the event bars thin and clear: Double click any of the bars to open their properties, and then set the border to “No line” and change the Fill to a dark gray. Under the “Series Options” tab in the same property box, move the “Gap Width” slider to maximum value (500%). Now click on the X-axis to show its properties, and under the “Axis Options”, change the “Base” to Days. This squishes the bars even further to create the appearance of a thin line.
At this point, your chart should be close to this:
• Adding event labels: Right click on any of the bars and click “Add Data Labels”. Now double click any of the data labels shown to open their properties box. Choose the “Label Options” tab, deselect “Value” and instead choose “Value From Cells”. At this point you will see a small dialogue box asking for range of cells to pick values from. We need to click and drag to select the column containing the names of the events (Column E in this example). Then click OK to close this box. Now you should see the event names as data labels. They might be jumbled up and you can tweak their position either by choosing various label position options in the property box or manually adjusting them one by one. Finally, I like to give the labels a white fill so that any lines in the background don’t interfere with the text. You can also add a title to the chart now.
• Some more tweaking (optional): For both X and Y axes, I like to fade the text color to a lighter shade of gray but it depends on purpose of chart. In Edward Tufte’s spirit of reducing chart junk, I would also remove the horizontal grid lines by clicking them and pressing delete. However, this is also something that depends on the purpose of chart.
Congratulations! Now you have a refined, timeline chart with event labels nicely showing. I hope you found it useful. Please do leave feedback and comments. | 1,921 | 8,811 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-26 | latest | en | 0.918217 |
https://ehomedecor.net/how-many-square-feet-is-10x16/ | 1,686,127,852,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653631.71/warc/CC-MAIN-20230607074914-20230607104914-00259.warc.gz | 271,626,434 | 26,879 | # How many square feet is 10×16?
10×16 Sheds
Square footage: 160 sq. ft.
Just so, Is 12×12 bedroom small?
A 12-by-12 bedroom’s small, boxy footprint seems even more awkward and tighter when you consider standard furniture sizes; a queen-sized bed, for instance, measures 60-by-80 inches or 5-by-6.7 feet, consuming a big chunk of floor space.
How many square feet is 5×5? How big is a 5×5 storage unit? A 5×5 self storage unit is 5 feet wide and 5 feet long, totaling 25 square feet — comparable to a large closet.
Similarly, How many feet is a 10×12 room?
Area of the floor or ceiling: Multiply the length by the width (10 feet x 12 feet = 120 square feet of area). Area of a wall: Multiply the width of the wall by its height. So one of the walls is 80 square feet (10 feet wide x 8 feet high) and the other is 96 square feet (12 feet x 8 feet).
## How many sq ft is my house?
If you live in a property that’s a perfect rectangle, simply measure the length and width and multiply the two numbers. For example, if your one-story house is 60 feet by 40 feet long, then your property is 2,400 square feet (60 x 40 = 2,400).
## How many sq ft is 5X6?
Measuring Vaulted Ceilings
For example, a triangle that is 5 feet tall and 6 feet long has 15 square feet, (5X6 divided by 2). Measure each triangle you have in this manner.
## How many square feet is a 5×10 room?
5×10 Storage Unit Size
A 5×10 self storage unit is a 5 feet wide and 10 feet long small storage solution, totaling 50 square feet. For comparison, a 5×10 space is a small storage unit that’s about the size of an average walk-in closet. Many units have an 8-foot ceiling, giving you up to 400 cubic feet of storage space.
## How many square feet is 8 ft by 8 ft?
x 8 ft. = 72 sq. ft. For a more complicated room, divide it into smaller sections and multiply the length and width of each section.
## How many square feet is an 8×12 room?
Multiply 12 x 8 = 96 square feet for each wall, then multiply 96 x 4 (since there are four walls with 96 square feet each)= 384 total square feet for the room. 25 sq. ft.
## How many square feet is my bedroom?
To calculate the square footage, you will multiply the length of the room by the width of the room. You will measure this from the inside of the room. We will use a room that is 10 ft x 15 ft as an example in all of our room scenarios: This is the simple equation: Length (ft) x Width (ft) = Square Footage.
## How do you calculate the size of a room?
Calculate the Area as Square Footage
If you are measuring a square or rectangle area, multiply length times width; Length x Width = Area.
## How many 12×12 tiles do I need for 100 square feet?
For a 100 square foot tile project, you will need to install a total of 100 12×12 tiles.
## How many square feet is 50×10?
Now you have the amount of mulch needed in cubic yards.
For example, if your bed is 50 feet long by 10 feet wide, the square footage is 500 square feet. If you want the mulch to be 2 inches deep, you’ll need 3 cubic yards of mulch. 50 x 10 x 2 = 1000. Then 1000 divided by 324 = 3.08 cubic yards.
## What is the difference between 5×10 and 10×5?
10×5 Unit (50 sq. ft.): Identical to a 5×10 unit just with a wider door and shallower storage-depth. Tends to make accessing your items slightly easier than a 5×10.
## How do I calculate 2m of a room?
In order to calculate the size of a room or space in m2, you simply multiply the length of the space (in metres) by the width of the space (in metres).
## What is the square footage of an 8×12 room?
Multiply 12 x 8 = 96 square feet for each wall, then multiply 96 x 4 (since there are four walls with 96 square feet each)= 384 total square feet for the room.
## How much is 32 linear feet?
2×4 boards: 8 feet times 4 boards = 32 linear feet.
## How many square feet is a board?
By definition, a board foot is one square foot, one inch thick. The most common mistake made in calculating board footage is forgetting to multiply by the thickness.
## What is the meaning of 1 square feet?
noun. a unit of area measurement equal to a square measuring one foot on each side; 0.0929 square meters. Abbreviation: ft2, sq.
## How do I figure out area?
Area is calculated by multiplying the length of a shape by its width. In this case, we could work out the area of this rectangle even if it wasn’t on squared paper, just by working out 5cm x 5cm = 25cm² (the shape is not drawn to scale).
Also read : How do you measure for a runner?
154 Points
Upvote Downvote | 1,220 | 4,523 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2023-23 | longest | en | 0.917973 |
http://book.caltech.edu/bookforum/printthread.php?s=53905b580fcd8b5fa856c8509a8da749&t=1464 | 1,585,779,932,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506121.24/warc/CC-MAIN-20200401192839-20200401222839-00140.warc.gz | 29,526,588 | 3,600 | LFD Book Forum (http://book.caltech.edu/bookforum/index.php)
- The Final (http://book.caltech.edu/bookforum/forumdisplay.php?f=138)
victe 09-09-2012 04:01 PM
Question 20
I think there is two possible true answers to this question.
yaser 09-09-2012 08:43 PM
Re: Question 11
Quote:
Originally Posted by victe (Post 5070) I think there is two possible true answers to this question.
Could you pmail me why you think so?
the cyclist 09-10-2012 08:32 AM
Re: Question 11
Quote:
Originally Posted by victe (Post 5070) I think there is two possible true answers to this question.
Hm. I don't think so. I can say the following without giving away too much info:
Out of the first four statements, I am sure three of them are false, because I can create simple counterexamples for them. For the remaining statement, I am not sure yet. (I am neither sure it is true, nor do I have a counterexample.)
So, I have narrowed it down to that one statement or "none of the above". Regardless, I am very confident that there are not two true statements in the list.
TonySuarez 09-10-2012 08:38 AM
Re: Question 11
:D
Couldn't resist (but in the end perhaps I'll have to swallow...)
feallertmalty 09-10-2012 08:47 AM
Re: Question 11
Quote:
Couldn't resist (but in the end perhaps I'll have to swallow...)
You narrated . On no other word
victe 09-12-2012 05:20 AM
Re: Question 11
Quote:
Originally Posted by TonySuarez (Post 5091) :D Couldn't resist (but in the end perhaps I'll have to swallow...)
In the end I was wrong.
I derived mathematically the true solution, and there is only one correct answer. Only, the statement should indicate some property on the functions to be met, but with the roles that are often used is understood.
Indeed, there are very simple counterexamples for all incorrect answers.
TonySuarez 09-14-2012 10:05 AM
Re: Question 11
Quote:
Originally Posted by TonySuarez (Post 5091) :D Couldn't resist (but in the end perhaps I'll have to swallow...)
I just jumped in, and delivered the final.
Worst than I expected :-(, lowered my percentile somewhat... but that's life...
So, in the end I had to swallow Q11, FYI :-(
fgpancorbo 09-14-2012 07:25 PM
Re: Question 11
Quote:
Originally Posted by the cyclist (Post 5090) Hm. I don't think so. I can say the following without giving away too much info: Out of the first four statements, I am sure three of them are false, because I can create simple counterexamples for them. For the remaining statement, I am not sure yet. (I am neither sure it is true, nor do I have a counterexample.) So, I have narrowed it down to that one statement or "none of the above". Regardless, I am very confident that there are not two true statements in the list.
I find myself in a similar situation! I spent a couple of hours trying to come up with the counter example to no avail. I then went analytical and I see no easy way to prove it. Still 4 days, so there is time for inspiration :D.
yaser 09-14-2012 08:42 PM
Re: Question 11
Quote:
Originally Posted by fgpancorbo (Post 5292) I find myself in a similar situation! I spent a couple of hours trying to come up with the counter example to no avail. I then went analytical and I see no easy way to prove it. Still 4 days, so there is time for inspiration :D.
Hint (from Mr. Hints himself :)): Cauchy.
I hope this helps more than it confuses.
fgpancorbo 09-14-2012 09:01 PM
Re: Question 11
Quote:
Originally Posted by yaser (Post 5295) Hint (from Mr. Hints himself :)): Cauchy. I hope this helps more than it confuses.
:D, sure, Cauchy like in "Cauchy-Schwarz"? I don't want to sound cocky but I thought about bringing that ammunition myself except that I was unable to identify the inner product but now that I think about it... there is that cross term that looks very much like one... Interesting. Correction; I said earlier that I had the proof, but I still have to work out the details on paper. I just did it in my mind; the actual details are a bit more elaborated but I don't see a problem getting there :D.
All times are GMT -7. The time now is 03:25 PM. | 1,090 | 4,104 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.924758 |
http://www.calculators4u.com/circle-area-calculator | 1,618,235,072,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038067400.24/warc/CC-MAIN-20210412113508-20210412143508-00311.warc.gz | 117,475,577 | 4,033 | # Circle Area Calculator
Circle area calculator tool is used to calculate circumference, diameter, area of the circle, if the radius is given. Here, in the tool enter the radius of the circle to calculate all parameters.
In a circle, all the points are equidistant from the given point called the centre of the circle. The distance from the centre of the circle to the longest chord (diameter) is zero.
## Formulas
The formula for the area of a circle is: π r2
The formula for the circumference of a circle is: 2 x π x r
The formula for the diameter of a circle is : 2 x r
### What is a Circle?
A round plane figure whose boundary (the circumference) consists of points equidistant from a fixed point (the centre).
#### Center of a Circle
A point in the circle and is at equal distance from all points on the circumference.
#### Radius
A line segment with one endpoint on the center and the other end point on the circle or circumference.
#### Diameter
A line segment that passes through the center of a circle and has endpoints on the circle.
#### π (Pi)
In math, there are numbers that come up so often, we give them their own symbol. One such special number is π(pi).
The symbol for pi is π. And the value of π is 3.14 (approx).
The number shows up as the ratio of a circle’s circumference to its diameter. The circumference is the distance around the circle and the diameter is the distance across the circle through the center. Another dimension we will often mention is the radius of a circle. The radius is half the length of the diameter. This means the diameter is twice the length of the radius. | 359 | 1,625 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2021-17 | latest | en | 0.892416 |
https://www.geeksforgeeks.org/count-numbers-upto-n-which-are-both-perfect-square-and-perfect-cube/?ref=rp | 1,685,913,962,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650264.9/warc/CC-MAIN-20230604193207-20230604223207-00762.warc.gz | 867,076,180 | 37,073 | GeeksforGeeks App
Open App
Browser
Continue
# Count numbers upto N which are both perfect square and perfect cube
Given a number N. The task is to count total numbers under N which are both perfect square and cube of some integers.
Examples:
```Input: N = 100
Output: 2
They are 1 and 64.
Input: N = 100000
Output: 6```
Naive Approach: The idea is to use the power, square_root and cube_root functions from the math library.
## C++
`// C++ program for the above approach``#include ``#include ` `// Function to count the perfect square``// and cubes``int` `countPerfectSquaresCubes(``int` `N)``{`` ``int` `count = 0;` ` ``// Iterate over the range [1, N]`` ``for` `(``int` `i = 1; i <= N; i++) {`` ``if` `(std::``pow``((``int``)std::``sqrt``(i), 2) == i`` ``&& std::``pow``((``int``)std::cbrt(i), 3) == i) {`` ``count++;`` ``}`` ``}` ` ``// Return the resultant count`` ``return` `count;``}` `// Driver Code``int` `main()``{`` ``int` `count = countPerfectSquaresCubes(100000);`` ``std::cout << count << std::endl;`` ``return` `0;``}` `// This code is contributed by aeroabrar_31`
## Java
`// Java program for the above approach` `import` `java.util.*;``class` `GFG {` ` ``// Function to count the perfect square`` ``// and cubes`` ``public` `static` `int` `countPerfectSquaresCubes(``int` `N)`` ``{`` ``int` `count = ``0``;`` ``for` `(``int` `i = ``1``; i <= N; i++) {`` ``if` `(Math.pow((``int``)Math.sqrt(i), ``2``) == i`` ``&& Math.pow((``int``)Math.cbrt(i), ``3``) == i) {`` ``count++;`` ``}`` ``}` ` ``// Return the resultant count`` ``return` `count;`` ``}` ` ``// Driver Code`` ``public` `static` `void` `main(String[] args)`` ``{`` ``int` `count = countPerfectSquaresCubes(``100000``);`` ``System.out.println(count);`` ``}``}`
## C#
`// C# program for the above approach``using` `System;` `public` `class` `GFG {` ` ``// Function to count the perfect square`` ``// and cubes`` ``public` `static` `int` `countPerfectSquaresCubes(``int` `N)`` ``{`` ``int` `count = 0;`` ``for` `(``int` `i = 1; i <= N; i++) {`` ``if` `(Math.Pow((``int``)Math.Sqrt(i), 2) == i`` ``&& Math.Pow((``int``)Math.Cbrt(i), 3) == i) {`` ``count++;`` ``}`` ``}` ` ``// Return the resultant count`` ``return` `count;`` ``}` ` ``// Driver Code`` ``public` `static` `void` `Main()`` ``{`` ``int` `count = countPerfectSquaresCubes(100000);`` ``Console.WriteLine(count);`` ``}``}``// This code is contributed by aeroabrar_31`
Output
`6`
Time Complexity: O(N*(logN))
Space Complexity: O(1)
Method 2 : Optimal
Approach: For a given positive number N to be a perfect square, it must satisfy P2 = N Similarly, Q3 = N for a perfect cube where P and Q are some positive integers.
N = P2 = Q3
Thus, if N is a 6th power, then this would certainly work. Say N = A6 which can be written as (A3)2 or (A2)3.
So, pick 6th power of every positive integers which are less than equal to N.
Below is the implementation of the above approach:
## C++
`// C++ implementation of the above approach``#include ``using` `namespace` `std;` `// Function to return required count``int` `SquareCube(``long` `long` `int` `N)``{` ` ``int` `cnt = 0, i = 1;` ` ``while` `(``int``(``pow``(i, 6)) <= N) {`` ``++cnt;`` ``++i;`` ``}` ` ``return` `cnt;``}` `int` `main()``{`` ``long` `long` `int` `N = 100000;` ` ``// function call to print required answer`` ``cout << SquareCube(N);`` ``return` `0;``}`
## Java
`// Java implementation of the above approach` `public` `class` `GFG{` ` ``// Function to return required count`` ``static` `int` `SquareCube(``long` `N)`` ``{`` ` ` ``int` `cnt = ``0``, i = ``1``;`` ` ` ``while` `((``int``)(Math.pow(i, ``6``)) <= N) {`` ``++cnt;`` ``++i;`` ``}`` ` ` ``return` `cnt;`` ``}`` ` ` ``public` `static` `void` `main(String []args)`` ``{`` ``long` `N = ``100000``;`` ` ` ``// function call to print required answer`` ``System.out.println(SquareCube(N)) ;`` ``}`` ``// This code is contributed by Ryuga``}`
## Python3
`# Python3 implementation of the``# above approach` `# Function to return required count``def` `SquareCube( N):` ` ``cnt, i ``=` `0``, ``1` ` ``while` `(i ``*``*` `6` `<``=` `N):`` ``cnt ``+``=` `1`` ``i ``+``=` `1` ` ``return` `cnt` `# Driver code``N ``=` `100000` `# function call to print required``# answer``print``(SquareCube(N))` `# This code is contributed``# by saurabh_shukla`
## C#
`// C# implementation of the above approach``using` `System;` `public` `class` `GFG{`` ` ` ``// Function to return required count`` ``static` `int` `SquareCube(``long` `N)`` ``{`` ` ` ``int` `cnt = 0, i = 1;`` ` ` ``while` `((``int``)(Math.Pow(i, 6)) <= N) {`` ``++cnt;`` ``++i;`` ``}`` ` ` ``return` `cnt;`` ``}`` ` ` ``public` `static` `void` `Main()`` ``{`` ``long` `N = 100000;`` ` ` ``// function call to print required answer`` ``Console.WriteLine(SquareCube(N)) ;`` ``}``}` `/*This code is contributed by 29AjayKumar*/`
## PHP
``
## Javascript
``
Output:
`6`
Time Complexity: O(N1/6)
Auxiliary Space: O(1), since no extra space has been taken.
My Personal Notes arrow_drop_up | 1,971 | 5,612 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2023-23 | latest | en | 0.496385 |
https://discuss.codechef.com/t/kingbob-editorial/14308 | 1,601,194,224,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400265461.58/warc/CC-MAIN-20200927054550-20200927084550-00187.warc.gz | 340,080,003 | 4,049 | # KINGBOB - Editorial
Author: Varun Singh
### DIFFICULTY:
CAKEWALK - SIMPLE
Math
### PROBLEM:
Given a positive integer N, find the total number of factors of N.
### QUICK EXPLANATION:
Loop till square root of the N, to count both factors in a single iteration. To take care of the corner case when N is a perfect square, increment the count only by 1.
### EXPLANATION:
Counting the factors of the given integer using Brute Force will result in TLE, so we will exploit an interesting property of numbers. For a number N, it can be factored into two factors a and b (in case of prime, 1 and N itself). One of both factors is always less than or equal to the square root of N and the other is always greater than or equal to the square root. If both a and b were greater than the square root of N, (a*b) would be greater than N.
So we run our loop only till square root of N, and if i completely divides N, then i and N/i, both are factors, hence we increase the count by 2. If N is a perfect square, i and N/i will be equal, so for this iteration, we only increase the counter by 1.
### AUTHOR’S SOLUTIONS:
Author’s solution can be found here. | 293 | 1,158 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2020-40 | latest | en | 0.896377 |
http://mathhelpforum.com/discrete-math/70222-math-question-inclusion-exclusion-principle.html | 1,527,056,075,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865450.42/warc/CC-MAIN-20180523043959-20180523063959-00491.warc.gz | 187,197,841 | 11,993 | # Thread: math question - inclusion exclusion principle
1. ## math question - inclusion exclusion principle
x1+x2+x3+x4+x5=32
3 of the variable are naturals and even
2 of the variable are naturals and odd
nobody of the variables equals to 1 or 0
get the final solution
i think i need to use the inclusion exclusion principle
someone can help me with this please
2. Originally Posted by tukilala
x1+x2+x3+x4+x5=32
3 of the variable are naturals and even
2 of the variable are naturals and odd
nobody of the variables equals to 1 or 0
get the final solution
i think i need to use the inclusion exclusion principle
someone can help me with this please
tukilala,
I don't understand your problem statement.
What do you mean by "get the final solution"? Do you mean "Find the number of solutions to the equation"?
3. Originally Posted by tukilala
x1+x2+x3+x4+x5=32
3 of the variable are naturals and even
2 of the variable are naturals and odd
nobody of the variables equals to 1 or 0
get the final solution
i think i need to use the inclusion exclusion principle
someone can help me with this please
Let $\displaystyle a,b,c,d,e$ be natural numbers.
Let $\displaystyle 2a = x_1$
$\displaystyle 2b = x_2$
$\displaystyle 2c = x_3$
$\displaystyle 2d+1 = x_4$
$\displaystyle 2e+1 = x_5$
Then your equation can be expressed:
$\displaystyle 2a+2b+2c +2d+1+2e+1 = 32$
$\displaystyle 2a+2b+2c +2d+2e + 2 = 32$
$\displaystyle 2a+2b+2c +2d+2e = 30$
$\displaystyle 2(a+b+c +d+e) = 30$
$\displaystyle a+b+c +d+e = 15$
There are many solutions. To get one, just try some numbers. As long as they're natural, it doesn't matter.
Since none are 0 or 1, I'd just put all 2s until the last one, and then see what you need to make up 15.
$\displaystyle 2+2+2+2+7 = 15$
Hence
$\displaystyle a = 2$
$\displaystyle b = 2$
$\displaystyle c = 2$
$\displaystyle d = 2$
$\displaystyle e = 7$
$\displaystyle x_1 = 4$
$\displaystyle x_2 = 4$
$\displaystyle x_3 = 4$
$\displaystyle x_4 = 5$
$\displaystyle x_5 = 15$
Is a solution.
4. ## i dont understand somthing...
i dont understand why u deside that
2a=x1
2b=x2
2c=x3
2d+1=x4
2e+1=x5
you dont know wich of the variable is odd and wich is even,
you just know that there are 3 even and 2 odd
so i think that you cant decide that x4 and x5 are odd by give them the value 2d+1 and 2e+1
am i right? of not? why?
thanks
5. Originally Posted by tukilala
i dont understand why u deside that
2a=x1
2b=x2
2c=x3
2d+1=x4
2e+1=x5
you dont know wich of the variable is odd and wich is even,
you just know that there are 3 even and 2 odd
so i think that you cant decide that x4 and x5 are odd by give them the value 2d+1 and 2e+1
am i right? of not? why?
thanks
Yes.
Any odd natural number can be expressed in the form 2k+1, where k is any natural number. Any even natural number can be expressed in the form 2k, where k is any natural number. That is why i choose these.
6. but if x1 is odd? so x1 cant be equal to 2a
my point is that you dont know wich number is odd and wich is even
so how you decide that x1 is even and x5 is odd??????
so i think something missing here, no?
7. Originally Posted by tukilala
but if x1 is odd? so x1 cant be equal to 2a
my point is that you dont know wich number is odd and wich is even
so how you decide that x1 is even and x5 is odd??????
so i think something missing here, no?
It doesn't matter what you decide.
The question doesn't specify WHICH variables are odd or even, it just says that 3 are even, and 2 are odd. You are at liberty to decide which ones you want to make.
The fact of the matter is that there are hundreds of different solutions to the equation under the given requirements. It doesn't really matter what one you choose.
8. Originally Posted by Mush
Let $\displaystyle a,b,c,d,e$ be natural numbers.
Let $\displaystyle 2a = x_1$, $\displaystyle 2b = x_2$, $\displaystyle 2c = x_3$
$\displaystyle 2d+1 = x_4$, $\displaystyle 2e+1 = x_5$
So $\displaystyle a+b+c +d+e = 15$
Mush’s solution is interesting and solves the problem if we specify that $\displaystyle x_1 ,\,x_2 ,\;\& \,x_3$ are the even variables and the others are odd. However, there are more solutions that Mush seems to indicate.
For example: $\displaystyle a = 1,\,b = 3,\,c = 4,\,d = 6\;\& \,e = 1$ gives $\displaystyle x_1 = 2,\,x_2 = 6,\,x_3 = 8,\,x_4 = 13\;\& \,x_5 = 3$ which also works.
So we must count the number of solutions to $\displaystyle a+b+c +d+e = 15$ in the positive integers (each variable at least 1).
That number is $\displaystyle {{10+5-1}\choose {10}}$.
As has been pointed out, we do not know which of the variables the odd ones are.
So we to account for that consider that the two odds may be in $\displaystyle {5\choose 2}$ places.
The total then is $\displaystyle {{10+5-1}\choose {10}}$$\displaystyle {5\choose 2}$
9. but why u did: (10+5-1)C(10)
there are 15 digits
so it is not sepose to be (15+5-1)C(15)
????
10. Do you have a textbook that contains this material?
Placing N identical objects into k different cells can be done in $\displaystyle {{N+k-1} \choose {N}}$ ways.
However, some of the cells may be empty.
Like $\displaystyle a = 1,\,b = 0,\,c = 8,\,d = 6\;\& \,e = 0$ is a solution to $\displaystyle a+b+c +d+e = 15$ in the nonnegative integers.
Placing N identical objects into k different cells, with no empty cell, can be done in $\displaystyle {{N-1} \choose {N-k}}$ ways.
The solutions to $\displaystyle a+b+c +d+e = 15$ in the positive integers number $\displaystyle {{15-1} \choose {15-5}}$. | 1,749 | 5,508 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2018-22 | latest | en | 0.824518 |
http://openstudy.com/updates/504c1258e4b0925ffedd234c | 1,448,560,334,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398447769.81/warc/CC-MAIN-20151124205407-00057-ip-10-71-132-137.ec2.internal.warc.gz | 172,414,062 | 18,369 | ## rgarcia876 3 years ago How do I do this problem? 7+3 * [ 8+8 *( 5+9)] they say the answer is 367 but I can't get it right? Thanx
1. EulerGroupie
I agree with "they"
2. Chlorophyll
Work from the inner most bracket out!
3. Chlorophyll
@rgarcia876 Can you start???
4. rgarcia876
Yeah but the answer doesn't come out right for me. This is how I've done it 5+9 = 14 = 7+3 * [8+8 * 14] then I do 8+8 = 16 7+3 * [16 *14] then 16*14 = 224 7 + 3 * 224 and then i get stuck?! Can someone explain a little how they do it?
5. EulerGroupie
Nope... do 8 times 14 after 5+9... PEMDAS within each set of parenthesis
6. Anikate
do inner brackets then multiply by 8. then add 8. then times that by 3. then add 7.
7. Denebel
|dw:1347163260137:dw|
8. Anikate
garcia do u get what i'm saying?
9. rgarcia876
Yep! I get what you all are saying Thanks everyone for the responses I appreciate it I got my Order of Ops confused!
10. Anikate
np anytime | 327 | 948 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2015-48 | longest | en | 0.847053 |
https://learn.careers360.com/school/question-need-solution-for-rd-sharma-maths-class-12-chapter-differential-equation-exercise-21-point-2-question-16-subquestion-iii/?question_number=16.3 | 1,718,359,908,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861545.42/warc/CC-MAIN-20240614075213-20240614105213-00484.warc.gz | 308,875,201 | 40,998 | #### Need solution for RD Sharma maths class 12 chapter Differential Equation exercise 21.2 question 16 subquestion (iii)
The required equation is
$y-2x\frac{\mathrm{d} y}{\mathrm{d} x}=0$
Hint:
The equation is the equation of parabola
Given:
$y^{2}=4ax$
Solution:
$y^{2}=4ax$
\begin{aligned} &2y\frac{\mathrm{d} y}{\mathrm{d} x}=4a\\ &\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2a}{y}\\ &\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{y}\left ( \frac{y^{2}}{4x} \right )\\ &\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{y}{2x}\\ &2x\frac{\mathrm{d} y}{\mathrm{d} x}=y \end{aligned}
The required equation is
$y-2x\frac{\mathrm{d} y}{\mathrm{d} x}=0$ | 258 | 647 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.473764 |
http://www.stevekrenzel.com/articles/simple-statistical-substring-searching | 1,561,564,605,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000367.74/warc/CC-MAIN-20190626154459-20190626180459-00221.warc.gz | 306,506,915 | 3,272 | 2. Statistical Substring
Previous Index Next
Substring Searching at Scale
Here is an extremely simple implementation of a statistical full text substring search which has a runtime independent of the corpus size. Recall is 100% and precision is adjustable.
We use the sample index from http://www.dotnetdotcom.org/ for the test corpus (2MB in size) and break the source into lowercased n-grams. In this case of length 5. Given the string "The dog ran.", we'd get 8 5-grams: "the ", "he d ", "e dog", " dog ", "dog r", "og ra", "g ran", and " ran."
After all grams have been enumerated, the amount of space required increases roughly 5 fold (or n for an n-gram). Then we just iterate through each gram and insert it into a hashmap where the key is the gram and the value is the set of URLs which have contained that gram.
Searching consists of breaking the search string into grams, looking up each set of URLs for each gram, and intersecting all the sets. Whatever URLs are in the resultant set will contain the search string (with a high probability, which will be defined in another post).
The following code implements what we've described here. The script accepts two arguments, the file with the web content (from http://www.dotnetdotcom.org) and the gram length. You can run it with "python script.py web_index 5". The first loop does the indexing, and the second does the searching.
```#!/usr/bin/python
from sys import argv
fin = open(argv[1])
index = {}
pairs = [(content[i], content[i+2]) for i in xrange(0, len(content)-1, 3)]
min_tok = int(argv[2])
fin.close()
for url, html in pairs:
for i in range(len(html)-min_tok+1):
key = html[i:i+min_tok].lower()
value = index.setdefault(key, set())
inp = raw_input("Search: ").lower()
while inp != "":
results = index.get(inp[:min_tok], set())
for i in range(len(inp)-min_tok+1):
results = results.intersection(index.get(inp[i:i+min_tok], set()))
for i, result in enumerate(results):
print "%d) %s"%(i+1, result)
print ""
inp = raw_input("Search: ").lower()
```
There are two big drawbacks with this approach: 1) The search string can't be shorter than the gram length and 2) The index becomes huge, in this case 13MB from 2MB. But the benefit is that you can do full text searching extremely quickly, even for very large sets of corpuses, and recall is always 100%. You can also increase or decrease the size of the index by adjusting the size of the grams and the frequency of hash collisions. By hash collisions, I'm referring to two different grams pointing to the same set of urls even if all of the urls in that set don't contain both of the grams. By adjusting these parameters, we can finely tune the precision of the search and the size of the index. I plan on visiting these tradeoffs in a follow up post.
About the author I'm Steve Krenzel, a software engineer and co-founder of Thinkfuse. Contact me at steve@thinkfuse.com.
Previous Next | 705 | 2,927 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-26 | latest | en | 0.80271 |
https://www.teacherspayteachers.com/Browse/Search:perimeter%20estimation%20and%20actual | 1,571,235,843,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986668994.39/warc/CC-MAIN-20191016135759-20191016163259-00228.warc.gz | 1,120,259,424 | 75,128 | # Results forperimeter estimation and actual
by
Estimation can be a hard concept to teach. The best thing I have found to help students' estimation skills improve is practice, practice, practice! This product can be used for students to estimate and find the perimeter or inside area of the included pictures. With this purchase you will receive 4
Subjects:
Types:
\$6.00
\$5.00
27 Ratings
3.9
PDF (6.46 MB)
Skills and Activities 3rd Grade Mathematics / 3ero Matemáticas Great for Mathematics Practice for whole groups, warm-up, stations, intervention groups or individual work. Just print on card stock or construction paper then laminate for durability. 143 Pages with practice 1-27 Concepts below, picture
Subjects:
Types:
\$2.50
5 Ratings
3.8
PDF (4.18 MB)
Here is a fun math activity with a pumpkin theme! Students use chips, beans, candy corns, or other markers to find the area and perimeter of four different pumpkins. Students first estimate how many markers it will take to fill the area or perimeter of each pumpkin. They then use the markers to f
Subjects:
Types:
\$1.00
4 Ratings
3.9
PDF (627.46 KB)
This is a math activity that has students practice the concept of estimation. It is an outline of a bunny face. First students make an estimation as to how many marshmallows they can fit around the perimeter of the bunny face. Then glue marshmallows around the bunny face. Count the marshmallows
Subjects:
Types:
\$1.00
1 Rating
4.0
DOC (42.5 KB)
Students can struggle with the idea that the perimeter of a shape is the sum of its sides – get your students understanding and thinking of perimeter and units with this fun measuring and graphing worksheet suitable for high school students! Open the preview to check out the product! Contents This
Subjects:
Types:
\$1.00
1 Rating
4.0
PDF (698.99 KB)
This activity teaches perimeter and area in the real world. Students are asked to work in pairs to find the perimeter and area of items around the classroom or around the school. They must first think about the best unit of measurement for each object they choose. Then, they must make an estimate
Subjects:
Types:
\$2.99
not yet rated
N/A
DOCX (14.35 KB)
When I began teaching it quickly became apparent to me that many of my computationally most proficient students, my students with the greatest prior knowledge, and my students who did the best on in-class work were lacking a basic understanding of common units of measure. I had students who could do
Subjects:
Types:
\$3.00
not yet rated
N/A
ZIP (545.98 KB)
This PDF can be shown on the board like a PPT to use with the whole class or printed to use individually or in groups. There are 9 multiple choice questions over reasonable estimates of measurements and finding the perimeter of shapes (including a square with just one side labeled). Some questions a
Subjects:
\$2.00
not yet rated
N/A
PDF (513.44 KB)
Formula 1 Math Powerpac F Lesson 4, Perimeter, Area, and Metric Systems Approximate working time is 20 minutes per lesson. With Formula I Math Intervention students will find success and enjoyment in these highly motivating math lessons. Powerpac lessons concentrate on basic operations with whole n
Subjects:
\$9.99
not yet rated
N/A
ZIP (42.06 MB)
by
Estimation can be a hard concept to teach. The best thing I have found to help students' estimation skills improve is practice, practice, practice! These FREE estimation worksheets will allow early learners to estimate and find the perimeter and area of the included pictures. With this FREE download
Subjects:
Types:
FREE
13 Ratings
4.0
PDF (796.58 KB)
This resource includes 104 middle school math anchor charts based on the CCSS of 6th grade math! The standards are NOT posted on the actual anchor charts. This resource is great for enhancing classroom instruction, aiding in individual or group work, and offers a "go to" for students when the teache
Subjects:
Types:
CCSS:
\$25.75
\$20.00
158 Ratings
4.0
Bundle
This No Prep Notebook thoroughly covers ALL of the 4th grade TEKS. Simply print and hole punch and this notebook is ready for your students to use. This notebook is an excellent way to introduce new concepts in whole class or small group guided math lessons and is the perfect resource for students
Subjects:
\$32.00
\$25.50
163 Ratings
4.0
Bundle
Are you looking for premium quality math resources that are uniquely aligned to the TEKS Math Standards for 3rd Grade? Perhaps you are hopeful to find the perfect math stations to add to your centers this year? Let NUMBEROCK’s professionally edited task card bundle be your one-stop solution to con
Subjects:
Types:
\$160.00
\$89.95
31 Ratings
4.0
Bundle
End the year strong with a math scavenger hunt!At the end of the school year it's harder than ever to keep students motivated, engaged, and learning-so why not take them outside? This end of the year math activity is designed to be completed outdoors and incorporates multiple learning styles to keep
Subjects:
Types:
\$3.50
648 Ratings
4.0
PDF (19.24 MB)
This bundle includes every 3rd grade math Boom Card Deck that is in my Teachers pay Teachers store, at about 25% below the actual cost!Formerly a "growing" bundle, this bundle includes at least one Boom Card Deck for each Virginia math SOL standard 3.1 through 3.17.Virginia math standards covered by
Subjects:
CCSS:
VA SOL:
\$100.65
\$75.00
18 Ratings
4.0
Bundle
Also use as spiral math review all year long. This packet contains more than 124 fourth grade math problems covering the common core state standards (CCSS). Use it to: - Review problems at the end of the year to find out what skills to revisit - Spiral review throughout the year - Sub packets that
Subjects:
Types:
\$7.00
115 Ratings
4.0
PDF (1.46 MB)
This 457 page file includes station activities which cover the ENTIRE YEAR. They are designed to align with common core standards for seventh grade math. You save 25% when buying the bundle as compared to all the activities purchased individually. As I add more activities, I will increase the price
Subjects:
Types:
CCSS:
\$116.00
\$87.00
50 Ratings
4.0
PDF (77.41 MB)
PLEASE be sure to download the preview to see all of the goodies included in this packet. This bundle contains 200 printables - August, September, October, November, and December. All are for third grade and get progressively more difficult with each set. Each month contains twenty literacy and t
Subjects:
Types:
CCSS:
\$25.00
\$20.00
14 Ratings
4.0
Bundle
The Gingerbread Measurement Unit is an ideal set of activities for students to practice their measurement skills by measuring gingerbread men, cookies and houses. This is a perfect addition to a Gingerbread thematic unit or to enhance learning during the Christmas season. ••••••••••••••••••••••••••
Subjects:
Types:
CCSS:
\$5.00
58 Ratings
4.0
PDF (14.46 MB)
Third Grade Common Core Math Worksheets Measurement and Data 3.MD Over 120 worksheets with most answer keys covering ALL Third Grade Measurement and Data Standards Measurement and Data Solve problems involving measurement and estimation. CCSS.MATH.CONTENT.3.MD.A.1 Tell and write time to the near
Subjects:
Types:
CCSS:
\$7.00
\$5.00
31 Ratings
4.0
PDF (2.04 MB)
Over 400 pages covering ALL THIRD GRADE MATH STANDARDS Operations and Algebraic Thinking Represent and solve problems involving multiplication and division. CCSS.MATH.CONTENT.3.OA.A.1 Interpret products of whole numbers, e.g., interpret 5 × 7 as the total number of objects in 5 groups of 7 object
Subjects:
Types:
CCSS:
\$26.00
\$15.00
17 Ratings
4.0
PDF (5.02 MB)
Complete Fourth Grade Common Core Math Worksheets All Standards Over 260 Fourth Grade Common Core Worksheet with Most Answer Sheets Operations and Algebraic Thinking Use the four operations with whole numbers to solve problems. CCSS.Math.Content.4.OA.A.1 Interpret a multiplication equation as a
Subjects:
Types:
CCSS:
\$24.00
\$14.00
13 Ratings
4.0
PDF (8.45 MB)
This is an entire NYS Integrated Geometry Curriculum. All Units are broken down into separate PDF files. The Units are broken down as follows: Unit 1: Basics of Geo Unit 2: Logic Unit 3: Parallel and Perpendicular Lines Unit 4: Triangle Thm and Intro to Congruence Unit 5: Triangle Properties and Ri
Subjects:
\$29.99
\$24.99
10 Ratings
3.9
ZIP (66.79 MB)
This 70 page file includes station activities focused on expressions, equations, and inequalities. They are designed to align with common core standards for seventh grade math. You save 15% when buying the bundle as compared to all the activities purchased individually. As I add more activities, I w
Subjects:
Types:
CCSS:
\$13.60
8 Ratings
4.0
PDF (12.23 MB)
showing 1-24 of 29 results
Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 2,132 | 8,807 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | longest | en | 0.88612 |
https://www.lotterypost.com/thread/213553/11 | 1,485,032,104,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281202.94/warc/CC-MAIN-20170116095121-00105-ip-10-171-10-70.ec2.internal.warc.gz | 947,060,726 | 17,061 | Welcome Guest
You last visited January 21, 2017, 2:27 pm
All times shown are
Eastern Time (GMT-5:00)
# P-3 Bet generator 8 combos
Topic closed. 265 replies. Last post 7 years ago by threetwowin.
Page 11 of 18
United States
Member #75358
June 1, 2009
5345 Posts
Offline
Posted: May 6, 2010, 3:36 pm - IP Logged
Hello Joker,
Georgia midday 5/6/10 ......538. Do you have time to do the magic? Thanks! maggie
048, 154, 910, 942, 985, 821, 637, 502
Hyattsville, MD
United States
Member #39877
May 23, 2006
74 Posts
Offline
Posted: May 6, 2010, 3:43 pm - IP Logged
Md got 835 today, isn't that something
middle Georgia
United States
Member #1893
July 20, 2003
1868 Posts
Offline
Posted: May 6, 2010, 3:43 pm - IP Logged
048, 154, 910, 942, 985, 821, 637, 502
Wow! That was quick..... Thank you. maggie
colonial heights
United States
Member #31712
February 2, 2006
12265 Posts
Offline
Posted: May 6, 2010, 3:47 pm - IP Logged
802
507, 490, 345, 306, 379, 764, 218, 956
034
740, 634, 967, 948, 903, 086, 125, 378
wow mirror the 764 7-2 6-1 4 214 first try way to go...........
when you do it do it right the first time...........
New Mexico
United States
Member #86099
January 29, 2010
11166 Posts
Offline
Posted: May 6, 2010, 4:00 pm - IP Logged
Last night's pick 3 in new mexico was 990. What can you tell me about tonights #'s? Thanks!
Michigan
United States
Member #62830
July 10, 2008
497 Posts
Offline
Posted: May 6, 2010, 4:32 pm - IP Logged
Hi joker i have a question does it matter what order the number fall? Like would 082 and 802 generate the same number. I have another question for example you gave me the numbers for 761 which generated these
379, 467, 243, 270, 296, 904, 185, 630
Is there a way to determine another number that has these numbers in it's run down?
United States
Member #75358
June 1, 2009
5345 Posts
Offline
Posted: May 6, 2010, 5:42 pm - IP Logged
Last night's pick 3 in new mexico was 990. What can you tell me about tonights #'s? Thanks!
392, 409, 643, 695, 620, 254, 781, 035
United States
Member #75358
June 1, 2009
5345 Posts
Offline
Posted: May 6, 2010, 5:43 pm - IP Logged
Hi joker i have a question does it matter what order the number fall? Like would 082 and 802 generate the same number. I have another question for example you gave me the numbers for 761 which generated these
379, 467, 243, 270, 296, 904, 185, 630
Is there a way to determine another number that has these numbers in it's run down?
Ist question, yes, same numbers no matter what order. 2nd question, no.
Pennsylvania
United States
Member #74096
May 2, 2009
23238 Posts
Offline
Posted: May 6, 2010, 5:44 pm - IP Logged
joker, I'm curious and if you don't mind, PA had 666 last night..does that change or complicate things?
What do you see from that number?
Thank you!
United States
Member #75358
June 1, 2009
5345 Posts
Offline
Posted: May 6, 2010, 5:57 pm - IP Logged
joker, I'm curious and if you don't mind, PA had 666 last night..does that change or complicate things?
What do you see from that number?
Thank you!
069, 176, 310, 362, 397, 921, 458, 702
Pennsylvania
United States
Member #74096
May 2, 2009
23238 Posts
Offline
Posted: May 6, 2010, 6:00 pm - IP Logged
Thank you joker!
United States
Member #14
November 9, 2001
31542 Posts
Offline
Posted: May 6, 2010, 7:25 pm - IP Logged
Only p-3 folks...c'mon.....
486, 598, 254, 283, 269, 635, 170, 943
Bingo!!!! ga. night 486
love to nibble those micey feet.
wisconsin
United States
Member #49379
January 28, 2007
2391 Posts
Offline
Posted: May 6, 2010, 7:27 pm - IP Logged
The generator on lp I can't put the drawing number in. Are you using that? Thanks
United States
Member #81314
October 16, 2009
19438 Posts
Online
Posted: May 6, 2010, 7:49 pm - IP Logged
055
257, 865, 382, 351, 376, 718, 940, 621
I"m putting all my money on, 823, 382, 283, 328, 832, 238.
The Struggle is real!
United States
Member #75358
June 1, 2009
5345 Posts
Offline
Posted: May 6, 2010, 8:37 pm - IP Logged
Bingo!!!! ga. night 486
Wow...Isn't that 2 nights in a row?
Page 11 of 18 | 1,504 | 4,160 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-04 | latest | en | 0.710107 |
http://gmatclub.com/forum/ds-geometry-12265.html?fl=similar | 1,484,592,243,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279224.13/warc/CC-MAIN-20170116095119-00477-ip-10-171-10-70.ec2.internal.warc.gz | 121,102,122 | 43,577 | DS: Geometry : DS Archive
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack
It is currently 16 Jan 2017, 10:44
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# DS: Geometry
Author Message
Director
Joined: 07 Nov 2004
Posts: 689
Followers: 6
Kudos [?]: 142 [0], given: 0
### Show Tags
14 Dec 2004, 07:12
00:00
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 0 sessions
### HideShow timer Statistics
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
Attachments
q1.GIF [ 5.04 KiB | Viewed 962 times ]
Director
Joined: 21 Sep 2004
Posts: 610
Followers: 1
Kudos [?]: 33 [0], given: 0
### Show Tags
14 Dec 2004, 07:22
i think one of the properties states that if it makes a right angled triangle then it woul dbe semi circle ..(that half of it ) which would make AC is the diameter..
answer A? not too sure though..
CIO
Joined: 09 Mar 2003
Posts: 463
Followers: 2
Kudos [?]: 58 [0], given: 0
### Show Tags
14 Dec 2004, 09:21
agree with A. when a right triangle is inscribed in a circle, the hypotenuse is the diameter.
Director
Joined: 19 Nov 2004
Posts: 559
Location: SF Bay Area, USA
Followers: 4
Kudos [?]: 195 [0], given: 0
### Show Tags
14 Dec 2004, 10:30
1) Angle of a triangle inscribed in a semicircle is 90. Here, the angle is 90, so it is a semi circle (corollary) => AC is a diameter
Ok
2) AB/BC = 3/4. Perhaps they are baiting that we might assume AB:BC:AC = 3:4:5 and therefore conclude angle ABC is 90*. But that is not the case. It does not say BC:AC is 4:5.
A it is
Director
Joined: 07 Nov 2004
Posts: 689
Followers: 6
Kudos [?]: 142 [0], given: 0
### Show Tags
14 Dec 2004, 15:01
OA is A.
Good explanation nocilis.
14 Dec 2004, 15:01
Display posts from previous: Sort by | 725 | 2,369 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2017-04 | latest | en | 0.881373 |
http://pakturkmaths.com/debt-snowball-worksheet/ | 1,566,434,597,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316555.4/warc/CC-MAIN-20190822000659-20190822022659-00150.warc.gz | 149,610,782 | 29,052 | # Debt Snowball Worksheet
Debt Snowball Worksheet – In terms of achieving a purpose you would simply put down on paper what it is you want. Nevertheless many people discover that we often lose these items of paper, or we don’t use a great format just like the SMART objectives format. Therefore it would truly be significantly better should you had been to make use of one of the many objective setting Debt Snowball Worksheet on provide at present.
However with so many different types of aim setting Debt Snowball Worksheet templates on provide discovering the right one can prove a challenge in itself. What makes it easier is these templates are very often built-in right into a aim setting software.
I’d advise studying as many evaluations as you may in relation to the totally different setting objectives Debt Snowball Worksheet available. If there is a chance to try any of these out then accomplish that – you will find most objective setting software give a free trial interval. By making an attempt out just a few even if just for just a few days can help you to raised understand how they work and which of them you’re feeling most comfortable using.
## Using Microsoft Excel to Create a Monetary Worksheet Template
Excel users can make the most of texts, formulation, and double click adjustments to create a template Debt Snowball Worksheet for any residence, business, or church. We’ll define here how you can set up a worksheet template in Microsoft Excel. This primary template can then be used for primary document keeping or modified for a lot of different makes use of.
For this text we are going to use four words to manage our spreadsheet: Description, Expense, Deposit, and Stability. Enter the phrase ‘Description’ at A1, ‘Expense’ at F1, ‘Deposit’ at H1, and ‘Stability’ at J1. Proceed by coming into formulas into the text containers beginning with ‘Expense’ in F 11, and ‘Deposit’ in H 11. For F 11 enter within the formulation, =sum(f2:f10) and for H11 enter =sum(h2:h10). Remember to include the entire formula which starts with the equals signal ‘=’ and ends with the last parenthesis ‘)’. A pleasant function of Excel is its capacity to adjust formulas when copied and pasted into one other cell. In other words, in case you were to enter the first method above then copy that cell and paste it into H11, then Excel will routinely alter the method from F’s to H’s. What these formulation will do is take the numbers you’ll enter and routinely add the worth in all cells between H2 and H10 and display the whole in H11 in addition to the same within the F column. For those who need more space you can substitute H11 for any number of slots. For example H2:H100 will add up from H2-H100. That is additionally one other space where Excel will mechanically regulate formulation for you. Should you insert rows anywhere between your formulas range (H2 via H10 for instance) then your formula in the final cell will automatically change to include all rows utilizing the appropriate cell location. Also, it is very important be aware that when getting into an expense to you’ll want to embrace the unfavourable sign so it’s subtracted from the total quite than added.
The following step is to create a method to calculate your total balance of all columns. In the H13 textbox enter the system =sum(f11:h11), what this will do is total the unfavorable expenses and the constructive deposits, creating a grand whole quantity. Additionally, you will wish to create a beginning steadiness (start of the month balance) at J2. If you’re utilizing this template for a brand new undertaking, then your beginning steadiness shall be zero.
## A Budgeting Worksheet Gets You Started With a Finances
You must start someplace and a budgeting Debt Snowball Worksheet could make the dreaded task of budgeting a lot easier. Whether you utilize computer based mostly software or plan to keep your funds on paper, a worksheet may also help you to brainstorm the classes you’ll need to finances for. While most of them will likely be slightly different than the finances you eventually come up with, they can function a useful instrument that will make your entire process much simpler.
You will discover a number of types of a budgeting worksheet on-line that you can print or use as a template in considered one of your current applications. Even if you just use it to get concepts, it may be an enormous assist while you sit all the way down to create a cash administration plan. While not everybody has the same earnings and expense circumstances, this kind of worksheet is a superb place to start.
In the event you do discover a budgeting worksheet that you simply like, you can merely add classes to it or substitute the ones you will not use with your individual items. For instance, many of these worksheets that one can find on-line have a spot for funding income but when you have no investments to trace either ignore that category or substitute it with one of your personal.
## Utilizing Math Worksheets
What are math worksheets and what are they used for? These are math types which might be utilized by mother and father and teachers alike to assist the younger kids learn primary math akin to subtraction, addition, multiplication and division. This tool is very important and when you’ve got a small child and you don’t have a worksheet, then its time you bought your self one or created one to your child. There are a number of sites over the internet that supply free worksheets which might be downloadable and printable to be used by mother and father and lecturers at dwelling or at school.
When you can not purchase a math work sheet because you think it’s possible you’ll not have time to, then you can create on using your private home computer and customise it in your child. Doing this is simple. All you need is Microsoft phrase software in your pc to achieve this. Just open the phrase software in your laptop and start a brand new doc. Be sure that the new document you are about to create is based on a template. Then, ensure that your web connection is on earlier than you possibly can search the term “math worksheet” from the internet. You’ll get templates of all types to your worksheet. Choose the one you want after which download.
As soon as downloaded, you’ll be able to customise the mathematics worksheet to fit your kid. The level of the child at school will decide the look and content material of the worksheet. Use the college textbook that your little one uses in school as a reference information to help you in the creation of the math worksheet. This may be certain that the worksheet is completely relevant to the kid and can assist the child improve his or her grades at school.
Thanks for reading my article about Debt Snowball Worksheet, i hope you take pleasure in it. Please come again to read my another article
## Related posts of "Debt Snowball Worksheet"
#### Vietnam War Worksheets
Vietnam War Worksheets - With regards to reaching a aim you can simply put down on paper what it is you want. Nonetheless many people find that we often lose these pieces of paper, or we do not use a superb format like the SMART targets format. Due to this fact it might really be...
#### Ionic Compound Formula Writing Worksheet
Ionic Compound Formula Writing Worksheet - Relating to attaining a purpose you could just put down on paper what it is you want. Nonetheless many of us find that we regularly lose these pieces of paper, or we do not use format like the SMART targets format. Therefore it would actually be much better when...
#### Free Name Tracing Worksheets For Preschool
Free Name Tracing Worksheets For Preschool - On the subject of attaining a aim you possibly can just put down on paper what it is you want. Nevertheless many of us discover that we frequently lose these pieces of paper, or we don't use a very good format just like the SMART goals format. Therefore...
#### Prokaryotic And Eukaryotic Cells Worksheet Answer Key
Prokaryotic And Eukaryotic Cells Worksheet Answer Key - With regards to achieving a goal you may just put down on paper what it's you need. However many of us find that we regularly lose these items of paper, or we do not use a superb format just like the SMART targets format. Subsequently it would... | 1,701 | 8,323 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2019-35 | latest | en | 0.922947 |
http://functions.wolfram.com/HypergeometricFunctions/ChebyshevTGeneral/06/04/01/04/0009/ | 1,519,075,338,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812841.74/warc/CC-MAIN-20180219211247-20180219231247-00652.warc.gz | 138,911,038 | 9,365 | html, body, form { margin: 0; padding: 0; width: 100%; } #calculate { position: relative; width: 177px; height: 110px; background: transparent url(/images/alphabox/embed_functions_inside.gif) no-repeat scroll 0 0; } #i { position: relative; left: 18px; top: 44px; width: 133px; border: 0 none; outline: 0; font-size: 11px; } #eq { width: 9px; height: 10px; background: transparent; position: absolute; top: 47px; right: 18px; cursor: pointer; }
ChebyshevT
http://functions.wolfram.com/07.04.06.0096.01
Input Form
ChebyshevT[\[Nu], z] == ((\[Nu] Cos[Pi \[Nu]]^2 Sin[Pi \[Nu]])/Pi^(3/2)) (Sum[Residue[((Gamma[s] Gamma[1/2 + s] Gamma[-s + \[Nu]])/((1 - z)/2)^s) Gamma[-s - \[Nu]], {s, j - \[Nu]}], {j, 0, Infinity}] + Sum[Residue[((Gamma[s] Gamma[1/2 + s] Gamma[-s - \[Nu]])/((1 - z)/2)^s) Gamma[-s + \[Nu]], {s, j + \[Nu]}], {j, 0, Infinity}]) - ((Sqrt[1 + z] Cos[Pi \[Nu]] Sin[Pi \[Nu]]^2)/(Sqrt[2] Pi^(3/2))) (Sum[Residue[((Gamma[s] Gamma[1/2 + s] Gamma[1/2 - s + \[Nu]])/ ((1 - z)/2)^s) Gamma[1/2 - s - \[Nu]], {s, 1/2 + j - \[Nu]}], {j, 0, Infinity}] + Sum[Residue[((Gamma[s] Gamma[1/2 + s] Gamma[1/2 - s - \[Nu]])/ ((1 - z)/2)^s) Gamma[1/2 - s + \[Nu]], {s, 1/2 + j + \[Nu]}], {j, 0, Infinity}]) /; Abs[z - 1] > 2
Standard Form
Cell[BoxData[RowBox[List[RowBox[List[RowBox[List["ChebyshevT", "[", RowBox[List["\[Nu]", ",", "z"]], "]"]], "\[Equal]", RowBox[List[RowBox[List[FractionBox[RowBox[List["\[Nu]", " ", SuperscriptBox[RowBox[List["Cos", "[", RowBox[List["\[Pi]", " ", "\[Nu]"]], "]"]], "2"], " ", RowBox[List["Sin", "[", RowBox[List["\[Pi]", " ", "\[Nu]"]], "]"]]]], SuperscriptBox["\[Pi]", RowBox[List["3", "/", "2"]]]], " ", RowBox[List["(", RowBox[List[RowBox[List["Sum", "[", RowBox[List[RowBox[List["Residue", "[", RowBox[List[RowBox[List[RowBox[List["(", RowBox[List[RowBox[List["Gamma", "[", "s", "]"]], " ", RowBox[List["Gamma", "[", RowBox[List[FractionBox["1", "2"], "+", "s"]], "]"]], " ", RowBox[List["Gamma", "[", RowBox[List[RowBox[List["-", "s"]], "+", "\[Nu]"]], "]"]], SuperscriptBox[RowBox[List["(", FractionBox[RowBox[List["1", "-", "z"]], "2"], ")"]], RowBox[List["-", "s"]]]]], ")"]], RowBox[List["Gamma", "[", RowBox[List[RowBox[List["-", "s"]], "-", "\[Nu]"]], "]"]]]], ",", RowBox[List["{", RowBox[List["s", ",", RowBox[List["j", "-", "\[Nu]"]]]], "}"]]]], "]"]], ",", RowBox[List["{", RowBox[List["j", ",", "0", ",", InterpretationBox["\[Infinity]", DirectedInfinity[1]]]], "}"]]]], "]"]], "+", RowBox[List["Sum", "[", RowBox[List[RowBox[List["Residue", "[", RowBox[List[RowBox[List[RowBox[List["(", RowBox[List[RowBox[List["Gamma", "[", "s", "]"]], " ", RowBox[List["Gamma", "[", RowBox[List[FractionBox["1", "2"], "+", "s"]], "]"]], " ", RowBox[List["Gamma", "[", RowBox[List[RowBox[List["-", "s"]], "-", "\[Nu]"]], "]"]], SuperscriptBox[RowBox[List["(", FractionBox[RowBox[List["1", "-", "z"]], "2"], ")"]], RowBox[List["-", "s"]]]]], ")"]], " ", RowBox[List["Gamma", "[", RowBox[List[RowBox[List["-", "s"]], "+", "\[Nu]"]], "]"]]]], ",", RowBox[List["{", RowBox[List["s", ",", RowBox[List["j", "+", "\[Nu]"]]]], "}"]]]], "]"]], ",", RowBox[List["{", RowBox[List["j", ",", "0", ",", InterpretationBox["\[Infinity]", DirectedInfinity[1]]]], "}"]]]], "]"]]]], ")"]]]], "-", RowBox[List[FractionBox[RowBox[List[SqrtBox[RowBox[List["1", "+", "z"]]], " ", RowBox[List["Cos", "[", RowBox[List["\[Pi]", " ", "\[Nu]"]], "]"]], " ", SuperscriptBox[RowBox[List["Sin", "[", RowBox[List["\[Pi]", " ", "\[Nu]"]], "]"]], "2"]]], RowBox[List[SqrtBox["2"], " ", SuperscriptBox["\[Pi]", RowBox[List["3", "/", "2"]]]]]], RowBox[List["(", RowBox[List[RowBox[List["Sum", "[", RowBox[List[RowBox[List["Residue", "[", RowBox[List[RowBox[List[RowBox[List["(", RowBox[List[RowBox[List["Gamma", "[", "s", "]"]], " ", RowBox[List["Gamma", "[", RowBox[List[FractionBox["1", "2"], "+", "s"]], "]"]], " ", RowBox[List["Gamma", "[", RowBox[List[FractionBox["1", "2"], "-", "s", "+", "\[Nu]"]], "]"]], SuperscriptBox[RowBox[List["(", FractionBox[RowBox[List["1", "-", "z"]], "2"], ")"]], RowBox[List["-", "s"]]]]], ")"]], RowBox[List["Gamma", "[", RowBox[List[FractionBox["1", "2"], "-", "s", "-", "\[Nu]"]], "]"]]]], ",", RowBox[List["{", RowBox[List["s", ",", RowBox[List[FractionBox["1", "2"], "+", "j", "-", "\[Nu]"]]]], "}"]]]], "]"]], ",", RowBox[List["{", RowBox[List["j", ",", "0", ",", InterpretationBox["\[Infinity]", DirectedInfinity[1]]]], "}"]]]], "]"]], "+", RowBox[List["Sum", "[", RowBox[List[RowBox[List["Residue", "[", RowBox[List[RowBox[List[RowBox[List["(", RowBox[List[RowBox[List["Gamma", "[", "s", "]"]], " ", RowBox[List["Gamma", "[", RowBox[List[FractionBox["1", "2"], "+", "s"]], "]"]], " ", RowBox[List["Gamma", "[", RowBox[List[FractionBox["1", "2"], "-", "s", "-", "\[Nu]"]], "]"]], " ", SuperscriptBox[RowBox[List["(", FractionBox[RowBox[List["1", "-", "z"]], "2"], ")"]], RowBox[List["-", "s"]]]]], ")"]], RowBox[List["Gamma", "[", RowBox[List[FractionBox["1", "2"], "-", "s", "+", "\[Nu]"]], "]"]]]], ",", RowBox[List["{", RowBox[List["s", ",", RowBox[List[FractionBox["1", "2"], "+", "j", "+", "\[Nu]"]]]], "}"]]]], "]"]], ",", RowBox[List["{", RowBox[List["j", ",", "0", ",", InterpretationBox["\[Infinity]", DirectedInfinity[1]]]], "}"]]]], "]"]]]], ")"]]]]]]]], "/;", RowBox[List[RowBox[List["Abs", "[", RowBox[List["z", "-", "1"]], "]"]], ">", "2"]]]]]]
MathML Form
T ν ( z ) ν cos 2 ( π ν ) sin ( π ν ) π 3 / 2 ( j = 0 res s ( ( Γ ( s ) Γ ( s + 1 2 ) Γ ( ν - s ) ( 1 - z 2 ) - s ) Γ ( - s - ν ) ) ( j - ν ) + j = 0 res s ( ( Γ ( s ) Γ ( s + 1 2 ) Γ ( - s - ν ) ( 1 - z 2 ) - s ) Γ ( ν - s ) ) ( j + ν ) ) - z + 1 cos ( π ν ) sin 2 ( π ν ) 2 π 3 / 2 ( j = 0 res s ( ( Γ ( s ) Γ ( s + 1 2 ) Γ ( - s + ν + 1 2 ) ( 1 - z 2 ) - s ) Γ ( - s - ν + 1 2 ) ) ( j - ν + 1 2 ) + j = 0 res s ( ( Γ ( s ) Γ ( s + 1 2 ) Γ ( - s - ν + 1 2 ) ( 1 - z 2 ) - s ) Γ ( - s + ν + 1 2 ) ) ( j + ν + 1 2 ) ) /; "\[LeftBracketingBar]" z - 1 "\[RightBracketingBar]" > 2 Condition ChebyshevT ν z ν ν 2 ν 3 2 -1 j 0 DirectedInfinity 1 Subscript res s Gamma s Gamma s 1 2 Gamma ν -1 s 1 -1 z 2 -1 -1 s Gamma -1 s -1 ν j -1 ν j 0 DirectedInfinity 1 Subscript res s Gamma s Gamma s 1 2 Gamma -1 s -1 ν 1 -1 z 2 -1 -1 s Gamma ν -1 s j ν -1 z 1 1 2 ν ν 2 2 1 2 3 2 -1 j 0 DirectedInfinity 1 Subscript res s Gamma s Gamma s 1 2 Gamma -1 s ν 1 2 1 -1 z 2 -1 -1 s Gamma -1 s -1 ν 1 2 j -1 ν 1 2 j 0 DirectedInfinity 1 Subscript res s Gamma s Gamma s 1 2 Gamma -1 s -1 ν 1 2 1 -1 z 2 -1 -1 s Gamma -1 s ν 1 2 j ν 1 2 z -1 2 [/itex]
Rule Form
Cell[BoxData[RowBox[List[RowBox[List["HoldPattern", "[", RowBox[List["ChebyshevT", "[", RowBox[List["\[Nu]_", ",", "z_"]], "]"]], "]"]], "\[RuleDelayed]", RowBox[List[RowBox[List[FractionBox[RowBox[List[RowBox[List["(", RowBox[List["\[Nu]", " ", SuperscriptBox[RowBox[List["Cos", "[", RowBox[List["\[Pi]", " ", "\[Nu]"]], "]"]], "2"], " ", RowBox[List["Sin", "[", RowBox[List["\[Pi]", " ", "\[Nu]"]], "]"]]]], ")"]], " ", RowBox[List["(", RowBox[List[RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["j", "=", "0"]], "\[Infinity]"], RowBox[List["Residue", "[", RowBox[List[RowBox[List[RowBox[List["(", RowBox[List[RowBox[List["Gamma", "[", "s", "]"]], " ", RowBox[List["Gamma", "[", RowBox[List[FractionBox["1", "2"], "+", "s"]], "]"]], " ", RowBox[List["Gamma", "[", RowBox[List[RowBox[List["-", "s"]], "+", "\[Nu]"]], "]"]], " ", SuperscriptBox[RowBox[List["(", FractionBox[RowBox[List["1", "-", "z"]], "2"], ")"]], RowBox[List["-", "s"]]]]], ")"]], " ", RowBox[List["Gamma", "[", RowBox[List[RowBox[List["-", "s"]], "-", "\[Nu]"]], "]"]]]], ",", RowBox[List["{", RowBox[List["s", ",", RowBox[List["j", "-", "\[Nu]"]]]], "}"]]]], "]"]]]], "+", RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["j", "=", "0"]], "\[Infinity]"], RowBox[List["Residue", "[", RowBox[List[RowBox[List[RowBox[List["(", RowBox[List[RowBox[List["Gamma", "[", "s", "]"]], " ", RowBox[List["Gamma", "[", RowBox[List[FractionBox["1", "2"], "+", "s"]], "]"]], " ", RowBox[List["Gamma", "[", RowBox[List[RowBox[List["-", "s"]], "-", "\[Nu]"]], "]"]], " ", SuperscriptBox[RowBox[List["(", FractionBox[RowBox[List["1", "-", "z"]], "2"], ")"]], RowBox[List["-", "s"]]]]], ")"]], " ", RowBox[List["Gamma", "[", RowBox[List[RowBox[List["-", "s"]], "+", "\[Nu]"]], "]"]]]], ",", RowBox[List["{", RowBox[List["s", ",", RowBox[List["j", "+", "\[Nu]"]]]], "}"]]]], "]"]]]]]], ")"]]]], SuperscriptBox["\[Pi]", RowBox[List["3", "/", "2"]]]], "-", FractionBox[RowBox[List[RowBox[List["(", RowBox[List[SqrtBox[RowBox[List["1", "+", "z"]]], " ", RowBox[List["Cos", "[", RowBox[List["\[Pi]", " ", "\[Nu]"]], "]"]], " ", SuperscriptBox[RowBox[List["Sin", "[", RowBox[List["\[Pi]", " ", "\[Nu]"]], "]"]], "2"]]], ")"]], " ", RowBox[List["(", RowBox[List[RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["j", "=", "0"]], "\[Infinity]"], RowBox[List["Residue", "[", RowBox[List[RowBox[List[RowBox[List["(", RowBox[List[RowBox[List["Gamma", "[", "s", "]"]], " ", RowBox[List["Gamma", "[", RowBox[List[FractionBox["1", "2"], "+", "s"]], "]"]], " ", RowBox[List["Gamma", "[", RowBox[List[FractionBox["1", "2"], "-", "s", "+", "\[Nu]"]], "]"]], " ", SuperscriptBox[RowBox[List["(", FractionBox[RowBox[List["1", "-", "z"]], "2"], ")"]], RowBox[List["-", "s"]]]]], ")"]], " ", RowBox[List["Gamma", "[", RowBox[List[FractionBox["1", "2"], "-", "s", "-", "\[Nu]"]], "]"]]]], ",", RowBox[List["{", RowBox[List["s", ",", RowBox[List[FractionBox["1", "2"], "+", "j", "-", "\[Nu]"]]]], "}"]]]], "]"]]]], "+", RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["j", "=", "0"]], "\[Infinity]"], RowBox[List["Residue", "[", RowBox[List[RowBox[List[RowBox[List["(", RowBox[List[RowBox[List["Gamma", "[", "s", "]"]], " ", RowBox[List["Gamma", "[", RowBox[List[FractionBox["1", "2"], "+", "s"]], "]"]], " ", RowBox[List["Gamma", "[", RowBox[List[FractionBox["1", "2"], "-", "s", "-", "\[Nu]"]], "]"]], " ", SuperscriptBox[RowBox[List["(", FractionBox[RowBox[List["1", "-", "z"]], "2"], ")"]], RowBox[List["-", "s"]]]]], ")"]], " ", RowBox[List["Gamma", "[", RowBox[List[FractionBox["1", "2"], "-", "s", "+", "\[Nu]"]], "]"]]]], ",", RowBox[List["{", RowBox[List["s", ",", RowBox[List[FractionBox["1", "2"], "+", "j", "+", "\[Nu]"]]]], "}"]]]], "]"]]]]]], ")"]]]], RowBox[List[SqrtBox["2"], " ", SuperscriptBox["\[Pi]", RowBox[List["3", "/", "2"]]]]]]]], "/;", RowBox[List[RowBox[List["Abs", "[", RowBox[List["z", "-", "1"]], "]"]], ">", "2"]]]]]]]]
Date Added to functions.wolfram.com (modification date)
2007-05-02 | 3,830 | 10,436 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-09 | latest | en | 0.211971 |
https://zh-tw.coursera.org/learn/data-science-project/?authMode=signup | 1,582,593,297,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145989.45/warc/CC-MAIN-20200224224431-20200225014431-00463.warc.gz | 954,563,628 | 148,897 | 25,545 次近期查看
## 19%
### 您將學到的內容有
• Apply your exploratory data analysis skills
• Build an efficient and accurate prediction model
• Create a useful data product for the public
• Produce a presentation deck to showcase your findings
### 您將獲得的技能
Data ScienceMachine LearningR ProgrammingNatural Language Processing
## 19%
### Jeff Leek, PhD
Associate Professor, Biostatistics
Bloomberg School of Public Health
895,478 個學生
20 門課程
### Roger D. Peng, PhD
Associate Professor, Biostatistics
Bloomberg School of Public Health
886,577 個學生
22 門課程
### Brian Caffo, PhD
Professor, Biostatistics
Bloomberg School of Public Health
901,944 個學生
21 門課程
1
## Overview, Understanding the Problem, and Getting the Data
7 個視頻 （總計 20 分鐘）, 6 個閱讀材料, 1 個測驗
7 個視頻
Welcome from SwiftKey1分鐘
You Are a Data Scientist Now1分鐘
Introduction to Task 0: Understanding the Problem1分鐘
Introduction to Task 1: Getting and Cleaning the Data1分鐘
Regular Expressions: Part 1 (Optional)5分鐘
Regular Expressions: Part 2 (Optional)8分鐘
6 個閱讀材料
A Note of Explanation2分鐘
Project Overview10分鐘
Syllabus10分鐘
Task 0 - Understanding the problem10分鐘
Task 1 - Getting and cleaning the data10分鐘
1 個練習
Quiz 1: Getting Started12分鐘
2
## Exploratory Data Analysis and Modeling
2 個視頻 （總計 3 分鐘）, 2 個閱讀材料, 1 個測驗
2 個視頻
2 個閱讀材料
Task 2 - Exploratory Data Analysis10分鐘
3
## Prediction Model
1 個視頻 （總計 1 分鐘）, 1 個閱讀材料, 1 個測驗
1 個視頻
1 個閱讀材料
1 個練習
Quiz 2: Natural language processing I20分鐘
4
## Creative Exploration
1 個視頻 （總計 1 分鐘）, 1 個閱讀材料, 1 個測驗
1 個視頻
1 個閱讀材料
1 個練習
Quiz 3: Natural language processing II20分鐘
4.5
252 條評論
### 來自数据课程毕业项目的熱門評論
Capstone did provide a true test of Data Analytics skills. Its like a being left alone in a jungle to survive for a month. Either you succumb to nature or come out alive with a smile and confidence.
Wow i finally managed to finish the specialization!! definitely learned a lot and also found out difficulties in building predictors by trying to balancing speed, accuracy and memory constraints!!!
## 常見問題
• 注册以便获得证书后，您将有权访问所有视频、测验和编程作业（如果适用）。只有在您的班次开课之后，才可以提交和审阅同学互评作业。如果您选择在不购买的情况下浏览课程，可能无法访问某些作业。
• 您注册课程后，将有权访问专项课程中的所有课程，并且会在完成课程后获得证书。您的电子课程证书将添加到您的成就页中，您可以通过该页打印您的课程证书或将其添加到您的领英档案中。如果您只想阅读和查看课程内容，可以免费旁听课程。 | 811 | 2,211 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-10 | latest | en | 0.466732 |
https://betterlesson.com/lesson/resource/2766783/addressing-prior-knowledge-gaps-fraction-sense-wmv | 1,510,949,604,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934803906.12/warc/CC-MAIN-20171117185611-20171117205611-00133.warc.gz | 576,516,637 | 20,340 | ## Addressing prior knowledge gaps fraction sense.wmv - Section 2: Warm up
Addressing prior knowledge gaps fraction sense.wmv
# Is that a Coincidence?
Unit 8: Exploring Rational Numbers
Lesson 9 of 20
## Big Idea: Because of the commutative property 50% of 60 is the same as 60% of 50.
Print Lesson
Standards:
Subject(s):
Math, Number Sense and Operations, white boards, box diagram, student led inquiry, equivalent percent equations, rational numbers
35 minutes
### Erica Burnison
##### Similar Lessons
###### Ratios and Percents
6th Grade Math » Unit Rate Applications and Percents
Big Idea: Percent means ‘per 100.’ A percent can be represented using a ratio of ‘a’ to 100.
Favorites(55)
Resources(17)
New Haven, CT
Environment: Urban
###### Budgets
6th Grade Math » The College Project - Working with Decimals
Big Idea: What is a budget? How do we use it and why is it important? Students work on these questions in the context of the College Project.
Favorites(15)
Resources(33)
Somerville, MA
Environment: Urban
###### Converting Fractions and Decimals
6th Grade Math » Fractions, Decimals, and Percents
Big Idea: Students develop the relationship between fractions and decimals.
Favorites(13)
Resources(12)
Brooklyn, NY
Environment: Urban | 307 | 1,255 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | latest | en | 0.811421 |
http://hrtanswers.com/which-glide-reflection-describes-the-mapping-abc-def-solved/ | 1,680,028,854,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948868.90/warc/CC-MAIN-20230328170730-20230328200730-00109.warc.gz | 21,779,190 | 12,967 | # Which glide reflection describes the mapping abc def [Solved]
1. Replicate ABC in regards to the line AC after which translate 1 unit to the correct. 2. Translate ABC 1 unit to the correct after which replicate it in regards to the line AC. Step-by-step rationalization: Were provided that, ABC is reworked utilizing glide reflection to map onto DEF. Since, we all know, Glide Reflection is the transformation involving translation and reflection. So, we are able to see that, ABC could be mapped onto DEF by any of the next glide reflections: 1. Replicate ABC in regards to the line AC after which translate 1 unit to the correct. 2. Translate ABC 1 unit to the correct after which replicate it in regards to the line AC. Therefore, any of the 2 glide reflection will map ABC onto DEF.
1. Replicate ABC in regards to the line AC after which translate 1 unit to the correct. 2. Translate ABC 1 unit to the correct after which replicate it in regards to the line AC. Step-by-step rationalization: Were provided that, ABC is reworked utilizing glide reflection to map onto DEF. Since, we all know, Glide Reflection is the transformation involving translation and reflection. So, we are able to see that, ABC could be mapped onto DEF by any of the next glide reflections: 1. Replicate ABC in regards to the line AC after which translate 1 unit to the correct. 2. Translate ABC 1 unit to the correct after which replicate it in regards to the line AC. Therefore, any of the 2 glide reflection will map ABC onto DEF.
Get Answer for Mastering Physics Problem:What is the rms speed of a molecule in Marss atmosphere Assume R = 8.315 J/molK. [Solved]
1. Replicate ABC in regards to the line AC after which translate 1 unit to the correct. 2. Translate ABC 1 unit to the correct after which replicate it in regards to the line AC. Step-by-step rationalization: Were provided that, ABC is reworked utilizing glide reflection to map onto DEF. Since, we all know, Glide Reflection is the transformation involving translation and reflection. So, we are able to see that, ABC could be mapped onto DEF by any of the next glide reflections: 1. Replicate ABC in regards to the line AC after which translate 1 unit to the correct. 2. Translate ABC 1 unit to the correct after which replicate it in regards to the line AC. Therefore, any of the 2 glide reflection will map ABC onto DEF.
This post is last updated on hrtanswers.com at Date : 1st of September – 2022 | 561 | 2,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.900516 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.