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 |
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http://mathhelpforum.com/trigonometry/15701-very-simple-trig-problem-print.html | 1,508,349,670,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823016.53/warc/CC-MAIN-20171018161655-20171018181436-00055.warc.gz | 216,550,182 | 4,024 | # Very simple trig problem
• Jun 6th 2007, 05:39 PM
redeemys24
Very simple trig problem
Hey, I need some help with this VERY basic trig/geometry problem. I just had my first trig class yesterday and we got this problem that is fairly basic, yet I can't seem to be able to do it without more advanced trig. We haven't learned anything with sin, cosin, or tangent yet and although I've already learned that in another class we're supposed to be able to solve this in a very basic way. Here's the problem:
An astronomical object has an actual diameter of 128 km and an apparent size of 24'16" as viewed from earth. How far is this object from earth?
So, I converted 24'16" into DD format and got .40444444 (with the last four repeating). I assume that the diagram should look something like this:
math
How should I find the lengths of the sides? Am I just going about this completely wrong? I wish we were allowed to use more advanced techniques but there's got to be a way to do this without them, or he wouldn't have assigned it. Thanks in advance.
• Jun 6th 2007, 09:03 PM
Soroban
Hello, redeemys24!
Quote:
An astronomical object has an actual diameter of 128 km
and an apparent size of 24'16" as viewed from earth.
How far is this object from earth?
So, I converted $24'16''$ into DD format and got $0.40444...^o$
Bisect the vertex angle of that isosceles triangle.
You will have a right triangle.
One side runs from earth (E) to the center of the object (C).
. . Let $d = EC$
From the center directly to the right, CA is the radius of the object: 64 km.
. . And EA is the hypotenuse.
Let $\theta \:= \:\angle CEA \:=\:\frac{1}{2}(0.40444...^o) \:=\:0.20222...^o$
We have: . $\tan\theta \:=\:\frac{64}{d}\quad\Rightarrow\quad d \:=\:\frac{64}{\tan\theta}$
Hence: . $d \;=\;\frac{64}{\tan(0.20222...)} \:=\:18133.09449 \:=\:18133\text{ km}$
• Jun 6th 2007, 09:17 PM
earboth
Quote:
Originally Posted by redeemys24
...We haven't learned anything with sin, cosin, or tangent yet .... Here's the problem:
An astronomical object has an actual diameter of 128 km and an apparent size of 24'16" as viewed from earth. How far is this object from earth?
So, I converted 24'16" into DD format and got .40444444 (with the last four repeating). I assume that the diagram should look something like this:
math
...
Hello,
this is only a guess:
With very small angles the diameter of the object and the corresponding arc are nearly the same and the distance to the object and the length of the legs of the angle are nearly the same too.
Let a be the arc corresponding with the angle of 0°24'16'' = 91/225°. Then you get:
$\frac{a}{2 \pi} = \frac{\frac{91}{360}}{360}\ \Longrightarrow \ a \approx 0.007058887...$
Let d be the diameter of the object and R the length of one leg of the angle, then the arc a is calculated by:
$\frac{d}{R} = a\ \Longrightarrow \ R = \frac{d}{a}$
Plug in the values you know and you'll get $R \approx 18.100 \ km$
(Personal remark: If I haven't made some ridiculous mistake then this object flies between the surface of the earth and the cloud of geo-stationary satelites and it is remarkably big: It nearly can cover the sun completly (31') so there will happen some additional eclipses of the sun every day) | 900 | 3,250 | {"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": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2017-43 | longest | en | 0.920637 |
https://www.albert.io/ie/act-science/stopping-distance-and-sleep-deprivation | 1,485,300,652,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560285315.77/warc/CC-MAIN-20170116095125-00003-ip-10-171-10-70.ec2.internal.warc.gz | 883,435,220 | 22,473 | Free Version
Difficult
# Theoretical Impact of Velocity Changes on Stopping Distances
ACTSCI-KR22IS
Although motorcyclists have an inherent disadvantage when colliding with larger vehicles, their smaller mass also means that they have the ability accelerate and respond to potential collisions faster. The Kinetic Energy-Work Theorem describes some of the relationships between the mass of a moving object and the distance required for it to stop. In this theorem the stopping distance, $d$, which can be achieved with a uniform force, $F$, is determined by both the mass, $m$, and the original velocity, ${v}_{i}$, of the cycle according to $F d = \frac{1}{2} {m {v}_{i}}^{2}$.
A study was conducted to analyze driver awareness and the effects on stopping distance. In this study participants partook in two days of test sessions in which their stopping distances were recorded at various times throughout the course of the day. One of the two days was arranged to occur after a normal night of sleep and the other day was arranged to occur after a night of total sleep deprivation (TSD) as shown in Figure 1. The order of the two days of testing was randomly assigned and separated by a period of one week to allow for recovery from the night of TSD.
Figure 1
To examine the impact of ${v}_{i}$, the stopping distances which the motorcyclists were able to achieve throughout the course of a day were also examined for two initial traveling velocities. Comparisons can be made with riders who slept a normal night of sleep and those who underwent total sleep deprivation (TSD). Figure 2 shows the entire range of stopping distances for all participants as well as the average for each group.
Figure 2
Which of the following figures most accurately illustrates the theoretically determined impact of velocity on stopping distance?
A
B
C
D | 392 | 1,847 | {"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} | 2.875 | 3 | CC-MAIN-2017-04 | longest | en | 0.971817 |
https://www.dataperspective.info/2015/10/introduction-to-logistic-regression.html?showComment=1569306926680 | 1,670,055,962,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710926.23/warc/CC-MAIN-20221203075717-20221203105717-00324.warc.gz | 766,749,125 | 23,243 | # Introduction to Logistic Regression with R
In my previous blog I have explained about linear regression. In today’s post I will explain about logistic regression.
Consider a scenario where we need to predict a medical condition of a patient (HBP) ,HAVE HIGH BP or NO HIGH BP, based on some observed symptoms – Age, weight, Issmoking, Systolic value, Diastolic value, RACE, etc.. In this scenario we have to build a model which takes the above mentioned symptoms as input values and HBP as response variable. Note that the response variable (HBP) is a value among a fixed set of classes, HAVE HIGH BP or NO HIGH BP.
### Logistic regression – a classification problem, not a prediction problem:
In my previous blog I told that we use linear regression for scenarios which involves prediction. But there is a check; the regression analysis cannot be applied in scenarios where the response variable is not continuous. In our case the response variable is not a continuous variable but a value among a fixed set of classes. We call such scenarios as Classification problem rather than prediction problem. In such scenarios where the response variables are more of qualitative nature rather than continuous nature, we have to apply more suitable models namely logistic regression for classification.
Definition:
Assume we have a binary category output variable Y and a vector of p input variables X. Rather than modeling this response Y directly, logistic regression models the conditional probability, Pr(Y = 1|X = x) as a function of x, that Y belongs to a particular category.
Mathematically, logistic regression is expressed as:
Estimating Coefficients – Maximum Likelihood function:
The unknown parameters, β0/ β1, in the function are estimated by maximum likelihood method using available input training data. The Maximum likelihood function expresses the probability of the observed data as a function of the unknown parameters. The maximum likelihood estimators of these parameters are chosen to be those values that maximize this function. Thus, the estimators are those which agree most closely with the observed data.
For now we assume that solving the above equation can be used to estimate the unknown parameters.
In R, we glm() which takes training data as input and gives us the fitted model with estimated parameters as output, which we will see in the later section.
#### Making Predictions:
Once the coefficients have been estimated, it is a simple matter to compute the probability of response variable for any given input values by putting values of β0/ β1/X in the below equation.
Note: we have predict() in R which takes fitted model, input parameters as input values to predict the response variables.
#### Use case – Classify if a person may have HBP/not HBP :
Let us take a use case and implement logistic regression in R. Let us classify/predict if a person suffers with High Blood Pressure (HPB) given input predictors AGE, SEX, IsSmoking, Avg Systolic BP, Average diastolic BP, RACE, Body Weight, Height, etc. Access the data set from here. The dataset is NHANES III dataset. In order to implement logistic regression, we need to follow the below steps:
str(data)
data.frame': 15643 obs. of 10 variables:
\$ HSAGEIR : int 21 32 48 35 48 44 42 56 82 44 ...
\$ HSSEX : int 1 0 0 1 1 1 0 0 0 1 ...
\$ BMPWTLBS: num 180 136 150 204 155 ...
\$ BMPHTIN : num 70.4 63.9 61.8 69.8 66.2 70.2 62.6 67.6 59.5 71.1 ...
\$ TCP : int 268 160 236 225 260 187 216 156 179 162 ...
\$ HBP : int 0 0 0 0 0 0 0 0 1 0 ...
\$ RACE_2 : num 0 0 0 0 0 1 1 0 0 0 ...
\$ RACE_3 : num 0 0 0 0 0 0 0 0 0 0 ...
\$ SMOKE_2 : num 0 0 1 0 1 0 0 1 0 0 ...
\$ SMOKE_3 : num 0 0 0 0 0 1 1 0 0 1 ...
#removing SEQN,SDPPSU6,SDPSTRA6 from the dataset, from description of the datset,
#we concluded that it is not related to data.
data = data[,-c(1:3)]
#excluding HAR3 for now.
data = data[,-c(1,10)]
# removing entire row if there is any single missing column in the data.
#check for NA's or missing values in the data
#removing all the na rows from data
data = na.omit(data)
# from the description we consider HBP as response variable.
names(data)
'[1] "HSAGEIR" "HSSEX" "DMARACER" "BMPWTLBS" "BMPHTIN" "PEPMNK1R" "PEPMNK5R" "HAR1" "SMOKE" "TCP" "HBP" '
summary(data)
Data Transformations:
'Creating dummy variables/design variables for DMARACER/HSSEX create number of dummy variables based on the levels -1, i.e if if there 3 levels create 3-1 levels of dummy variables x= data
for RACE create 2 dummy variables DMRACE1, DMRACE2 where White(1) - (0,0) black(2) (1,0),other (3) - (1,1)'
for(level in unique(data\$DMARACER)){
data[paste("RACE", level, sep = "_")] <- ifelse(data\$DMARACER == level, 1, 0)
}
#removing race_1
data = data[,-12]
for(level in unique(data\$SMOKE)){
data[paste("SMOKE", level, sep = "_")] <- ifelse(data\$SMOKE == level, 1, 0)
}
#removing SMOKE_1
data = data[,-14]
table(data\$HSSEX,data\$HBP)
` 0 1`
` 0 6576 1702`
` 1 5817 1548`
Fem_HBP = (1702/6576+1702)) = 0.2056052
Mal_HBP = (1548/(1548+5817)) = 0.2101833
Modelling:
#remove HAR1,PEPMNK1R as SMOKE & HBP are created from these variables.
data = data[,-c(3,6,7,8,9,16)]
#create training & testing data
train_x = data
smp_size <- floor(0.75 * nrow(train_x))
train_ind <- sample(seq_len(nrow(train_x)), size = smp_size)
train <- train_x[train_ind, ]
test <- train_x[-train_ind, ]
glm2 = glm(train\$HBP~train\$HSAGEIR+train\$HSSEX+train\$BMPWTLBS+train\$BMPHTIN+train\$TCP+train\$RACE_2+train\$RACE_3,data=train,family=binomial)
The image shows us the summary of the model. In my next post let us evaluate the logistic regression model and let us consider few other models and choose better model among them.
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9. i believe you have a nice page here these days was my initial time coming here.. i just happened to discover it doing a google search. anyway, great post.. ill be bookmarking this page for certain. | 2,003 | 7,376 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.876312 |
http://metamath.tirix.org/mo2.html | 1,675,521,784,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500140.36/warc/CC-MAIN-20230204142302-20230204172302-00262.warc.gz | 26,878,184 | 8,456 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > mo2 Unicode version
Theorem mo2 2293
Description: Alternate definition of "at most one." (Contributed by NM, 8-Mar-1995.) Restrict dummy variable z. (Revised by Wolf Lammen, 28-May-2019.)
Hypothesis
Ref Expression
mo2.1
Assertion
Ref Expression
mo2
Distinct variable group: ,
Proof of Theorem mo2
Dummy variable is distinct from all other variables.
StepHypRef Expression
1 mo2v 2289 . 2
2 mo2.1 . . . . 5
3 nfv 1707 . . . . 5
42, 3nfim 1920 . . . 4
54nfal 1947 . . 3
6 nfv 1707 . . 3
7 equequ2 1799 . . . . 5
87imbi2d 316 . . . 4
98albidv 1713 . . 3
105, 6, 9cbvex 2022 . 2
111, 10bitri 249 1
Colors of variables: wff setvar class Syntax hints: ->wi 4 <->wb 184 A.wal 1393 E.wex 1612 F/wnf 1616 E*wmo 2283 This theorem is referenced by: mo3 2323 mo3OLD 2324 mo 2325 eu5OLD 2330 morimvOLD 2342 rmo2 3427 nmo 27384 bj-eu3f 34413 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1618 ax-4 1631 ax-5 1704 ax-6 1747 ax-7 1790 ax-10 1837 ax-11 1842 ax-12 1854 ax-13 1999 This theorem depends on definitions: df-bi 185 df-an 371 df-ex 1613 df-nf 1617 df-eu 2286 df-mo 2287
Copyright terms: Public domain W3C validator | 585 | 1,290 | {"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} | 3.03125 | 3 | CC-MAIN-2023-06 | latest | en | 0.147617 |
https://artofproblemsolving.com/wiki/index.php?title=1974_AHSME_Problems/Problem_20&oldid=56272 | 1,712,979,962,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816535.76/warc/CC-MAIN-20240413021024-20240413051024-00425.warc.gz | 95,503,132 | 11,793 | 1974 AHSME Problems/Problem 20
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Problem
Let
$T=\frac{1}{3-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{5}}+\frac{1}{\sqrt{5}-2}.$ (Error making remote request. Unexpected URL sent back)
Then
$\mathrm{(A)\ } T<1 \qquad \mathrm{(B) \ }T=1 \qquad \mathrm{(C) \ } 12 \qquad$
$\mathrm{(E) \ }T=\frac{1}{(3-\sqrt{8})(\sqrt{8}-\sqrt{7})(\sqrt{7}-\sqrt{6})(\sqrt{6}-\sqrt{5})(\sqrt{5}-2)}$
Solution
Let's try to rationalize $\frac{1}{\sqrt{n+1}-\sqrt{n}}$. Multiplying the numerator and denominator by $\sqrt{n+1}+\sqrt{n}$ gives us $\frac{1}{\sqrt{n+1}-\sqrt{n}}=\sqrt{n+1}+\sqrt{n}$.
Therefore,
$$T=(3+\sqrt{8})-(\sqrt{8}+\sqrt{7})+(\sqrt{7}+\sqrt{6})-(\sqrt{6}+\sqrt{5})+(\sqrt{5}+2)=5.$$
Hence the answer is $\boxed{\text{D}}$. | 351 | 854 | {"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": 7, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-18 | latest | en | 0.446351 |
https://www.eurocode.us/structural-design-eurocode-7/footings-subject-to-vertical-actions.html | 1,560,798,706,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998558.51/warc/CC-MAIN-20190617183209-20190617205209-00536.warc.gz | 743,343,062 | 6,204 | ## Footings subject to vertical actions
For a spread foundation subject to vertical actions, Eurocode 7 requires the design vertical action Vd acting on the foundation to be less than or equal to the design bearing resistance Rd of the ground beneath it:
Vd should include the self-weight of the foundation and any backfill on it. This equation is merely a re-statement of the inequality:
Ed < Rd discussed at length in Chapter 6. Rather than work in terms of forces, engineers more commonly consider pressures and stresses, so we will rewrite this equation as:
fwhich also appear in BS 8004.
qEd - qRd where qEd is the design bearing pressure on the ground (an action effect), and qRd is the corresponding design resistance.
Figure 10.4 shows a footing carrying characteristic vertical actions VGk (permanent) and VQk (variable) imposed on it by the super-structure. The characteristic self-weights of the footing and of the backfill upon it are both permanent actions (WGk). The following sub-sections explain how qEd and qRd are obtained from VGk, VQk, WGk, and ground properties.
Figure 10.4. Vertical actions on a spread foundation
10.4.1 Effects of actions
The characteristic bearing pressure qEk shown in Figure 10.4 is given by:
0 0 | 285 | 1,250 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2019-26 | latest | en | 0.929005 |
https://www.coursehero.com/file/6619113/CSE331-Lecture-20/ | 1,495,545,089,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607636.66/warc/CC-MAIN-20170523122457-20170523142457-00118.warc.gz | 844,189,922 | 23,269 | CSE331 Lecture 20
# CSE331 Lecture 20 - Lecture 20 CSE 331 Temp letter grades...
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Unformatted text preview: Lecture 20 CSE 331 Oct 17, 2011 Temp letter grades assigned See the blog post for more details A feedback… Time pressure on mid-term favors “quick thinkers” Tested different skills than those on HWs Mid-term has more weight than the homeworks Need another scribe for today Any volunteer? Algorithm for Interval Scheduling R: set of requests Set A to be the empty set While R is not empty Choose i in R with the earliest finish time Add i to A Remove all requests that conflict with i from R Return A*=A A* is optimal Run time analysis O(n log n) time sort intervals such that f(i) ≤ f(i+1) O(n) time build array s[1..n] s.t. s[i] = start time for i Set A to be the empty set While R is not empty Choose i in R with the earliest finish time Add i to A Remove all requests that conflict with i from R Return A*=A Do the removal on the fly Algorithm implementation Go through the intervals in order of their finish time 3 2 1 4 In general, if jth interval is the last one chosen Pick smallest i>j such that s[i] ≥ f(j) How can you tell Check if ts[i] <iff(1) in O(1) ime any of 2,3 or 4 conflict with 1? O(n log n) run time The final algo O(n log n) time sort intervals such that f(i) ≤ f(i+1) O(n) time build array s[1..n] s.t. s[i] = start time for i Add 1 to A and set f = f(1) For i = 2 .. n If s[i] ≥ f Add i to A Set f = f(i) Return A* = A Reading Assignment Sec 4.1of [KT] Questions? The “real” end of Semester blues There are deadlines and durations of tasks Write up a term paper Party! Exam study 331 HW Project Monday Tuesday Wednesday Thursday Friday The “real” end of Semester blues There are deadlines and durations of tasks Write up a term paper Exam study Party! 331 HW Project Monday Tuesday Wednesday Thursday Friday The algorithmic task YOU decide when to start each task Write up a term paper You have to do ALL the tasks Exam study Party! 331 HW Project Monday Tuesday Wednesday Thursday Friday Scheduling to minimize lateness All the tasks have to be scheduled GOAL: minimize maximum lateness Write up a term paper Exam study Party! 331 HW Project Monday Tuesday Wednesday Thursday Friday One possible schedule All the tasks have to be scheduled GOAL: minimize maximum lateness Lateness = 0 331 HW Monday Tuesday Lateness = 2 Party! Exam study Wednesday Thursday Write up a term pa Friday ...
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## This note was uploaded on 12/11/2011 for the course CSE 331 taught by Professor Rudra during the Fall '11 term at SUNY Buffalo.
Ask a homework question - tutors are online | 688 | 2,755 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2017-22 | longest | en | 0.841235 |
https://mathsciencesgto.cimat.mx/node/611 | 1,596,646,709,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735963.64/warc/CC-MAIN-20200805153603-20200805183603-00022.warc.gz | 417,956,728 | 6,936 | # Modeling with Linear Algebra
COURSE DESCRIPTION This course is about different aspects of linear algebra with emphasis on its applications. Linear algebra unifies the study of linear equations with the geometry of lines and planes. It has great theoretical significance in mathematics and rich applications. The course will start with real-life problems modeled with linear equations, which we will represent using matrices and vectors. Later on, we will deal with topics such as determinants, vector spaces, eigenvalues, and least-square problems. Although the course is mainly focused on modeling, the students will learn the formality and rigor that form a crucial part of algebra. Practical aspects of the course and applications will be emphasized throughout the course by the use of the computer language Python. Prerequisites The common requirements for the semester in Mathematical Tools for Modeling. Faculty Fall 2019 Manuel GONZÁLEZ VILLA
COURSE GOALS On completion of the course, students will be able to: understand concepts and techniques in linear algebra used for mathematical modeling; identify real-life problems that can be modeled with linear algebra; understand several applications and uses of linear algebra; employ mathematical formality and rigor.
COURSE CONTENT 1. Uses of linear equations and matrices (4 weeks) Review of linear equations and matrix algebra. Linear equations in economics, chemistry, and engineering. Adjacency matrix of a graph. Invertible matrices. Linear algebra and cryptography. Interpolating polynomials. Matrix factorizations. Determinant and its properties. Determinant and volume. 2. Vector Spaces and linear transformations (4 weeks) Vector spaces and subspaces. Spanning sets and linear independence. Basis and dimension of a vector space. Coordinate systems. Linear transformations. Geometry of linear transformations and computer graphics. Kernel and image of a linear transformation. The dimension theorem. Linear isomorphisms. The matrix of a linear transformation. Change of basis and similarity. 3. Inner product (3 weeks) Inner product and length. Orthogonality. Projections. Applications to machine learning. The Gram-Schmidt process. Least-squares solutions. 4. Eigenvalues and canonical forms (3 weeks) Review of complex numbers. Eigenvalues and eigenvectors of a matrix. The characteristic equation. Diagonalization of symmetric matrices and the spectral theorem. The Page Rank algorithm. Markov chains. Bibliography Kwak, Jin Ho; Hong, Sungpyo (2004). Linear Algebra, 2nd edition, Springer. Lay, David C.; Lay, Steven R.; McDonald, Judi J. (2016). Linear Algebra and Its Applications, 5th edition, Pearson. [Main textbook] Support Sessions 2 hours per week with a teaching assistant Grading Midterm exam (25%), final exam (35%), quizzes (15%), homework (10%), a simulation project to be submitted/presented by the end of the course (15%) | 590 | 2,918 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.887711 |
https://workforce.libretexts.org/Bookshelves/Electronics_Technology/Book%3A_Electric_Circuits_III_-_Semiconductors_(Kuphaldt)/08%3A_Operational_Amplifiers/8.08%3A_Averager_and_Summer_Circuits | 1,726,063,427,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651387.63/warc/CC-MAIN-20240911120037-20240911150037-00572.warc.gz | 593,517,120 | 30,734 | # 8.8: Averager and Summer Circuits
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If we take three equal resistors and connect one end of each to a common point, then apply three input voltages (one to each of the resistors’ free ends), the voltage seen at the common point will be the mathematical average of the three.
This circuit is really nothing more than a practical application of Millman’s Theorem:
This circuit is commonly known as a passive averager, because it generates an average voltage with non-amplifying components. Passive simply means that it is an unamplified circuit. The large equation to the right of the averager circuit comes from Millman’s Theorem, which describes the voltage produced by multiple voltage sources connected together through individual resistances. Since the three resistors in the averager circuit are equal to each other, we can simplify Millman’s formula by writing R1, R2, and R3 simply as R (one, equal resistance instead of three individual resistances):
If we take a passive averager and use it to connect three input voltages into an op-amp amplifier circuit with a gain of 3, we can turn this averaging function into an addition function. The result is called a noninverting summer circuit:
With a voltage divider composed of a 2 kΩ / 1 kΩ combination, the noninverting amplifier circuit will have a voltage gain of 3. By taking the voltage from the passive averager, which is the sum of V1, V2, and V3divided by 3, and multiplying that average by 3, we arrive at an output voltage equal to the sum of V1, V2, and V3:
Much the same can be done with an inverting op-amp amplifier, using a passive averager as part of the voltage divider feedback circuit. The result is called an inverting summer circuit:
Now, with the right-hand sides of the three averaging resistors connected to the virtual ground point of the op-amp’s inverting input, Millman’s Theorem no longer directly applies as it did before. The voltage at the virtual ground is now held at 0 volts by the op-amp’s negative feedback, whereas before it was free to float to the average value of V1, V2, and V3. However, with all resistor values equal to each other, the currents through each of the three resistors will be proportional to their respective input voltages. Since those three currents will add at the virtual ground node, the algebraic sum of those currents through the feedback resistor will produce a voltage at Vout equal to V1 + V2 + V3, except with reversed polarity. The reversal in polarity is what makes this circuit an inverting summer:
Summer (adder) circuits are quite useful in analog computer design, just as multiplier and divider circuits would be. Again, it is the extremely high differential gain of the op-amp which allows us to build these useful circuits with a bare minimum of components.
## Review
• A summer circuit is one that sums, or adds, multiple analog voltage signals together. There are two basic varieties of op-amp summer circuits: noninverting and inverting.
This page titled 8.8: Averager and Summer Circuits is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Tony R. Kuphaldt (All About Circuits) via source content that was edited to the style and standards of the LibreTexts platform. | 2,479 | 7,620 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-38 | latest | en | 0.19793 |
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# Day Problems 9/12/12 - PowerPoint PPT Presentation
Day Problems 9/12/12. Name the intersection of plane AEH and plane GHE. What plane contains points B, F, and C? What plane contains points E, F, and D?. 1.3 Measuring Segments 9/12/12. Postulate 1.5 – Ruler Postulate Every point on a line can be paired with a real number.
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## PowerPoint Slideshow about ' Day Problems 9/12/12' - lionel-glass
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• Name the intersection of plane AEH and plane GHE.
• What plane contains points B, F, and C?
• What plane contains points E, F, and D?
• Postulate 1.5 – Ruler Postulate
• Every point on a line can be paired with a real number.
• The real number that corresponds to a point is called the coordinate of the point.
• Allows you to measure lengths of segments and will allow you to find the distances between points.
• The distance between points A and B is the absolute value of the difference of their coordinates.
• What is ST?
• What is SV?
• What is UV?
• What is TV?
• ST = | -4 – 8 | = | -12| = 12
• SV = |-4 – 14 | = | -18| = 18
• UV = | 10 – 14| = | -4| = 4
• TV = |8 – 14| = | -6 | = 6
• If three points A, B, and C are collinear and B is between A and C, then AB + BC = AC.
• If EG = 59, what are EF and FG?
EF + FG = EG
(8x – 14) + (4x + 1) = 59
12x – 13 = 59
12x = 72
x = 6
EF = 8(6) – 14 = 34
FG = 4(6) + 1 = 25
• The midpoint of a segment is a point that divides the segment into two congruent segments.
• A point, line, ray, or other segment that intersects a segment at its midpoint is said to bisect the segment.
• The point, line, ray, or segment is called a segment bisector.
A
M
B
• Q is the midpoint of PR. What are PQ, QR, and PR?
• Find x.
6x – 7 = 5x + 1
x – 7 = 1
x = 8
Find PQ and QR.
PQ = 6x – 7
= 6 ( 8 ) – 7
= 41
Find PR.
PR = PQ + QR = 41 + 41 = 82.
• Classwork – Textbook p. 24 #7 – 21 ODD.
• Homework – Textbook p. 24 #8 – 22 EVEN. | 793 | 2,495 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2017-34 | longest | en | 0.894774 |
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# Uniform Particle Motion - PowerPoint PPT Presentation
Isfahan University of Technology Department of Mechanical Engineering. Uniform Particle Motion. Mohammad Reza Tavakoli. Outline. Newton’s Resistance Law Stokes’s Law Settling Velocity & Mechanical Mobility Slip Correction Factor Nonspherical Particles Aerodynamic Diameter
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## PowerPoint Slideshow about ' Uniform Particle Motion' - thane-nash
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### Uniform Particle Motion
Outline
• Newton’s Resistance Law
• Stokes’s Law
• Settling Velocity & Mechanical Mobility
• Slip Correction Factor
• Nonspherical Particles
• Aerodynamic Diameter
• Settling at High Reynolds Number
Newton’s Resistance Law
• Resistance is a result of the acceleration of the air that has to be pushed aside to allow the sphere to pass through.
• In 1 sec. a sphere of diameter d will push aside a volume of gas equal to the projected area of the sphere times its velocity V.
Newton’s Resistance Law
• Change of momentum= Force required to move the sphere through the gas
CD= Constant for Re>1000: CD=0.44
Newton’s Resistance Law
Stokes region: Re<1
Transition region: 1< Re< 1000 ( 3 < Re < 400 error < 2% )
( 400 < Re < 1000 error < 10%)
Newton region: Re>1000
Cd=0.44
Stokes’s Law
In general Navier-Stokes equation
Stokes assumptions:
Inertial forces are negligibly small compare to viscous forces
Fluid is incompressible
There are no wall or other particle nearby
Motion is constant
Particle is rigid sphere
No slip condition at particle’s surface
Net force= normal force + tangential force
Both of the forces acting in a direction opposite to particle motion
Stokes’s Law
Normal component:
Tangential component:
Total resisting force on a spherical particle due to its velocity V relative to the fluid:
(Re<1 & Err<10%)
See Appendix
Newton’s Resistance Law & Stokes’s Law
Compare drag forces:
Stokes’s law contains viscosity but NOT inertia factors
like rho
Newton’s law contain rho but NOT viscosity.
Newton’s Resistance Law & Stokes’s Law
Stokes’s law :
Newton’s law :
Flow in tubes: No normal force cd=16/Re
Validation of Stokes assumptions:
• Fluid is incompressible
• There are no wall or other particle nearby
• Particle is rigid sphere ( 0.7% error for water drops)
• No slip condition at particle’s surface (Slip Correction Factor)
• Nonspherical particles (shape factor)
Slip Correction Factor
The No Slip Condition is not valid for small particle whose size approaches the mean free path of the gas.
1910- Cunningham Correction factor (Cc):
Cc >1 so, reduces the Stokes drag force by:
For Particle of 0.1 micron.
Slip Correction Factor
For particle to below 0.01 micron: (2.1% error for all particle sizes)
Terminal Velocity:
Slip Correction Factor
Slip Correction factor increases when the particle size decreases.
Slip Correction Factor
Slip Correction factor increases when pressure decreases because the mean free path increases.
Pd: multiplying particle diameter by the pressure in atmospheres gives diameter of the particle that has the same slip correction factor at 1 atm pressure.
Look at A12: compare particle of 1 micron at 2 atm pressure vs. particle at 2 micron and 1 atm pressure.
Slip Correction Factor
Slip Correction factor increases when pressure decreases because the
mean free path increases.
Nonspherical Particle (Dynamic shape factor)
actual resistance force of nonspherical particle
Dynamic shape factor = ----------------------------------------------------------------------------------
resistance force of the sphere with same volume and velocity
de= equivalent volume diameter (diameter of the sphere having the same volume as the irregular particle)
Nonspherical Particle (Dynamic shape factor)
The Dynamic shape factor >1:
nonspherical particle settle more slowly than their equivalent volume spheres
Aerodynamic Diameter
• Stokes diameter (ds): diameter of the sphere that has the same density and settling velocity as the particle
• Aerodynamic diameter (da): diameter of the unit density sphere that has the same settling velocity as the particle
Aerodynamic Diameter
• If a particle has an aerodynamic diameter 1 micron it behaves in an aerodynamics sense like a 1 micron water droplet (density=1 g/cm^3) regardless of its shape, density or physical size
• Rho_b: Bulk material
• Rho_p: stokes particle
• Rho_b==Rho_p
Settling at high Reynolds Number
- Table: 3.4 (d is known and V is unknown)
-(1945) Davies (up to Re=4):
Settling at high Reynolds Number
• If settling velocity is known(V is known and d is unknown) (Table 3.5)
• Empirical Equation: | 1,202 | 5,280 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2017-43 | latest | en | 0.796732 |
https://communities.sas.com/t5/SAS-Statistical-Procedures/How-do-I-adjust-for-clustered-data-in-logistic-regression/td-p/359048?nobounce | 1,516,217,980,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886964.22/warc/CC-MAIN-20180117193009-20180117213009-00151.warc.gz | 670,768,154 | 27,779 | New Contributor
Posts: 3
# How do I adjust for clustered data in logistic regression?
I am using proc logistic to investigate the association between the variables laek and pv (indexar, alder, arv, and koen are confounders). The model looks something like this:
proc logistic data=dataset;
class indexar (ref='2010') koen (ref='K') / param = ref ;
model laek(EVENT='1') = pv indexar alder arv koen;
run;
The individual level dataset that I am using is obtained from patients who are grouped within different clinics. I believe that the clinic effect is a mediator rather than a confunder. How can I adjust for clustering? Should I use a "sandwich" estimator and in that case, how do I do that in SAS?
Super User
Posts: 10,200
## Re: How do I adjust for clustered data in logistic regression?
Try Generalize Linear Mixed Model.
Check PROC GLIMMIX .
Frequent Contributor
Posts: 98
## Re: How do I adjust for clustered data in logistic regression?
Depending on the exact type of inference you are interested in, you can account for such clustering in a number of ways. The two simplest ways are probably in GENMOD or GLIMMIX (though, depending on the details of the analysis you can also use PROC SURVEYLOGISTIC or even PROC PHREG, or reparameterize your data to use a conditional maximum likelihood approach in PROC LOGISTIC. But these are less intuitive and more complicated approaches, I think).
For example, you can fit a generalized estimating equation (GEE):
PROC GENMOD data=dataset;
class indexar (ref='2010') koen (ref='K') / param = ref ;
model laek(EVENT='1') = pv indexar alder arv koen;
repeated subject=patient_id/ type=un;
run;
You can fit just fit a normal logistic model but with the empirical ("sandwich") variance estimators:
PROC GLIMMIX data=dataset empirical;
class indexvar koen;
model laek(event='1') = pv indexar alder arv koen;
run;
You can fit a mixed effects model:
PROC GLIMMIX data=dataset;
class indexvar koen;
model laek(event='1') = pv indexar alder arv koen;
random intercept / subject=patient_id;
run;
For both the RANDOM statement in GLIMMIX and the REPEATED statement in GENMOD, you can also specify a hierarchical (nested) structure (e.g. "random intercept / subject=patient_id(clinic_id)" for patients nested within clinics), though I would consult this note (http://support.sas.com/kb/24/200.html) and/or this document (https://support.sas.com/resources/papers/proceedings14/SAS026-2014.pdf) to make sure you understand exactly what it is you are specifying and how to interpret it.
New Contributor
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https://justaaa.com/accounting/558465-sales-158000-710000-46000-265000-sales-discounts | 1,726,470,211,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651676.3/warc/CC-MAIN-20240916044225-20240916074225-00708.warc.gz | 298,018,848 | 10,025 | Question
# Sales \$ 158,000 \$ 710,000 \$ 46,000 \$ 265,000 Sales discounts 5,000 16,000 600 4,200 Sales...
Sales \$ 158,000 \$ 710,000 \$ 46,000 \$ 265,000 Sales discounts 5,000 16,000 600 4,200 Sales returns and allowances 16,000 6,000 4,700 800 Cost of goods sold 92,064 442,384 31,339 131,040
Compute net sales, gross profit, and the gross margin ratio for each of the four separate companies. (Round your gross margin ratio to 1 decimal place; i.e.; 0.2367 should be entered as 23.7%.)
Required:
Computation of net sales:
1 2 3 4 Sales \$158,000 \$710,000 \$46,000 \$265,000 (-) Sales discount (\$5,000) (\$16,000) (\$600) (\$4,200) (-) Sales return and allowance (\$16,000) (\$6,000) (\$4,700) (\$800) Net sales \$137,000 \$688,000 \$40,700 \$260,000
Computation of gross profit:
1 2 3 4 Net sales \$137,000 \$688,000 \$40,700 \$260,000 (-) Cost of goods sold (\$92,064) (\$442,384) (\$31,339) (\$131,040) Gross profit \$44,936 \$245,616 \$9,361 \$128,960
Computation of gross margin ratio:
1 2 3 4 Gross profit \$44,936 \$245,616 \$9,361 \$128,960 (/) Net sales \$137,000 \$688,000 \$40,700 \$260,000 Gross margin ratio 32.8% 35.7% 23% 49.6%
#### Earn Coins
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\$30.00
## Description
5/5 - (1 vote)
Part I. The attached data set was reported by an article in Technometrics on the selling
price, y, and the annual taxes, x (local, school, county) for 24 houses. By using R (or any
appropriate software you prefer), answer questions 1–5 and submit the relevant outputs.
1. Construct and submit a scatter plot of y versus x. Does a simple linear regression
model seem appropriate here?
2. Fit the simple linear regression model using the method of least squares, i.e., find
the least squares line, ˆy = βˆ
0 + βˆ
1x by using the software. Submit your solution
(output).
3. In plain English, interpret the meaning of the slope parameter β1.
4. In plain English, interpret the meaning of the intercept β0. Does it have a practical
meaning here?
5. Report the value of s; and then calculate s
2 and SSE.
Part II. Suppose that you obtained the following summary quantities to estimate the
parameters in a regression study. Assume that x and y are related according to the
simple linear regression model ˆy = βˆ
0 + βˆ
1x.
n = 14,
Xn
i=1
yi = 572,
Xn
i=1
y
2
i = 23530,
Xn
i=1
xi = 43,
Xn
i=1
x
2
i = 157.42, and Xn
i=1
xiyi = 1697.80. | 358 | 1,202 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-38 | latest | en | 0.832436 |
https://ecce216.com/mixed/how-much-was-100-worth-in-1897/ | 1,720,785,947,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514387.30/warc/CC-MAIN-20240712094214-20240712124214-00531.warc.gz | 141,232,650 | 11,458 | # How much was \$100 worth in 1897?
## How much was \$100 worth in 1897?
\$100 in 1897 is equivalent in purchasing power to about \$3,483.24 today, an increase of \$3,383.24 over 125 years. The dollar had an average inflation rate of 2.88% per year between 1897 and today, producing a cumulative price increase of 3,383.24%.
## How much was \$10000 1897?
When \$10,000 is equivalent to \$321,204.40 over time, that means that the “real value” of a single U.S. dollar decreases over time….Buying power of \$10,000 in 1890.
Year Dollar Value Inflation Rate
1895 \$9,230.77 -2.33%
1896 \$9,230.77 0.00%
1897 \$9,120.88 -1.19%
1898 \$9,120.88 0.00%
How much is \$1 in the 1890s?
Value of \$1 from 1890 to 2022 \$1 in 1890 is equivalent in purchasing power to about \$32.56 today, an increase of \$31.56 over 132 years. The dollar had an average inflation rate of 2.67% per year between 1890 and today, producing a cumulative price increase of 3,156.16%.
What would a dollar in 1896 be worth today?
Value of \$1 from 1896 to 2022 \$1 in 1896 is equivalent in purchasing power to about \$34.42 today, an increase of \$33.42 over 126 years. The dollar had an average inflation rate of 2.85% per year between 1896 and today, producing a cumulative price increase of 3,341.77%.
### How much was 100k worth in 1900?
Value of \$100,000 from 1900 to 2022
Cumulative price change 3,379.71%
Average inflation rate 2.95%
Converted amount (\$100,000 base) \$3,479,714.29
Price difference (\$100,000 base) \$3,379,714.29
CPI in 1900 8.400
### How much was 50 cents 1890?
\$0.50 in 1890 is equivalent in purchasing power to about \$16.28 today, an increase of \$15.78 over 132 years. The dollar had an average inflation rate of 2.67% per year between 1890 and today, producing a cumulative price increase of 3,156.16%.
How much was 50 cents in the 70s?
\$0.50 in 1970 is equivalent in purchasing power to about \$3.82 today, an increase of \$3.32 over 52 years.
How much was 1000 in the 1800s?
\$1,000 in 1800 is equivalent in purchasing power to about \$23,198.10 today, an increase of \$22,198.10 over 222 years. The dollar had an average inflation rate of 1.43% per year between 1800 and today, producing a cumulative price increase of 2,219.81%.
#### How much was \$200 a day in 1966?
\$200 in 1966 is equivalent in purchasing power to about \$1,804.30 today, an increase of \$1,604.30 over 56 years. The dollar had an average inflation rate of 4.01% per year between 1966 and today, producing a cumulative price increase of 802.15%.
#### How much did a house cost in 1800?
What was the average pay in the 1800’s? What was the average wage in 1840? What was the average salary in 1790? How much did paper cost in the 1800s?…What was the average pay in the 1800’s?
Median home price March May
2019 \$259 400 \$278 200
How much was 50 cents 1850?
The inflation rate in 1850 was 1.30%. The current inflation rate compared to last year is now 9.06%….Value of \$50 from 1850 to 2022.
Cumulative price change 3,698.86%
Converted amount (\$50 base) \$1,899.43
Price difference (\$50 base) \$1,849.43
CPI in 1850 7.800
How much was \$200000 in the Gilded Age?
Value of \$200,000 from 1890 to 2022 \$200,000 in 1890 is equivalent in purchasing power to about \$6,424,087.91 today, an increase of \$6,224,087.91 over 132 years.
## How much was \$1000 dollars worth in 1890?
\$1,000 in 1890 is equivalent in purchasing power to about \$31,770.22 today, an increase of \$30,770.22 over 132 years. The dollar had an average inflation rate of 2.65% per year between 1890 and today, producing a cumulative price increase of 3,077.02%.
Posted in Mixed | 1,121 | 3,652 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-30 | latest | en | 0.949527 |
http://www.weknowtheanswer.com/q/is-2-miles-larger-than-3333-yards | 1,495,990,955,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463610342.84/warc/CC-MAIN-20170528162023-20170528182023-00195.warc.gz | 883,339,876 | 9,042 | # Is 2 miles larger than 3333 yards?
### Answer this question
• Is 2 miles larger than 3333 yards?
Anonymous30003 | 07/03 2017 06:06
Hddhdhdjfvjhjk
Positive: 50 %
... the circle’s diameter will be 27.3% larger than the ... Ra yards. 43200000/2 ... being 864575.9 miles or 1,521,653,584 British yards, ...
Positive: 50 %
How Many Feet Are There in a Mile? ... Miles to Yards: Miles: ... or about 3.2 mm longer than the International Mile. ...
Positive: 45 %
from a smaller unit to a larger unit 6 feet = yards ... 7. 64 in. 5 ft 8. 2 mi 3,333 yd 9. ... right arm in yards? 12. Which measure is less than 435 inches? | 205 | 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} | 2.796875 | 3 | CC-MAIN-2017-22 | latest | en | 0.868288 |
https://www.accountingformanagement.org/straight-line-method-of-depreciation/ | 1,726,814,177,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652138.47/warc/CC-MAIN-20240920054402-20240920084402-00173.warc.gz | 589,774,318 | 11,384 | # Straight-line method of depreciation
The straight-line method of depreciation attempts to allocate equal portion of depreciable cost to each period of the asset’s useful life. This method assumes that the depreciation is a function of the passage of time rather than the actual productive use of the asset. Under straight line method, the depreciation expense for a period is calculated by dividing the depreciable cost of the asset by the years of its useful. Depreciable cost is arrived at by deducting salvage or residual value from the original cost of the asset. For example, if an asset is purchased for \$10,000 and its salvage value is expected to be \$800 at the end of its useful life, the depreciable cost would be computed as follows:
Depreciable cost = \$10,000 – \$800
= \$9,200
Other names used for straight-line method are original cost method or fixed installment method of depreciation.
## Formula:
The following formula is used to calculate depreciation under straight line method:
## Example:
The Eastern company provides the following information regarding one of its fixed assets that has been purchased on January 1, 2022:
• Cost of the asset: \$35,000
• Salvage value: \$3,000
• Useful life: 10 years
Required: Calculate annual depreciation expense of this asset using straight line method.
### Solution:
= (\$35,000 – \$3,000)/10 years
= \$3,200
The Eastern Company will allocate a depreciation of \$3,200 to all the years of the useful life of the fixed asset.
## Straight-line depreciation rate
Alternatively, we can calculate a depreciation rate by dividing 1 by the years of useful life of the asset. This is known as straight line depreciation rate that can be applied to the total depreciable cost to calculate the depreciation expense for the period. In our example, the useful life of the asset is 10 years and the depreciation rate is 10%, as computed below:
1/10 = 0.1 or 10%
It means the depreciation expense to be recognized in each period of the useful life of the asset is 10% of \$32,000 (depreciable cost), as computed below:
(\$35,000 – \$3,000) × 0.1 = \$3,200
The straight-line method of depreciation is popular among companies world wide because it is more conceptual and simple to employ.
## Limitations:
The major limitations/disadvantages of straight-line method of depreciation are given below:
1. It assumes that the useful economic life of the asset is same each year that may not always be true.
2. It assumes that the annual repair and maintenance expenses are the same during the useful life of the asset. In practice, an asset may require different repair and maintenance expenses in different years.
3. It creates a distortion in the rate of return analysis.
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One Comment on Straight-line method of depreciation
1. kumarperumal
nice | 630 | 2,847 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-38 | latest | en | 0.878552 |
http://math.stackexchange.com/questions/91875/computing-the-union-closure-of-a-finite-discrete-set | 1,467,184,879,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783397636.15/warc/CC-MAIN-20160624154957-00168-ip-10-164-35-72.ec2.internal.warc.gz | 199,223,228 | 16,763 | # Computing the union-closure of a finite discrete set
You must excuse my notation if it is not stndard, I have my background in Computer Science and not in Math. Hope this topic still is in the right forum.
Let $U = {1,2,...,n}$ be our finite universe. Let $\mathcal{F}$ be a family of subsets of $U$.
The union-closure of $\mathcal{F}$ is the unique minimal family $\mathcal{C}_{\mathcal{F}}$ such that:
1. Every element in $\mathcal{F}$ is in $\mathcal{C}$.
2. For every pair of elements in $\mathcal{C}$ their union is also in $\mathcal{C}$.
I have a couple of questions regarding the union-closure.
1. How fast can we compute the union-closure?
It can be done in $O(|\mathcal{C}| \cdot \log{|\mathcal{C}|} \cdot n^2)$.
2. Can we bound $\log_2{|\mathcal{C}|}/n$ as $n$ goes to infinite?
The set system can be modelled as a bipartite graph (sets on one side and U on the other edges)
Then we can make a lower and upper bound based on the Maximum Matching and Maximum Induced Matching (equivalent to VC-dimension).
Does anyone know of work related to this?
-
Even if it has sets, it's not really set theory. I removed the tag. – Asaf Karagila Dec 15 '11 at 23:33
Without conditions on $\mathcal{F}$, $\log_2|\mathcal{C}|/n$ can range from $0$ if, say, $\mathcal{F}=\{U\}$, to $1-\epsilon$, if $\mathcal{F}=\{\{k\}:k\in U\}$, where $\epsilon\to 0$ as $n\to\infty$. – Brian M. Scott Dec 15 '11 at 23:47
Interesting question: you might consider asking it (if you don't get answers here) on cstheory.stackexchange.com – Suresh Venkat Dec 21 '11 at 0:31 | 487 | 1,558 | {"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} | 3.265625 | 3 | CC-MAIN-2016-26 | longest | en | 0.824025 |
https://byjus.com/question-answer/quadrilateral-efgh-is-a-rectangle-in-which-j-is-the-point-of-intersection-of-the-1/ | 1,642,735,211,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302715.38/warc/CC-MAIN-20220121010736-20220121040736-00358.warc.gz | 219,153,294 | 20,931 | Question
# Quadrilateral EFGH is a rectangle in which J is the point of intersection of the diagonals. If JF = 8x+4 and EG = 24x–8, find the value of x.
A
1.5
B
2
C
3
D
4
Solution
## The correct option is C 2 Given, JF =8x+4 and EG =24x−8 Quadrilateral is a rectangle. So, ∠EJF=90∘, EJ=JF and EJ=JG=24x−82 ⇒24x−82=8x+4 ⇒24x–8=2×(8x+4) ⇒24x–8=16x+8 ⇒24x−16x=8+8 ⇒8x=16 ⇒x=2 Mathematics
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View More | 198 | 451 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2022-05 | latest | en | 0.582676 |
https://mathathome.mathlearningcenter.org/kindergarten | 1,591,497,771,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348523476.97/warc/CC-MAIN-20200607013327-20200607043327-00232.warc.gz | 436,705,889 | 17,844 | # Kindergarten Math at Home
Welcome! We’ll update this page every weekday with a new math activity. Many of the activities will be fun and worthwhile for students of all ages so that you can do them as a family. The following day, we’ll share examples of how other people did the activity so you can check your work and stretch your thinking.
The resources on this page focus on the major work for kindergarten:
• counting to 100 by 1 and 10, and reading and writing numbers to 20
• solving problems that involve adding and subtracting within 10
• understanding that teen numbers always have a group of 10 and some more ones
We’ll do a lot of work with shapes and patterns, too!
## Math & Game Tools
DIY spinners, origami dice, and more!
# Activity of the Day
Check the archive to see past activities with sample solutions.
## Math in Our World: How Many Marbles?
Hayden and Charlotte love to build tracks for their marbles. We wondered how many marbles we had in our home, so we decided to count all the marbles we could find.
1. What do you notice? What do you wonder?
1. Show or tell what is happening using pictures, numbers, or words.
### How could you get started?
1. Draw a number line starting at 64 and count on.
2. Draw a ten-frame to organize and count the number of marbles outside the jar.
### Challenge
1. Show your work to someone else, and explain your thinking. Did they understand?
### Follow-Up
We have a lot of marbles at our house so we made a counting collection. Can you make your own counting collection? You could use rocks, toys, or even food!
## Activities of the Day
Want more Activities of the Day? Check out the archive for past activities, including sample responses.
## Practice Pages
Collections of practice pages organized by math skill. In English and Spanish with answer keys.
## Online Games
Digital math games to play independently or together online.
## Family Games
Math games for families to play together at home with materials and tools easily made, downloaded, or found online.
## Behind the Scenes
A special message to you! | 461 | 2,099 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-24 | latest | en | 0.937412 |
http://math.stackexchange.com/questions/366355/linearly-independent-functions-not-solutions-of-ode | 1,449,027,394,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398538144.64/warc/CC-MAIN-20151124205538-00202-ip-10-71-132-137.ec2.internal.warc.gz | 143,847,538 | 19,495 | # Linearly independent functions not solutions of ODE
If I have a set of $N$ linearly independent functions $f_1,\dots,f_N$, that may NOT be the solutions of a differential equation, and I impose initial conditions $f(0)=K_0,\dots,D^{N-1}f(0)=K_{N-1}$, is it true that we will always find coefficients $a_1,\dots,a_N$ such that the function $$f(x)=a_1f_1(x)+\dots +a_Nf_N(x)$$ satisfies the initial conditions imposed? Or is only true if the linearly independent functions are solutions of a differential equation?
(Of course assuming this functions are differentiable up to order $N$)
My problem raises from thinking in $$\{sin(x),cos(x)\}$$ In this case is true the statement becasue both functions are never zero for the same value $x$. On the other side $$\{x,x^2\}$$Are also linearly independent but at $x=0$ both functions are zero so the statement is false. I don't understand why the fact that they the first set is an independent set of solutions and the second not, makes a difference.
-
Presumably you meant to impose general initial conditions for $f(0),\ldots,f^{(N)}(0)$ since $a_i=0$ is always a valid solution otherwise. – Sharkos Apr 19 '13 at 12:02
Also, I assume you only meant to impose $N$ initial conditions, in which case you should stop at $f^{(N-1)}(0)$. – Sharkos Apr 19 '13 at 12:05
Sorry, Already fixed. – Ambesh Apr 19 '13 at 12:16
Linear independence as functions doesn't tell you much about pointwise behaviour in general.
The set of functions $f_i(x)$ in this case are fairly irrelevant - all that matters is the set $f_i^{(k)}(0)$ which forms an $N\times N$ matrix $A_i\,^k$ if we assume we impose initial conditions $f(0)=b_0,\ldots,f^{(N-1)}(0)=b_{N-1}$. (Sorry about the weird mixture of 0-based and 1-based indices.)
Then the question is simply this:
Does there exist a vector $\mathbf a$ such that $A \mathbf a = \mathbf b$?
or in general,
Is $A$ invertible?
In general, regardless of whether the functions are linearly independent as functions, the rows of this matrix may or may not be independent. (For instance, the functions can all be piecewise defined as identically $0$ for $x\in[0,\frac{1}{2}]$ and then become linearly independent just by having different forms in $[1,2]$, say.)
The reason why a basis of solutions to a differential equations are special is because - provided they are nice and nonsingular $N$th order equations, they give a prescription for evolving any initial condition $\mathbf b$ forwards in $x$. Therefore, you can choose the rows to be independent just by choosing each $f_i$ to have initial conditions such that e.g. $A$ is the identity matrix! ($f_i^{(k)}(0) = \delta_k^i$.) This is just what we mean by choosing a basis of equations.
Note that $x,x^2$ are a basis of solutions for a differential equation of the form $$x^2f''+bxf'+cf$$ but that this is a singular equation, since you have to divide through by $x^2$.
-
+1 nice explanation – Babak S. Apr 19 '13 at 12:16
@Sharkos I don't see why is the case of differential equations is not possible $f_i^{(i)}=0$ ? Thank you – Ambesh Apr 19 '13 at 12:20
If I understand you right, then you're thinking about this the wrong way: when you take a basis of solutions to a differential equation, it must be the case that from whatever point you choose ($x_0$) it is possible to construct any possible initial condition at that point. This is because a basis by definition spans the set of all possible solutions, and the DE has a solution with any initial conditions you set. On the other hand, you can choose a set of solutions stupidly, like $\sin x,2\sin x$. These don't have the property you want! – Sharkos Apr 19 '13 at 12:25
@Sharkos Yes I understand. My problem is that I have a independent set of solutions of a fractional differential equation. I can't use the theorem that grants me that there is a solution to proof that the do satisfy the initial conditions. So I have N solutions, they are independent...and I'm stuck there. – Ambesh Apr 19 '13 at 12:36
Well, I've never played with fractional differential equations, but if they're at least as singular as $x^n y^{(n)}$-type equations, you're likely out of luck, since these produce $x,x^2,\ldots,x^k, x\ln x$ type solutions. – Sharkos Apr 19 '13 at 12:39
Let $f_1,\ldots,f_n$ be linearly independent functions and define $$D\colon f_i\mapsto \left(f_i(0),f^\prime_i(0),\ldots,f^{(n-1)}_i(0)\right).$$ For solving your initial value problem you need $Df_1,\ldots,Df_n$ to span $\mathbb{R}^n$. This is true in the case of differential equations, but not for general systems of linearly independent functions.
Since $D\colon\mathbb{R}^{[0,1]}\to\mathbb{R}^n$ its actually fairly easy to find independent functions in $\mathbb{R}^{[0,1]}$ such that their projection is no longer independent like for instance your example $x,x^2$. You just need some $g\in\mathrm{span}(f_1,\ldots,f_n)$ such that $Dg=0$, which can be done even for $g\neq 0$.
- | 1,386 | 4,943 | {"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} | 3.625 | 4 | CC-MAIN-2015-48 | longest | en | 0.89718 |
https://spiral.ac/sharing/sgjv5ps/rnt22-principal-ideal-domains | 1,618,694,260,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038464045.54/warc/CC-MAIN-20210417192821-20210417222821-00377.warc.gz | 622,636,521 | 9,737 | rnt22-principal-ideal-domains
# Interactive video lesson plan for: RNT2.2. Principal Ideal Domains
#### Activity overview:
Ring Theory: We define PIDs and UFDs and describe their relationship. Prime and irreducible elements are defined, and conditions for implication are given.
(second version: corrections to definition of prime and irreducible; comment should be 'R UFD implies R[x] UFD, improved proof that, in UFD, irreducible implies prime).
Tagged under: mathematics,ring,theory,abstract,algebra,Ring (mathematics),integral,domain,principal,ideal,unique,factorization,prime,irreducible,unit,Gaussian,integer,polynomial,field,degree,PID,UFD
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Absolutely amazing collaboration from year 10 today. 100% engagement and constant smiles from all #lovetsla #spiral
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Students show better Interpersonal Writing skills than Speaking via @SpiralEducation Great #data #langchat folks! | 480 | 2,193 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | latest | en | 0.833689 |
https://www.rjwala.com/2023/05/a-spinning-cricket-ball-in-air-does-not.html | 1,686,446,873,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646652.16/warc/CC-MAIN-20230610233020-20230611023020-00238.warc.gz | 1,083,078,947 | 68,619 | # A spinning cricket ball in air does not follow a parabolic trajectory?
Question: A spinning cricket ball in air does not follow a parabolic trajectory?
A spinning cricket ball in air does not follow a purely parabolic trajectory due to the effect of the Magnus force.
When a cricket ball is released with backspin or topspin, it experiences a force perpendicular to its velocity, known as the Magnus force. This force is a result of the difference in air pressure on the top and bottom of the spinning ball.
If a cricket ball is released with backspin, the air pressure on the top of the ball is lower than the air pressure on the bottom, resulting in an upward force on the ball. This Magnus force causes the ball to curve downwards less steeply than it would if it were not spinning. Similarly, if the ball is released with topspin, the air pressure on the top of the ball is higher than the air pressure on the bottom, resulting in a downward force on the ball. This Magnus force causes the ball to curve downwards more steeply than it would if it were not spinning.
The combination of the Magnus force and the force due to gravity results in the cricket ball following a curved trajectory instead of a purely parabolic one. The amount of curvature depends on the spin rate, speed of the ball, and the air density, among other factors.
Therefore, a spinning cricket ball in air does not follow a parabolic trajectory but rather a curved path due to the effect of the Magnus force.
Rjwala is an educational platform, in which you get many information related to homework and studies. In this we also provide trending questions which come out of recent recent exams. | 341 | 1,676 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2023-23 | latest | en | 0.940744 |
http://lists.openscad.org/pipermail/discuss_lists.openscad.org/2019-November/028992.html | 1,586,270,309,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371799447.70/warc/CC-MAIN-20200407121105-20200407151605-00457.warc.gz | 114,624,653 | 2,587 | Alex Gibson alex at alexgibson.net
Fri Nov 22 09:02:23 EST 2019
```Example:
mount_width = 15;
difference()
{
union()
{
//plate
translate([0,0,-5])
cube([200,100,10],center=true);
//mount
translate([0,0,25])
}
hull()
{
rotate([90,0,0])
rotate([90,0,0])
}
hull()
{
rotate([90,0,0])
rotate([90,0,0])
}
}
Alex Gibson
edumaker limited
. Project management
. Operations & Process improvement
. 3D Printing
-----Original Message-----
From: Discuss [mailto:discuss-bounces at lists.openscad.org] On Behalf Of Alex
Gibson
Sent: 22 November 2019 13:44
I think you mean that you want to re-create a part like this one, and you
want to be able to do the curved 'fillet' between the main plate and the
raised mount.
The way you can do this is to make the plate and the mount so that they
overlap, and put both inside a union() {.....}
Make the mount wider than you want it to be, by 2x the radius of the fillet
you want.
Translate the parts so that the TOP surface of the plate on Z=0.
Then, you want to take a cylinder with the radius of the fillet you want,
make it easily long enough to be wider than the whole part. Rotate it over
so that it is in line with the fillet you want to make.
Translate it up to Z=fillet radius, so it is lying on the surface of the
plate. Translate it horizontally so its centre overlaps the EXTRA width of
the mount, and its edges are at the correct width.
Now, copy that cylinder, paste it and translate the copy in Z so it is way
above the parts. Then put both inside a hull() {.....}
Use difference() to cut the new cylinder hull out of the union above. Then
copy-paste the cylinder hull over to the other side to do the same on the
other side.
This should get you what you are after!
Alex Gibson
edumaker limited
. Project management
. Operations & Process improvement
. 3D Printing
-----Original Message-----
From: Discuss [mailto:discuss-bounces at lists.openscad.org] On Behalf Of
ats3788
Sent: 22 November 2019 12:37
How an we do The Soft connection to the plate in OpenSCad
--
_______________________________________________
_______________________________________________
-------------- next part --------------
An HTML attachment was scrubbed... | 577 | 2,243 | {"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} | 2.640625 | 3 | CC-MAIN-2020-16 | latest | en | 0.856943 |
https://mathoverflow.net/questions/374071/computation-of-the-groups-kbu-times-mathbbz-and-hbu-times-mathbbz/374112 | 1,607,080,789,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141735600.89/warc/CC-MAIN-20201204101314-20201204131314-00537.warc.gz | 378,115,347 | 29,713 | # Computation of the groups $K(BU \times \mathbb{Z})$ and $H^*(BU \times \mathbb{Z})$
Let $$U$$ denote the limiting group of the chain $$U(1) \to U(2) \to U(3) \to \cdots$$
I wish to compute the group $$K^{-1}\mathbb{C}/\mathbb{Z}(BU \times \mathbb{Z})$$. For this, we have the long exact sequence
$$\cdots \to K^{-1}(M) \to H^{odd}(M;\mathbb{C}) \to K^{-1}\mathbb{C}/\mathbb{Z}(M) \to K(M) \xrightarrow{ch \otimes \mathbb{C}} H^{even}(M;\mathbb{C}) \to \cdots$$ .
It therefore helps to know the groups $$K^{\pm 1}(BU \times \mathbb{Z})$$ and $$H^*(BU \times \mathbb{Z};\mathbb{C})$$.
However I am unable to compute these groups. Could anyone please suggest some references where these groups may have been explicitly computed ? Or may be some hints as to how I may proceed ? Thanks so much !
• $H^\ast(BU;\mathbb Z) = \mathbb Z[c_1,c_2,\dots]$ is given by the Chern classes. A great place to learn about this is Milnor and Stasheff's Characteristic Classes. I should know something about $K^\ast(BU)$ off the top of my head, but I don't. But googling "K-theory of BU" leads to this wikipedia page, which calculates both your groups in section 3 and 4 respectively, referring to Adams' Stable Homotopy and Generalized Homology for the $K$-theory computation. I'm still confused by the notation $K^n\mathbb C / \mathbb Z$... – Tim Campion Oct 14 at 19:10
• Ah, and as wikipedia notes, since $BU$ has even cells, its odd-dimensional $K$-theory vanishes. – Tim Campion Oct 14 at 19:13
• The odd cohomology vanishes, and then since $KU_\ast$ is also concentrated in even degrees the Atiyah-Hirzebruch spectral sequence collapses (everything is in even bidegree so there is no room for differentials) -- then since everything is in even bidegree, it's in even total degree, so the $K$-theory is also concentrated in even degrees. – Tim Campion Oct 14 at 19:41
• I feel like anything I know about this stuff I learned through some weird osmosis at some point, but I think the Adams book mentioned above might be a good resource. Just try not to get too hung up on whatever point-set formalism he's using. – Tim Campion Oct 14 at 19:47
• Yes Adams book is going to be one of the best places to learn this. Be warned: his book is written in reverse order of how one learning the subject should read it. You should read what he has to say about the AH spectral sequence (I think in the third part of the book), and then look at his computations of complex oriented cohomology using it (K-theory is complex oriented). This will be the second part of the book. – Connor Malin Oct 14 at 19:59
First, you just have $$K^n(\mathbb{Z}\times BU)=\text{Map}(\mathbb{Z},K^n(BU))$$, so you can work with $$K^*(BU)$$, which is technically more convenient. In particular, this is the inverse limit of the rings $$K^*(BU(n))$$. If $$V$$ is a complex vector bundle of dimension $$n$$ over a base space $$X$$, we can consider the polynomial $$g_V(u)=\sum_{k=0}^n[\lambda^k(V)]u^{n-k}\in K^0(X)[u]$$. We can then define $$c^{KU}_k(V)\in K^0(X)$$ to be the coefficient of $$t^{n-k}$$ in $$g_V(t-1)$$. By considering the universal case, we get classes $$c^{KU}_0=1,c^{KU}_1,\dotsc,c^{KU}_n\in K^0(BU(n))$$. It is a standard fact that $$K^0(BU(n))=\mathbb{Z}[[c^{KU}_1,\dotsc,c^{KU}_n]]$$, with $$c^{KU}_n$$ mapping to zero in $$K^0(BU(n-1))$$. One way to prove this is to use the general framework of complex orientable cohomology theories. Another is to use the Atiyah-Hirzebruch spectral sequence, which is in (even,even) bidegrees and so collapses. By passing to the limit we get $$K^0(BU) = \mathbb{Z}[[c^{KU}_1,c^{KU}_2,c^{KU}_3,\dotsc]].$$ We also have $$H^*(BU) = \mathbb{Z}[[c^{H}_1,c^{H}_2,c^{H}_3,\dotsc]],$$ with $$c_k^H\in H^{2k}(BU)$$ being the Chern classes in ordinary cohomology. | 1,202 | 3,774 | {"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": 23, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2020-50 | latest | en | 0.886053 |
https://www.compadre.org/PQP/quantum-need/section5_10.cfm?NOH=1 | 1,716,165,988,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058009.3/warc/CC-MAIN-20240519224339-20240520014339-00800.warc.gz | 634,356,235 | 4,431 | ## Section 5.10: Exploring the Uncertainty Principle
One way to begin to understand wave-particle duality is to think of a particle as a wave packet, constructed from a superposition of waves. Consider the wave given by the equation y(x, t) = A cos(kx − ωt) or for one instant in time (picking t = 0 for convenience), y(x) = A cos(kx). Restart.
1. Look at this wave. Where it is located?
2. Keep k2 the same and choose k1 = 8.0 rad/m. Now, there is a localized packet. What is the uncertainty in x (measure the distance from one zero amplitude to the next)? What is the uncertainty in kk = |k1 k2|). What is the uncertainty in x and the uncertainty in k for part (a)?
3. Pick several more values of the wave number, k1. As the uncertainty in kk) increases, what happens to the uncertainty in xx)?
In general, as the uncertainty in the x position decreases, the uncertainty in the k value increases so that ΔxΔk ~ 1. Note that this is just an approximate relationship. In fact, ΔxΔk ≥ 1/2.
Instead of simply adding two waves together with different k values (and possibly different amplitudes), at a time t = 0, we can add a group of waves together. The wave given by the sum (or in the case of a continuous distribution of k values, an integral) of all the superimposed waves is
y(x) = ∫ A(k) cos(kx) dk, (5.10)
where the amplitude of each individual wave added together can depend on the k value. Consider the simplest case where the amplitude is equal to 1 over a range of k values and is zero otherwise.
1. Integrate the above equation and show that (again this is at t = 0): y(x) = (2/x) sin(Δkx/2) cos(k0x), where the amplitude is 1 for k0 − Δk/2 < k < k0 + Δk/2.
2. Use the "Wave Packet: set values" button above to see this wave packet (here, k0 is the average of k1 and k2 and Δk = |k1 k2|) and measure Δx for different Δk values.
Sections 8.4-8.7 discuss constructing quantum-mechanical wave packets and the time evolution of such wave packets. | 539 | 1,986 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.865439 |
https://communities.sas.com/t5/SAS-Statistical-Procedures/How-to-do-multivariate-analysis-in-SAS-proc-logistic/td-p/69905?nobounce | 1,516,418,721,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888878.44/warc/CC-MAIN-20180120023744-20180120043744-00798.warc.gz | 675,989,068 | 29,183 | ## How to do multivariate analysis in SAS (proc logistic)
Occasional Contributor
Posts: 18
# How to do multivariate analysis in SAS (proc logistic)
I've been reading about multivariate analysis and proc logistic, and although there are some online descriptions of multivariate analysis there are few that describe how to do it in SAS. I need something that takes me step by step through the output to determine what adjustments I need to make (i.e. When to exclude a given independent variable).
From what I've read and been told, it's my interpretation that if the p-value of any independent variable is above .25, I should exclude the variable with the highest p-value until all p-values are are below .25. Is that a standard and accepted approach?
Any help is greatly appreciated.
Thanks.
Super Contributor
Posts: 281
## Re: How to do multivariate analysis in SAS (proc logistic)
Hard to answer any of this without a more detailed description of what your predictors are and what your dependent variables are, and what you hope to learn from this analysis.
Also, based on my understanding of the word "multivariate", PROC LOGISTIC does not do multivariate analyses. To me, multivariate means multiple response variables, analyzed with respect to their joint (correlated) distributions. Maybe you are using this word to mean something than what I think it means? Message was edited by: Paige
Occasional Contributor
Posts: 18
## Re: How to do multivariate analysis in SAS (proc logistic)
I'm probably using the work multivariate incorrectly.
This is the code I wrote to test the relationship of some binary (1=Yes, 2=No) independent variables on the dependent variable BreastFeeding (binary as well).
proc logistic data=nbscrBirthVars;
class NoCollege (ref="1") cesarean (ref="1") PreTerm (ref="1") LBW (ref="1") NICU (ref="1") TenStep (ref="1")/ param=ref;
model BreastFeeding (event="2")= NoCollege cesarean PreTerm LBW NICU TenStep;
run;
The output is below. So, my understanding is that I would remove Macrosomia from the model because the Pr > Chisq in the Type 3 analysis is greater than 0.25 (0.6956). Is that the standard way of determining what to remove?
Thanks.
The LOGISTIC Procedure
Model Information
Data Set WORK.NBSCRBIRTHVARS
Response Variable FormulaSupp
Number of Response Levels 2
Model binary logit
Optimization Technique Fisher's scoring
Number of Observations Used 99826
Response Profile
Ordered Formula Total
Value Supp Frequency
1 1 18503
2 2 81323
Probability modeled is FormulaSupp=2.
NOTE: 6875 observations were deleted due to missing values for the response or explanatory variables.
Class Level Information
Design
Class Value Variables
NoCollege 1 0
2 1
cesarean 1 0
2 1
PreTerm 1 0
2 1
LBW 1 0
2 1
NICU 1 0
2 1
Macrosomia 1 0
2 1
TenStep 1 0
2 1
------------------------------------------------------------------------------------------------------
The LOGISTIC Procedure
Model Convergence Status
Convergence criterion (GCONV=1E-8) satisfied.
Model Fit Statistics
Intercept
Intercept and
Criterion Only Covariates
AIC 95717.853 93430.154
SC 95727.364 93506.243
-2 Log L 95715.853 93414.154
Testing Global Null Hypothesis: BETA=0
Test Chi-Square DF Pr > ChiSq
Likelihood Ratio 2301.6993 7 <.0001
Score 2338.5007 7 <.0001
Wald 2265.3540 7 <.0001
Type 3 Analysis of Effects
Wald
Effect DF Chi-Square Pr > ChiSq
NoCollege 1 462.7169 <.0001
cesarean 1 47.0002 <.0001
PreTerm 1 13.8791 0.0002
LBW 1 3.6452 0.0562
NICU 1 229.8353 <.0001
Macrosomia 1 0.1531 0.6956
TenStep 1 1166.5014 <.0001
Analysis of Maximum Likelihood Estimates
Standard Wald
Parameter DF Estimate Error Chi-Square Pr > ChiSq
Intercept 1 0.7874 0.0964 66.7686 <.0001
NoCollege 2 1 0.3688 0.0171 462.7169 <.0001
cesarean 2 1 0.1185 0.0173 47.0002 <.0001
PreTerm 2 1 0.1106 0.0297 13.8791 0.0002
LBW 2 1 0.0706 0.0370 3.6452 0.0562
NICU 2 1 0.5175 0.0341 229.8353 <.0001
Macrosomia 2 1 0.0348 0.0890 0.1531 0.6956
TenStep 2 1 -0.5757 0.0169 1166.5014 <.0001
------------------------------------------------------------------------------------------------------
9
The LOGISTIC Procedure
Odds Ratio Estimates
Point 95% Wald
Effect Estimate Confidence Limits
NoCollege 2 vs 1 1.446 1.398 1.495
cesarean 2 vs 1 1.126 1.088 1.165
PreTerm 2 vs 1 1.117 1.054 1.184
LBW 2 vs 1 1.073 0.998 1.154
NICU 2 vs 1 1.678 1.569 1.794
Macrosomia 2 vs 1 1.035 0.870 1.233
TenStep 2 vs 1 0.562 0.544 0.581
Association of Predicted Probabilities and Observed Responses
Percent Concordant 56.6 Somers' D 0.232
Percent Discordant 33.3 Gamma 0.259
Percent Tied 10.1 Tau-a 0.070
Pairs 1504719469 c 0.616
Super Contributor
Posts: 281
## Re: How to do multivariate analysis in SAS (proc logistic)
While I am not familiar with the advice to use 0.25 as your cutoff, I would use 0.05 as the cutoff. In any event, it seems reasonable to remove Macrosomia from the model.
Valued Guide
Posts: 684
## Re: How to do multivariate analysis in SAS (proc logistic)
There are many stepwise variable-selection options in proc logistic. Check out the documentation for the model statement. But note: one should be cautious with all of these methods. Use them as an exploratory guide, not as a final model-selection method.Model selection (i.e., variable selection in a model) is a complex endeavor.
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https://www.maplesoft.com/support/help/errors/view.aspx?path=GroupTheory/IsSubnormal&L=E | 1,726,287,205,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651548.18/warc/CC-MAIN-20240914025441-20240914055441-00354.warc.gz | 801,259,953 | 21,873 | GroupTheory/IsSubnormal - Maple Help
For the best experience, we recommend viewing online help using Google Chrome or Microsoft Edge.
# Online Help
###### All Products Maple MapleSim
Home : Support : Online Help : GroupTheory/IsSubnormal
GroupTheory
IsSubnormal
test whether one group is contained as a subnormal subgroup of another
Calling Sequence IsSubnormal( H, G )
Parameters
H - a group G - a group
Description
• A group $H$ is a subnormal subgroup of a group $G$ if $H$ is a subgroup of $G$, and if there is a chain
$G={G}_{0}▹{G}_{1}▹\dots ▹H$
such that ${G}_{k}$ is normal in ${G}_{k-1}$, for each $i$. Every normal subgroup of a group is subnormal, but not conversely.
• The IsSubnormal( H, G ) command tests whether the group H is a subnormal subgroup of the group G. It returns true if H is subnormal in G, and returns false otherwise. For some pairs H and G of groups, the value FAIL may be returned if IsSubnormal cannot determine whether H is a subnormal subgroup of G.
Examples
> $\mathrm{with}\left(\mathrm{GroupTheory}\right):$
> $G≔\mathrm{Group}\left(\mathrm{Perm}\left(\left[\left[1,2,3,6,4,5,7,8\right]\right]\right),\mathrm{Perm}\left(\left[\left[2,5\right],\left[6,8\right]\right]\right)\right)$
${G}{≔}⟨\left({1}{,}{2}{,}{3}{,}{6}{,}{4}{,}{5}{,}{7}{,}{8}\right){,}\left({2}{,}{5}\right)\left({6}{,}{8}\right)⟩$ (1)
> $\mathrm{GroupOrder}\left(G\right)$
${16}$ (2)
> $H≔\mathrm{Subgroup}\left(\left\{\mathrm{Perm}\left(\left[\left[2,5\right],\left[6,8\right]\right]\right)\right\},G\right)$
${H}{≔}⟨\left({2}{,}{5}\right)\left({6}{,}{8}\right)⟩$ (3)
> $\mathrm{IsSubnormal}\left(H,G\right)$
${\mathrm{true}}$ (4)
Every normal subgroup of a group is subnormal.
> $\mathrm{andmap}\left(\mathrm{IsSubnormal},\mathrm{NormalSubgroups}\left(G\right),G\right)$
${\mathrm{true}}$ (5)
Compatibility
• The GroupTheory[IsSubnormal] command was introduced in Maple 2018.
• For more information on Maple 2018 changes, see Updates in Maple 2018.
See Also | 679 | 2,013 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 19, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.427826 |
https://timestablesworksheets.com/rocket-math-multiplication-worksheets/ | 1,722,829,250,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640427760.16/warc/CC-MAIN-20240805021234-20240805051234-00575.warc.gz | 470,548,547 | 18,913 | # Rocket Math Multiplication Worksheets
Rocket Math Multiplication Worksheets – One of the more difficult and challenging things that you can do with basic school college students is purchase them to enjoy math. Addition worksheets and subtraction worksheets aren’t what most kids need to be carrying out in their day time. Math in simple terminology is normally all of that simple to train, but when it comes to training the greater complicated ideas of mathematics, around 3rd quality, it may be a little more hard to maintain the college students intrigued. It can be even more complicated to put into action ways of training which will help the students understand the ideas of multiplication without the need of getting puzzled.
Multiplication tables are a fun way to get rid of into instructing multiplication. The five times tables along with other little numbers confirm quicker to educate most primary school pupils. College students may find it easy to just maintain adding yet another digit to their final final result to get the next number from the time tables. By teaching this primary, you can get the students from the mood and cozy with multiplying. The down sides might happen, however, once you’ve reached a higher times tables than five. The trick of merely adding an additional digit to the very last final result, to obtain the after that number within the table becomes much more difficult.
There comes a position in which pupils will need to commence memorizing the times tables to be able to remember them, rather than having the capability to utilize a distinct technique. Here you should be capable of put into practice enjoyable strategies for training to produce memorizing the times tables a lot much less difficult job. This is an integral part of mathematics. If your pupils fail to turn out to be efficient with their times tables soon they’ll have difficulties within the later levels of school. So what can you truly do in order to improve their capability within this field?
## Rocket Math Multiplication Worksheets is Important to Improve your Math Skills
There is a number of stuff that both parents and instructors as well are capable of doing to assist the pupil become successful. For their parents, they need to execute a number of stuff. To begin with, a due diligence system must be applied in the home to ensure the child is not really disregarding their studies in the home. Overlooking their work at home and attempting to perform every thing at school by no means helps! By seated with the children through the evening, transforming from the t . v . and aiding them come to be acquainted with multiplication you are going to eventually be assisting them conduct much better in school. I frequently advise flash cards for multiplication in your own home.
If you believe your kids will benefit, you can find issues you can get to enhance their research. It is possible to certainly get guides to assist them, however, many kids see studying being a laborious task nowadays. You can decide on a number of pc software Compact disks and math sites that make multiplication fun. These math games enable your little one to obtain exciting playing games while learning multiplication tables at the same time.
Professors need to clearly established correct homework, and make sure that it’s not very very much. Research is an essential part of a student’s studies, but placing an excessive amount of can decrease morale and make a little one feel overloaded. Setting fun and sensible amounts of homework can significantly aid a young child on their own approach to memorizing and learning the multiplication tables.he most significant factor to remember when attemping to assist your youngster is to try to ensure that it stays fun. Math should always be shown to kids inside a entertaining way. They must not really understand they are studying. | 726 | 3,892 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-33 | latest | en | 0.94927 |
http://forums.wolfram.com/mathgroup/archive/2009/Jul/msg00222.html | 1,638,118,746,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358570.48/warc/CC-MAIN-20211128164634-20211128194634-00210.warc.gz | 25,468,014 | 7,610 | Re: Is it possible to impose a condition on an iterator of
• To: mathgroup at smc.vnet.net
• Subject: [mg101548] Re: [mg101499] Is it possible to impose a condition on an iterator of
• Date: Fri, 10 Jul 2009 06:42:10 -0400 (EDT)
• References: <200907090552.BAA16854@smc.vnet.net>
```In[6]:= r1 = Sum[a^2 - 3*(a/(4*a^2)), {a, {1, 2, 3, 4, 5, 13, 14, 15, 16,
17,
18, 19, 20}}];
In[7]:= r2 = Sum[a^2 - 3*(a/(4*a^2)), {a, DeleteCases[Range[20],
i_ /; i >= 6 && i <= 12]}];
In[8]:= r3 = Sum[If[a >= 6 && a <= 12, 0, a^2 - 3*(a/(4*a^2))], {a, 1,
20}];
In[9]:= r1 === r2 === r3
Out[9]= True
In[10]:= r1
Out[10]= 64135524211/28217280
2009/7/9 Mauro <mauro at nonoyahoo.com>
> I would like to perform a summation of this type,
>
> Sum[a^2 - 3 a/(4 a^2), {a, 1, 20}]
>
> excluding however some values of the iteratore, I would for instance
> like to exclude the values than to greater of 6 and inferior to 12. Is
> it possible?
>
> Thanks
>
>
```
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• Next by thread: Re: Is it possible to impose a condition on an iterator of | 464 | 1,214 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2021-49 | longest | en | 0.794914 |
https://www.codesdope.com/practice/java-decide-if-or-else/ | 1,716,816,207,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059040.32/warc/CC-MAIN-20240527113621-20240527143621-00267.warc.gz | 612,078,705 | 14,646 | # Practice questions on Decide if or else
## Level 1
1.
Take values of length and breadth of a rectangle from user and check if it is square or not.
```import java.util.Scanner;
class Ans{
public static void main(String[] args){
Scanner s = new Scanner(System.in);
System.out.println("Enter length");
int x = s.nextInt();
int y = s.nextInt();
if(x==y){
System.out.println("Square");
}
else{
System.out.println("Rectangle");
}
}
}
```
2.
Take two int values from user and print greatest among them.
3.
A shop will give discount of 10% if the cost of purchased quantity is more than 1000.
Suppose, one unit will cost 100.
Judge and print total cost for user.
```import java.util.Scanner;
class Ans{
public static void main(String[] args){
Scanner s = new Scanner(System.in);
System.out.println("Enter quantity");
int x = s.nextInt();
if((x*100)>1000){
System.out.println("You get a discount of "+(.1*x*100)+" and your total cost is "+(x*100-(.1*x*100)));
}
else{
System.out.println("No discount");
}
}
}
```
4.
A company decided to give bonus of 5% to employee if his/her year of service is more than 5 years.
Ask user for their salary and year of service and print the net bonus amount.
5.
A school has following rules for grading system:
a. Below 25 - F
b. 25 to 45 - E
c. 45 to 50 - D
d. 50 to 60 - C
e. 60 to 80 - B
f. Above 80 - A
```import java.util.Scanner;
class Ans{
public static void main(String[] args){
Scanner s = new Scanner(System.in);
int x = s.nextInt();
if(x<25){
System.out.println("F");
}
else if((x>=25)&&(x<45)){
System.out.println("E");
}
else if((x>=45)&&(x<50)){
System.out.println("D");
}
else if((x>=50)&&(x<60)){
System.out.println("C");
}
else if((x>=60)&&(x<80)){
System.out.println("B");
}
else if((x>=80)&&(x<=100)){
System.out.println("A");
}
else{
System.out.println("Not correct marks");
}
}
}
```
6.
Take input of age of 3 people by user and determine oldest and youngest among them.
7.
Write a program to print absolute vlaue of a number entered by user. E.g.-
INPUT: 1 OUTPUT: 1
INPUT: -1 OUTPUT: 1
```import java.util.Scanner;
class Ans{
public static void main(String[] args){
Scanner s = new Scanner(System.in);
System.out.println("Enter number");
int x = s.nextInt();
if(x<0){
System.out.println("Absolute value : "+(-1*x));
}
else{
System.out.println("Absolute value : "+x);
}
}
}
```
8.
A student will not be allowed to sit in exam if his/her attendence is less than 75%.
Take following input from user
Number of classes held
Number of classes attended.
And print
percentage of class attended
Is student is allowed to sit in exam or not.
9.
Modify the above question to allow student to sit if he/she has medical cause. Ask user if he/she has medical cause or not ( 'Y' or 'N' ) and print accordingly.
10.
If
x = 2
y = 5
z = 0
then find values of the following expressions:
a. x == 2
b. x != 5
c. x != 5 && y >= 5
d. z != 0 || x == 2
e. !(y < 10)
```class Ans{
public static void main(String[] args){
int x = 2;
int y = 5;
int z = 0;
System.out.println(x==2);
System.out.println(x != 5);
System.out.println(x != 5 && y >= 5);
System.out.println(z != 0 || x == 2);
System.out.println(!(y < 10));
}
}
```
11.
Write a program to check whether a entered character is lowercase ( a to z ) or uppercase ( A to Z ).
You can use ASCII value to do so.
## Level 2
1.
Write a program to check if a year is leap year or not.
If a year is divisible by 4 then it is leap year but if the year is century year like 2000, 1900, 2100 then it must be divisible by 400.
2.
Ask user to enter age, sex ( M or F ), marital status ( Y or N ) and then using following rules print their place of service.
if employee is female, then she will work only in urban areas.
if employee is a male and age is in between 20 to 40 then he may work in anywhere
if employee is male and age is in between 40 t0 60 then he will work in urban areas only.
And any other input of age should print "ERROR".
3.
A 4 digit number is entered through keyboard. Write a program to print a new number with digits reversed as of orignal one. E.g.-
INPUT : 1234 OUTPUT : 4321
INPUT : 5982 OUTPUT : 2895
```import java.util.Scanner;
class Ans{
public static void main(String[] args){
Scanner s = new Scanner(System.in);
System.out.println("Enter number");
int x = s.nextInt();
int first_digit = x%10;
int second_digit = (x/10)%10;
int third_digit = (x/100)%10;
int fourth_digit = (x/1000)%10;
int new_number = (first_digit*1000)+(second_digit*100)+(third_digit*10)+(fourth_digit*1);
System.out.println(new_number);
}
}
``` | 1,282 | 4,558 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-22 | latest | en | 0.613592 |
https://vanityfairrewards.com/skin-diseases/your-question-how-many-moles-of-naoh-are-contained-in-30ml-of-its-0-15-m-solution.html | 1,660,853,174,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573399.40/warc/CC-MAIN-20220818185216-20220818215216-00462.warc.gz | 537,745,790 | 18,276 | # Your question: How many moles of NaOH are contained in 30ml of its 0 15 M solution?
Contents
## How many moles of NaOH are contained in 30ml of its 0.15 M solution?
= 0.15 / 1000 X 27 moles. = 0.004 Moles.
=0.045 (ans).
## How many moles are in 15 mL of NaOH?
Equation 1
How many moles of NaOH would be present in 15.0 mL? 5.00 moles/L X 0.0150 L= 7.50 X102 moles of NaOH.
1 mol of NaOH .
## How many moles of NaOH are present in its 3m solution?
To calculate the moles of solute, substitute the values of molarity and volume in the given formula. Therefore, the 0.03 moles of NaOH are present in 10.0 mL of a 3.00 M solution. Note: Make sure to convert the volume given in milliliter into liters as the molarity is calculated in terms of liters.
IT IS INTERESTING: How many atoms are there in 0 5 moles of ammonia?
## How many moles of NaOH are present in 90g of NaOH?
There are 2.25 moles of NaOH in 90g of NaOH.
## What volume of 0.15 m NaOH will contain 0.184 g NaOH?
∴ Volume of solution =(0.184g)(0.15mol L-1)×(40g mol-1)=0.0307L=30.7mL.
## How many moles are in 20 ml of NaOH?
You are given 1 Molar NaOH. This is in moles per liter. However, if you consider the density approach, 1M of NaOH has a density of 1.04 g/ml so 20 ml is 20 x 1.04 g = 20.8 g/40 g/mol (MW of NaOH) that is going to yield 520 mmol of NaOH in 20 mll of 1 M NaOH.
## How do you make a 15% NaOH solution?
15 grams of sodium hydroxide is present in 100 grams of solution. So, to make 50 g of 15%(w/w) sodium hydroxide solution, 47.2 mL of water is needed under experimental conditions.
## How many moles are in NaOH?
First, the molar mass of NaOH is 39.997 g/mol. Divide the mass by the molar mass and you’ll have the equation: 60 g / 39.997 g/mol. The grams cancel and you are left with 1.50 moles of NaOH.
## How many grams of NaOH are present in 250 mL of NaOH solution?
So 250mL of a 0.5m solution is 5.0g of NaOH.
## How many grams of NaOH are there in 500 mL of a 0.125 M NaOH solution?
The mass of NaOH needed to prepare 500 mL of 0.125 M NaOH is 2.5 g.
## How do you prepare 500.00 mL of a 0.1 m NaOH solution from NaOH s )?
Dissolve 20 gms of NaOH in 500 ml of water to prepare 500 ml 1N NaOH soln.
IT IS INTERESTING: How long does a skin cancer screening take? | 769 | 2,279 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.916887 |
https://mathoverflow.net/questions/381368/is-there-a-concrete-application-of-topos-theory | 1,726,608,498,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.53/warc/CC-MAIN-20240917204739-20240917234739-00560.warc.gz | 345,686,409 | 25,380 | # Is there a concrete application of topos theory?
The notion of topos was originally formulated in SGA 4 in the context of attacking the Weil conjectures. This formalism turned out to be unnecessary for the purposes of proving those conjectures. But the Weil conjectures do give a good example of a theorem where other abstract things, such as Grothendieck topologies and etale cohomology, are used in an essential way.
Is there any example of a concrete result in which the usage of topos theory is essential?
• A concrete result in which field? Algebraic geometry? Differential geometry? Any field that's not category theory? (Genuinely curious, since you haven't used a top-level tag) Commented Jan 16, 2021 at 13:42
• I'm confused by how you consider that topos theory is not used in the proof of the Weil conjecture. You seem to be saying that there use can be replaced by Grothendieck topologies and étale cohomology (which I agree with) but topos theory as introduce in SGA 4 is really nothing more than the theory of Grothendieck topologies. So I wouldn't consider this as getting ride of topos theory. I'm pointing this out, because the same thing apply to all application of Grothendieck toposes : you can always write everything in terms of Sites and Grothendieck topologies. Commented Jan 16, 2021 at 14:58
• Yes "Grothendieck topos theory" and "site theory" are two different point of view on exactly the same thing. Anything you can do with one you can also do it with the other. Of course there are indeed a lot of technical and conceptual advantages of working with toposes rather than sites, but at the end of the day you can always translate everything in terms of sites. Commented Jan 16, 2021 at 15:13
• Since many different sites can give the same topos of sheaves, I'm inclined to view a topos as capturing the "important" aspects of a site and discarding irrelevant details. I could describe this as an analogy "topos : site :: group : group-presentation. In principle, group theory could be developed entirely in terms of presentations, never mentioning the groups themselves, and in some situations that's useful, but in most situations it just makes things less clear. Commented Jan 16, 2021 at 15:50
• @Kim I think you can't lift the functoriality of the crystalline topos to sites so that might be an example. Commented Jan 16, 2021 at 16:12 | 572 | 2,373 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-38 | latest | en | 0.945832 |
https://viewer.wu.ac.at/viewer/fulltext/AC14446373/12/ | 1,702,059,924,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100769.54/warc/CC-MAIN-20231208180539-20231208210539-00271.warc.gz | 667,296,520 | 18,648 | # Full text: The Personal Distribution of Income
```9
<l( y)
<3 (.7)
f(w-y) e” dw
ft; •-* 7
ior
w ^ C
(5)
or w^ 7 ,
where is the Laplace transform of f(w)
The above mixture is. a Laplace transform of f ( w) shifted to the
right by 7.
The Laplace transform requires that the argument of the
function f be non-begative. We have therefore to assume that
j ( we shall further below haw this restriction can be relaxed ).
Equation (5) shows that the Pareto form of the wealth
distribution is reproduced in the income distribution, provided
the independence condition is fulfilled and y w.
We have now to face the fact that the rate of" return on wealth
will in reality not be independent of wealth* -he cross-classifications
of wealth and income of wealth owners for Holland and Sweden
show that mean income is a linear function of wealth, the regression
coefficient being smaller than unity.
For the decline of the rata of return with increasing
The earned income will be/less important the greater
the wealth. In particular the income from ( non-corporate) business
will be higher in relation to wealth in the lower wealth classes.
Further, capital gains are not counted as income, but they afffict
wealth, and they will be more important for lagge wealth, because
the proportion of shares held increases with wealth. The internal
accumulationof firms will not find expression in the income,
but quite probably in the wealth of the share holders. Also
appreciation of real estate may affect the large wealth proportionately
more.
Wow the rate of return is independent of we a lth if
its conditional distribution is the same whatever the size of
wealth. It would seem that we might perhaps restore the
condition of independence simply by turning the system of coordinates
in the appropriate way, so that we would reduce the present to the
former case. If we man make the covariance of w and w-y zero
then the ceefficient of regression of y on w should be one,
as in zhe former case:
weauixn
various reasons are res
rela
oasiblej
ively
```
### Note to user
Dear user,
In response to current developments in the web technology used by the Goobi viewer, the software no longer supports your browser. | 516 | 2,196 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.940603 |
http://mathematicssolution.com/wilsons-theorem-of-number-theory/ | 1,501,209,270,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549436321.71/warc/CC-MAIN-20170728022449-20170728042449-00011.warc.gz | 202,793,068 | 11,086 | BIGtheme.net http://bigtheme.net/ecommerce/opencart OpenCart Templates
Friday , July 28 2017
Home / Elementary Number Theory / Wilson’s theorem of number theory
# Wilson’s theorem of number theory
Proof: Dismissing the case p = 2 and p = 3 as being evident, let us take p > 3. Suppose that a is any of the p – 1 positive integers
1, 2, 3, … , p – 1
and consider the linear congruence ax ≡ 1 (mod p) . Then gcd(a, p) = 1. By linear congruence theorem , this congruence admits a unique solution modulo p; hence, there is a unique integer aʹ , with 1 ≤ aʹ ≤ p – 1 , satisfying aaʹ ≡ 1 (mod p).
Because p is prime, a = a ʹ if and only if a = 1 or a = p – 1 . Indeed, the congruence a 2 ≡ 1 (mod p) is equivalent to (a – 1). (p + 1) ≡ 0 (mod p). Therefore either a – 1 ≡ 0 (mod p), in which case a = 1 or a + 1 ≡ 0 (mod p), in which case
a = p – 1.
If we omit the numbers 1 and p – 1 , the effect is to group the remaining integers 2, 3, … , p – 1 into pairs a, aʹ, where a ≠ aʹ, such that there product aaʹ ≡ 1 (mod p). When these (p – 3 ) / 2 congruence are multiplied together and factors rearranged, we get
1. 3….(p – 2 ) ≡ 1 (mod p)
Or rather
(p – 2 ) ! ≡ 1(mod p)
Now multiply by p – 1 to obtain the congruence
(p – 1 ) ! ≡ p – 1 ≡ –1 (mod p).
as was to be proved.
Example: A concrete example should help to clarify the proof Wilson’s theorem. Specifically, let us take p = 13. It is possible to divide the integers 2, 3, …, 11into (p – 3) / 2 = 5 pairs, each product of which is congruent to 1 modulo13. To write this congruences out explicitly:
2 . 3 ≡ 1 (mod 13)
3 . 9 ≡ 1 (mod 13)
1. 10 ≡ 1 (mod 13)
2. 8 ≡ 1 (mod 13)
6 . 11 ≡ 1 (mod 13)
Multiplying these congruences gives the result
11 ! = (2.7) (3.9)(4.10)(5.8)(6.11) ≡ 1 (mod 13)
and so
12 ! ≡ 12 ≡ – 1 (mod 13)
Thus, (p – 1 ) ! ≡ – (mod p), with p = 13.
Converse of Wilson’s theorem: If (p – 1)! + 1 ≡ 0, then p is prime.
Proof: Suppose that p is not prime. Then p is composite and so there exist a divisor d where 1 < d < p. Furthermore since d is a factor of 1.2.3…(p – 1), so we can write
(p – 1) ! ≡ 0 (mod d)
(p – 1) ! + 1 ≢ 0 (mod d)
⇒ (p – 1) ! + 1 ≢ 0 (mod p)
This is a contradiction of the converse statement.
It follows that p must be a prime. (Proved)
## Application of Euclidian’s algorithm in Diophantine equation
Problem-1: Which of the following Diophantine equations cannot be solved – a) 6x + ... | 881 | 2,407 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2017-30 | latest | en | 0.814176 |
https://ch.mathworks.com/matlabcentral/cody/problems/105-how-to-find-the-position-of-an-element-in-a-vector-without-using-the-find-function/solutions/38546 | 1,596,468,434,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735812.88/warc/CC-MAIN-20200803140840-20200803170840-00180.warc.gz | 255,986,354 | 15,431 | Cody
# Problem 105. How to find the position of an element in a vector without using the find function
Solution 38546
Submitted on 12 Feb 2012 by Roberto Montagna
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = [1 3 5 4 2]; y = 2 posX_correct = 5; assert(isequal(findPosition(x,y),posX_correct))
y = 2 s = 1 2 3 4 5
2 Pass
%% x = [1 5 8 6 7 6 0]; y = 8 posX_correct = 3; assert(isequal(findPosition(x,y),posX_correct))
y = 8 s = 1 2 3 4 5 6 7 | 201 | 577 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.623019 |
http://71.cialisewq.top/surface-area-worksheet/ | 1,607,143,374,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141746320.91/warc/CC-MAIN-20201205044004-20201205074004-00058.warc.gz | 1,851,698 | 29,449 | # Surface Area Worksheet
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Surface Area Worksheet Learn how to calculate surface area with this sixth grade geometry worksheet introduce your students to the concept of surface area with this concrete teaching tool in which learners will practice The third dimension is arguably the coolest let your fifth graders explore 3d shapes in this worksheet that asks them to find the surface area of items with a cube shape after reading a description Now watch this video to explore how to show the surface area of a sphere is 4 r 178 activities to help you understand practise and test your knowledge you can print out the worksheets below or.
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## Surface Area Worksheets Math Worksheets 4 Kids
Surface Area Of L Shaped Prisms Worksheets Lay A Strong Foundation In Decomposing Shapes With These Printable Surface Area Of L Shaped Rectangular Prism Worksheets Solve For Surface Area Of Independent 3d Shapes Add Them And Subtract The Area Of The Face That Connects The Rectangular Blocks Surface Area Of Triangular Prisms Worksheets
### Surface Area Worksheets Tutoringhour
The Surface Area Is The Total Area That The Surface Of An Object Occupies These Thorough Paced Free Surface Area Worksheets Include The First Stage Which Is Counting Unit Squares To Determine The Surface Area Followed By Finding The Surface Area Of Solid Shapes Such As Rectangular Prisms Cylinders Cones Spheres Hemispheres Square Pyramids And Compound Shapes
#### Surface Area Worksheets Super Teacher Worksheets
Surface Area 2 Basic This Worksheet Has Three Solid Shapes With The Dimensions Labeled Students Determine The Sa Of Each Includes Plenty Of Space To Show Work
##### Surface Area Worksheets Printable At Rudolph Academy
These Surface Area Worksheets Are Printable Find Surface Area Worksheets For Rectangular Prisms Cylinders Cones Spheres And Triangular Prisms Teachers Parents And Students Can Print These Worksheets And Make Copies Surface Area Of Rectangular Prism Worksheets
###### Volume And Surface Area Worksheets Dsoftschools
Some Of The Worksheets Below Are Volume And Surface Area Worksheets Surface Area Objectives To Find The Surface Of A Cube To Find The Surface Of A Cuboid Volume And Surface Area Of Rectangular Prisms And Cylinders Solid Geometry Calculate The Volume Of Prisms And Cylinders Calculate The Volume Of Pyramids Cones And Spheres Calculate The Surface Area Of Prisms And
Surface Area Using Nets Worksheets
Finding Total Surface Area Remain Brimming With Energy And Enthusiasm Throughout These Printable Worksheets Showing 3d Shapes Along With Their Nets Use The Dimensions And Find The Area Of Each Region On The Net To Compute The Surface Area Of The Given Solid Shape Draw The Net And Find Its Surface Area
Area And Surface Area Worksheets Teacher Worksheets
Area And Surface Area Showing Top 8 Worksheets In The Category Area And Surface Area Some Of The Worksheets Displayed Are Surface Area Surface Area Surface Area Rectangular Prism Surface Area Of Solids Formulas For Perimeter Area Surface Volume Surface Area Math Resource Studio
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Top | 1,889 | 9,727 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2020-50 | latest | en | 0.918315 |
https://www.doubtnut.com/rd-sharma-solutions/class-8-chapter-squares-and-square-roots-1 | 1,621,393,111,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991562.85/warc/CC-MAIN-20210519012635-20210519042635-00557.warc.gz | 750,997,974 | 49,953 | # RD Sharma Class 8 and Chapter Squares And Square Roots Maths Solutions
## Maths Class 8 and Chapter Squares And Square Roots RD Sharma Solutions
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### RD SHARMA Class 8 | SQUARES AND SQUARE ROOTS
Is 225 a perfect square ? if so, find the number whose square is 225.
### RD SHARMA Class 8 | SQUARES AND SQUARE ROOTS
Show that 63504 is perfect square. Also, find the number whose square is 63504.
### RD SHARMA Class 8 | SQUARES AND SQUARE ROOTS
Find the smallest number by which 180 must be multiplied so that the product is a perfect square.
### RD SHARMA Class 8 | SQUARES AND SQUARE ROOTS
Find the smallest number by which 25200 should be divided so that the result is a perfect square.
The Central Board of Education is a high-level acclaimed education board. The syllabus of the board recommends NCERT books to its students since class 8th. NCERT had a very simplified pattern of theory and examples in its books.
This pattern is good for the students to understand the concept for their school exams, but to ace the exams and strengthen their concepts further students require more than just the NCERT books.
To ace their exams and the relatively higher competitive exams, students need to have a better command on Math. Math is a part of almost all competitive exams and therefore holds quite importance.
Students struggle in such situations to completely master the subject just with the help of NCERT books. This is when they look to reference books for help, to understand the concept in more detail and get more exercises to solve. The best reference book for Math is RD SHARMA, which not only helps students strengthen their concepts and understand them but also provides a whole lot of exercises and examples. The students can master the subject only by practicing regularly.
The only part where it lacks is providing a detailed stepwise solution to those problems. Doubtnut helps the students in that case, where subject experts put their knowledge together and prepare a solution guide which is 100% accurate and is comprehended to help the students understand the topic better.
The third chapter in the book is Square and Square Roots. The book does perfect justice by explaining each concept clearly and providing with illustrative examples and exercises for practice.
## Square and Square Roots Class 8th Solutions
The third chapter of Square and square roots is a very basic and important topic for future use in math. It explains the students the basic method of finding squares and their square roots respectively.
A square of any number can be calculated by multiplying that number to itself.
The square root of the other hand is quit inverse of what we do to find a square of any number. Symbolic representation of square roots is (√). For example, √16 = 4, because 4×4 = 16.
The chapter lists all the methods, algorithms to find out square roots of numbers.
In recent times with the changes in education system a new concept of providing solutions and guides to students with the help of online experts has emerged. The solutions provided by these online experts help the students to get a better understanding of each chapter and topic. The solutions and guides that are provided for books like RD Sharma ensure that the students get a better chance to practice questions and exercises.
The solutions help the student understand how a problem is solved in a stepwise manner so as they do not get confused during the exams. The answers to exercises in the solution guide are in the same order as the exercises in the book itself.
With this guide and Solutions provided to the students they perform better in the exams with much more knowledge of the topics and the chapter.
## Topics and Subtopics of Rd Sharma Class 8th Chapter 3
The book RD Sharma has a very elaborate layout for each chapter. Every chapter no matter how small or lengthy is distributed effectively into various topics and subtopics for maintaining clear guideline for the students.
The writers and experts meticulously create the book and ensure that the material provided in the book is easily understood by students of the concerned class. Every topic is explained in detail both theoretically and with the help of illustrative examples. All of this explanation and theory is then followed by exercise with questions for the students to practice.
Since the writers divide every topic into smaller subtopics, it gets easier for the student who understands the basics and concept of the entire chapter. The solutions provided to the students follow the pattern as it is according to the book, just for the students.
The following topics and sub-topics are covered by the book RD Sharma:
### Introduction
The concept follows the lead of Powers. Any number that has the power of 2, is called a square or a perfect square and is written as 22, 32, etc. The power of 2 or a square means the number is multiplied by itself, once. The chapter entails the whole concept of squares for one, two and three digit numbers (or even more) and also to find Square roots of both perfect and non-perfect squares.
### Square Of A Number And Square Numbers
Square of a number is the number raise to the power 2. For any number n it can be written and understood as, n2 = n×n. Square numbers or perfect square is a natural number which is the square of a natural number.
### Properties and Patterns of Some Square Numbers
This section explains the various properties and patterns that are used to determine whether or not the given square is a perfect square. These properties and patterns are simple tricks to verify the square number, without the tedious work of finding its Square roots. The book gives a detailed explanation with illustrative examples to clear the concept for the students.
### Some Shortcuts to Find Squares
The concept of finding squares by simple multiplication is easy to use in small numbers. But to find the square of a bigger number, simple multiplication gets complicated and a tedious method. In that case the book provides students with some shortcuts to find squares. These methods are • column method • visual method • diagonal method etc.
### Square Roots
The concept of square roots is rather very easy and qualitatively explained in RD Sharma. Square root is the number which is multiplied by it to give the square.
### Properties of Square Roots
Like any other concept in Math, the properties of square roots help understand the topic in depth. The book details these properties and their use in a very student friendly language with examples.
### Various Algorithm and Methods for Deriving Square Roots
This topic is divided into major subtopics in the book. RD Sharma covers all the methods and algorithms in an illustrative manner and shows the students how to solve the problems instead of elongated theory. The major methods covered in the book are • Square root of a perfect square by prime factorization • Relation between the digits of a perfect square and its square roots • Square roots of perfect square by the method of Long division • Square roots of rational numbers in the form of fractions • Square root of rational numbers in decimal form. These concepts are detailed and provided with examples to help students understand their implication in exercises and questions.
## Importance of Rd Sharma Class 8 Math
The book is a major help to students of every class and also the students who prepare for higher competitive exam. RD Sharma is stated to be one of the best reference books formats available to the students. Not only it provides the students with help in the studies but also ensures a confidence boost by the amount of practice.
• In depth explanations to topics: RD Sharma follows the pattern of in depth knowledge and explanation for each chapter which is divided into various topics and subtopics. With this pattern students get to understand every concept in a very effective and easy manner.
• Number of practice questions & exercises: RD Sharma does not just focus on the theory of a concept but also provides the students with many examples for understanding the basics and how question is to be solved. Each topic or concept is followed by a number of illustrative examples which is then followed by the exercise on the same topic for the students to practice and master the concept. This is done to ensure that the student get to practice a lot, because math is not a subject of memorizing theory.
• Prepared by a team of experts: RD Sharma is regularly edited to keep it up to date according to the patterns of exams and the needs of students. This is overseen by a group of experts who entered the questions are accurately prepared and in accordance to recent changes in exam patterns so as the students are aware of it all.
• Aided with online solutions: Students can easily find online solutions and guides to the book which helps them to solve the questions and any difficulties they encounter while practicing. This makes RD Sharma a great book for practicing as students have a backup to look up to in case they get stuck on some question.
Math is a subject of great weight age in every competitive exam and therefore needs to be practiced to boost the confidence and score of the students. RD Sharma aids the students in this aim and therefore is recommended to them as a reference book. | 1,880 | 9,454 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2021-21 | latest | en | 0.945653 |
https://justaaa.com/finance/33722-beverly-hills-started-a-paper-route-on-january-1 | 1,723,531,950,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641063659.79/warc/CC-MAIN-20240813043946-20240813073946-00470.warc.gz | 267,262,543 | 10,596 | Question
# Beverly Hills started a paper route on January 1. Every three months, she deposits \$1,000 in...
Beverly Hills started a paper route on January 1. Every three months, she deposits \$1,000 in her bank account, which earns 8 percent annually but is compounded quarterly. Four years later, she used the entire balance in her bank account to invest in an investment at 12 percent annually. How much will she have after three more years? Use Appendix A and Appendix C for an approximate answer, but calculate your final answer using the formula and financial calculator methods. (Do not round intermediate calculations. Round your final answer to 2 decimal places.)
Future value:
In the first half of the question, 1 Period = 3 Months.
Periodic Payments (A) = \$1000
Rate = 8% annually (compounded quarterly). Therefore, Rate per Period (r) = 8% / 4 = 2%
Number of Periods (n) = 4 years x 4 = 16 periods
Using Appendix C,
FVA = A x FVIFA ( 2% , 16 periods)
FVA = 1000 x 18.639 = \$18639
Using Formula,
FVA = \$18639.29
Using Calculator,
Enter N = 16, I/Y = 2, PV = 0, PMT = 1000
Press CPT, then FV.
FV = \$18639.29
This is the Value she will have in her bank account after 4 years. After this, she invests the entire amount for 3 years @ 12 percent annually.
Using Appendix A,
FV = PV x FVIF ( 12%, 3 Periods)
FV = 18639 x 1.405 = \$26,187.80
Using Formula,
FV = PV x (1 + r)n
FV = 18639.29 (1.12)3 = \$26186.86
Using Financial Calculator,
Enter N = 3, I/Y = 12, PV = 18639.29, PMT = 0
Press CPT and then FV.
FV = 26186.86
#### Earn Coins
Coins can be redeemed for fabulous gifts. | 482 | 1,615 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-33 | latest | en | 0.913015 |
https://testbook.com/question-answer/a-researcher-selects-a-probability-sample-of-100-o--61791455b5e01413a92a8ba5 | 1,675,497,785,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500095.4/warc/CC-MAIN-20230204075436-20230204105436-00736.warc.gz | 582,270,238 | 74,739 | # A researcher selects a probability sample of 100 out of the total population. It is _____.
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1. a cluster sample
2. a random sample
3. a stratified sample
4. a systematic sample
Option 2 : a random sample
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## Detailed Solution
To draw valid conclusions from your results, you have to carefully decide how you will select a sample that is representative of the group as a whole. There are two types of sampling methods:
• Probability sampling involves random selection, allowing you to make strong statistical inferences about the whole group.
• Non-probability sampling involves non-random selection based on convenience or other criteria, allowing you to easily collect data.
A researcher selects a probability sample of 100 out of the total population. It is a random sample.
Key Points
Simple random sampling:
• In a simple random sample, every member of the population has an equal chance of being selected.
• Your sampling frame should include the whole population.
• To conduct this type of sampling, you can use tools like random number generators or other techniques that are based entirely on chance.
For example, You want to select a simple random sample of 100 employees of Company X. You assign a number to every employee in the company database from 1 to 1000, and use a random number generator to select 100 numbers.
Stratified random sampling
• Stratified sampling is a method of sampling from a population that can be partitioned into subpopulations.
• When subpopulations within an overall population vary, it could be advantageous to sample each subpopulation independently.
Systematic sampling
• Systematic sampling is the selection of Participants from an ordered sampling frame.
• The most common form of systematic sampling is an equiprobability method.
Cluster sampling
• Cluster sampling is a sampling used when in a statistical population there are mutually homogeneous yet internally heterogeneous groups.
• In this sampling plan, the total population is divided into these groups and a simple random sample of the groups is selected.
• The researcher divides the population into separate groups, called clusters. | 493 | 2,363 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.898591 |
https://estebantorreshighschool.com/quadratic-equations/energy-of-electron-equation.html | 1,675,712,420,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500357.3/warc/CC-MAIN-20230206181343-20230206211343-00717.warc.gz | 256,971,968 | 11,403 | ## How do you calculate the energy of an electron?
A simple expression for the energy of an electron in the hydrogen atom is_E=−13.6n2 where the energy is in electron volts.n is the principle quantum number.So for an electron in n=1 :E=−13.6eV.To convert to joules you can x this by 1.6×10−19.
## What is the energy of an electron?
The energy of an electron is of the same order of magnitude (is in the same range) as the energy of light. The lines in the spectrum of an element represent changes in the energy of electrons within the atoms of that element.
## What is Bohr’s equation for energy levels?
By keeping the electrons in circular, quantized orbits around the positively-charged nucleus, Bohr was able to calculate the energy of an electron in the nth energy level of hydrogen: E ( n ) = − 1 n 2 ⋅ 13.6 eV E(n)=-dfrac{1}{n^2} cdot 13.6,text{eV} E(n)=−n21⋅13.
## How are energy levels calculated?
When you look at the Periodic Table of the Elements, the energy levels of the atoms correspond to the rows of the table. The two elements in the top row, hydrogen and helium, are filling their first energy level with their final electrons. The eight elements of the second row are filling their second energy level.
## Do electrons have energy?
There is also a maximum energy that each electron can have and still be part of its atom. Beyond that energy, the electron is no longer bound to the nucleus of the atom and it is considered to be ionized. When an electron temporarily occupies an energy state greater than its ground state, it is in an excited state.
## Why is the energy of electron negative?
When a stable atom is formed, the electron is attracted to the nucleus, r is less than infinity, and the energy will be negative. A negative value for the energy implies that energy must be supplied to the system if the electron is to overcome the attractive force of the nucleus and escape from the atom.
## Which electron has highest energy?
Valence electrons
You might be interested: Differentiate equation
## What is ground state energy?
The ground state of an electron, the energy level it normally occupies, is the state of lowest energy for that electron. There is also a maximum energy that each electron can have and still be part of its atom.
## Why is energy quantized?
Energy is quantized in some systems, meaning that the system can have only certain energies and not a continuum of energies, unlike the classical case. This would be like having only certain speeds at which a car can travel because its kinetic energy can have only certain values.
## What are Bohr’s 4 postulates?
Postulates of Bohr’s Model of an Atom The energy levels are represented by an integer (n=1, 2, 3…) known as the quantum number. This range of quantum number starts from nucleus side with n=1 having the lowest energy level. The orbits n=1, 2, 3, 4… are assigned as K, L, M, N….
## Why was Bohr rejected?
The moment that the Bohr model was applied to an element with more than one electron (which, unfortunately, includes every element except hydrogen), the Bohr model failed miserably. Bohr’s model failed because it treated electrons according to the laws of classical physics.
## How do you find Bohr radius?
Bohr Radius Formulaao is the Bohr radius.me is the rest mass of electron.εo is the permittivity of the free space.left ( frac{h}{2pi } right ) = ħ is the reduced Planck constant.c is the velocity of light in vacuum.α is the fine structure constant.e is the elementary charge.
## What are the 4 energy levels?
There are four types of orbitals that you should be familiar with s, p, d and f (sharp, principle, diffuse and fundamental). Within each shell of an atom there are some combinations of orbitals.
## How many electron energy levels are there?
Thus, the first level can contain up to 2 electrons, 2(12) = 2; the second up to 8 electrons, 2(22) = 8; the third up to 18, 2(32) = 18; and so on. Only seven energy levels are needed to contain all the electrons in an atom of any of those elements now known.
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#### Characteristic equation complex roots
What are roots of characteristic equations? discussed in more detail at Linear difference equation#Solution of homogeneous case. The characteristic roots (roots of the characteristic equation) also provide qualitative information about the behavior of the variable whose evolution is described by the dynamic equation. How do I know if my roots are complex? When graphing, if […]
#### Free fall time equation
What is the formula for time in free fall? Free fall means that an object is falling freely with no forces acting upon it except gravity, a defined constant, g = -9.8 m/s2. The distance the object falls, or height, h, is 1/2 gravity x the square of the time falling. Velocity is defined as […] | 1,111 | 4,818 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2023-06 | latest | en | 0.912869 |
bim4u.nl | 1,618,727,791,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038468066.58/warc/CC-MAIN-20210418043500-20210418073500-00443.warc.gz | 252,255,831 | 8,839 | ball mill kw calculator
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## ball mill kw calculator
• ###### Ball Mill Design/Power Calculation
The basic parameters used in ball mill design (power calculations), rod mill or any tumbling mill sizing are; material to be ground, characteristics, Bond Work Index, bulk density, specific density, desired mill tonnage capacity DTPH, operating % solids or pulp density, feed size as F80 and maximum ‘chunk size’, product size as P80 and maximum and finally the type of circuit open/closed ...
• ###### Ball Nose Finishing Mills Speed Feed Calculator - DAPRA
The calculator will automatically provide the necessary speed and feed in the green fields. For assistance setting up your milling program, contact a Dapra applications specialist or call (800) 243-3344. Click here to download a chart of recommended Ball Nose cutting speeds and feeds (PDF).
• ###### Bond formula for the grinding balls size calculation
Oct 19, 2017 The enterprises consumers grinding media have a question about right choise the grinding ball size (diameter) for the mill in order to achieve the required grinding quality. We noted earlier, this information can be obtained from several sources: — Technical documentation. It attached to the milling equipment (mill).
• ###### Ball Mill Calculation Kw And Rpm - Stone Crushing Machine
ball mill, cement ball mill, energy-saving ball mill, cone ball As a ball mill manufacturer,We The grinding efficiency means the consumption 1 kw when 3 according to the surface when electricity to calculate tons Click Chat Now
• ###### AMIT 135: Lesson 7 Ball Mills Circuits – Mining Mill ...
The ball size factor, F B, is added to the mall shaft power for mill sizes greater than 3.3 meters:F↓B – 1.102 (d↓max – 12.5D / 50.8) kW/t Other Rowland Kjos Power Draw Corrections
• ###### motor kw calculation for ball mill
ball mil design calculation? yahoo answers the power (in kw) required to turn a ball mill is approximated by: p = 0.285 d (1.073 j) m n where d is the internal diameter in metres, j is the fraction of the mill volume occupied by media, m is the total mass of media in the mill, in tonnes, and n is the mill speed, rpm. the actual motor energy ...
• ###### Calculate and Select Ball Mill Ball Size for Optimum Grinding
In Grinding, selecting (calculate) the correct or optimum ball size that allows for the best and optimum/ideal or target grind size to be achieved by your ball mill is an important thing for a Mineral Processing Engineer AKA Metallurgist to do. Often, the ball used in ball mills is oversize “just in case”. Well, this safety factor can cost you much in recovery and/or mill liner wear and ...
• ###### BALL MILL POWER - Page 1 of 2
Nov 12, 2013 ball mill power . dear experts. we have a ball mill in combi mode for cement grinding . roller press before the ball mill. the supplier has given 28% grinding media filling , with shaft power 2270 kw only. ball mill size is 4.4*11 m. grinding media size is max 30mm . it is a monochamber mill designed to grind the roller press product.
• ###### TECHNICAL NOTES 8 GRINDING R. P. King
the mill is used primarily to lift the load (medium and charge). Additional power is required to keep the mill rotating. 8.1.3 Power drawn by ball, semi-autogenous and autogenous mills A simplified picture of the mill load is shown in Figure 8.3 Ad this can be used to establish the essential features of a model for mill
• ###### A Method of C alculating Autogenous/ Semi-Autogenous ...
mills with the rod mill and ball mill laboratory work indices. Note, in Figure. 1, that the rod mill product slope is less than 0.5 due to an extra amount of nes present being fi fi ner than 650 μm. These fi nes proceed to the ball mill improving the ball mill effi ciency. Also, the plotted rod mill P80 value, as shown in Figure 1, is 2900 ...
• ###### Ball Nose Finishing Mills Speed Feed Calculator - DAPRA
The calculator will automatically provide the necessary speed and feed in the green fields. For assistance setting up your milling program, contact a Dapra applications specialist or call (800) 243-3344. Click here to download a chart of recommended Ball Nose cutting speeds and feeds (PDF).
• ###### Speeds and Feeds Calculator - Good Calculators
How to use: Choose a type of operation (drilling, reaming, boring, counterboring, face milling, slab milling/side milling, end milling, or turning), select your stock material, choose a material for the tool (high-speed steel or carbide), input the quantity of teeth for the tool and the diameter of the workpiece/cutter. Hit the "Calculate ...
• ###### Ball Mill Torque Calculation - Krosline
Ball Mill Torque Calculation. The horsepower at both the spindle and the motor are shown, as well as the spindle torque for a given spindle speed rpmhe motor horsepower required for the milling operation can then be compared to the horsepower capability of the machineearn more about milling.
• ###### How To Calculate Rpm Of A Motor To Ball Mill
ball mill power - cemnet. if we calculate the shaft power with 28% filling it comes to more than 2600 kw with 210 t grinding media. ... supplier has given a ball mill motor installed kw of 2600 only. ... n, mill speed, rpm. read more
• ###### motor kw calculation for ball mill
ball mil design calculation? yahoo answers the power (in kw) required to turn a ball mill is approximated by: p = 0.285 d (1.073 j) m n where d is the internal diameter in metres, j is the fraction of the mill volume occupied by media, m is the total mass of media in the mill, in tonnes, and n is the mill speed, rpm. the actual motor energy ...
• ###### Ball Mill Operating Speed - Mechanical Operations Solved ...
At what speed will the mill have to be run if the 100 mm balls are replaced by 50 mm balls, all the other conditions remaining the same? Calculations: The critical speed of ball mill is given by, where R = radius of ball mill; r = radius of ball. For R = 1000 mm and r = 50 mm, n c = 30.7 rpm. But the mill is operated at a speed of 15 rpm.
• ###### Ball mill media optimization - Metcom Tech
plant ball mill’s grinding efficiency (Fig. 1). The functional performance ... (kW) x Mill grinding rate of coarse material (t/kWh) (1) The power applied to ... mill speed are used to calculate the energy input. The specific energy input (E, in kWh/t) is the
• ###### how to estimate the wear rate for Ball mill - Page 1 of 10
Dec 08, 2009 Re: how to estimate the wear rate for Ball mill. Most practical way to by measuring empty height and calculating the % volumetric filling of grinding media.Alternative way to make track of mill main drive kW consumption reduction and add makeup charge according to the extent of fall in power consumption.
• ###### BALL MILL POWER - Page 1 of 2
Nov 12, 2013 ball mill power . dear experts. we have a ball mill in combi mode for cement grinding . roller press before the ball mill. the supplier has given 28% grinding media filling , with shaft power 2270 kw only. ball mill size is 4.4*11 m. grinding media size is max 30mm . it is a monochamber mill designed to grind the roller press product.
• ###### A Method of C alculating Autogenous/ Semi-Autogenous ...
mills with the rod mill and ball mill laboratory work indices. Note, in Figure. 1, that the rod mill product slope is less than 0.5 due to an extra amount of nes present being fi fi ner than 650 μm. These fi nes proceed to the ball mill improving the ball mill effi ciency. Also, the plotted rod mill P80 value, as shown in Figure 1, is 2900 ...
• ###### Ball Mill Ball Consumption Calculation-ball Mill
Mill Ball Mill Ball Consumption Calculation. Calculations for ball mill calculation capacity ball mill xball mill the specific power consumption of clinker gypsum grinding in the ball mill should be about 3000 cm2 g 244 kwht 3200 cm2 g 268 kwh t 3400 cm2 g 294 kwh t 3600 cm2 g 320 kwh t 3800 cm2 g 347 kwh t 4000 cm2 g 375 kwt note the specific power requirement of large pulverizer is 1015 ...
• ###### Cement mill notebook - SlideShare
Jan 07, 2015 Raw mills usually operate at 72-74% critical speed and cement mills at 74-76%. 3.2 Calculation of the Critical Mill Speed: G: weight of a grinding ball in kg. w: Angular velocity of the mill tube in radial/second. w = 2*3.14*(n/60) Di: inside mill diameter in meter (effective mill diameter). n: Revolution per minute in rpm. 7.
• ###### Ball Mill Finish Calculator - Martin Chick Associates
The Ball Mill Finish Calculator can be used when an end mill with a full radius (a ball mill) is used on a contoured surface. The tool radius on each side of the cut will leave stock referred to as a scallop. The finish of the part will be determined by the height of the scallop, amd the scallop will be determined by the stepover distance ...
• ###### Ball Mill Ball Consumption Calculation-ball Mill
Mill Ball Mill Ball Consumption Calculation. Calculations for ball mill calculation capacity ball mill xball mill the specific power consumption of clinker gypsum grinding in the ball mill should be about 3000 cm2 g 244 kwht 3200 cm2 g 268 kwh t 3400 cm2 g 294 kwh t 3600 cm2 g 320 kwh t 3800 cm2 g 347 kwh t 4000 cm2 g 375 kwt note the specific power requirement of large pulverizer is 1015 ...
• ###### Ball Nose Finishing Mills Speed Feed Calculator - DAPRA
The calculator will automatically provide the necessary speed and feed in the green fields. For assistance setting up your milling program, contact a Dapra applications specialist or call (800) 243-3344. Click here to download a chart of recommended Ball Nose cutting speeds and feeds (PDF).
• ###### Ball Mill Torque Calculation - Krosline
Ball Mill Torque Calculation. The horsepower at both the spindle and the motor are shown, as well as the spindle torque for a given spindle speed rpmhe motor horsepower required for the milling operation can then be compared to the horsepower capability of the machineearn more about milling.
• ###### How To Calculate Rpm Of A Motor To Ball Mill
ball mill power - cemnet. if we calculate the shaft power with 28% filling it comes to more than 2600 kw with 210 t grinding media. ... supplier has given a ball mill motor installed kw of 2600 only. ... n, mill speed, rpm. read more
• ###### GROSS POWER CALCULATOR - SMC Testing
base are up to 40ft in diameter and ball mills are up to 26ft in diameter. The measured gross power draws from these mills are plotted against the model predictions in Figure 1. As indicated by this data base the accuracy of the model (1 standard deviation) is 4.6%. How to Use the Model . Inputs to the model are made in three windows, viz: Mill ...
• ###### Speeds and Feeds Calculator - Good Calculators
How to use: Choose a type of operation (drilling, reaming, boring, counterboring, face milling, slab milling/side milling, end milling, or turning), select your stock material, choose a material for the tool (high-speed steel or carbide), input the quantity of teeth for the tool and the diameter of the workpiece/cutter. Hit the "Calculate ...
• ###### Ball mill media optimization - Metcom Tech
plant ball mill’s grinding efficiency (Fig. 1). The functional performance ... (kW) x Mill grinding rate of coarse material (t/kWh) (1) The power applied to ... mill speed are used to calculate the energy input. The specific energy input (E, in kWh/t) is the
• ###### Ball Mill Operating Speed - Mechanical Operations Solved ...
At what speed will the mill have to be run if the 100 mm balls are replaced by 50 mm balls, all the other conditions remaining the same? Calculations: The critical speed of ball mill is given by, where R = radius of ball mill; r = radius of ball. For R = 1000 mm and r = 50 mm, n c = 30.7 rpm. But the mill is operated at a speed of 15 rpm.
• ###### how to estimate the wear rate for Ball mill - Page 1 of 10
Dec 08, 2009 Re: how to estimate the wear rate for Ball mill. Most practical way to by measuring empty height and calculating the % volumetric filling of grinding media.Alternative way to make track of mill main drive kW consumption reduction and add makeup charge according to the extent of fall in power consumption.
• ###### Cement mill notebook - SlideShare
Jan 07, 2015 Raw mills usually operate at 72-74% critical speed and cement mills at 74-76%. 3.2 Calculation of the Critical Mill Speed: G: weight of a grinding ball in kg. w: Angular velocity of the mill tube in radial/second. w = 2*3.14*(n/60) Di: inside mill diameter in meter (effective mill diameter). n: Revolution per minute in rpm. 7.
• ###### An innovative approach for determining the grinding media ...
(4) The calculation of motor power of ball mill. The media addition was calculated via equality (17) . (17) N = K ωb × G. Where N (kW) represented the power of ball mill pinion shaft; K ωb (kW/t) was the power for per ton ball medium; G (t) represented the weight of media loaded in the ball mill. Usually, the power of the ball mill pinion ...
• ###### Mill (grinding) - Wikipedia
The power predictions for ball mills typically use the following form of the Bond equation: = (−) where E is the energy (kilowatt-hours per metric or short ton) W is the work index measured in a laboratory ball mill (kilowatt-hours per metric or short ton) P 80 is the mill circuit product size in micrometers
• ###### Milling formulas and definitions - Sandvik Coromant
The milling process – definitions Cutting speed,v c Indicates the surface speed at which the cutting edge machines the workpiece. Effective or true cutting speed, v e Indicates the surface speed at the effective diameter (DC ap).This value is necessary for determining the true cutting data at the actual depth of cut (a p).This is a particularly important value when using round insert cutters ... | 3,341 | 13,922 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | longest | en | 0.785073 |
https://sergworks.wordpress.com/2012/09/ | 1,623,862,190,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487625967.33/warc/CC-MAIN-20210616155529-20210616185529-00380.warc.gz | 470,081,226 | 17,437 | # Big Integer redux
9
I started a project that will include Big Integer math implementation (common Delphi&FPC codebase) based on interfaces without objects. Right now the project reached the stage where I am able to compile (and run!) code like that:
```{
Usage example: BinomCoff 120 42
Demonstrates how to use BigCardinal type
http://rosettacode.org/wiki/Evaluate_binomial_coefficients#Delphi
}
program BinomCoff;
{\$APPTYPE CONSOLE}
uses
SysUtils, tfNumerics;
function BinomialCoff(N, K: Cardinal): BigCardinal;
var
L: Cardinal;
begin
if N < K then
Result:= 0 // Error
else begin
if K > N - K then
K:= N - K; // Optimization
Result:= 1;
L:= 0;
while L < K do begin
Result:= Result * (N - L);
Inc(L);
Result:= Result div L;
end;
end;
end;
var
A: BigCardinal;
M, N: Cardinal;
begin
ReportMemoryLeaksOnShutdown:= True;
try
if ParamCount <> 2 then begin
Writeln('Usage example: BinomCoff 120 42');
Exit;
end;
N:= StrToInt(ParamStr(1));
M:= StrToInt(ParamStr(2));
A:= BinomialCoff(N, M);
Writeln('C(', N, ', ', M, ') = ', A.AsString);
A:= BigCardinal(nil); // A is global var and should be freed explicitely
// to prevent memory leak on shutdown
except
on E: Exception do
Writeln(E.ClassName, ': ', E.Message);
end;
end.
```
Output:
A lot still yet to be done, but it works! 🙂
# Why StackOverflow sucks
906
StackOverflow is a monster Q&A machine. If you have a programming question, StackOverflow is probably the best place to ask – you have a better chance to get an answer on SO than anywhere else.
The paradox is that SO is not interested in users getting answers to their questions. Usually Q&A sites want their questioners to be happy, but not SO. SO wants great questions and great answers. Hence the reputation system and an army of Nazi retards moderating everything they can see.
If a question is considered poor by the user with moderating privileges, it will be downvoted, closed and finally deleted. But that is not all – SO has an automatic ban system. Users providing questions & answers that received low marks can be banned by robots.
One of the first questions I answered on SO more that 2 years ago was:
Here is a task:
“3 brothers have to transfer 9 boxes from one place to another.
These boxes weigh 1, 2, 4, 5, 6, 8, 9, 11, 14 kilos.
Every brother takes 3 boxes.
So, how to make a program that would count the most optimal way to take boxes?
The difference between one brother’s carrying boxes weight and other’s has to be as small as it can be.”
Can you just give me an idea or write a code with any programming language you want ( php or pascal if possible? ).
Thanks.
I thought the question was interesting and after spending some time found a solution based on checking all permutations of 9 weight numbers, it appeared to be blazingly fast. I posted an answer, and my answer was accepted by the questioner.
Sure that was not a great question. Also the question was not properly tagged – with ‘php’ and ‘pascal’. I guess ‘php’ tag was a mistake; the questioner got a whole army of moderating idiots attacking his question.
The question received 17 downvotes. The question got the comment `Smells like homework to me` and the comment got 14 upvotes; the presumption of innocence does not work on SO, and a guy with the editing privileges tagged the question as ‘homework’. Later on it was closed by the moderator called Bill the Lizard with the following resolution:
It’s difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form.
Strange resolution, isn’t it? The question was answered…
The strange guy Bill the Lizard did not stop after closing the question. More than 1.5 year (!) after the question was answered he returned to it and deleted both the question and my answer (my answer probably because it contradicted his resolution).
If you think your post was not well accepted on SO, just think of the whole picture.
[Updated]
I was saying in this post “If you have a programming question, StackOverflow is probably the best place to ask”. It was when SO started some years ago; not now.
Today your chances to get a useful answer to your question on SO are close to zero. Instead you get a lot of comments arguing for example that your question does not fit SO or wrongly worded or else and nothing useful.
SO today is yet another trolls&noobs zoo.
# Nice feature of Lazarus IDE
4
Lazarus treats include files (*.inc) as a part of a project. That means you should not bother where you store include files in a project – just
```{\$I Smth.inc}
```
is enough.
Not so in Delphi (Delphi XE to be exact). You can add *.inc file to a Delphi project, but it does not matter for Delphi, you still need to write paths to include files like that
```{\$I ..\..\Source\Common\Smth.inc}
```
and sure the paths are different for units in different project folders. | 1,244 | 4,966 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.784891 |
http://v8doc.sas.com/sashtml/stat/chap13/sect6.htm | 1,553,066,015,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202303.66/warc/CC-MAIN-20190320064940-20190320090940-00057.warc.gz | 230,059,588 | 2,827 | Chapter Contents Previous Next
Introduction to Nonparametric Analysis
## Comparing Two Independent Samples
SAS/STAT software provides several nonparametric tests for location and scale differences.
When you perform these tests, your data should consist of a random sample of observations from two different populations. Your goal is either to compare the location parameters (medians) or the scale parameters of the two populations. For example, suppose your data consist of the number of days in the hospital for two groups of patients: those who received a standard surgical procedure and those who received a new, experimental surgical procedure. These patients are a random sample from the population of patients who have received the two types of surgery. Your goal is to decide whether the median hospital stays differ for the two populations.
### Tests in the NPAR1WAY Procedure
The NPAR1WAY procedure provides the following location tests: Wilcoxon rank sum test (Mann-Whitney U test), Median test, Savage test, and Van der Waerden test. Also note that the Wilcoxon rank sum test can be obtained from the FREQ procedure. In addition, PROC NPAR1WAY produces the following tests for scale differences: Siegel-Tukey test, Ansari-Bradley test, Klotz test, and Mood test.
When data are sparse, skewed, or heavily tied, the usual asymptotic tests may not be appropriate. In these situations, exact tests may be suitable for analyzing your data. The NPAR1WAY procedure can produce exact p-values for all of the two-sample tests for location and scale differences.
Chapter 47, "The NPAR1WAY Procedure," provides detailed statistical formulas for these statistics, as well as examples of their use.
### Tests in the FREQ Procedure
This procedure provides a test for comparing the location of two groups and for testing for independence between two variables.
The situation in which you want to compare the location of two groups of observations corresponds to a table with two rows. In this case, the asymptotic Wilcoxon rank sum test can be obtained by using SCORES=RANK in the TABLES statement and by looking at either of the following:
• the Mantel-Haenszel statistic in the list of tests for no association. This is labeled as "Mantel Haenszel Chi-square" and PROC FREQ displays the statistic, the degrees of freedom, and the p-value.
• the CMH statistic 2 in the section on Cochran-Mantel-Haenszel statistics. PROC FREQ displays the statistic, the degrees of freedom, and the p-value. To obtain this statistic, specify the CMH2 option in the TABLES statement.
When you test for independence, the question being answered is whether the two variables of interest are related in some way. For example, you might want to know if student scores on a standard test are related to whether students attended a public or private school. One way to think of this situation is to consider the data as a two-way table; the hypothesis of interest is whether the rows and columns are independent. In the preceding example, the groups of students would form the two rows, and the scores would form the columns. The special case of a two-category response (Pass/Fail) leads to a 2 ×2 table; the case of more than two categories for the response (A/B/C/D/F) leads to a 2 ×c table, where c is the number of response categories.
For testing whether two variables are independent, PROC FREQ provides Fisher's exact test. For a 2 ×2 table, PROC FREQ automatically provides Fisher's exact test when you use the CHISQ option in the TABLES statement. For a 2 ×c table, use the EXACT option in the TABLES statement to obtain the test.
Chapter Contents Previous Next Top | 783 | 3,664 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-13 | latest | en | 0.903047 |
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Sunday, February 25 2018 @ 11:35 AM CST
Select a Forum » General Chat » Primary 1 Matters » Primary 2 Matters » Primary 3 Matters » Primary 4 Matters » Primary 5 Matters » Primary 6 Matters » Question, Feedback and Comments » Education Classified
Forum Index > Test Paper Related > Primary 6 Matters Ratio Problem Sum
| Printable Version
By: rpgwwk (offline) Tuesday, March 06 2012 @ 12:12 AM CST (Read 15263 times)
rpgwwk
A square is divided into four parts W, X, Y and Z.
The ratio of the area of W to X is 7 : 8.
The ratio of the area of Y to Z is 3 : 2.
(a) What fraction of the square is part Y? (Give your answer in the simplest form)
(b) If part Z is smaller than part X by 10 cm2, find the area of the square.
Newbie
Registered: 09/27/11
Posts: 6
By: awyw1201 (offline) Tuesday, March 06 2012 @ 01:46 AM CST
awyw1201
Is there a diagram for this question?
Regular Member
Registered: 07/08/11
Posts: 87
By: rpgwwk (offline) Tuesday, March 06 2012 @ 11:02 AM CST
rpgwwk
Yes.
Thank you.
Newbie
Registered: 09/27/11
Posts: 6
By: awyw1201 (offline) Tuesday, March 06 2012 @ 08:28 PM CST
awyw1201
In the diagram, the diagonal cuts it into two equal parts.
Thus, W + X = Y + Z
So looking at the ratios, we need to equalise the total number of units.
L.C.M of 15 and 5 is 15. Therefore, we need to multiply Y : Z by 3.
W : X = 7 : 8
Y : Z = 9 : 6
8-6 = 2u----10
1u---5
30u---150
Ans: 150 sqaure cm.
Regular Member
Registered: 07/08/11
Posts: 87
By: awyw1201 (offline) Tuesday, March 06 2012 @ 08:29 PM CST
awyw1201
Sorry, ans to (a) is 9/30
Regular Member
Registered: 07/08/11
Posts: 87
All times are CST. The time is now 11:35 am.
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New Post Sticky Topic w/ New Post Locked Topic w/ New Post
View Anonymous Posts Able to Post HTML Allowed Censored Content | 632 | 1,864 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2018-09 | latest | en | 0.783635 |
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posted by .
The U.S. Weather Service reports that in a certain northern city it rains 35 days and snows 40 days in the winter. However, it rains and snows on only 10 of those days. Based on this information, what is the probability that it will rain or snow in that city on a particular winter day? Assume that there are 90 days of winter.
• algebra -
P(A) + P(B) - P(A and B)
35/90 + 40/90 -10/90 = 13/18
## Similar Questions
1. ### math
at a mountain village in new guinea it rains on average 6 days a week. determine the probability that it rains on 1. any one day 2. two successive days 3. three sucessive days. what is the sample space/
2. ### math
at a mountain village in new guinea it rains on average 6 days a week. determine the probability that it rains on 1. any one day 2. two successive days 3. three sucessive days. what is the sample space/
3. ### math
In a mountain village it rains on average 5 days a week. Find the probability that over a weekend it rains on exactly one of the two days.
4. ### algebra
The U.S. Weather Service reports that in a certain northern city it rains 35 days and snows 40 days in the winter. However, it rains and snows on only 10 of those days. Based on this information, what is the probability that it will …
5. ### English
If it snows tomorrow, what will you do? 1. If it snows tomorrow, I will go out with my dog. 2. If it snows tomorrow, I will make a snow man at the back yard. 3. If it snows tomorrow, I will meet my girl friend. 4. If it snows tomorrow,
6. ### Math
Ansgar is writing a novel. He writes seven days a week. On each of those days he write for at least 4 hours but never more than 8 hours. Last week, he wrote for exactly 46 hours. What is the maximum number of days on which he could …
7. ### statistics
in a beach side suburb it rains on 20% of days and is windy on 30% of days. if rain and wind are independent, on any particular day find the probability that: (a) it rains and is windy (b) it does not rain and is not windy (c) it rains …
8. ### statistic
in a beacside it rains on 20percent of days and is windy on 30percent a days.if rain and wind are independent on any particular days find the probability that it rains and it windy
9. ### English
(1) It rains. (2) It is cold. (3) It is growing dark. (4) It seems that there is no end to this. (5) It is unclear why he cut the rope. (6) It snows a lot. (7) It is snowing a lot. ------------------------------ "Rains" and "snows" …
10. ### math
During the winter it snows 3 out of every 10 days what is the probability that it will snow, in any given week 7 days, more than 3 times?
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```NATIONAL INSTITUTE OF TECHNOLOGY NAGALAND
CHUMUKEDIMA, DIMAPUR – 797 103.
B.Tech Degree Programme
Curriculum
Regulations – 2012
Bachelor of Technology in Computer Science and Engineering
Semester III
Course
Code
MA202
Course Title
L
T
P
C
Discrete Mathematics
3
1
0
4
CS201
Data Structures
3
0
0
3
CS202
Computer Architecture
3
0
0
3
CS203
Database Management Systems
3
0
0
3
EE208
Electrical Science
3
0
2
4
CS204
3
0
0
3
CS205
Data Structures Laboratory
0
0
3
2
CS206
0
0
3
2
CS207
Database Management Systems Laboratory
0
0
3
2
TOTAL
18 1 11 26
MA202 DISCRETE MATHEMATICS
PROPOSITIONAL CALCULUS
L T P C
3 1 0 4
12
Propositions − Logical Connectives − Compound Propositions − Conditional and
Biconditional Propositions − Truth Tables − Tautologies and Contradictions −
Contrapositive − Logical Equivalences and Implications − Demorgan’s Laws −
Normal Forms − Principal Conjunctive and Disjunctive Normal Forms − Rules of
Inference − Arguments − Validity of Arguments
PREDICATE CALCULUS
12
Predicates − Statement Function − Variables − Free and Bound Variables −
Quantifiers − Universe of Discourse − Logical Equivalences and Implications for
Quantified Statements − Theory of Inference − The Rules of Universal Specification
and Generalization −Validity of Arguments
SET THEORY
12
Basic Concepts − Notations − Subset − Algebra of Sets − The Power Set − Ordered
Pairs and Cartesian Product − Relations on Sets − Types of Relations and their
Properties − Relational Matrix and the Graph of a Relation − Partitions − Equivalence
Relations − Partial Ordering − Poset − Hasse Diagram − Lattices and their Properties
− Sublattices − Boolean Algebra – Homomorphism
FUNCTIONS
12
Definition of functions − Classification of Functions − Type of Functions − Examples
− Composition of Functions − Inverse functions − Binary and n-ary Operations −
Characteristic Function of a Set − Hashing Functions − Recursive Functions −
Permutation Functions
GROUPS
12
Algebraic Systems − Definitions − Examples − Properties − Semigroups − Monoids
− Homomorphism − Sub Semigroups and Submonoids − Cosets and Lagrange’s
Theorem − Normal Subgroups − Normal Algebraic System with two Binary
Operations − Codes and Group Codes − Basic Notions of Error Correction − Error
Recovery in Group Codes
Total: 60 Periods
TEXT BOOKS
1.
Trembly J .P and Manohar R, “Discrete Mathematical Structures with
Applications to Computer Science”, TMH Pub. Co. Ltd, New Delhi, 2003.
2.
Ralph P. Grimaldi, “Discrete and Combinatorial Mathematics: An Applied
Introduction”, Fourth Edition, Pearson Education Asia, Delhi, 2002.
REFERENCES
1.
2.
3.
Bernard Kolman, Robert C. Busby and Sharan Cutler Ross, “Discrete
Mathematical Structures”, Fourth Indian reprint, Pearson Education Pvt Ltd.,
New Delhi, 2003.
Kenneth H. Rosen, “Discrete Mathematics and its Applications”, Fifth Edition,
TMH Pub. Co. Ltd., New Delhi, 2003.
Richard Johnsonbaugh, “Discrete Mathematics”, Fifth Edition, Pearson
Education Asia, New Delhi, 2002.
CS201 DATA STRUCTURES
L T P C
3 0 0 3
LINEAR DATA STRUCTURES
9
Abstract Data Types – Algorithm Notations - Basic data structures – Arrays – Lists
– Singly linked lists – Doubly linked lists – Circular lists - Stacks and Queues –
Applications of Stack and Queues
NON-LINEAR DATA STRUCTURES
9
Trees – Binary Trees – Binary tree representation and traversals – Threaded binary
trees – Binary tree representation of general trees – Application of trees: Set
representation – Graph and its representations – Graph Traversals
SEARCH TREES AND PRIORITY QUEUES
9
AVL Trees – Red-Black Trees – Splay Trees – Binary Heap – Leftist Heap
SORTING
9
Insertion sort – Merge sort – Quick sort – Heap sort – Sorting with disks – K-way
Merging – Sorting with Tapes – Polyphase Merge
SEARCHING AND INDEXING
9
Linear Search – Binary Search - Hash Tables – Overflow Handling – Cylinder
Surface Indexing – Hash Index – B-Tree Indexing.
Total: 45 Periods
TEXT BOOKS
1.
Ellis Horowitz and Sartaj Sahni, “Fundamentals of Data Structures”,
Computer Press, 1983.
2.
Sartaj Sahni, “Data Structures, Algorithms, and Applications in C++”,
Second Edition, McGraw Hill NY, Silicon Press, 2005.
REFERENCE BOOKS
1.
Goodrich, Michael T., Roberto Tamassia, David Mount. “Data Structures and
Algorithms in C++”, Seventh Edition, Wiley Publishers, 2004.
2.
Mark Allen Weiss, Data Structures and Algorithm Analysis in C++, 3rd
edition, Pearson Education India, 2007.
3.
Jean-Paul Tremblay and Paul G. Sorenson, An Introduction to Data
Structures with Applications, Second Edition, Tata McGraw-Hill, New Delhi,
1991.
4.
Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, Clifford Stein,
"Introduction to Algorithms", Third Edition, Prentice Hall of India, , 2010.
CS202 COMPUTER ARCHITECTURE
L T P C
3 0 0 3
FUNDAMENTAL STRUCTURE OF COMPUTERS
9
Functional units – Basic operational concepts – Bus structures – Software
Performance – Memory Locations and Addresses – Memory Operations –
Instruction and Instruction Sequencing
–
Classification of Computers (SISD, MISD, MIMD) – RISC – CISC – ALU
Design – Fixed and Floating Point Operations
ARITHMETIC FOR COMPUTERS
9
Division – Floating Point Numbers – Representation – Arithmetic Operators
BASIC PROCESSING UNIT
9
Fundamental Concepts – Multiple Bus Organization – Execution of Complete
Instruction – Hardwired Control – Micro Programmed Control – Instruction Level
Parallelism
PIPELINING AND ILP
9
Basic Concepts – Data Hazards – Instruction Hazards – Influence on Instruction
Sets – Data Path and Control Considerations – Performance Considerations –
Exception Handling – Advanced Concepts in Pipelining – Exploitation of more
ILP – Hardware and Software Approaches – Dynamic Scheduling – Speculation –
Compiler Approaches – Multiple Issue Processors
MEMORY SYSTEM AND I/O
9
Basic Concepts – Semiconductor RAM – ROM – Speed – Size and Cost –
Cache Memories – Improving Cache Performance – Virtual Memory – Memory
Management requirements – Associative Memories – Secondary Storage Devices –
Accessing I/O Device – Interrupts – DMA – Buses – Interface Circuits – Standard
I/O Interfaces (PCI, SCSI, USB) – I/O Devices and Processors
Total: 45 Periods
TEXT BOOKS
1. Carl Hamacher, Zvonko Vranesic and Safwat
Organization”, Fifth Edition, Tata McGraw Hill, 2002.
Zaky,
“Computer
2. David A. Patterson and John L. Hennessy, “Computer Organization and
Design: The Hardware/Software Interface”, Third Edition, Elsevier, 2005.
REFERENCE BOOKS
1. William Stallings, “Computer Organization and Architecture: Designing for
Performance”, Sixth Edition, Pearson Education, 2003.
2. John P. Hayes, “Computer Architecture and Organization”, Third Edition,
Tata McGraw Hill, 1998.
3. V.P. Heuring, H.F. Jordan, “Computer Systems Design and Architecture”,
Second Edition, Pearson Education, 2004.
4.
Behrooz Parhami, “Computer Architecture”, Oxford University Press, 2007.
CS203 DATABASE MANAGEMENT SYSTEMS
L T P C
3 0 0 3
DATABASE FUNDAMENTALS
9
Introduction and Need for Database Systems – Database Vs File systems –
Database – DBMS distinction – Approaches to build a Database – Database System
Architecture – Data Modeling and Languages – Entity-Relationship Model – Weak
Entity Sets – Mapping ER Model to Relational Model
RELATIONAL DATA MODEL
9
Concept of Relations – Schema-Instance distinction – Integrity Constraints –
Relational Algebra – Tuple Relation Calculus – Domain Relational Calculus –
Overview of QBE – SQL Queries – Nested Queries – Aggregate Operators – Null
values – Embedded SQL – Database Security – Views
DATABASE DESIGN
9
Importance of a Good Schema Design – Problems encountered with Bad Schema
Designs – Functional Dependencies – Normalization – Decomposition –
Armstrong’s Axioms – First, Second, Third Normal Forms – Dependency
Preservation – Boyce/Codd Normal Form – Multi-valued Dependencies and Fourth
Normal Form – Join Dependencies and Fifth Normal Form
DATA STORAGE AND INDEXES
9
File Organizations – Primary and Secondary Index Structure – Various Index
Structures – Hash-based – Dynamic Hashing Techniques – Multi-level Indexes – B
Trees – B+ Trees
9
Transaction Concepts – ACID Properties – Recovery and Concurrency Control –
Locking Protocols – Recovery Methods – Object Oriented and Object Relational
Databases – Logical Databases – Web Databases – Distributed Databases – Data
Warehousing and Data Mining
Total: 45 Periods
TEXT BOOKS
1. Abraham Silberschatz, Henry F. Korth, S. Sudharshan, “Database System
Concepts”, Fifth Edition, Tata McGraw Hill, 2006.
2. Ramez Elmasri, Shamkant B. Navathe, “Fundamentals of Database Systems”,
Fourth Edition, Pearson / Addision Wesley, 2007.
REFERENCE BOOKS
1. Raghu Ramakrishnan, “Database Management Systems”, Third Edition, Tata
McGraw Hill, 2003.
2. S.K.Singh, “Database Systems Concepts, Design and Applications”, First
Edition, Pearson Education, 2006.
3. C.J.Date, A.Kannan, S.Swamynathan, “An Introduction to Database
Systems”, Eighth Edition, Pearson Education, 2006.
EE208 ELECTRICAL SCIENCE
BASIC CIRCUIT CONCEPTS
L T P C
3 0 2 4
9
Lumped circuits – Circuits elements – V-I relationships of R, L and C – Independent
sources – Dependent sources – simple resistive circuits – Kirchhoff’s Laws –
Analysis of series and parallel circuits – Network reduction – Voltage division –
Current division – Source transformation – Star delta transformation.
9
Concepts of phasor and complex impedance and admittance – Analysis of simple
series and parallel circuits – Active power, Reactive power and Power factor – Series
resonance and parallel resonance – Bandwidth and Q factor – Solution of three phase
balanced circuits – Power measurement by two- wattmeter method – Solution of three
phase unbalanced circuits.
MAGNETIC CIRCUITS
9
Self and Mutual Inductances – Leakage Reactance - Magnetic circuits with DC and
AC excitation - Components of Exciting Current - B-H curve - Hysteresis and Eddy
Currents and losses - Separation of Hysteresis and eddy current losses - Dynamic BH curve - Determination of Magnetic Force and Torque from Energy and Co-energy
– Single and Multiple Excited Systems - Simple designs
INTRODUCTION TO ROTATING MACHINES
9
Introduction to AC and DC machines – MMF of distributed windings – Magnetic
fields in rotating machines – Rotating MMF in AC machines – Generated voltage in
AC and DC machines – Electro-magnetic Torque in Non-Salient pole machines –
Linear machines – Magnetic saturation – Flux Leakages
DC MACHINES AND TRANSFORMERS
9
Electromagnetic Torque and Generated Voltage in a DC Machine - Effect of
Armature MMF – Electric Circuit Aspects – Magnetic Circuit Aspects – Testing of
Machines - Transformers : Principle of operation – EMF Equation – No load
Conditions – Effect of Secondary Current – Equivalent Circuit – Phasor Diagrams –
Open Circuit, Short Circuit, Load, Sumpner’s Tests – Separation of No load Losses –
Voltage regulation and efficiency
Total: 45 Periods
TEXT BOOKS
1. M. Nahvi and J.A. Edminister, “Electric Circuits”, Fifth Edition, Schaum's
Outline Series, McGraw-Hill, 2011
2. W H Hayt, J E Kemmerly, S M. Durbin “Engineering Circuit Analysis”,
Seventh Edition, Tata McGraw-Hill, International Editions, 2010
3. A Fitzgerald, Charles Kingsley, Stephen Umans, “Electric Machinery”, 6th
Edition, Tata McGraw Hill, 2010
REFERENCE BOOKS
1. Donald Fink and H. Wayne Beaty “Standard Handbook for Electrical
Engineers”, Tata Mc-Graw Hill, 2006.
2. Van Vallenburg, "Network Analysis", Prentice Hall India.
3. Nilsson and Riedel, “Electric Circuits”, Nineth Edition, Pearson Education,
2011.
4. Vincent Del Toro," Electrical Engineering Fundamentals", Prentice Hall India
5. Cotton, H., “Electrical Technology”, CBS Publishers, New Delhi, 6th edition
1984
L T P C
3 0 0 3
9
Object Oriented Programming Concepts – Classes – Objects – Methods and
Messages – Abstraction and Encapsulation – Inheritance – Abstract Classes –
Polymorphism – C++ Fundamentals – I/O operations – Constructors – Destructors –
9
INHERITANCE AND POLYMORPHISM
Inheritance – Public, Private and Protected Derivations – Multiple Inheritance –
Virtual Base Class – Abstract Class – Virtual Functions – Pure Virtual
Functions – Templates
JAVA FUNDAMENTALS
9
Java Virtual Machine – Reflection – I/O Streaming – Filter And Pipe Streams –
Byte Codes – Byte Code Interpretation – Dynamic Reflexive Classes – Threading –
Java Native Interfaces – GUI Applications
NETWORK PROGRAMMING IN JAVA
9
Socket Programming - Stream Customization – Secure Sockets – Custom Sockets –
UDP Datagrams – Multicast Sockets – URL Classes – Reading Data from the
Server – Writing Data – Configuring The Connection – Reading The Header –
Content Handlers – Telnet Application – Java Messaging Services – Remote
Method Invocation – Activation Models – JAR File Creation – JDBC –Multimedia
Data Handling
WEB PROGRAMMING ELEMENTS
9
Browser configuration – Plug-in components – Web standards and validation – Web
Quality – XML standards – HTML – CSS – DHTML – Java Script – Functions –
Events – Document Object Model
Total: 45 Periods
TEXT BOOKS
1.
Bjarne Stroustrup, “The C++ Programming Language”, Third Edition,
Pearson Education, 2000.
2.
Cay S. Horstmann, Gary Cornell, “Core JAVA Volume 1”, Eighth Edition,
Pearson Education, 2008.
3.
P.J. Deitel and H.M. Deitel, “Internet & World Wide Web: How to Program”,
Fourth Edition, Pearson Education, 2009.
REFERENCE BOOKS
1.
S. B. Lippman, Josee Lajoie, Barbara E. Moo, “C++ Premier”, Fourth
Edition, Pearson Education, 2005.
2.
Robert Lafore, “Object Oriented Programming in C++”, Fourth Edition, Sams
Publishers, 2001.
3.
K. Arnold and J. Gosling, “The JAVA programming language”, Third
Edition, Pearson Education, 2000.
4.
Robert W. Sebesta, “Programming the World Wide Web”, Sixth Edition,
CS205 DATA STRUCTURES LABORATORY
L T P C
0 0 3 2
1. Implementation of Abstract Data Type – Lists – Array based - Linked List – adding,
inserting, deleting and sorting
2. Application of Lists – Polynomial Manipulation
3. Implementation of Stack ADT – Evaluating Arithmetic Expressions – Convert Infix
expression to Postfix expression
4. Implementation of Queue ADT – Circular Queue and Double ended Queue
5. Priority Queue using Heaps
6. Implementation of Trees – Binary Trees – Tree Traversals
7. Application of Trees – Hashing – Hash Functions – Rehashing – Extendible Hashing
8. Insertion sort – Shell sort – Bubble sort – Heap sort – Quick sort – Merge Sort
9. Linear Search and Binary Search Techniques
10. Implementation of AVL Trees
11. Implementation of B Trees and B+ Trees
12. Implementation of Binomial Heaps and Fibonacci Heaps
13. Representation of Graphs – Breadth-first Search – Depth-first Search
14. Implementation of Kruskal’s, Prim’s, Sollin’s Algorithm
L T P C
0 0 3 2
1. Simple C++ programs to understand the concepts of user defined types
(classes) and predefined objects.
2. Implementation of String class in C++ (length, reverse, uppercase and
lowercase, copy operations)
3. Implementation of C++ application to demonstrate explicit handling of this
pointer, copy constructor, constant members in a class, static member
functions and function pointers for member functions
representation of complex numbers, simplification of fractional expressions
etc.,)
5. C++ applications for demonstrating compile time and run time polymorphism
6. Implementation of class and function templates in C++.
7. Implementation of Multi-level Inheritance in C++.
8. Simple Java applications to demonstrate default inheritance, pre-defined
classes and packages
9. Implementation of Java applications to demonstrate reflection for method
invocation
10. Simple Java applications using Threads and Simple Networking applications
using Sockets
11. Implement multi-threaded echo server and a corresponding GUI client using
sockets in Java.
12. Develop a RMI application to call a remote method to retrieve a video content.
13. Develop a java application to the retrieve the student’s records from the
database.
14. Design a simple Web page using DHTML and CSS.
15. Design a student course registration form with appropriate validation. Use
DOM objects and Java Script.
CS207 DATABASE MANAGEMENT SYSTEMS LABORATORY L T P C
0 0 3 2
1. Experiment with SQL basics and commands
• Understanding Database Schema and Table Definition
• Create a Database, ALTER TABLE, Updating the Table, Dropping a
Table
• Data Definition Language commands
• Data Manipulation Language and transaction control commands
• Adding, inserting, editing and deleting records in the database,
generating SQL Queries and Sub Queries, Retrieval of data, Printing of
• Modifying the Structure of Tables,
• Use of logical operators, BETWEEN AND, IN Function, LIKE
operator in queries
• Implement the concept of JOINS with Single Table and Multi Tables
• Use of built-in functions, ordering and CONCAT operations,
AGGREGATE and GROUPING functions, SET operations
3. Understanding E-R diagrams, entities, relationships and mapping constraints
4. Implement the concept of Indexes and Views
5. Experiment with Normalization
6. Database Connectivity with front end tools and Embedded SQLs
7. Use of Forms and Reports Generation
8. Use of Control Structures, Procedures, Functions and View in PL/SQL
9. Implement Triggers and Cursors
10. Experiment on Networked Databases
12. Database Design and Implementation (Case Study)
NATIONAL INSTITUTE OF TECHNOLOGY NAGALAND
CHUMUKEDIMA, DIMAPUR – 797 103.
B.Tech Degree Programme
Curriculum
Regulations – 2012
Bachelor of Technology in Computer Science and Engineering
Semester V
Course
Code
CS301
Course Title
L
T
P
C
Computer Graphics
3
0
0
3
CS302
Digital Signal Processing Fundamentals
3
0
0
3
CS303
Formal Languages and Automata Theory
3
0
0
3
CS304
Computer Networks
3
0
0
3
CS903
Data Mining and Data Warehousing
3
0
0
3
CS910
XML and Web Services
3
0
0
3
CS305
Computer Graphics Laboratory
0
0
3
2
CS306
Digital Signal Processing Laboratory
0
0
3
2
CS307
Computer Networks Laboratory
0
0
3
2
TOTAL
18 0
9 24
CS301 COMPUTER GRAPHICS
L T P C
3 0 0 3
2D PRIMITIVES
9
Coordinate Systems - Elements of pictures created in computer graphics – Graphics
input primitives and devices – OpenGL basic Graphics primitives – Output
primitives – Line, Circle and Ellipse drawing algorithms – Attributes of output
primitives – Line drawings in OpenGL – Display Technologies
2D GEOMETRIC TRANSFORMATIONS
9
2D Viewing – Window-Viewport Transformation – Two dimensional Geometric
transformations – Line, Polygon, Curve and Text clipping algorithms – 2D
Geometric Transformations using OpenGL
3D GRAPHICS
9
Parallel and Perspective projections – Three dimensional object representation –
Polygons, Curved lines, Splines, Quadric Surfaces – Visualization of data sets – 3D
Transformations – Viewing – Visible surface identification – Color Models –
Graphics Programming
MULTIMEDIA BASICS
9
Introduction and Definitions – Applications – Elements – Animations –
Compression – Types of Compressions: Lossless – Lossy – Video compression –
Image Compression – Audio compression – Media Representation and File format –
Multimedia data structures: KD Trees – R trees
MULTIMEDIA AUTHORING AND APPLICATIONS
9
Creating interactive multimedia – Multimedia Authoring Systems – Video on
Demand – Virtual Reality – Augmented Reality – Content based retrieval –
Multimedia for portable devices
Total: 45 Periods
TEXT BOOKS
1. Donald D. Hearn, M. Pauline Baker and Warren Carithers, “Computer
Graphics with OpenGL”, Fourth Edition, Pearson Education, 2010.
2. Ze-Nian Li and Mark S.Drew, “Fundamentals of Multimedia”, First Edition,
Pearson Education, 2007.
REFERENCE BOOKS
1. F.S.Hill and Stephen M Kelley, “Computer Graphics using OPENGL”, Third
Edition, Prentice Hall, 2007.
2. Prabhat K Andleigh, Kiran Thakrar, “Multimedia Systems Design”, First
Edition, PHI, 2007.
3. Ralf Steinmetz and Klara, “Multimedia Computing, Communications and
Applications”, Pearson Education, 2004.
4. Peter Shirley, “Fundamentals of Computer Graphics”, Third Edition, A K
Peters, 2009.
CS302 DIGITAL SIGNAL PROCESSING FUNDAMENTALS
L T P C
3 0 0 3
INTRODUCTION
9
Classification of Systems: Continuous, discrete, linear, causal, stable, dynamic,
recursive, time variance; classification of signals: continuous and discrete, energy and
power; mathematical representation of signals; spectral density; sampling techniques,
quantization, quantization error, Nyquist rate, aliasing effect. Digital signal
representation.
DISCRETE TIME SYSTEM ANALYSIS
9
Z-transform and its properties, inverse z-transforms; difference equation – Solution
by z-transform, application to discrete systems - Stability analysis, frequency
response – Convolution – Introduction to Fourier Transform– Discrete time Fourier
transform.
DISCRETE FOURIER TRANSFORM & COMPUTATION
9
DFT properties, magnitude and phase representation - Computation of DFT using
FFT algorithm – DIT & DIF - FFT using radix 2 – Butterfly structure
DESIGN OF DIGITAL FILTERS
9
FIR & IIR filter realization – Parallel & cascade forms. FIR design: Windowing
Techniques – Need and choice of windows – Linear phase characteristics. IIR design:
Analog filter design - Butterworth and Chebyshev approximations; digital design
using impulse invariant and bilinear transformation - Warping, prewarping Frequency transformation.
DIGITAL SIGNAL PROCESSORS
9
Introduction – Architecture of one DSP processor– Features – Addressing Formats –
Functional modes - Introduction to Commercial Processors
TOTAL: 45 PERIODS
TEXT BOOKS
1.
J.G. Proakis and D.G. Manolakis, ‘Digital Signal Processing Principles,
Algorithms and Applications’, Pearson Education, New Delhi, 2009.
2.
S.K. Mitra, ‘Digital Signal Processing – A Computer Based Approach’, Tata
McGraw Hill, New Delhi, 2006.
REFERENCES
1.
Alan V. Oppenheim, Ronald W. Schafer and John R. Buck, ‘Discrete – Time
Signal Processing’, Pearson Education, New Delhi, 2003.
2.
Emmanuel C Ifeachor and Barrie W Jervis ,”Digital Signal Processing – A
Practical approach” Pearson Education, Second edition, 2002.
3.
B. Venkataramani, M. Bhaskar, ‘Digital Signal Processors, Architecture,
Programming and Applications’, Tata McGraw Hill, New Delhi, 2003.
4.
S.Salivahanan, A.Vallavaraj, C.Gnanapriya, "Digital Signal Processing", Tata
McGraw-Hill Publishing, 2000.
CS303 FORMAL LANGUAGES AND AUTOMATA THEORY
L T P C
3 0 0 3
REGULAR LANGUAGES
9
Finite Automata (FA) – Deterministic Finite Automata (DFA) – Non-deterministic
Finite Automata (NFA) – Finite Automata with Epsilon transitions - Regular
Expression – FA and Regular Expressions – Pumping lemma for Regular
languages - Equivalence and minimization of Finite Automata
CONTEXT FREE LANGUAGES
9
Context-Free Grammar (CFG) – Parse Trees – Ambiguity in grammars and
languages – Equivalence of Parse trees and derivation - Normal forms for CFG Definition of the Pushdown automata – Languages of a Pushdown Automata –
Equivalence of Pushdown automata and CFG – Pumping lemma for CFL
CLOSURE PROPERTIES AND TURING MACHINES
9
Closure properties of Regular Sets: Complement and Intersection – Closure
properties of CFL: Union, Concatenation, Kleene Closure, Intersection and
Complement – Turing Machines – Language of a Turing machine – Turing
machine as a computing device - Various techniques for construction of TMs –
Equivalence of one tape and multi-tape Turing machines - Halting Problem Stack Automata
UNDECIDABILITY
9
A language that is not Recursively Enumerable (RE) – An undecidable problem
that is RE – Undecidable problems about Turing Machine – Rice theorem for
Recursive and Recursively enumerable languages – Post’s Correspondence
Problem
RECENT TRENDS & APPLICATIONS
9
Matrix grammar – Programmed grammar – Random context grammar – Regular
Control grammar – Lindenmayer systems – A glance on DNA computing and
Membrane computing
Total: 45 Periods
TEXT BOOKS
1. E.Hopcroft and J.D.Ullman, "Introduction to Automata Theory", Languages
and Computation, Second Edition, Pearson Education, 2003.
2. Peter Linz, "An Introduction to Formal Language and Automata", Narosa
Pub. House, Reprint 2000.
REFERENCE BOOKS
1. H.R. Lewis and C.H. Papadimitriou, “Elements of the theory of
Computation”, Second Edition, Pearson Education, 2003.
2. J. Martin, “Introduction to Languages and the Theory of Computation”, Third
Edition, Tata Mc Graw Hill, 2003.
3. Micheal Sipser, “Introduction of the Theory and Computation”, Thomson
Brokecole, 1997.
CS304 COMPUTER NETWORKS
L T P C
3 0 0 3
INTRODUCTION
9
Layering and protocols – Internet Architecture – Networking devices – Modems,
Routers, Switches, Gateways – Needs – Data Communication concepts – Data
Transmission – Transmission media – Signal encoding techniques – Multiplexing
– Spread spectrum and Channel access techniques – TDM – FDM
9
Link layer services – Framing – Error control – Flow control – Media access control
– Ethernet – CSMA/CD – Token Ring – FDDI – Wireless LANs – CSMA/CA
NETWORK ROUTING
9
Circuit switching – Packet switching – Virtual circuit switching – Routing – IP –
Global Address – Datagram Forwarding – Subnetting – CIDR – ARP – DHCP –
RIP – OSPF – BGP – ICMP – IPv6 – Multicasting – PIM
TRANSPORT LAYER
9
Overview of Transport layer – UDP – TCP – Reliable byte stream – Connection
management – Flow control – Retransmission – Congestion control – Congestion
avoidance
APPLICATION LAYER
9
Principles of Application Layer Protocols – Web and HTTP – FTP – Telnet –
Electronic Mail (SMTP, POP3, IMAP, MIME) – DNS – SNMP
Total: 45 Periods
TEXT BOOKS
1. William Stallings, “Data and Computer Communications”, Ninth Edition,
Pearson Education, 2010.
2. Larry L. Peterson, Bruce S. Davie, “Computer Networks: A Systems
Approach”, Fifth Edition, Morgan Kaufmann Publishers, 2010.
REFERENCE BOOKS
1. Andrew S. Tanenbaum, “Computer Networks”, Fourth Edition, Pearson
Education, 2003.
2. James F. Kurose, Keith W. Ross, “Computer Networking - A Top-Down
Approach Featuring the Internet”, Fifth Edition, Pearson Education, 2009.
3. Nader. F. Mir, “Computer and Communication Networks”, Pearson Prentice
Hall Publishers, 2010.
CS903 DATA WAREHOUSING AND DATA MINING
L T P C
3 0 0 3
INTRODUCTION TO DATA WAREHOUSING
9
Evolution of Decision Support Systems - Data warehousing Components – Building a
Data warehouse - Data Warehouse and DBMS Data Marts - Metadata Multidimensional Data Model - OLAP Vs OLTP - OLAP operations - Data Cubes Schemas for Multidimensional Database: Stars, Snowflakes and Fact Constellations
DATA WAREHOUSE PROCESS AND ARCHITECTURE
9
Types of OLAP Servers, 3–Tier Data Warehouse Architecture - Distributed And
Virtual Data Warehouses - Data Warehouse Implementation, Tuning and Testing of
Data Warehouse - Data Staging (ETL) Design and Development - Data Warehouse
Visualization - Data Warehouse Deployment, Maintenance, Growth - Business
Intelligence Overview- Data Warehousing and Business Intelligence Trends Business Applications - Tools - SAS
INTRODUCTION TO DATA MINING
9
KDD Vs. Data Mining - Stages of the Data Mining Process - Task primitives - Data
Mining Techniques - Data Mining Knowledge Representation – Data Mining Query
Languages - Integration of a Data Mining System with a Data Warehouse – Issues:
Data preprocessing – Data cleaning - Data transformation - Feature selection Dimensionality reduction - Discretization - Mining Frequent Patterns - Association Correlation
CLASSIFICATION AND CLUSTERING
9
Decision Tree Induction - Bayesian Classification – Rule Based Classification –
Classification by Back propagation – Support Vector Machines – Associative
Classification – Lazy Learners – Clustering - Partitioning methods - k-means Hierarchical Methods - distance-based agglomerative and divisible clustering Density-based Methods – expectation maximization - Grid Based Methods – ModelBased Clustering Methods – Constraint-based Cluster Analysis – Outlier Analysis
DATA MINING SOFTWARE AND APPLICATIONS
9
Mining complex data objects - Spatial Databases - Temporal Databases - Multimedia
Databases - Time series and Sequence data; Text Mining – Graph Mining - Web
Mining - Application and trends in Data Mining
TOTAL: 45 PERIODS
TEXT BOOKS
1. Jiawei Han and Micheline Kamber, "Data Mining: Concepts and Techniques",
Morgan Kaufmann Publishers, third edition, 2011, ISBN: 1558604898.
2. Alex Berson and Stephen J. Smith, “Data Warehousing, Data Mining &
OLAP”, Tata McGraw-Hill Edition, Tenth Reprint 2007.
3. G. K. Gupta, “Introduction to Data Mining with Case Studies”, Easter
Economy Edition, Prentice Hall of India, 2006.
REFERENCE BOOKS
1. Mehmed kantardzic, “Data Mining Concepts, Models, Methods and
Algorithms”, Wiley Interscience, 2003.
2. Ian Witten, Eibe Frank, "Data Mining; Practical Machine Learning Tools and
Techniques", third edition, Morgan Kaufmann, 2011.
3. George M Marakas, "Modern Data Warehousing, Mining and Visualization",
Prentice Hall, 2003.
CS910 XML AND WEB SERVICES
L T P C
3 0 0 3
XML Fundamentals
9
Basics – XML Tree – Syntax – XML Elements – Attributes – Validation – XML
Viewing – CSS – XSLT – XML Namespace – CDATA - XML Parser – DOM –
XML to HTML – Applications
XML Technology
9
XML Essentials – Schema – XML Design Techniques – Security – Transformation –
Query – XML Components – XML Processing – XML Publishing
Web Services Fundamentals
9
RPC concepts – RMI Implementation – Concepts and Use of Web Services - Web
Service Architecture - JAX-RPC – XML-RPC - Web Services Platform: SOAP –
UDDI – WSDL – Simple Web Service Creation – Deployment
Web Services Development and Deployment
9
XML Web Services Standards – AXIS2 Framework - SOAP Messages – Life Cycle
of a Message – Message Exchange Patterns – Handling of SOAP Messages - AXIS2
Clients and Services – SOAP Messages with Attachments – Applications
DEPLOYMENT PLATFORM ARCHITECTURAL MODELS
9
AXIS2 Requirements – Architecture – Information Model – XML Processing Model
– SOAP Message Processing Model – Deployment Model – Client Communication
with Web Services – Transports – Code Generation – Data Binding – Modules –
Handlers – SOAP Faults
Total: 45 Periods
Text Books
1. Gustavo Alonso, Fabio Casati, Harumi Kuno, Vijay Machiraju, “Web Services
- Concepts, Architectures and Applications”, Springer Verlag, 2010.
2. Ron Schmelzer, “XML and Web Services Unleashed”, Sams, 2002.
Reference Books
1. Frank. P. Coyle, XML, Web Services And The Data Revolution, Pearson
Education, 2002.
2. Eric Newcomer, “Understanding Web Services: XML, WSDL, SOAP and
3. David A. Chappell and Tyler Jewell, “Java Web services”, O'Reilly Media,
Inc., 2002.
4. Anne Thomas Mannes, “Web Services: A Manger’s Guide”, Addison Wesley
Professional, 2003.
5. http://www.w3schools.com
6. http://www.w3.org
7. http://axis.apache.org/axis2/java/core/docs/toc.html
CS305 COMPUTER GRAPHICS LABORATORY
L T P C
0 0 3 2
Implement the exercises from 1 to 4 using C/OpenGL/Java
1. Implementation of Algorithms for drawing 2D Primitives Line (DDA, Bresenham) – all slopes
Circle (Midpoint)
2. 2D Geometric transformations Translation
Rotation
Scaling
Reflection
Shear
Window-Viewport
3. Composite 2D Transformations
4. Liang - Barsky Line Clipping
Implement the exercises from 5 to 7 using OpenGL
5. 3D Transformations - Translation, Rotation, Scaling
6. 3D Projections – Parallel, Perspective
7. Creating 3D Scenes
8. Compression Algorithms - To implement text and image compression algorithms
9. Image Editing and Manipulation - Basic Operations on image using any image
editing software, Creating gif animated images, Image optimization
10. 2D Animation – To create Interactive animation using any authoring tool
CS306 DIGITAL SIGNAL PROCESSING LABORATORY
L T P C
0 0 3 2
USING DSP TRAINER
1. Study of various Addressing modes of DSP with simple programming
examples using TMS320C5X, TMS320C67XX, ADSP 21XXX, BF53X
2. Implementation of Linear and Circular Convolution
3. Sampling of Input Signal and Display
4. Waveform Generation
5. Calculation of FFT
6. Implementation of FIR and IIR Filters
USING MATLAB
1. Linear and Circular convolution of two sequences
2. Noise cancellation of Signal
3. Long Sequence convolution (Overlap add & save method)
4. Design of FIR Filters
5. Design of IIR Filters
6. Calculation of FFT of a Signal
CS307 COMPUTER NETWORKS LABORATORY
L T P C
0 0 3 2
1. Familiarization with configuring and installing a LAN
2. Learn to use basic networking commands like ipconfig, ping, arp, rarp, TCP
Dump, Netstat, TraceRoute
3. Simple Chat Program using TCP Sockets
4. Simulation of HTTP Protocol using TCP Sockets
5. Simulation of Sliding Window Protocol using TCP Sockets
6. Simulation of DNS using UDP Sockets
7. Simulation of Ping using Raw Sockets
8. Learn to use commands like Develop applications and understand the behavior
of TCP Options
9. Study of TCP/UDP performance using OPNET tool
10. Performance comparison of MAC protocols using OPNET tool
11. Performance comparison of Routing protocols using OPNET tool
12. Study and configure functionalities of a router and switches (or by simulation)
Semester VII (Regulations 2008)
Course
Code
CS401
L
T
P
C
Mobile and Pervasive Computing
3
0
0
3
EE956
Wireless and Sensor Networks
3
0
0
3
CS911
Cloud Computing and Services
3
0
0
3
CS961
Cryptography and Network Security
3
0
0
3
CS403
Mobile and Pervasive Computing Laboratory
0
0
3
2
CS404
Project Phase - I
0
0
8
4
Course Title
TOTAL
12 0 11 18
CS401 MOBILE AND PERVASIVE COMPUTING
L T P
C
3 0 0
3
MOBILE NETWORKS
9
Overview of Wireless Communication – Media Access Control – SDMA, FDMA,
TDMA, CDMA – Generation of Cellular Wireless Networks – GSM – Architecture
– Protocols – Connection Establishment – Frequency Allocation – Localization –
Handover – Security – GPRS
WIRELESS NETWORKS
9
Wireless LANs and PANs – IEEE 802.11 Standard – Architecture – Services –
Network – Emerging technologies – Piconet - Bluetooth, Wi-Fi, WiMAX, 3G, WATM
– Mobile IP protocols – WAP push architecture – WML Scripts and Applications
ROUTING AND MOBILITY MANAGEMENT
9
Mobile IP – DHCP – Handoff in wireless mobile networks – Handoff schemes –
Location management in cellular networks – Mobility models – Location and Tracking
management schemes – Time, Movement, Profile and Distance based update strategies
– ALI technologies - Routing in Mobile Ad-hoc Networks - TCP Improvements - TCP
over 2.5/3G
TRANSPORT AND APPLICATION LAYERS
9
Mobile TCP – WAP – Architecture – WWW Programming Model – WDP – WTLS –
WTP – WSP – WAE – WTA Architecture – WML – WML Scripts
PERVASIVE COMPUTING
9
Pervasive computing Principles – Characteristics – Infrastructure –Interaction
Transparency – Context Aware – Applications – Device Technology – Hardware,
Human-machine Interfaces, Biometrics, and Operating systems – Device Connectivity
– Protocols, Security, and Device Management – Pervasive Web Application
architecture –Access from PCs and PDAs – Access via WAP
Total: 45 Periods
TEXT BOOKS
1. Jochen Schiller, “Mobile Communications”, Second Edition, Pearson, 2009.
2. Frank Adelstein, Sandeep KS Gupta, Golden Richard, “Fundamentals of Mobile
and Pervasive Computing”, Tata McGraw-Hill Edition, 2005.
3. Asoke Talukder, Hasan Ahmed, Rupa Yavagal, “Mobile Computing:
Technology, Applications and Services Creation”, Second Edition, TMH, 2010.
4. Jochen Burkhardt, “Pervasive Computing: Technology and Architecture of
Mobile Internet Applications”, Addison- Wesley Professional; 3rd edition, 2007.
REFERENCE BOOKS
1. William Stallings, “Wireless Communication and Networks”, Pearson
Education, 2009.
2. Uwe Hansmann, Lothar Merk, Martin S. Nicklons and Thomas Stober,
Principles of Mobile Computing”, Springer, New York, 2003.
3. Ivan Stojmenovic, “Handbook
Computing”, Wiley, 2006.
of
Wireless
Networks
and
Mobile
4. Stefan Poslad, “Ubiquitous Computing: Smart Devices, Environments and
Interactions”, Wiley, 2009.
EE956 WIRELESS AND SENSOR NETWORKS
Fundamentals
L T P C
3 0 0 3
9
Introduction to Sensor Networks - Unique constraints and challenges - Advantage of
Sensor Networks - Applications of Sensor Networks - Cellular and Mobile Adhoc
NETworks (MANETs) - Enabling technologies for Wireless Sensor Networks – Key
Definitions of Sensor Networks – Collaborative Processing.
Architectures
9
Sensor Node Hardware and Network Architecture: Single-Node Architecture Hardware components & Design constraints - Operating Systems and Execution
Environments - Network architecture - Physical Layer and Transceiver Design
Considerations - Optimization goals and figures of merit - Design principles for
WSNs - Service interfaces of WSNs - Gateway concepts.
Networking Sensors
9
Sensor Management Network Protocols - MAC Protocols - Issues in designing MAC
Protocol for WSNs - Classification of MAC Protocols - S-MAC Protocol - B-MAC
protocol - IEEE 802.15.4 standard - Zig Bee - Dissemination protocol for large sensor
network - Routing protocols: Issues in designing routing protocols - Classification of
routing protocols - Energy-efficient routing - Unicast - Broadcast - Multicast Geographic routing.
Infrastructure Establishment
9
Topology control - Clustering - Time Synchronization - Deployment and
Configuration: Localization and positioning - Coverage and connectivity - Single-hop
and multi-hop localization - Self configuring localization systems - Roles of Sensor
Nodes and Utilities – Sensor Tasking and Control.
Sensor Network Platforms and Tools
9
Data Storage and Manipulation: Data Centric and Content Based Routing Compression Technologies for WSN - Data Aggregation Technique - Applications:
Detecting unauthorized activity using a Sensor Network - WSN for Habitat
Monitoring - Operating Systems for Wireless Sensor Networks - Introduction to
TinyOS and nesC - Berkeley Motes - Programming Challenges - Node Level
Software Platforms - Node Level Simulators - State-centric Programming.
Total: 45 Periods
Text Books
1. Holger Kerl, Andreas Willig, “Protocols and Architectures for Wireless Sensor
Networks”, John Wiley & Sons, 2005.
2. Feng Zhao, Leonidas Guibas, “Wireless Sensor Networks: An Information
Processing Approach”, Elsevier, 2007.
Reference Books
1. Anna Hac, “Wireless Sensor Network Designs”, John Wiley, 2003.
2. Bhaskar Krishnamachari, “Networking Wireless Sensors”, Cambridge
University Press, 2005.
3. C. Siva Ram Murthy and B. S. Manoj, “Adhoc Wireless Networks:
Architectures and Protocols”, Prentice Hall, 2004.
4. Kazem Sohraby, Daniel Minoli and Taieb Znati, “Wireless Sensor Networks Technology, Protocols, and Applications”, John Wiley, 2007.
5. Wayne Tomasi, “Introduction to Data Communication and Networking”,
Pearson Education, 2007.
CS911 CLOUD COMPUTING AND SERVICES
INTRODUCTION
L T P C
3 0 0 3
9
Cloud definition, benefits, usage scenarios - History of Cloud Computing - Cloud
Architecture - Types of Clouds - Business Models around Clouds – Major Players in
Cloud Computing - Issues in Clouds - Eucalyptus - Nimbus - Open Nebula - Cloud
Sim
CLOUD SERVICES
9
Types of Cloud services: Software as a Service - Platform as a Service –
Infrastructure as a Service - Database as a Service - Monitoring as a Service –
Communication as a Service - Service providers - Google, Amazon, Microsoft Azure,
IBM, Sales Force
COLLABORATION USING CLOUD SERVICES
9
Email Communication over the Cloud - CRM Management - Project Management Event Management - Task Management – Calendar - Schedules - Word Processing –
Presentation – Spreadsheet - Databases – Desktop - Social Networks and Groupware
VIRTUALIZATION FOR CLOUD
9
Need for Virtualization – Pros and cons of Virtualization – Types of Virtualization –
System Vm, Process VM, Virtual Machine monitor – Virtual Machine properties Interpretation and binary translation, HLL VM - Hypervisors – Xen, KVM,
VMWare, Virtual Box - Hyper-V
SECURITY, STANDARDS AND APPLICATIONS
9
Security in Clouds: Cloud security challenges – Software as a Service Security,
Common Standards: The Open Cloud Consortium – The Distributed management
Task Force – Standards for application Developers – Standards for Messaging –
Standards for Security, End user access to cloud computing, Mobile Internet devices
and the Cloud
Total: 45 Periods
TEXT BOOKS
1. John Rittinghouse & James Ransome, "Cloud Computing, Implementation,
Management and Strategy", CRC Press, 2010.
2. Michael Miller, "Cloud Computing: Web-Based Applications That Change the
Way You Work and Collaborate". Que Publishing, August 2008.
3. James E Smith, Ravi Nair, "Virtual Machines", Morgan Kaufmann Publishers,
2006.
REFERENCE BOOKS
1. David E.Y. Sarna, "Implementing and Developing Cloud Application", CRC press
2011.
2. Lee Badger, Tim Grance, Robert Patt-Corner, Jeff Voas, " Draft cloud computing
synopsis and recommendations", NIST, May 2011.
3. Anthony T Velte, Toby J Velte, Robert Elsenpeter, "Cloud Computing : A
Practical Approach", Tata McGraw-Hill 2010.
4. Haley Beard, "Best Practices for Managing and Measuring Processes for Ondemand Computing, Applications and Data Centers in the Cloud with SLAs",
Emereo Pty Limited, July 2008.
5. G.J.Popek, R.P. Goldberg, "Formal Requirements for Virtualizable Third
Generation Architectures", Communications of the ACM, No.7 Vol.17, July 1974.
CS961 CRYPTOGRAPHY AND NETWORK SECURITY
L T P C
3 0 0 3
INTRODUCTION
9
Security trends – Attacks and Services – Classical Crypto Systems – Different
types of Ciphers – LFSR sequences – Basic Number theory – Congruence
Modulo – Chinese Remainder theorem – Modular exponentiation – Fermat and
Euler's theorem – Secure programs – Non-malicious Program Errors – Viruses ––
Controls against Program Threat – Control of Access to General Objects – User
Authentication – Good Coding Practices – Open Web Application Security Project
ENCRYPTION TECHNIQUES
9
Simple DES – Differential Crypto Analysis – DES – Modes of operation –
Triple DES – AES – RC5, RC4 – RSA – Attacks – Primality test – Factoring
KEY EXCHANGE AND AUTHENTICATION TECHNIQUES
9
Discrete Logarithms – Computing discrete logarithms – Diffie–Hellman key
exchange – Elliptic curve cryptography Key exchange – Elgamal Public Key
Cryptosystems – Message Authentication codes – Hash functions – Hash
algorithms – Secure Hash – Birthday attacks – MD5 – Authentication protocols
– Digital signatures – RSA, DSA
NETWORK SECURITY AND STANDARDS
9
Public Key Infrastructure – Kerberos – X.509 – IPSec – Virtual Private Networks –
E–Mail Security – PGP and PEM – Web Security – Secured DNS – SSL, TLS and
SET – CoBIT Framework – Compliances – Credit Card Applications – GLBA –
Standards – ISO 27000
OPERATING SYSTEMS AND DATABASE SECURITY
9
Trusted Operating systems – Security models – Designing trusted OS – Assurance
– Database Security – Multi-level databases – Multi-level security
Total: 45 Periods
TEXT BOOKS
1. Wade Trappe, Lawrence C Washington, “Introduction to Cryptography with
Coding Theory”, Second Edition, Pearson Education, 2007.
2.
William Stallings, “Cryptography and Network Security Principles and
Practice”, Fifth Edition, Prentice Hall, 2010.
3.
Matt Bishop, “Computer Security: Art and Science”, Pearson Education Inc.,
2003.
REFERENCE BOOKS
1.
Charles Pfleeger, Shari Lawrence Pfleeger, Devin N Paul, “Security in
Coding”, Pearson, 2007.
2.
Wenbo Mao, “Modern Cryptography Theory and Practice”, Pearson, 2004.
CS403 MOBILE AND PERVASIVE COMPUTING LABORATORY L T P C
0 0 3 2
1. Develop mobile applications using J2ME environment and test it
2. Simulation of applications to access web sites using Microsoft Windows
Mobile .net environment
3. Implementation of playing games and photo sharing applications using
BREW (Binary Runtime Environment for Wireless Toolkit)
4. Simulation of Infotainment (news, weather forecasts etc) using WAP, WML
Scripts
5. Pervasive devices connectivity – Using of server side programming in Java
6. Write web application via WAP phones
7. Develop simple applications for Android mobile devices
8. Develop an android application to access a Web service
```
Related documents | 10,857 | 44,124 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.55097 |
https://metanumbers.com/582493 | 1,624,162,550,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487655418.58/warc/CC-MAIN-20210620024206-20210620054206-00006.warc.gz | 360,748,379 | 10,922 | ## 582493
582,493 (five hundred eighty-two thousand four hundred ninety-three) is an odd six-digits composite number following 582492 and preceding 582494. In scientific notation, it is written as 5.82493 × 105. The sum of its digits is 31. It has a total of 2 prime factors and 4 positive divisors. There are 580,320 positive integers (up to 582493) that are relatively prime to 582493.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 6
• Sum of Digits 31
• Digital Root 4
## Name
Short name 582 thousand 493 five hundred eighty-two thousand four hundred ninety-three
## Notation
Scientific notation 5.82493 × 105 582.493 × 103
## Prime Factorization of 582493
Prime Factorization 313 × 1861
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 582493 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 582,493 is 313 × 1861. Since it has a total of 2 prime factors, 582,493 is a composite number.
## Divisors of 582493
4 divisors
Even divisors 0 4 4 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 584668 Sum of all the positive divisors of n s(n) 2175 Sum of the proper positive divisors of n A(n) 146167 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 763.212 Returns the nth root of the product of n divisors H(n) 3.98512 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 582,493 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 582,493) is 584,668, the average is 146,167.
## Other Arithmetic Functions (n = 582493)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 580320 Total number of positive integers not greater than n that are coprime to n λ(n) 48360 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 47658 Total number of primes less than or equal to n r2(n) 16 The number of ways n can be represented as the sum of 2 squares
There are 580,320 positive integers (less than 582,493) that are coprime with 582,493. And there are approximately 47,658 prime numbers less than or equal to 582,493.
## Divisibility of 582493
m n mod m 2 3 4 5 6 7 8 9 1 1 1 3 1 2 5 4
582,493 is not divisible by any number less than or equal to 9.
## Classification of 582493
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (582493)
Base System Value
2 Binary 10001110001101011101
3 Ternary 1002121000211
4 Quaternary 2032031131
5 Quinary 122114433
6 Senary 20252421
8 Octal 2161535
10 Decimal 582493
12 Duodecimal 241111
20 Vigesimal 3cg4d
36 Base36 chgd
## Basic calculations (n = 582493)
### Multiplication
n×i
n×2 1164986 1747479 2329972 2912465
### Division
ni
n⁄2 291246 194164 145623 116499
### Exponentiation
ni
n2 339298095049 197638765279377157 115123197303880238312401 67058456567129111655305395693
### Nth Root
i√n
2√n 763.212 83.5148 27.6263 14.2252
## 582493 as geometric shapes
### Circle
Diameter 1.16499e+06 3.65991e+06 1.06594e+12
### Sphere
Volume 8.27867e+17 4.26375e+12 3.65991e+06
### Square
Length = n
Perimeter 2.32997e+06 3.39298e+11 823770
### Cube
Length = n
Surface area 2.03579e+12 1.97639e+17 1.00891e+06
### Equilateral Triangle
Length = n
Perimeter 1.74748e+06 1.4692e+11 504454
### Triangular Pyramid
Length = n
Surface area 5.87682e+11 2.3292e+16 475604
## Cryptographic Hash Functions
md5 74e3afbb6e39c77a5d4a1c9aea16ae3d 7313c05e42b680231b6c6644558a6f360106e7c1 6617d3bb9128608cebb3b6cf2374b40439e7cda82d4c7aac283e2aa6ed3d353a c93c19d46dc2556a8b3602cbf199507943168e82433bcc0aa32c6a1e3ec58788b3538261dac569061edab2af430c9228646dbc5074732ab1c83a4f448bc77d67 79c6aa5b61677498c66d687fe2253c3e7ec61817 | 1,479 | 4,272 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.815366 |
https://virtualpsychcentre.com/characteristics-of-visible-waves/ | 1,726,487,122,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.32/warc/CC-MAIN-20240916112213-20240916142213-00556.warc.gz | 575,859,988 | 12,293 | ## What are the properties of visible waves?
Visible light makes up a large proportion of the electromagnetic spectrum and has a wide range of frequencies. Visible light is detected by the eye and makes up a limited range of frequencies in the electromagnetic spectrum. Visible light is a longitudinal wave that travels at the speed of light.
## What are characteristics of visible wavelengths?
WAVELENGTHS OF VISIBLE LIGHT
As the full spectrum of visible light travels through a prism, the wavelengths separate into the colors of the rainbow because each color is a different wavelength. Violet has the shortest wavelength, at around 380 nanometers, and red has the longest wavelength, at around 700 nanometers.
## What are the characteristics of visible light radiation?
Light waves have wavelengths between about 400 and 700 nanometers (4,000 to 7,000 angstroms). Our eyes perceive different wavelengths of light as the rainbow hues of colors. Red light has relatively long waves, around 700 nm long. Blue and purple light have short waves, around 400 nm.
## What are 3 characteristics of light waves?
Light has the usual attributes of waves, namely frequency, wavelength and speed. The speed of light (c) in a vacuum has been measured at 3.00 x 108 ms–1 .
## Which is a difference between visible light and radio waves?
In that section, it was pointed out that the only difference between radio waves, visible light and gamma rays is the energy of the photons. Radio waves have photons with the lowest energies. Microwaves have a little more energy than radio waves.
## What are the 5 main characteristics of light?
Photography is “writing with light.” So, let’s build our conversation about lighting on five fundamental characteristics of light: Direction, Intensity, Color, Contrast, and Hardness.
## What are the two characteristics of light wave?
Light has a wavelength and a frequency.
## Which of the following is not a characteristic of visible light?
Answer and Explanation: Frequency is not a characteristic of light as it does not change when light goes from one medium to another. Sun and stars are source of light.
## What are the characteristics of infrared?
Infrared waves have longer wavelengths than visible light and can pass through dense regions of gas and dust in space with less scattering and absorption. Thus, infrared energy can also reveal objects in the universe that cannot be seen in visible light using optical telescopes.
## What is the range of visible light wavelengths?
approximately 0.4 to 0.7 µm
The visible wavelengths cover a range from approximately 0.4 to 0.7 µm. The longest visible wavelength is red and the shortest is violet. Common wavelengths of what we perceive as particular colours from the visible portion of the spectrum are listed below.
## What makes the visible spectrum of light visible quizlet?
The visible spectrum is the portion of the electromagnetic spectrum that is visible to the human eye. Electromagnetic radiation in this range of wavelengths is called visible light or simply light.
## How visible rays are produced?
Relatively long radio waves are produced by electrical current flowing through huge broadcast antennas, while much shorter visible light waves are produced by the energy state fluctuations of negatively charged electrons within atoms.
## Is infrared light visible?
Infrared waves are a portion of the light spectrum that follows red. They have longer wavelengths than visible light, ranging from 700 nanometers to one millimeter. This renders them invisible to humans in almost all conditions.
## What are the three types of radio waves?
Radio waves have the longest wavelengths in the EM spectrum, according to NASA, ranging from about 0.04 inches (1 millimeter) to more than 62 miles (100 kilometers).
BandFrequency rangeWavelength range
High Frequency (HF)3 to 30 MHz10 to 100 m
Very High Frequency (VHF)30 to 300 MHz1 to 10 m
•
27 feb 2019
## What are examples of radio waves?
AM and FM Radio Broadcasting which involves transmitting sound to a wide audience. Radar is a detection system that uses radio waves to get information about objects. Bluetooth and wireless communication use radio waves to create connections between devices.
## What is visible light used for?
Visible light is the light we can see, so is used in photography and illumination. It is also used in fibre optic communications, where coded pulses of light travel through glass fibres from a source to a receiver.
## Is Bluetooth a radio wave?
Devices connected in a Bluetooth network communicate with each other using ultra-high frequency (UHF) radio waves. These are electromagnetic waves with frequencies around 2.4 gigahertz (2.4 billion waves per second).
## Do phones use radio waves?
Mobile phones communicate by transmitting radio waves through a network of fixed antennas called base stations. Radiofrequency waves are electromagnetic fields, and unlike ionizing radiation such as X-rays or gamma rays, can neither break chemical bonds nor cause ionization in the human body.
## What is a visible wave?
Visible light waves are the only wavelengths of the electromagnetic spectrum that humans can see. The different wavelengths of visible light are seen as the colors of the rainbow: red, orange, yellow, green, blue, indigo, and violet. | 1,080 | 5,356 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-38 | latest | en | 0.941519 |
https://sitemate.com/resources/articles/safety/ltifr-calculation/ | 1,713,978,152,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819668.74/warc/CC-MAIN-20240424143432-20240424173432-00215.warc.gz | 465,719,845 | 29,444 | # LTIFR Calculation: Here's how to calculate and use LTIFR
## LTIFR calculation formula
The formula for calculating lost time injury frequency rate is the number of lost time injuries multiplied by 1,000,000, divided by the employee total hours worked.
As you can see, there are just three parts of the LTIFR formula, two of which need to be reconciled on your end, and then the 1,000,000 figure which remains the same across all LTIFR calculations and is used to standardise the LTIRF 'score' into per million hours worked.
In terms of what constitutes a lost time injury, a lost time injury is any injury which results in a fatality, permanent disability or time lost from work (must be at least 24 hours, one day or one shift).
## LTIFR calculation examples
Almost all companies today - especially those in heavy industries where safety has been a concern - calculate their safety key performance indicators on a regular basis.
These calculations have become important for internal purposes, and have become increasingly important for external purposes too, with authorities and other stakeholders using the indicators as a tool for making important decisions themselves.
Below is an LTIFR calculation example which helps to illustrate exactly what an LTIFR calculation does look like.
In this example, the company calculating their LTIFR has these numbers:
• 10 lost time injuries
• 500,000 hours worked on the project
From these two numbers, which would have been documented and collected during the phase of works, the company can calculate their LTIFR:
(10 lost time injuries x 1,000,000) / 500,000 = 20 lost time injuries per million hours worked
You can see some LTIFR industry averages here.
## Calculating your lost time injury frequency rate
Obviously you need to use the above LTIFR calculation formula to calculate LTIFR properly, but the tool you use for calculating your lost time injury rate is up to you - and can have a big impact on the amount of time and effort required to get these safety metrics into a safety scorecard or other useful format.
Some companies still use paper log forms and other sheets, and then reconcile the data before placing it into a calculator.
Other companies rely on updating a tracking spreadsheet with data from the field at regular intervals.
And some companies leave the heavy lifting to real-time safety softwares which aggregate all of this safety data in real-time until it needs to be calculated.
## Using your safety calculation results effectively
Calculating LTIFR isn't supposed to be the last step in the process. In fact, it's about the second step of the process after tracking and compiling the individual results.
The idea behind any measurement - and the purpose of measuring - is to understand how you are performing and make well-informed decisions about how to improve.
Once you have your LTIFR score, you, management or your safety teams will want to use that information along with other leading and lagging indicators to create actions and next steps for making positive change.
A bad lost time injury rate might expedite the corrective actions or require some additional resources or external help, but a good result should also create further analysis and action.
If specific projects or teams are performing particularly well and keeping that score low, then you will want to pull high performing or positive safety culture elements from the site to other sites.
There should be a constant learning process and cross-pollination of positive safety behaviours, as well as the intentions to rectify the bad ones.
People in 100+ countries use this safety management system to improve how they track and calculate safety KPIs. | 747 | 3,724 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-18 | latest | en | 0.95389 |
https://grandyang.com/leetcode/861/ | 1,723,665,800,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641121834.91/warc/CC-MAIN-20240814190250-20240814220250-00711.warc.gz | 223,326,825 | 34,098 | # 861. Score After Flipping Matrix
We have a two dimensional matrix `A` where each value is `0` or `1`.
A move consists of choosing any row or column, and toggling each value in that row or column: changing all `0`s to `1`s, and all `1`s to `0`s.
After making any number of moves, every row of this matrix is interpreted as a binary number, and the score of the matrix is the sum of these numbers.
Return the highest possible score.
Example 1:
``````Input: [[0,0,1,1],[1,0,1,0],[1,1,0,0]]
Output: 39
Explanation: Toggled to [[1,1,1,1],[1,0,0,1],[1,1,1,1]].
0b1111 + 0b1001 + 0b1111 = 15 + 9 + 15 = 39
``````
Note:
1. `1 <= A.length <= 20`
2. `1 <= A[0].length <= 20`
3. `A[i][j]` is `0` or `1`.
``````class Solution {
public:
int matrixScore(vector<vector<int>>& A) {
int m = A.size(), n = A[0].size(), res = (1 << (n - 1)) * m;
for (int j = 1; j < n; ++j) {
int cnt = 0;
for (int i = 0; i < m; ++i) {
cnt += (A[i][j] == A[i][0]);
}
res += max(cnt, m - cnt) * (1 << (n - 1 - j));
}
return res;
}
};
``````
Github 同步地址:
https://github.com/grandyang/leetcode/issues/861
https://leetcode.com/problems/score-after-flipping-matrix/
https://leetcode.com/problems/score-after-flipping-matrix/discuss/143722/C%2B%2BJavaPython-Easy-and-Concise
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Help us with donation | 525 | 1,342 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-33 | latest | en | 0.551158 |
https://tolstoy.newcastle.edu.au/R/e2/help/07/06/20047.html | 1,586,445,534,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371858664.82/warc/CC-MAIN-20200409122719-20200409153219-00537.warc.gz | 712,326,648 | 4,083 | # Re: [R] Matlab end operator
From: AA <aa2007r_at_gmail.com>
Date: Wed, 27 Jun 2007 17:18:04 -0400
Hi Markus, Christophe
I also use both matlab and R.
I agree with Christophe: you can define the 'end' functionality by nrow or length
also have a look at the following link that may be useful. http://mathesaurus.sourceforge.net/octave-r.html good luck
AA.
----- Original Message -----
From: "Christophe Pallier" <christophe_at_pallier.org> To: "Markus Loecher" <loecher_at_eden.rutgers.edu> Cc: <r-help_at_stat.math.ethz.ch>
Sent: Wednesday, June 27, 2007 2:43 PM
Subject: Re: [R] Matlab end operator
> Hello Markus,
>
> On 6/27/07, Markus Loecher <loecher_at_eden.rutgers.edu> wrote:
>>
>> Dear list members,
>> I use both R and Matlab and find that each has its own strengths. Matlab
>> definitely has the edge when it comes to the interactivity of its graphs.
>
>
> I also use both. R definitely has the edge when it comes to do perform
> statistical data analyses :)
> (and also when you consider the price...)
>
> In addition I find the little operator end extremely useful in indexing
>> arrays. (as in x(1:end,) )
>
>
> You mean 'x(1:end,1:end)' or 'x(:,:)' (':' is equivalent to "1:end")
>
> When I go from R to Matlab, I tend to forget to type the ':' ("a[,2]" in R
> is "a(:,2)" in Matlab.)
>
> The interest of 'end' is clearer when the starting index is larger than 1
> as
> in, e.g., 'x(2:end)'
>
> Yet note that in R, you can use negative indexes:
>
> x[-1] is the R equivalent of Matlab's x(2:end)
>
> x[-(1:(n-1))] is equivalent to x(n:end)
>
>
> I agree that R syntax may be a bit less "elegant" in this particular
> situation (but try to write the equivalent of a[-2,] in Matlab)
> Personally, I would stick to "x[n:length(x)]" (or "a[n:nrow(a),]" for a
> matrix). Anyway this kind of code would probably appear inside a loop and
> I
> would put the numbers of rows or columns in variables if there are needed
> more than once.
>
> Best,
>
> --
> Christophe Pallier
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help_at_stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
R-help_at_stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Wed 27 Jun 2007 - 22:03:58 GMT
Archive maintained by Robert King, hosted by the discipline of statistics at the University of Newcastle, Australia.
Archive generated by hypermail 2.2.0, at Wed 27 Jun 2007 - 23:32:38 GMT.
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# orie474sec6 - Fall 2006 ORIE474: Section 6 notes Nikolai...
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Fall 2006 ORIE474: Section 6 notes Nikolai Blizniouk The goal of these notes is to provide some guidance for the use of Regression node in SAS . The setup assumes that you have drawn a diagram similar to that from Section 5. Doing regression with categorical variables using SAS Suppose we have a population divided in J disjoint strata/categories and we make a measurement on a subject that belongs to one of these strata. Consider a simple regression model of the form Y i = β 0 + J s j =1 β 1 ,j · I ( i th subject belongs to j th category) + ǫ i , where j = 1 , . . . , J and ǫ i is a zero-mean noise term. 1 The coeFcient β 0 is treated as the overall population mean (aka “grand mean”) and β 1 ,j ’s measure the amount by which the mean of the j th stratum deviates from the grand mean. Given that we have the data Y 1 , . . . , Y n , the goal is to estimate the parameters ( β 0 , β 1 , 1 , ..., β 1 ,J ) of the model. It is known from which subpopulation each Y i comes from. Notice that the above model does not determine the parameters uniquely. In statistics, a formal statement is “parameterization is not identi±able”. This means that no matter how much data we have, it will not be possible to say for certain which parameter values actually were used to generate the data. Why? Because Y i = β 0 + β 1 ,j + ǫ i = ( β 0 α ) + ( β 1 ,j + α ) + ǫ i , for every value of α , so even if you knew the expectation (i.e., true mean) μ i of each of Y i ’s, you would still be unable to determine the β ’s uniquely. ²or example, suppose we have two subpopulations of ORIE graduates, one of which consists of individuals whose highest degree is Bachelor and the other of those with Masters. Let Y i denote the random variable for the current salary of the i th individual. Then β 1 ,j ’s will capture deviations of the subpopulation means ( μ j = β 0 + β 1 ,j ) from some “base level” mean β 0 . Obviously, unless β 0 is ±xed, one would be unable to determine β 1 ,j ’s even if Y i ’s did not include the error term ǫ i . It is thus desirable to avoid this kind of situation by putting constraints on pa-
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## This note was uploaded on 02/06/2011 for the course ORIE 474 taught by Professor Apanasovich during the Spring '07 term at Cornell.
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Ask a homework question - tutors are online | 740 | 2,782 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2017-17 | longest | en | 0.88844 |
https://i-alpha.it/quiz-2-compound-inequalities.html | 1,618,350,332,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038075074.29/warc/CC-MAIN-20210413213655-20210414003655-00429.warc.gz | 420,393,827 | 13,403 | # Quiz 2 compound inequalities
• This quiz and attached worksheet will help gauge your understanding of set notation, compound inequalities and systems of inequalities. Topics you will need to know to pass the quiz include types ...
• Preview this quiz on Quizizz. Which represents the word sentence as an inequality?A number n is at least 17. Quiz 2 Algebra 1 Simple and Compound Inequalities DRAFT
• Inequalities Determine the truth of an inequality Solving inequalities by adding or subtracting (whole numbers) Solving inequalities by adding or subtracting (fractions) Solving Inequalities by multiplying or dividing Solving Multi-Step Inequalities Inequalities Review Printable
• Compound Interest is one most important topic for bank exams, 1 to 2 questions have been seen in Bank PO Prelims exams. Here at Smartkeeda you will get Compound Interest PDF with Tricks to Solve Fast.
• Solving and Graphing Inequalities. If you are beginning your study of inequalities, I have a lot of lessons for you to study. As you check them out below, make sure you start right here on this page for a quick introduction to basic inequalities.
• -2- ©X s2l0 21A2X NKZurtRa7 DSborf stew8aFrxej NLvL KCm.a x KA8l PlW QrAiWgIh Rt3s q Cr 6e8s Neqrnv cekd y.1 R VM6aUdbeT 2w miGteh s KIpnDfSi YniktHeb 1AUlZgKepb 5rPaJ o1w.V Worksheet by Kuta Software LLC
• A compound inequality includes two inequalities in one statement. A statement such as $4<x\le 6$ means $4<x$ and $x\le 6$. There are two ways to solve compound inequalities: separating them into two separate inequalities or leaving the compound inequality intact and performing operations on all three parts at the same time.
• solution of equations, inequalities and systems (of both equations and inequalities) graphing equations and functions; operations with matrices; operations with functions . We hope you like our site (if you do we would appreciate you spreading the word via likes, digs, pins etc.).
• Ssa 3380 bk sample answers
• Apr 06, 2018 · 2. Solving Linear Inequalities. The procedure for solving linear inequalities in one variable is similar to solving basic equations. (See Solving Equations.). We need to be careful about the sense of the equality when multiplying or dividing by negative numbers.
• Graph each system of linear inequalities on the provided coordinate plane. Make sure to shade carefully and indicate the solution area. Remember that we have two special solutions- infinitely many or no solution. See if you end up with either special solution on any problems! 1. { <4 R−5 2. { +7>2 >−1 3 −5 3. { R +2 R −2 4. { R +2 Q −2
• This page contains a lot of printable two-step inequalities worksheets based on solving and graphing for 7th grade, 8th grade, and high school students. Easy level has positive integer coefficients with answers only in positive numbers.
• This is all explained on Solving Inequalities. Instead, bring "2" to the left: 3x−10x−4 − 2 > 0. Then multiply 2 by (x−4)/(x−4): 3x−10x−4 − 2 x−4x−4 > 0. Now we have a common denominator, let's bring it all together: 3x−10 − 2(x−4)x−4 > 0. Simplify: x−2x−4 > 0 . Second, let us find "points of interest".
• Solve the compound inequality. 2 See answers amna04352 amna04352 Answer: ... HELP PLEASE They are on - Quiz 5.1 - 5.2 - Quiz 5.3-5.4 - Chapter 5 Test - Applications ...
• In this activity, students playfully develop their skills using simple and compound inequalities to name intervals on the x-axis.
• Solving Absolute Value EquationsSolving Compound/Absolute Value Inequalities Lessons 1-4 and 1-6. Content Objectives • Solve Absolute Value Equations • Apply definition to solve, including ‘2’ possible equations • Recognize when there is no solution • Solve Compound Inequalities • Recognize unions versus intersections • Graph solutions/write solutions in Interval Notation ...
• Improve your math knowledge with free questions in "Solve two-step inequalities" and thousands of other math skills.
• Students begin to work with Graphing Linear Inequalities in a series of math worksheets, lessons, and homework. A quiz and full answer keys are also provided.
• Apr 8, 2016 - Explore Emily Wagner's board "Compound Inequalities" on Pinterest. See more ideas about compound inequalities, inequality, middle school math.
Phase 3 nyc gymsThe number 2 will work, as will –2. But 4 will not work, and neither will –4, because they are too far away from zero. Even 3 and –3 won't work (though they're right on the edge), because this is a "less than" (but not equal to) inequality. However, the number 2.99 will work, as will –2.99. Solve x 2 4. Test and worksheet generators for math teachers. Algebra 1 Worksheets Systems Of Equations And Inequalities Worksheets You can customize the worksheets to include one step two step or multi step equations variable on both sides parenthesis and more.
Apr 25, 2008 · 2. QUIZ 6.1 to 6.2 -Follow the directions -20 minutes -No talking -No questions When finished bring up paper and start to read pg. 289 to 293 3. 6.3 Compund Inequalities pg. 289 to 293 Inequalities describes expressions that are not equal COMPOUND inequalities-- combine 2 inequalities using the words AND and OR.
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• Check: p. 44 # 5-50(x5), 52 Algebra 2 Warm-up: * Solve and graph. Lesson Quiz 1.1-1.4(was Friday) Review special cases with compound inequalities. Classwork Puzzle D-27 Powerpoint 1.7 with practice Section 1.7 Objective: Solve and graph absolute value equations and inequalities.
• Chapter 2 Quiz For use after Section 2.3 Name _____ Date _____ Write the sentence as an inequality. 1. A number q plus 8 is less than or equal to 15. 2. The number 20 is no less than the difference of a number c and 8. Write an inequality that represents the graph. 3. 4. Solve the inequality. 5. 6 13 t + ≤ 6.
• Jan 29, 2015 - Are you looking for a fun and interesting application of compound inequalities to get your students thinking and applying the skills you have taught? In this activity students must graph and write compound inequalities to match parking signs. In all cases the students must show when a car MAY park...
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Compound Inequalities-Translate from words. 15 terms. hagadornl TEACHER. Driver's Ed Chp 1, 2, 3, 9. 16 terms. MarleePolevault. Farmer's Tax Guide Depreciation (ADS ... In today's lesson, we are going to be solving compound and absolute value inequalities.0002. A compound inequality consists of two or more inequalities combined by either "and" or "or."0009. To solve a compound inequality, solve each part.0016. For example, if you are given 4x - 2 < 10, and x - 1 ≥ 4, it is connected by the word and, so it is a compound inequality.0022. So, I am going to solve each part of that.0040
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To solve a compound inequality means to find all values of the variable that make the compound inequality a true statement. We solve compound inequalities using the same techniques we used to solve linear inequalities. We solve each inequality separately and then consider the two solutions.
Compound inequalities are the derived form of inequalities, which are very useful in mathematics whenever you are dealing with a range of possible values. For example, you get two solutions after solving a particular linear inequality , x > 3 and x < 12 .
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Solving Compound Inequalities. This is a game like "Who Wants to Be a Millionaire?" where you have to keep getting the answer right in order to move up in money amounts. I recommend that you work out the problem instead of making a guess.
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Solving Inequalities in One Variable Quiz Select the best answer. 1. The solutions for x 2 d 7 are all real numbers _____. A less than or equal to 5 B greater than or equal to 5 C less than or equal to 9 D greater than or equal to 9 2. Which is the graph of m 3? F B 5 G H J 3. Which inequality is shown by the graph below? A x! 2 C x 2
• The inequality is . The compound inequality is Module 2 , so the compound inequality involves AND. . The inequality is symbol and . 85 Lesson 5 Reflect 8. What is a short method to write the compound inequality x ≥ 0 AND x < 6? Your Turn Write the compound inequality shown by each graph. 9.
Mpi merge sort
• Solve the compound inequality. 2 See answers amna04352 amna04352 Answer: ... HELP PLEASE They are on - Quiz 5.1 - 5.2 - Quiz 5.3-5.4 - Chapter 5 Test - Applications ... Jan 29, 2015 - Are you looking for a fun and interesting application of compound inequalities to get your students thinking and applying the skills you have taught? In this activity students must graph and write compound inequalities to match parking signs. In all cases the students must show when a car MAY park... | 2,249 | 8,895 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 3.6875 | 4 | CC-MAIN-2021-17 | latest | en | 0.910091 |
http://www.perlmonks.org/?node_id=1006314 | 1,516,790,725,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084893629.85/warc/CC-MAIN-20180124090112-20180124110112-00521.warc.gz | 527,401,152 | 5,891 | Perl Monk, Perl Meditation PerlMonks
### Re^3: Determining if a rational number terminates
on Nov 29, 2012 at 18:12 UTC ( #1006314=note: print w/replies, xml ) Need Help??
Thanks for that, I didn't even consider that it wouldn't be in lowest form.
It's really clever to take advantage of the fact that powers of five always end with five, I suppose then it's better to divide out the powers of two first than the powers of five. Tomorrow I will ++ both of these :D
• Comment on Re^3: Determining if a rational number terminates
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Re^4: Determining if a rational number terminates
by tobyink (Abbot) on Nov 29, 2012 at 20:13 UTC
Meh... I was actually only checking multiples of five. D'oh! 1/15 was reporting "Y".
Fixed...
```@ARGV = (15, 40);
sub _{@i=sort{\$b-\$a}@_;return\$i[1]if\$i[0]==\$i[1];@i=(\$i[1],\$i[0]%\$i[1
])while\$i[1];\$i[0]}sub __{\$j=_@_;\$_/=\$j for@_;\$_[1]<0and do{\$_*=-1for
@_}}__((\$k,\$l)=@ARGV);for(2,5){\$l/=\$_ until\$l%\$_}print\$l eq 1?Y:N,\$/#
perl -E'sub Monkey::do{say\$_,for@_,do{(\$monkey=[caller(0)]->[3])=~s{::}{ }and\$monkey}}"Monkey say"->Monkey::do'
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https://www.physicsforums.com/threads/speeds-and-vectors.2338/ | 1,670,357,644,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711114.3/warc/CC-MAIN-20221206192947-20221206222947-00148.warc.gz | 985,642,952 | 16,908 | # Speeds and vectors
Not sure if this belongs here or in Maths. Just a question about convention and proper use of terms.
It seems to me that if something is moving, it is moving in relation to something else. So any object at any speed automatically has a velocity, unless the speed is 0 maybe. Is there any situation when you would use the term Speed and it would have nothing to do with Velocity?
This probably seems dumb, but I require an absolute definition that leaves no gaps.
Vector = (magnitude, direction)
Speed = Magnitude
Velocity = (Speed, Direction)
Meaningless content removed
Integral
Last edited by a moderator:
Is there any situation when you would use the term Speed and it would have nothing to do with Velocity?
No. Speed V by definition is simply a magnitude of velocity vector V
newton1
Originally posted by Alexander
No. Speed V by definition is simply a magnitude of velocity vector V
just like you have a velocity vector...you find the Norm of the vector, then it is a speed
It seems to me that if something is moving, it is moving in relation to something else.
you can pick whatever coordinate system you want, to get whatever velocity you want. The only time it matters is when the velocity is relative to some other coordinate system. You can have velocity magnitude without direction, but then it is not velocity in 3-d+t space, it's 1-d.
plus
A clear use of speed rather than velocity is to calculate the forces and power involved with a car traveling at a certain speed with respect to the earth. This is independent to velocity.
Incorrect for forces (force is vector quantity) but correct for power.
plus
Incorrect for forces (force is vector quantity) but correct for power.
But if the frame of reference is the car, then within the other frame of reference, of the earth, if it is traveling at a straight line, then you can compute the forces within the car. The vectors wrt car frame of references are known.
anil
Get this straight: Speed is just a magnitude with a unit of measurement but lacks direction in other words it's a scaler quanitity. If i say you are traveling @ 75 MPH all i know is you are traveling @ 75 miles per hour relative to the Earth but i don't know which direction you are moving relative to the earth. In order for a scaler quantity to qualify for a vector quantity it must have a direction relative to Earth or an object (usually frame of reference used is an object in Inertia since it's velocity is constant, that means it doesnot change direction). Earth is not in Inertia since it changes direction every singe moment because it rotates even though it's speed may be constant (also Earth's speed is not constant, Kepler's law). Let's say the car is moving east relative to earth. Speed is 75 MPH and Velocity is 75 MPH East.
When you say an object is moving it means object is moving compared to the relative of observer's body. Actually object can have a velocity of 0 compared to anybody in universe: (Object in Inertia at rest) So in reality object can have a velocity of 0. Object at any speed greater than zero has a velocity relative to any object. Object at rest has no velocity because it's velocity is zero and it lacks direction. Velocity is very much preferred in physics because it gives a brief description of a moving body unlike speed. Most physics problems require directions to solve projectile problems so speed is not a good choice. It's highly unlikely a physicist would prefer the term "speed" over "velocity". In dialy life using velocity would be very odd and speed seems to be preferred.
Last edited:
Work: W = (Fs)=Fscos([the])
Power: P= dW/dt=(Fv)=Fvcos([the])
Last edited by a moderator:
Singularity
I would say something moves relative to another object rather than in "relation" to another object. Oh, and yes : speed = scalar, velocity = vector =)
velocity of car from inside car =0
acceleration of car from inside car =0
acceleration of gravity from inside car > 0
acceleration of gravity from outside car >0
velocity of car wrt Earth >0
as you know, acceleration is absolute. velocity is relative.
plus
velocity of car from inside car =0
acceleration of car from inside car =0
acceleration of gravity from inside car > 0
acceleration of gravity from outside car >0
velocity of car wrt Earth >0
as you know, acceleration is absolute. velocity is relative.
If x is a vector, you cannot say
x>0
without giving some indication of the nature of the partial ordering.
newton1
Originally posted by plus
But if the frame of reference is the car, then within the other frame of reference, of the earth, if it is traveling at a straight line, then you can compute the forces within the car. The vectors wrt car frame of references are known.
you can not say the car is traveling at a straight line
you should say it travel a constant velocity
and not constant speed too, because the it can be constant speed but not at the same direction
and you must consider the gravity ...
newton1 | 1,101 | 5,002 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2022-49 | latest | en | 0.954638 |
https://people.richland.edu/james/spring15/m122/projects/project9.html | 1,513,277,586,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948550199.46/warc/CC-MAIN-20171214183234-20171214203234-00653.warc.gz | 603,870,838 | 1,460 | # Project 9: Parametric & Polar Equations
## Instructions
Use Maxima to work this project. Each group should email an annotated Maxima file to the instructor.
## Polar Functions
### What to do
Create a gallery of polar graphs. For each of function, find the following items.
1. Graph the function
2. Find the length of the curve
3. Find the area enclosed by the curve, unless otherwise indicated
Be sure to pay attention to domain.
### Functions to use
Here are the curves to use.
1. The circle $$r = 4$$
2. The odd petaled rose $$r = 4 \cos 3\theta$$
3. The even petaled rose $$r = 3 \cos 4\theta$$
4. The dimpled limaçon $$r = 4 +3\sin \theta$$
5. The cardiod $$r = 5 - 5 \cos \theta$$
6. The convex limaçon $$r = 5 - 2\sin \theta$$
7. The limaçon with an inner loop $$r = 3 + 4 \cos \theta$$; just find the area inside the loop
8. The lemniscate $$r^2 = 9 \sin 2\theta$$
9. The burst $$r = 5 + \sin (12 \theta )$$ | 298 | 926 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2017-51 | latest | en | 0.68995 |
https://fractad.wordpress.com/tag/math/ | 1,675,242,573,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499919.70/warc/CC-MAIN-20230201081311-20230201111311-00234.warc.gz | 288,674,241 | 34,810 | # 2013 Winterim: Light, Math & Color
I taught another three-week stained-glass mini-course this year. After my students learn the basic technique of copper-foil stained glass windows, they research a math topic, write a paper on it, and illustrate it with a window of their own design. Topics this year included systems of inequalities, the Fibonacci Sequence, corresponding angles formed by two lines and a transversal, the Four-Color Theorem, and the Pythagorean Theorem among others.
Here’s a gallery of their finished windows:
# Strange Attractors
This post isn’t about the strange attractors you hear of in chaos theory; rather, it’s about an excellent poetry anthology edited by Sarah Glaz and Joanne Growney. I’ve linked to Joanne’s blog, Intersections, before. After showing it to my school’s librarian, she ordered a copy of Strange Attractors, and it arrived a few days ago.
I haven’t had chance to really delve into it, but it looks wonderful. Glaz and Growney have selected poems from the beginning of recorded history up to the present. What holds it all together is their subject matter: love. Yes, mathematicians are susceptible to it, and these poems are ample illustration of the many ways math and poetry complement each other in expressing that emotion.
Most of the poets use plain words to get their point across, but there are several clever exceptions. I especially like this one by Kaz Maslanka, entitled “Sacrifice and Bliss”:
It reminds me of a Tolstoy quote I have in my classroom: “A man is like a fraction whose numerator is what he is, and whose denominator is what he thinks of himself. The larger the denominator, the smaller the fraction.”
Anyway, even if you’ve forgotten most of the math you learned in high school, you’ll find plenty of wonderful poetry to enjoy in this anthology. If you love math, you’ll derive (!) even more pleasure from it. If you teach math, Strange Attractors will be an invaluable resource for you and your students.
# A Really Useful Online Graphing Calculator
The good people at Desmos have made an excellent online graphing calculator even better. It’s incredibly fast and versatile. For example, if you input y = m*x + b, it will automatically ask you if you want to create sliders for “m” and “b”. Here’s an example of a quadratic function with sliders:
It will plot implicit relations:
As well as inequalities:
There’s even a “Share” button that lets you post your creation to Facebook, Twitter, or Google+. Once you create an account, you can save your graphs online, to be used later. There are countless applications of this product in the math classroom. | 571 | 2,637 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2023-06 | longest | en | 0.939972 |
https://www.thestudentroom.co.uk/showthread.php?t=6759960 | 1,718,888,058,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861940.83/warc/CC-MAIN-20240620105805-20240620135805-00312.warc.gz | 897,574,833 | 54,010 | # Equations of two tangents to a circle - A Level Maths
Hi,
Would someone be able to help with the following maths question?
A circle has equation x^2+y^2+4x+12y=-23
The lines l1 and l2 are both tangents to the circle, and they intersect at the point (5,0).
Find the equations of l1 and l2, giving your answers in the form y=mx+c.
Scroll to see replies
Here is a diagram for the question.
Original post by flappertard
Here is a diagram for the question.
What have you worked out so far?
Original post by laurawatt
What have you worked out so far?
I know that the circle equation in the form (x-a)^2+(y-b)^2=r^2 is (x+2)^2+(y+6)^2=17
Therefore the centre is (-2,-6) and the radius is √17
Using √(x1-x2)^2+(y1-y2)^2
I can find the length of the line from (5,0) to the centre which is √85
As the tangents meet the circle at 90 degrees you can use Pythagoras to work out the length of the tangent (from (5,0) to where it meets the circle) which is 2√17
Original post by flappertard
I know that the circle equation in the form (x-a)^2+(y-b)^2=r^2 is (x+2)^2+(y+6)^2=17
Therefore the centre is (-2,-6) and the radius is √17
Using √(x1-x2)^2+(y1-y2)^2
I can find the length of the line from (5,0) to the centre which is √85
As the tangents meet the circle at 90 degrees you can use Pythagoras to work out the length of the tangent (from (5,0) to where it meets the circle) which is 2√17
You've pretty much got most of it. What are you stuck on?
Original post by mqb2766
You've pretty much got most of it. What are you stuck on?
The question is asking for the equations of the tangents. Not quite sure how to get there.
I don't have the mark scheme but through a graphing tool, I think the points of intersection are (-3,-2) and (1.8,-7.6). However, I need to find this algebraically.
You could use the usual tangent - circle intersection where the line gradient is to be found using the quadratic discriminant.
Original post by mqb2766
You could use the usual tangent - circle intersection where the line gradient is to be found using the quadratic discriminant.
Thanks for your help so far. Would you be able to explain this method further? Thank you
Original post by flappertard
Thanks for your help so far. Would you be able to explain this method further? Thank you
y = m(x-5)
Sub that into the circle to get a quadratic in x. This must have a single solution if the line is a tangent, so solve the discriminant=0 and that should be a quadratic in m, giving the two gradients.
If you wanted to keep going with your approach, you could have got the gradients from the tan(x+y) formula with a few extra lines.
(edited 3 years ago)
Original post by mqb2766
y = m(x-5)
Sub that into the circle to get a quadratic in x. This must have a single solution if the line is a tangent, so solve the discriminant=0 and that should be a quadratic in m, giving the two gradients.
If you wanted to keep going with your approach, you could have got the gradients from the tan(x+y) formula with a few extra lines.
Hi, I just used the quadratic discriminant method and I ended up getting a polynomial to the power of 4! Upon solving for m I got 0.22 and -0.58. Any ideas where I am going wrong?
Original post by flappertard
Hi, I just used the quadratic discriminant method and I ended up getting a polynomial to the power of 4! Upon solving for m I got 0.22 and -0.58. Any ideas where I am going wrong?
Original post by mqb2766
This is my working.
Original post by flappertard
This is my working.
Will have a look, but it usually falls out a bit easier.
Can you do it using the tan(x+y) formula? You've got tan(x) and tan(y).
Original post by mqb2766
Will have a look, but it usually falls out a bit easier.
Can you do it using the tan(x+y) formula? You've got tan(x) and tan(y).
Hi, I've never come across this formula before so not sure what to do. When you say x and y are you referring to the point where the tangent meets the circle? Thanks
Original post by flappertard
Hi, I've never come across this formula before so not sure what to do. When you say x and y are you referring to the point where the tangent meets the circle? Thanks
For the circle - line - tangent one, I'd transform the origin to the point (5,0). That simplified the calculation a lot as the line is y=mx. The gradient is unchanged.
Will upload the other one in a few mins.
As you'd worked out these lengths already, it's far simpler to get m this way. The grads are
tan (x+/-y)
(edited 3 years ago)
Original post by flappertard
This is my working.
The coeff. of $x^2$ is incorrect in your quadratic - should be $1+m^2$
Original post by flappertard
Hi,
Would someone be able to help with the following maths question?
A circle has equation x^2+y^2+4x+12y=-23
The lines l1 and l2 are both tangents to the circle, and they intersect at the point (5,0).
Find the equations of l1 and l2, giving your answers in the form y=mx+c.
which paper was this from? I'm currently doing the same question but can't figure out the origin of the question
Original post by flappertard
This is my working.
I'm not saying this is *the* way to do it, but from (x+2)^2 + (m(x-5) +6)^2 = 17, substitute X = x-5 to get:
(X+7)^2+(mX+6)^2 = 17; if you carry on from here you will end up with b^2-4ac being a quadratic in m.
(My gut feeling is that you shouldn't need the substitution, but for sure the algebra gets messy and I didn't really want to spend the time trying to work it through).
Hi I don’t think it was from an exam. When my teacher gave the answers, they were from a question pack on the SaveMyExams website
Original post by DFranklin
I'm not saying this is *the* way to do it, but from (x+2)^2 + (m(x-5) +6)^2 = 17, substitute X = x-5 to get:
(X+7)^2+(mX+6)^2 = 17; if you carry on from here you will end up with b^2-4ac being a quadratic in m.
(My gut feeling is that you shouldn't need the substitution, but for sure the algebra gets messy and I didn't really want to spend the time trying to work it through).
where did you get x-5 from fro the substitution | 1,653 | 6,040 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-26 | latest | en | 0.952867 |
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t3(4)
t3(4) - (a Let p denote the price per job and x the number...
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Methods of Calculus MAC3233 Test 3 March 25, 1996 Instructions: This is a closed-book, closed-notes test. Do not write your answers on this paper, use separate answer sheets. No calculators are allowed to be used. To receive credit, you must show all work . For the optimization problems, you are required to show why your answer is a minimum or maximum. Some numerical answers may involve the number e . (1) (10 points each) Find f 0 ( x ) for the following (a) f ( x ) = (3 x 4 - 2 x + 1)( x + 4) 3 (b) f ( x ) = 3 x + 2 x 4 9 x 2 - 1 (c) f ( x ) = ( e x - 1 + x - 2 ) 4 (d) f ( x ) = e (3 x +5) (2) (15 points) Find ∂f ∂x and ∂f ∂y for f ( x ) = x + e xy (3) (15 points) f ( x ) = (4 x - 1) e 3 x . (a) f 0 ( x ) = ? (b) f 00 ( x ) = ? (c) Find the x coordinate of each possible extremum. (d) Use f 00 ( x ) to say whether each answer in (c) is a relative maximum or minimum. (4) (15 points) When the price for cleaning roofs was \$150, the Acme Roof Cleaning Service had 40 roof cleaning jobs per month. When the rate was raised to \$160, the number of jobs dropped to 38 per month.
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Unformatted text preview: (a) Let p denote the price per job, and x the number of jobs per month. Assuming a linear demand curve, write the formula for the demand curve. (b) Write the formula for the monthly revenue function. (c) Suppose the company has fixed costs of \$500 per month and the vari-able cost is \$30 per job. Write the formula for the monthly profit function. Find the price per job that will maximize the company’s monthly profit. (5) (15 points) Suppose a firm’s profit from producing and selling x units of a product is given by P ( x ) = 32 x 2-9 x 3 thousands of dollars. (a) Find the marginal profit, dP dx . (b) The production level at t weeks from the present is expected to be x = 4 + 3 t . Find the time rate of change of profit, dP dt . (c) How fast (with respect to time) are profits changing when t = 4? 1...
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http://www.evi.com/q/84_cm_equals_how_many_inches | 1,418,933,811,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802767843.95/warc/CC-MAIN-20141217075247-00117-ip-10-231-17-201.ec2.internal.warc.gz | 489,560,830 | 6,573 | # 84 cm equals how many inches
• 84 centimeters is 33.07 inches.
• tk10publ tk10ncanl
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• what is the equivalent of 84 cm to inches | 2,032 | 6,360 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-52 | latest | en | 0.877356 |
https://studysoup.com/tsg/589101/applied-calculus-5-edition-chapter-6-1-problem-6 | 1,582,746,778,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146485.15/warc/CC-MAIN-20200226181001-20200226211001-00363.warc.gz | 566,699,859 | 11,069 | ×
×
# Figure 6.6 shows the derivative g. If g(0) = 0, graph g. Give (x, y)-coordinates of all
ISBN: 9781118174920 268
## Solution for problem 6 Chapter 6.1
Applied Calculus | 5th Edition
• Textbook Solutions
• 2901 Step-by-step solutions solved by professors and subject experts
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3
Problem 6
Figure 6.6 shows the derivative g. If g(0) = 0, graph g. Give (x, y)-coordinates of all local maxima and minima. 1 2 3 4 5 6 1 1 g(x) x Figure 6.6
Step-by-Step Solution:
Step 1 of 3
COMM 215 EXAM 2 NOTES Social Penetration: o As a relationship develops o Not familiar/ intimate to intimate o Communication grows Social Exchange: the outcome in the relationship= rewardÂcost Comparison Level of Alternatives: o Deals with risk o Looking outside the relationship o Considering what can resolve a relationship o A constant...
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Step 3 of 3
#### Related chapters
Unlock Textbook Solution | 294 | 1,010 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2020-10 | latest | en | 0.722422 |
https://www.cleverism.com/lexicon/break-even/ | 1,719,186,465,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864968.52/warc/CC-MAIN-20240623225845-20240624015845-00773.warc.gz | 606,862,663 | 19,474 | Break-even is a term used in business and economics to describe a situation where the cost of the business is equal to the revenue received.
Break-even analysis is important in business and economics to determine profitability, the best unit price for a product or service and to develop company strategy in line with profitability. The users of the information provided by a break-even analysis are investors, managers and corporate strategists.
Supposing you have set up a business and it cost you \$X. After a period of time you find out that you have received sales worth \$X; which is the same amount of money you used to set up the business. This point in business where there isn’t any net loss or net gain is referred to as the break-even point.
The break-even point is usually measured in terms of sales – that is, how much you need to have sold before you recoup your investment. To calculate this value (X), we know that:
_ Total Revenue = Total Cost
but Total Revenue = Product Price (P) x Sales (X)
and Total Cost = Total Fixed Costs TFC + (the variable cost V x Sales X)
so P x X = TFC + (V x X)
_ P – (V x X) = TFC / X
Therefore, break-even point
_ X = TFC/ P – V
If it costs \$1 to buy an orange that you resell for \$2 and the cost of business is \$50 a day then you need to sell 50 oranges a day just to keep the business running. Anything below 50 oranges means you are making a loss and anything above that means that you are making a profit.
In general, the total cost can be divided into fixed and variable. The fixed costs represent costs that do not change regardless of the amount of sales the business makes or lack thereof. These costs do not change in the short term but have the ability to change in the long run. Examples of these costs are; rent charges, land and rates, marketing and advertisement costs, administration costs and depreciation.
The variable costs are costs that change depending on the amount of products produced and sold. Examples of variable costs include labor costs, cost of production materials and fuel costs and can be separated into two types: direct and indirect variable costs. Direct variable costs are those that have been expressly incurred in the production while indirect costs have not been expressly incurred in the production process.
Related topics: Accounting basics | 504 | 2,395 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.947077 |
https://fr.maplesoft.com/support/help/view.aspx?path=csgn&L=F | 1,670,419,993,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711162.52/warc/CC-MAIN-20221207121241-20221207151241-00259.warc.gz | 268,092,130 | 31,958 | csgn - Maple Help
# Online Help
###### All Products Maple MapleSim
csgn
sign function for real and complex expressions
Calling Sequence csgn(x) csgn(1, x) csgn(0, x, y)
Parameters
x - any algebraic expression y - any algebraic expression
Description
• The csgn function is used to determine in which half-plane ("left" or "right") the complex-valued expression or number x lies. With the exception described in the next bullet point, it is defined by
$\mathrm{csgn}\left(x\right)=\left\{\begin{array}{cc}1& 0<\mathrm{\Re }\left(x\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{or}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{\Re }\left(x\right)=0\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{and}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}0<\mathrm{\Im }\left(x\right)\\ -1& \mathrm{\Re }\left(x\right)<0\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{or}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{\Re }\left(x\right)=0\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{and}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{\Im }\left(x\right)<0\end{array}\right\$
• For the case of a complex number in which the real component is one of the floating point values -0. or +0., csgn returns the sign of the real part. For more information, see Numeric Computation in Maple.
• The value of csgn(0) is controlled by the environment variable _Envsignum0. The 3-argument calling sequence csgn(0, x, y) sets _Envsignum0 = y for the duration of the call to csgn. See signum for further information.
• The decision of whether or not to perform many of the automatic symmetry transformations in maple is based on the value of csgn. For example, if csgn(x) = -1, the transformation $\mathrm{sin}\left(x\right)\to -\mathrm{sin}\left(-x\right)$ is done.
• csgn uses signum to determine the signs of $\mathrm{\Re }\left(x\right)$ and $\mathrm{\Im }\left(x\right)$.
• The derivative of csgn is denoted by csgn(1, x). This is 0 for all non-purely-imaginary numbers, and is undefined otherwise.
• For mathematical consistency, the value of csgn(0), as determined either by the value of _Envsignum0 or by the third argument to csgn, should be either 0 (the default) or one of 1, -1, or undefined.
Examples
> $\mathrm{csgn}\left(1-\frac{2}{3}I\right)$
${1}$ (1)
> $\mathrm{csgn}\left(-1-\frac{2}{3}I\right)$
${-1}$ (2)
> $\mathrm{csgn}\left(-1+\frac{2}{3}I\right)$
${-1}$ (3)
> $\mathrm{csgn}\left(1+\frac{2}{3}I\right)$
${1}$ (4)
> $\mathrm{csgn}\left(-\frac{2}{3}\mathrm{\pi }I\right)$
${-1}$ (5)
> $\mathrm{csgn}\left(\mathrm{exp}\left(\frac{2\mathrm{\pi }}{3}I\right)\right)$
${-1}$ (6)
> $\mathrm{csgn}\left(\mathrm{\pi }\right)$
${1}$ (7)
> $\mathrm{diff}\left(\mathrm{csgn}\left(x\right),x\right)$
${\mathrm{csgn}}{}\left({1}{,}{x}\right)$ (8)
> $\mathrm{diff}\left(\mathrm{csgn}\left(x\right),x,x\right)$
${\mathrm{csgn}}{}\left({1}{,}{x}\right)$ (9)
> $\mathrm{csgn}\left(1,-3+I\right)$
${0}$ (10)
> $\mathrm{csgn}\left(0\right)$
${0}$ (11)
> $\mathrm{csgn}\left(0,0,-1\right)$
${-1}$ (12)
The following illustrates the exception for floating point complex numbers with real part equal to -0. or +0.:
> $\mathrm{csgn}\left(0+I\right)$
${1}$ (13)
> $\mathrm{csgn}\left(-0.+1.I\right)$
${-1}$ (14)
> $\mathrm{csgn}\left(0-I\right)$
${-1}$ (15)
> $\mathrm{csgn}\left(0.-1.I\right)$
${1}$ (16)
See Also | 1,247 | 3,340 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 36, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.559367 |
http://www.interfacebus.com/opamp-voltage-regulator-design-lm143.html | 1,656,364,789,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103341778.23/warc/CC-MAIN-20220627195131-20220627225131-00615.warc.gz | 89,009,862 | 4,989 | General Engineering Terms
"A" "B" "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M",
"N", "O", "P" "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z"
Voltage Regulator Op Amp Design
This design uses an LM143 operational amplifier. There's nothing special about this device, and was selected because it could be considered a replacement for the common LM741 Op Amp. In fact the LM143 is a higher voltage upgrade to the LM741 [uA741]. Although any other general purpose operational amplifier developed after the 741 could be considered an up-grade.
For simplicity the supply voltage to the operational amplifier is not shown in the schematic. However only an Op Amp which may use a single voltage supply should be considered. So in this case V+ is connected to a positive voltage and V- is connected to ground. A by-pass capacitor should also be used across the power rail, but is also omitted from the diagram. LM143 Voltage Regulator
The diode is a 1N4954 Zener Diode with a zener voltage [Vz] of 6.8 volts. Although almost any Zener diode could be used, and would depend on the actual voltage required on the output of the circuit. So the example diode used here is just a random example, just as the selection of the LM143 is random.
The output shown at the top of the diode is the zener voltage, and will be what ever diode voltage is used or required. So the point is fixed, depending on the Zener voltage of the diode selected. The [910] series resistor is used to select the current through the diode.
The input resistor [6.8k] and the feedback resistor [1.8k] are used to scale the output to the desired voltage:
Vo = 6.8 * (Ri + Rf)/ Ri.
In this case using the component values provided:
Vref = 6.8 volts & Vo = 8.6 volts.
Output Voltage Polarity
The polarity of the output voltage is positive, based on the direction of the zener diode. However if the zener diode is reversed, and the Op Amp is power from a negative voltage [and ground], than the circuit will be a negative voltage regulator.
Editor note; this topic is for informational purposes, to show how to develop a voltage regulator using an Op Amp. However; there is absolutely no need to design a voltage regulator circuit using an operational amplifier. There are many different styles of voltage regulator ICs available to cover any particular need. Along with all the different options available, come the many different packaging styles as well.
But there could be some obscure chance that a spare operational amplifier, already being used in another part of the design is free. Or that a particular Op-Amp is already being used in the parts list and could be re-purposed as a voltage regulator, there by not requiring a new line item on the PL. In any case the LM143 is offered here to represent one possible circuit design which utilizes an OpAmp.
The LM143 is depicted here to illustrate a general purpose operational amplifier used in this application. In this case the LM143 is being offered as a high voltage replacement for an LM741 Op-Amp, or in other words almost any other Op Amp.
Operational Amplifier Circuit Designs
LM386: Audio Mono Amplifier
LM741: Audio Stereo Amplifier
LM149: Audio Tone Control Filter
LM301: Audio 2-Band Tone Control Filter
TL082: Audio 3-Band Tone Control Filter
LM741: Audio Midrange Control Filter
LM381: Audio Mixer [Summer]
LM741: Audio Equalizer [Filter]
LM741: Resonant Equalizer [Filter]
LM741: All-pass Filter
LM101: Sine Wave Oscillator
LM107: Sine Wave Oscillator
LM108: Cosine Wave Oscillator
LM107: Triangle Wave Oscillator
LM3900: Triangle Wave Oscillator
LM124: Pulse Generator
The list of operational amplifier designs does not include all the different Op Amp circuits on this site. Nor does it include all of the different Op Amp devices that are used in the circuits. The list only serves as a starting point showing some common circuit examples and the Op Amp used to implement them. | 934 | 3,930 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.907547 |
https://www.physicsforums.com/threads/trigonometric-inverse-functions.234247/ | 1,582,220,092,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145260.40/warc/CC-MAIN-20200220162309-20200220192309-00148.warc.gz | 818,551,141 | 14,412 | # Trigonometric inverse functions
## Main Question or Discussion Point
i want the most general solution for
sin6x=sin4x-sin2x
rock.freak667
Homework Helper
sin6x=sin4x-sin2x
sin6x-sin4x+sin2x=0
Then remember that
$$sinP+sinQ=2sin(\frac{P+Q}{2})cos(\frac{P-Q}{2})$$
thanks | 104 | 278 | {"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} | 2.875 | 3 | CC-MAIN-2020-10 | longest | en | 0.502289 |
https://www.jiskha.com/display.cgi?id=1354040270 | 1,503,449,593,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886116921.70/warc/CC-MAIN-20170823000718-20170823020718-00215.warc.gz | 928,016,591 | 4,025 | # physics
posted by .
A car moving with a speed of 85 km/h is brought to rest in a didtance of 60 m. How much time did the car take to stop?
• physics -
Average speed x time = stopping distance
The average speed during stopping is half the starting speed, or 42.5 km/h
85 km/h = 23.6 m/s
42.5 km/h = 11.8 m/s
Time = 60m/11.8m/s = 5.1 seconds
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http://stackoverflow.com/questions/18075747/python-multiplying-a-3x3-matrix-to-3nx1-array-without-using-loops | 1,469,731,581,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828313.74/warc/CC-MAIN-20160723071028-00123-ip-10-185-27-174.ec2.internal.warc.gz | 239,619,367 | 17,176 | Dismiss
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Python: Multiplying a 3x3 matrix to 3nx1 array without using loops [closed]
I would like to do a matrix multiplication (a 3x3 matrix) with a vector (3x1). The "problem" ist that every component of the vector is taken each one of another matrix and I do not know how to proceed. Is there any way to do it?
``````import numpy as np
A = np.array([[1,1,1],[2,1,0],[1,0,1]])
v1 = np.array([[1,2,3,4]])
v2 = np.array([[5,6,7,8]])
v3 = np.array([[9,10,11,12]])
``````
And I would like to multiply: A x {1,5,9}.T and save the result. Then A x {2,6,10}.T, A x {3,7,11}.T and finally A x {4,8,12}.T. The lengths of arrays v1, v2, and v3 are the same.
Thank you in advance! Regards,
Xabi
-
closed as off-topic by msw, Hans Then, madth3, zhangyangyu, Antti HaapalaAug 7 '13 at 1:06
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "Questions asking for code must demonstrate a minimal understanding of the problem being solved. Include attempted solutions, why they didn't work, and the expected results. See also: Stack Overflow question checklist" – msw, Hans Then, madth3, zhangyangyu, Antti Haapala
If this question can be reworded to fit the rules in the help center, please edit the question.
Even showing your worst attempt gives us reason to think you've tried to solve it; it also helps us to understand what you've got in your mental toolbox which allows us to give you a more usable answer. – msw Aug 6 '13 at 9:07
do not see why you would want to do this with or without for loops. numpy has methods for matrix multiplication and Transposition. you are already importing numby why not just use it. it has C optimised, tested code exactly for that purpose. – Joop Aug 6 '13 at 10:10
2 Answers
You can do the operation you are after with a single matrix multiplication, if you first stack all your vectors together into a single array:
``````vectors = np.vstack((vv1, v2, v3))
products = np.dot(A, v)
``````
And now `products[:, i]` (or `products.T[i]`, if you prefer) has the product of `A` with the `i`-th vector.
-
Thank you for your response! It is a good way to solve the "problem"! – user2655987 Aug 7 '13 at 7:55
Numpy Arrays:
Using two numpy arrays; one 3 x 3 and one 3x1:
``````>>> import numpy as np
>>> a = np.ones((3,3))
>>> b=np.random.rand(3,1)
array([[ 0.08970952],
[ 0.56447089],
[ 0.57500698]])
``````
If you want matrix multiplication you can use `dot`
``````>>> np.dot(a,b)
array([[ 1.22918739],
[ 1.22918739],
[ 1.22918739]])
``````
If you want element wise muliplication you can use `*`
``````>>> a*b
array([[ 0.08970952, 0.08970952, 0.08970952],
[ 0.56447089, 0.56447089, 0.56447089],
[ 0.57500698, 0.57500698, 0.57500698]])
``````
Numpy Matrices:
Note that if you are using numpy matrices then the `*` operator can be used for matrix multiplication:
``````>>> c = np.mat(a) # converts from array to matrix
>>> d = np.mat(b)
>>> c*d
matrix([[ 1.22918739],
[ 1.22918739],
[ 1.22918739]])
``````
- | 993 | 3,234 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-30 | latest | en | 0.922157 |
https://www.askiitians.com/forums/Mechanics/10/54538/friction.htm | 1,714,054,500,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297295329.99/warc/CC-MAIN-20240425130216-20240425160216-00781.warc.gz | 577,724,419 | 44,076 | What is static friction and kinetic friction?
Vivek Joshi
20 Points
11 years ago
Static friction force is the force that resists the starting of motion of any mass. Eg: U just want to start the motion of the block. You need to apply minimum force to start the motion of the block. The force u applied is just equal to kinetic friction force.
While the kinetic friction force acts as a retarder. I opposes the motion of mass. Suppose any block is moving on a table, the friction force that is acting against the motion of the block is kinetic friction force.
Kinetic friction force is always less then or equal to static friction force.
Sourabh Meena
19 Points
11 years ago
Static Friction is friction which comes when a body is at rest and tries to move NOTE - IT IS SELF ADJUSTING FRICTION.
KINETIC Friction is the friction which comes into play when a body is actually moving on the surface of other body.
Mujahid Ahmed
42 Points
11 years ago
I''l make this Short and Easy.
Static Friction is the force acting between two contact surfaces that opposes the start of Relative Motion between the two Surfaces
Kinetic Friction is the force acting between two contact surfaces that opposes the Relative motion Once it has started.
Pay attention to the bolded part.
Click "yes" below if i helped :)
Gunna Laxmi Prasanna
32 Points
11 years ago
Frictional force is called a self adjusting force.
Frictional forces are divided into three types:1.Static friction2.Kinetic or Dynamic friction3.Rolling friction
Static friction:Frictional force acting on a body when it is under rest condition.The maximum value of this force is attained when the body is about to move.
Kinetic or Dynamic friction:-It is the frictional force acting on a body which is under motion.Generally kinetic or dynamic friction is less than static frictional force. | 410 | 1,848 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.92423 |
https://www.physicsforums.com/threads/rolling-disk-energy-before-pure-rotational.164127/ | 1,709,151,486,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474744.31/warc/CC-MAIN-20240228175828-20240228205828-00599.warc.gz | 951,044,727 | 15,263 | # Rolling disk, energy before pure rotational?
• kanki
In summary, the conversation discusses the distance traveled by a solid disk with a given angular speed when it is lowered onto a horizontal surface and released, with a specific coefficient of friction. Conservation of momentum is applied to find the work done by friction, which determines the distance traveled before pure rolling occurs. The concept of pure rolling is also defined as when the torque due to friction is less than a certain value, indicating a combination of translational and rotational motion.
#### kanki
Suppose a solid disk of radius R is given an angular speed about an axis through its center and then lowered to a horizontal surface and released, the coefficient of friction between disk and surface is u. What is the distance traveled before pure rolling occurs?
Conservation of Momentum is applied because no net torque.
The work done by the friction of the disk, fx is used to find the distance x travelled. however i do not know which energy has been changed into work done by friction. What is meant by pure rolling? Only translational and rotational? I'm stuck here.
The initial energy of the disc is rotational and is given by -
$$E = \frac{1}{2}I\omega ^2$$
Pure rolling starts when the slipping stops, that is, when the torque due to friction is less than a certain value.
## 1. What is the concept of rolling disk in physics?
The concept of a rolling disk in physics refers to an object that is moving both rotationally and translationally at the same time. This means that the disk is spinning on its own axis while also moving in a straight line.
## 2. How is energy related to pure rotational motion of a disk?
Energy is related to pure rotational motion of a disk through the conservation of energy principle. The kinetic energy of the disk is equal to the sum of its translational kinetic energy and its rotational kinetic energy. This means that as the disk rotates, its energy is conserved and remains constant.
## 3. What is the difference between kinetic energy and potential energy in a rolling disk?
Kinetic energy in a rolling disk refers to the energy of motion, specifically the energy of both rotation and translation. Potential energy, on the other hand, refers to the energy that an object has due to its position or configuration. In a rolling disk, potential energy is converted into kinetic energy as the disk moves.
## 4. How does the moment of inertia affect the energy of a rolling disk?
The moment of inertia, which is a measure of an object's resistance to rotational motion, affects the energy of a rolling disk in that objects with a higher moment of inertia require more energy to rotate at the same speed compared to objects with a lower moment of inertia.
## 5. Can a rolling disk have both kinetic and potential energy?
Yes, a rolling disk can have both kinetic and potential energy. As the disk rolls, it has both rotational and translational kinetic energy. Additionally, if the disk is on an inclined surface, it will also have potential energy due to its position in the gravitational field. | 632 | 3,136 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-10 | latest | en | 0.946742 |
http://www.thenakedscientists.com/forum/index.php?topic=39137.0 | 1,477,100,871,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718423.28/warc/CC-MAIN-20161020183838-00147-ip-10-171-6-4.ec2.internal.warc.gz | 745,464,293 | 14,871 | # The Naked Scientists Forum
### Author Topic: The Lorentz Factor Conflict MMXI (Read 4591 times)
#### butchmurray
• Sr. Member
• Posts: 194
• If I had my druthers, I would have druthers
##### The Lorentz Factor Conflict MMXI
« on: 07/05/2011 20:29:14 »
The Lorentz Factor Conflict MMXI
Thorntone E. ‘Butch’ Murray
May 2011
Problem
When the results of length contraction calculations are applied in the Pythagorean theorem, length perpendicular to the direction of motion is unique for different velocities.This conflicts with the Special Relativity stipulation to the contrary.
Hypothesis
“When one side of a triangle has a fixed length and the length of another side is changed the length of the remaining side must also change.” This property of triangles dictates that since the ‘light path’ the hypotenuse has a fixed length and the length of another side is changed with velocity the remaining side, length perpendicular to the direction of motion, must also change. This contradicts Special Relativity.
Additionally, at relative rest the length of the light path and length perpendicular to the direction of motion are equal. When in relative motion the light path becomes the hypotenuse of a right triangle. The hypotenuse of a right triangle is the longest side. The length of the light path and length perpendicular to the direction of motion are no longer equal. One or both of these two lengths has changed. Change of either of these lengths contradicts Special Relativity.
Data
The light path – c is fixed at 1
a2=c2-b2
a2=c2-.4352 a2=1-.189 a2=.811 a=.900
a2=c2-.52 a2=1-.25 a2=.75 a=.866
a2=c2-.9952 a2=1-.990 a2=.01 a=.1
The ‘light path’, the hypotenuse ‘c’, has a fixed length and the length of another side is changed with velocity the remaining side, length perpendicular to the direction of motion, must also change.
Conclusion
The Lorentz factor is a corner stone of Special Relativity. It, however, either conflicts with the Special Relativity tenet that length perpendicular to the direction of motion does not change OR that the speed of light is constant OR both.
Prediction
Chaos will ensue.
#### Phractality
• Hero Member
• Posts: 523
• Thanked: 1 times
##### The Lorentz Factor Conflict MMXI
« Reply #1 on: 07/05/2011 21:16:10 »
You are confused about the concept of frames of reference. The speed of light in a given reference frame is measured in m/s, where the meter sticks and clocks are stationary in that reference frame. If you measure the speed of light in two different reference frames, you get the same speed because you are measuring relative to a different set of meter sticks and clocks.
What you appear to be doing is measuring the speed of light in one reference frame by using meter sticks and clocks from a different reference frame.
#### imatfaal
• Neilep Level Member
• Posts: 2787
• rouge moderator
##### The Lorentz Factor Conflict MMXI
« Reply #2 on: 08/05/2011 10:36:11 »
Hi Butch. I think Fraktalitee is correct - but to get further into why you really need to provide more details on your thought experiment - preferably with a diagram. I reckon that once you do this you yourself will realise your error - BTW when going up against one of the most tested theorems of all time it is probably best to assume error until every possibly source is exhausted :-)
#### butchmurray
• Sr. Member
• Posts: 194
• If I had my druthers, I would have druthers
##### The Lorentz Factor Conflict MMXI
« Reply #3 on: 08/05/2011 14:24:07 »
Thank you both.
Everything referenced is as observed from relative rest.
Time is one unit. It doesn’t matter if it is seconds, hours, etc. but is the same unit throughout.
The body is one light unit square when at relative rest.
The right triangle is observed from relative rest with the body in uniform relative motion.
The right triangle:
The vertical side is the height of the body – one light unit in length
The horizontal side is the contracted length of the body
The hypotenuse is the light path for one unit of time - one light unit in length
SR dictates that the height of the body does not change. The speed of light is constant. Light therefore must traverse one light unit of length in one unit of time - one light unit of length.
That is the basic problem. The hypotenuse is the longest side of a right triangle. The vertical side and the hypotenuse cannot be the same length.
Thanks again,
Butch
#### butchmurray
• Sr. Member
• Posts: 194
• If I had my druthers, I would have druthers
##### The Lorentz Factor Conflict MMXI
« Reply #4 on: 08/05/2011 14:59:40 »
Forgot to mention:
By virtue of the constancy of the speed of light the hypotenuse will always be one light unit in length. In any triangle with one side of fixed length (the light path- hypotenuse), the change of length of another side (the horizontal leg that changes with relative velocity) will result in the change of length in the remaining side (the height of the body which according to SR does not change).
Butch
#### imatfaal
• Neilep Level Member
• Posts: 2787
• rouge moderator
##### The Lorentz Factor Conflict MMXI
« Reply #5 on: 09/05/2011 12:16:58 »
Butch - you are still not making a lot of sense. You need a diagram or a better description.
#### socratus
• Sr. Member
• Posts: 329
##### The Lorentz Factor Conflict MMXI
« Reply #6 on: 10/05/2011 12:24:10 »
The Lorentz Factor Conflict MMXI
Thorntone E. ‘Butch’ Murray
May 2011
Problem
When the results of length contraction calculations are applied in the Pythagorean theorem, length perpendicular to the direction of motion is unique for different velocities.This conflicts with the Special Relativity stipulation to the contrary.
Hypothesis
One of SRT’s postulate says that quantum of light moves
in a straight line with constant speed c=1 in the vacuum.
So, in SRT we have one reference frame and it is vacuum.
Not hypothetical problem:
Can quantum of light have geometrical form of a triangle?
===.
S.
#### imatfaal
• Neilep Level Member
• Posts: 2787
• rouge moderator
##### The Lorentz Factor Conflict MMXI
« Reply #7 on: 10/05/2011 12:46:26 »
Socratus - you are wrongly muddying the issue. SR does not deal with quantised light . SR is based on gedenken so it is acceptable to be hypothetical. Reference frame in these terms does not refer to whether it is a vacuum but to whether the observer is in relative motion or accelerating.
FYG a quick search on the entire original book give zero results for quantum, quanta, packet, discrete or wavelength. Einstein did publish his Nobel prize winning paper on Photoelectric Effect in same year as SR - but SR most certainly does not engage with quantum theory of light
#### butchmurray
• Sr. Member
• Posts: 194
• If I had my druthers, I would have druthers
##### The Lorentz Factor Conflict MMXI
« Reply #8 on: 10/05/2011 23:41:03 »
Unfortunately, I will probably not have time to dedicate to this project until this coming weekend.
Thank you,
Butch
#### butchmurray
• Sr. Member
• Posts: 194
• If I had my druthers, I would have druthers
##### The Lorentz Factor Conflict MMXI
« Reply #9 on: 13/05/2011 04:10:49 »
This revision (.01) addresses the issues of the previous version.
The Lorentz Factor Conflict MMXI.01
Problem
As viewed from rest, length perpendicular to the direction of motion is changed when the uniform relative velocity of the body is changed. This conflicts with the Special Relativity stipulation to the contrary.
Hypothesis
“When one side of a triangle has a fixed length and the length of another side is changed the length of the remaining side must also change.” This property of triangles dictates that since the ‘light path’ the hypotenuse has a fixed length and the length of another side is changed with velocity the remaining side, length perpendicular to the direction of motion, must also change.
Data
Fig. A, Fig. B & Fig. C
All are as viewed from relative rest.
Time duration for all (proper time) = 1 second.
IIA, IIB, & IIC light path: fixed length = 1 light second. (Light speed for 1 second.)
IIIB & IIIC the distance (length) this end of the light path has advanced in the direction of motion at a chosen velocity >0, <1c for 1 second. (For appearance IIIB < IIIC)
Fig. A: IA = height of body (length perpendicular to the direction of motion) at relative rest. Length = 1 light second
Fig. B: IIB length = 1 light second
IIIB length = .2 light seconds (.2c for 1 second)
Fig. C: IIC length = 1 light second
IIIC length = .6 light seconds (.6c for 1 second)
Fig B calculation:
a2+b2=c2 a2= c2- b2 IB2= IIB2- IIIB2 IB2=12-.22 IB2=1-.04
IB2=.96 IB=sqrt .96 IB=.98 light seconds
Fig C calculation:
a2+b2=c2 a2= c2- b2 IC2= IIC2- IIIC2 IC2=12-.62 IC2=1-.36
IC2=.64 IC=sqrt .64 IC=.8 light seconds
Length perpendicular to the direction of motion
IA at relative rest – 1 light second
IB at relative velocity .2c - .98 light seconds
IC at relative velocity .6c - .8 light seconds
Conclusion
Observed from relative rest, if the relative velocity of a body is changed the length of the body perpendicular to the direction of motion must also change.
Prediction
A new explanation for the results of the interferometer experiments by Michelson-Morley and others will be discovered.
#### yor_on
• Naked Science Forum GOD!
• Posts: 11582
• (Ah, yes:) *a table is always good to hide under*
##### The Lorentz Factor Conflict MMXI
« Reply #10 on: 15/06/2011 01:10:47 »
#### PhysBang
• Hero Member
• Posts: 579
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##### The Lorentz Factor Conflict MMXI
« Reply #11 on: 15/06/2011 22:44:19 »
This still doesn't make any sense whatsoever. This is not surprising for a claim that attempts to refute Special Relativity. There is little doubt that one must have a serious misunderstanding of the science and mathematics involved to put forward such a claim.
#### The Naked Scientists Forum
##### The Lorentz Factor Conflict MMXI
« Reply #11 on: 15/06/2011 22:44:19 » | 2,598 | 9,984 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2016-44 | longest | en | 0.856592 |
https://en.razmery.info/equipment/wheel_disks/dimensions-wheel-disks-lamborghini-gallardo-lp560-4.html | 1,624,043,071,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487640324.35/warc/CC-MAIN-20210618165643-20210618195643-00452.warc.gz | 219,253,664 | 7,441 | Disc sizes of Lamborghini Gallardo LP560-4
General parameters: Offset (ET): from 20 to x;Drilling (PCD): from 4x100 to x;Center hole (DIA): from 54.1 to x.Important: wheel disc parameter PCD indicates not only the diameter of circumference of the mounting holes, but also the number of mounting bolts or nuts (for example: 5x120, where 120 (mm) is the diameter of centers of the mounting holes, and 5 is the number of mounting bolts). Possible disk sizes for Lamborghini Gallardo LP560-4: 19x8.516x6.516x5.5 Procedure for calculating the disc offset: The distance between mating face of the wheel (the face with which the disk is pressed against the hub) and the middle of the width of disk is called disk offset (ET).The formula for calculating:ET = A - B/2, where:A is the distance between the inner plane of disk and the part in contact with the hub;B is disk width.The disk offset affects the width of the vehicle wheelbase, since the distance between the centers of symmetry (in width) of wheels on one axis directly depends on this parameter.Important: PCD of disks with four fixing bolts (or nuts) corresponds to the distance between the centers of opposite bolts (or nuts), PCD of disks with five fixing bolts (or nuts) corresponds to the distance between the centers of any non-adjacent bolts (or nuts) multiplied by the coefficient 1.051.
Wheel disks - full drawing
Possible disk parameters for Lamborghini Gallardo LP560-4
General view
General view
Disk
Possible disk sizes for Lamborghini Gallardo LP560-4 of different years of manufacture and modifications
Wheel
Possible wheel sizes for Lamborghini Gallardo LP560-4 of different years of manufacture and modifications
ET (mm)
Disk offset
Distance between the vertical plane of symmetry of the wheel and the plane of contact of the disk to the hub.
Measured in millimeters (mm)
(General view)
19''
16''
19x8.5
16x6.5
16x5.5
from 20 to x
Table of disc sizes of Lamborghini Gallardo LP560-4 modifications
Year
Model year of manufacture
Size
Disk size in format:
1. D - diameter of the disk in inches.
2. x - a sign indicating the inseparability of the disk.
3. B - disk width in inches.
ET (mm)
Disc offset:
Distance between the vertical plane of symmetry of the wheel and the plane of contact of the disc to the hub.
PCD (mm)
Disc drilling:
1. Number of bolt holes.
2. Diameter of the circle on which the mounting holes are located.
Tire
Tire size in format:
1. Width of the work face.
2. Percentage of profile height to width.
3. R - type of construction (radial) - inner diameter (in inches).
DIA (mm)
Center hole:
Diameter of center hole, another marking is CH.
5.2
201419x8.5425x112235/35R1957.1
201319x8.5425x112235/35R1957.1
201219x8.5425x112235/35R1957.1
201119x8.5425x112235/35R1957.1
201019x8.5425x112235/35R1957.1
200919x8.5425x112235/35R1957.1
200819x8.5425x112235/35R1957.1 | 777 | 2,870 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2021-25 | latest | en | 0.87093 |
http://fer3.com/arc/m2.aspx/Exmeridian-reasoning-muddle-LaPook-jan-2016-g34050 | 1,632,276,834,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057303.94/warc/CC-MAIN-20210922011746-20210922041746-00274.warc.gz | 23,627,478 | 7,334 | # NavList:
## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Message:αβγ
Message:abc
Add Images & Files Posting Code: Name: Email:
Re: Ex-meridian reasoning muddle?
From: Gary LaPook
Date: 2016 Jan 5, 05:35 +0000
Yes but here is nothing magical about having latitude, it is just another LOP that happens tor run straight east and west. to get a fix you still need to cross it with another LOP (perhaps advanced from the morning) and the east-west LOP is no more accurate or easy to use than any other LOP.
I think the reason for the ex-meridian process is that it dates from the days before LOPs in which latitude and longitude were determined separately and in the navigator's mind were also kept separately.
Just treat your sun shot near noon as a normal LOP shot. Any shot taken within two minutes of noon will produce the exact same Hc since the LHA of the sun will be the same whole number of degrees (since the AP will be within 30' of longitude of your DR) when you enter HO 249 or HO 229 or HO 214 or HO 218 or HO 208 or Weems LOP book. The Zn will always be 180 (or 360.) Even if you take the shot more than two minutes from noon (within reason) the resulting LOP will accurate enough for practical navigation, any inaccuracy lost in the noise of the observation.
gl
To: garylapook@pacbell.net
Sent: Monday, January 4, 2016 8:09 PM
Subject: [NavList] Re: Ex-meridian reasoning muddle?
Mark,
The ex-meridian needs knowledge of longitude as you say. So then what happens if the DR longitude is out by lets say 15' then how will this affect the ex-meridian correction. Example: Lat. 34° N, Dec. 22° S, LHA t 6°, and LHA t 6.25°
[1.964(.8290)(.9272)/.8290](.267)(6)(6) = 17.5' correction added to Ho
[1.964(.8290)(.9272)/.8290](.267)(6.25)(6.25) = 19' correction added to Ho
A difference of 1.5' only. Not much to worry about.
The benefit of the ex-meridian is that it is easier to calculate than the intercept method and gives latitude directly without the need of plotting.
Greg Rudzinski
Date: 2016 Jan 4, 18:36 -0800
I had managed to miss the ex-meridian exercise in my past midday shots education. I had learned a meridian shot is a nice way to get a good latitude and a so-so longitude without any sight reduction pain.
I understand the ex-meridian concept is if you miss your meridian averaging shots or "on the money” peak meridian shot, it gets you back to square one with a little mathematical hocus pocus.
The only thing is that bothers me is we are trying to get back to an accurate latitude using my estimated longitude to get LHA. In modern times with my accurate chronometers, if I have confidence in my longitude...I probably already have similar confidence in my latitude....so why not just take that sight as a LOP , bring up the morning LOP if available on the DR track, then shoot another in the afternoon, and have a 3 sight St Hilaire party with advanced DR’d LOP’s.
If we are doing this in olden days with more suspect chronometers..and thus suspect longitude....pre St Hilaire, wouldn't I play the safe game and use my tools to zero up the more sure latitude rather than use possibly tweaked longitude to calculate?
I mean latitude I might catch from other star or planet bodies. I understand St Hilaire wasn't around yet....If I understand correctly Sumner got a lat based LOP in 1848 by using three latitudes...I assume because he was more certain of Lat than Lon.
With the ex-meridian I feel like I am using a single suspect variable to calculate a suspect variable, rather than using multiple LOP’s to slowly reign in all the suspectness........if you get what I mean...
This probably means I missed some subtle mathematical elegance or the point completely.... I am trying to figure out when this is the right thing to do.
MC
PS I have my Fathers slide rule he used for navigation at the USCG. It is an intimidating looking Pickett N4-T, vector type LOG LOG. I remember him zooming through Nav problems with it when I was young in the 60's and early 70's. I always marveled at his ability to do seat of the pants navigation in his head. He told me it was because he grew up with the slide rule and manual calculation (he also had a near photographic memory). I treasure it for sentimental reasons, and would secretly like to learn to use it someday. I grew up at the very end of the slide rule era, and switched to the calculator too young to remember the slide rule much.
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https://community.fabric.microsoft.com/t5/DAX-Commands-and-Tips/Date-Range-based-on-Slicer-selection/m-p/3457325/highlight/true | 1,716,160,349,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058009.3/warc/CC-MAIN-20240519224339-20240520014339-00389.warc.gz | 166,762,848 | 117,713 | cancel
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Frequent Visitor
## Date Range based on Slicer selection
Hello - I'm looking to return a list of the previous "X" days based on a selected day from a slicer.
For example, once a day is selected in the 'Date' slicer, I want to return the date range in the visualization for the previous 180 days up to the selected day.
In simple words, the logic would be something like the below, but I cannot make it work:
IF ( Date >= SELECTEDVALUE ( Date )-180 || Date <= SELECTEDVALUE ( Date ) , Date , blank() )
And then use 'Date' in the visualization and exclude blanks.
Any recommendation is more than welcome, thanks!!
1 ACCEPTED SOLUTION
Frequent Visitor
Thanks @DataNinja777 . This can work if I want to have the cumulative "Daily Trends" up to the selected day.
For this purpose I find this other solution:
Measure =
if( MAX( 'Main_Table'[Date]) >= SELECTEDVALUE('Date'[Date],min('Date'[Date]) )-180 &&
MAX( 'Main_Table'[Date]) <= SELECTEDVALUE('Date'[Date],max('Date'[Date]) ),
"Yes", "No" )
And then use this 'Measure' as a filter in the visualization.
6 REPLIES 6
Frequent Visitor
Thanks @DataNinja777 . Your solutions provided a cumulative count for the last "X" days, while I was looking to include/exclude days based on the selection. I made it work with the following Measure and by adding a separate Table with just the 'Date' field, which is used as filter.
Measure_last_180_days =
IF (SELECTEDVALUE('Date'[Date])-180 <= MAX( 'Main Table'[Date]) &&
SELECTEDVALUE('Date'[Date]) >= MAX('Main Table'[Date]),"yes","no" )
Frequent Visitor
Thanks @DataNinja777 , this can help as it outputs the [Daily trends] for the previous 180 days, but it doesn't have any relation with the SELECTEDVALUE.
Below are two examples of how I would like this visualization to show based on the selected dates:
thanks!
Super User
Hi @ragnezza ,
Frequent Visitor
Thanks @DataNinja777 . This can work if I want to have the cumulative "Daily Trends" up to the selected day.
For this purpose I find this other solution:
Measure =
if( MAX( 'Main_Table'[Date]) >= SELECTEDVALUE('Date'[Date],min('Date'[Date]) )-180 &&
MAX( 'Main_Table'[Date]) <= SELECTEDVALUE('Date'[Date],max('Date'[Date]) ),
"Yes", "No" )
And then use this 'Measure' as a filter in the visualization.
Community Support
Hi @ragnezza ,
It seems that you have gotten a solution. Could you please mark the helpful post as Answered? It will help the others in the community find the solution easily if they face the same problem as yours. Thank you.
Best Regards
Community Support Team _ Rena
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Super User
Hi @ragnezza,
DATESINPERIOD function (DAX) - DAX | Microsoft Learn
Something like below is how it can be used:
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Top Kudoed Authors | 815 | 3,313 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.847066 |
econdojo.github.io | 1,686,030,154,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652235.2/warc/CC-MAIN-20230606045924-20230606075924-00687.warc.gz | 259,684,116 | 8,742 | This article is based on an old note I wrote in Spring 2014. I post it here to test the blog feature of this website. I review one of the most important notions in econometrics, namely the notion of identification, following the seminal paper Hurwicz (1962). In what follows, I begin with a general definition of identification.
#### 1 General Definition
In the context of econometrics, identifiability of the unknown quantity (e.g. parameters or functions of parameters) of our interest serves as the necessary condition for the existence of a consistent estimator for that quantity. The condition is only necessary because the estimator may not converge in a probabilistic sense to the true value of the quantity due, for example, to the existence of excessive dependency in the data. That is, the law of large numbers could fail even when the unknown quantity can be identified. If one could not logically deduce the value of the unknown quantity from a presample analysis, then the usual argument for proving the consistency of an estimator would fail, let alone the derivation of its asymptotic distribution.
To fix idea, we denote by $$P_X$$ the true distribution of the observed data, which is the probability measure on the state space of random element $$X$$. Also, let
$\mathcal{P}=\{P_{\theta}:\theta\in\Theta\}$
be our model for the distribution of the same data, which is indexed by the parameter $$\theta$$. The model $$\mathcal{P}$$ is of particular interest because we are intended to interpret it as a structural model for the distribution of the observed data that helps us not only summarize various kinds of statistics, but also understand the underlying mechanism that generates the data. We will introduce the notion of structural model in the next section. Also, there will be no need to introduce such a model if our interest lies only in the statistical characterization of the observed data. $$P_X$$ alone would suffice for that purpose.
We assume that the model $$\mathcal{P}$$ is complete in the following sense: there exists some $$\theta\in\Theta$$ such that $$P_{\theta}=P_X$$. Cautions are needed here because it is possible that there exists another $$\theta^*\in\Theta$$ such that $$\theta^*\neq\theta$$ and $$P_{\theta^*}=P_X$$ if, for example, the observed data is not sufficiently informative. Therefore, the only legitimate claim we can make from the knowledge of $$P_X$$ alone is that
$\theta\in\Theta_0(P_X):=\{\theta\in\Theta:P_{\theta}=P_X\}$
where $$\Theta_0(P_X)$$ is referred to as the identified set. We say that $$\theta$$ is identified if $$\Theta_0(P_X)$$ is a singleton. Identification in our general discussion so far is really a global concept. Rothenberg (1971) showed that the necessary and sufficient condition for local identification is the nonsingularity of the Fisher information matrix. Iskrev (2008) shows how the information matrix of linearized DSGE models can be evaluated analytically, which can be used to check the local identifiability in DSGE models. Thus, the natural question is that under what restrictions on $$\Theta_0(P_X)$$ can the identification of $$\theta$$ be achieved. We will explore these restrictions in the next section.
As an example, consider the following linear regression model $$Y=X'\beta+\epsilon$$, where $$X$$ is a $$k\times 1$$ vector and $$\theta=(P_X,\beta,P_{\epsilon|X})$$. A standard set of restrictions on $$\Theta_0(P)$$ under which $$\theta$$ can be identified are the following:
A1. $$\mathbb{E}_{P_{\theta}}[\epsilon X]=0$$.
A2. $$\mathbb{E}_{P_{\theta}}[XX']$$ is nonsingular.
Note that restriction (A1) is usually referred to as the orthogonality condition in the traditional framework of classical regression models. Park (2010) proposes a more general class of regression models, called the martingale regression since it is identified by the condition that the error process is a martingale. It allows for the presence of arbitrary time-varying and stochastic volatilities that are often quite persistent and strongly endogenous.
#### 2 A Priori Information and Prediction
Our general definition of identification has hid two essential points of identification: both the degree of and the need for identification are relative notions. First, we describe the identification issue one might encounter even in the simplest case. To fix idea, let $$x=[x_1,x_2,\ldots,x_n]'$$ denote the state of a configuration that takes the form of simultaneous equations system given by \begin{gather} f_i(x)=0,\qquad i=1,2,\ldots,m\label{configuration} \end{gather} Hurwicz refers to $$f_i$$ as the behavior pattern of the $$i$$-th component in the configuration. In the context of probabilistic models, the zero’s on the right hand side of \eqref{configuration} should be replaced by stochastic disturbance terms so that $$f_i$$’s take the form of, for example, first-order conditions derived from agents’ optimizing behavior in a dynamic stochastic general equilibrium (DSGE) model, or the classical linear regression model given in the previous section.
In what follows, we will focus only on the linear deterministic behavior pattern which, in matrix notation, can be compactly written as \begin{gather} Ax-b=0\label{linear} \end{gather} where the behavior pattern is now completely determined by the $$m\times n$$ matrix $$A$$ and the $$m\times 1$$ vector $$b$$. Note that \eqref{linear} imposes a restriction on the possible values that $$x$$ can take under the particular configuration specified by $$(A,b)$$. Let $$\mathcal{H}$$ be the state space of $$x$$ that is spanned by the true values of $$A$$ and $$b$$. Following the language and notation established in the previous section, we can take $$P_X$$ to be $$\mathcal{H}$$ here because the knowledge of $$P_X$$ boils down to that of $$\mathcal{H}$$ in a deterministic setting. Therefore, the identified set of $$(A,b)$$ can be written as
$\Theta_0(\mathcal{H})=\{(A,b):Ax-b=0,\ \ \ \forall\ x\in\mathcal{H}\}$
and the completeness assumption of our model $$\mathcal{P}$$ amounts to requiring that $$\Theta_0(\mathcal{H})$$ is a nonempty set. Cautions are needed again because it is straightforward to see that for any $$(A,b)\in\Theta_0(\mathcal{H})$$ and any $$m\times m$$ invertible matrix $$P$$, the combination \begin{gather} C=PA\ \ \ \text{and}\ \ \ d=Pb\label{transform} \end{gather} is also contained in $$\Theta_0(\mathcal{H})$$. Since these two behavior patterns both span the state space $$\mathcal{H}$$, we say that $$(A,b)$$ and $$(C,d)$$ are observationally equivalent with data. Therefore, the knowledge of $$\mathcal{H}$$ alone fails to single out the true behavior pattern specified by a particular combination of $$A$$ and $$b$$.
The above discussion still remains unclear about how much identification one is able to achieve and what purposes is identification intended for. Indeed, both the degree of and the need for identification are only defined as relative notions that are intimately connected.
##### 2.1 Identification is no free lunch
Because the knowledge of $$\mathcal{H}$$ alone does not suffice here, we may look for additional restrictions that are not contained in $$\mathcal{H}$$ for the purpose of identification. These extra restrictions are usually called a priori information, which comes from our economic theory or empirical experience, etc., and can be conveniently formulated in the Bayesian framework by imposing appropriate prior probability measures on $$(A,b)$$. Thus, the imposition of a priori information helps us rule out all those transformation matrices $$P$$ in \eqref{transform} that are incompatible with the restrictions provided by the a priori information. To see this more clearly, consider the following two extreme cases:
• If the a priori information is sufficiently restrictive about \eqref{transform} so that the class of all transformation matrices boils down to one containing only diagonal matrices, then we say that the behavioral pattern $$(A,b)$$ can be identified (up to normalization).
• If our economic theory or empirical experience remains silent, that is, there is no readily available a priori information, then all invertible transformation matrices are allowed and we say that the behavioral pattern $$(A,b)$$ cannot be identified.
Therefore, the cardinality of $$\mathcal{P}_I$$, which is the set of all transformation matrices that are compatible with a given a priori information $$I$$, measures exactly the degree of identification one is able to achieve relative to her a priori information $$I$$. That is, identification might be quite expensive and $$I$$ is the price that one must pay for it. But do we always need to pay a high price in order to buy identification?
##### 2.2 There is price discount on identification
As we shall see, the price of identification really depends on the specific purposes our behavioral patter is intended for. Here we closely follow Hurwicz and interpret the “need” for identification as the “need for purposes of prediction”. This requires a clear distinction between the true old behavioral pattern and all the possible modified behavioral patterns that we are intended to predict. For notational ease, let $$w$$ be a modifying variable that takes values in its domain $$\mathcal{W}$$. (Imagine that $$w$$ represents a particular outcome in our underlying “probability space” $$\mathcal{W}$$.) Let $$w^*$$ be the realized historical value of $$w$$. Then the true old behavioral pattern that generated $$\mathcal{H}$$ can be written as the combination of $$A^*=A(w^*)$$ and $$b^*=b(w^*)$$. Moreover, all the possible modified behavioral patterns, indexed by elements of $$W$$, can be written as
$A_0(w)=\phi[A^{*},b^{*},w],\ \ \ b_0(w)=\psi[A^{*},b^{*},w]$
where both $$\phi$$ and $$\psi$$ are known functions to us. That is, knowledge of the true old behavioral pattern is sufficient to determine the predicted behavioral pattern indexed by $$w$$, although the true old one itself remains undetermined. Now our need for identification of the true old behavioral pattern is for the purpose of finding the set of all predicted states of the configuration over the entire possibilities of our anticipated modifications
$X_0(\mathcal{\mathcal{W}})=\{x:A_0(w)x-b_0(w)=0,\ \ \ \forall\ w\in\mathcal{W}\}$
Similarly, the above set based on alternative old behavioral pattern (not necessarily the true one) can be written as
$X_P(\mathcal{\mathcal{W}})=\{x:A_P(w)x-b_P(w)=0,\ \ \ \forall\ w\in\mathcal{W}\}$
where
$A_P(w)=\phi[PA^*,Pb^*,w]\ \ \ b_P(w)=\psi[PA^*,Pb^*,w]$
for some transformation matrix $$P$$. Clearly, $$X_0(\mathcal{\mathcal{W}})$$ plays a similar role as the a priori information $$I$$ in ruling out all those transformation matrices $$P$$ in \eqref{transform} that are incompatible with the restrictions provided by $$X_0(\mathcal{\mathcal{W}})$$, i.e. all $$P$$’s such that $$X_P(\mathcal{W})\neq X_0(\mathcal{W})$$. In much of the structural vector autoregression (SVAR) literature, for example, $$X_0(\mathcal{\mathcal{W}})$$ corresponds to the restrictions imposed on the contemporaneous matrix with the interpretation that there are delays in agents’ reactions to the disturbances originating outside of their own sector. To see this more clearly, consider again the following two extreme cases:
• If $$\mathcal{W}$$ is sufficiently restrictive about \eqref{transform}, then the class of all transformation matrices compatible with $$X_0(\mathcal{\mathcal{W}})$$ boils down to one containing only diagonal matrices.
• If $$\mathcal{W}=\{w^*\}$$ so that the future configuration remains the same as the old one, then all invertible $$m\times m$$ transformation matrices are compatible with $$X_0(\mathcal{\mathcal{W}})$$.
Therefore, the elements of $$\mathcal{P}_{\mathcal{W}}$$, which is the set of all transformation matrices that are compatible with $$X_0(\mathcal{\mathcal{W}})$$, prescribe exactly those behavioral patterns that yield the same prediction. Now it becomes clear that, at least for the purpose of prediction, it suffices to impose a priori information enough to eliminate all those transformation matrices outside of $$\mathcal{P}_{\mathcal{W}}$$ so that
$\mathcal{P}_{\mathcal{I}}\subseteq\mathcal{P}_{\mathcal{W}}$
This is the main result of Hurwicz: the need for knowledge of the true old behavioral pattern is not absolute, but relative to the modification domain. Therefore, we will always get a price discount for buying identification if our anticipated modifications are not rich enough.
As a final remark, if $$\mathcal{P}_{\mathcal{W}}$$ contains only diagonal matrices, then we call $$(A^*,b^*)$$ the structure of the configuration with respect to the modification domain $$W$$. | 3,082 | 12,760 | {"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} | 3.1875 | 3 | CC-MAIN-2023-23 | longest | en | 0.880036 |
https://calculatorsbag.com/calculators/math/addition-calculator | 1,695,536,378,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506623.27/warc/CC-MAIN-20230924055210-20230924085210-00727.warc.gz | 185,427,911 | 8,522 | If you want to calculate the sum of two or more numbers, then you can use our addition calculator.
The addition is one of the arithmetic operations in mathematics. It can be defined as a mathematical process to sum a total quantity. Or in other words, an addition is a process of adding two or more numbers, which can be represented by “+”. For instance, if you have two numbers, 8 and 9, then the addition of these two numbers will be 8 + 9 = 17.
Suppose you want to find out the addition of two or more numbers. In that case, you can use our calculator, or to determine it manually, you can use the representation listed below:
Addition of number = a + b + .......
Where,
a and b can be any real number.
### Example
For a more precise understanding of a concept, let us solve an example below:
Suppose if you have two numbers, 11 and 8 then find the addition of two numbers.
Given data
x = 11
y = 8
To Find
Solution
To find out the addition of number, we will use the formula listed below:
Addition of number = x + y
Putting values in the formula:
Addition of number = 11 + 8 = 19
### How to use the Addition Calculator?
The steps to use the addition calculator are as follows:
Step 1: Enter any number in the first required input field.
Step 3: The calculator will automatically display an answer on the screen.
### Calculator use
You can use our addition calculator in every mathematical and arithmetic operation to instantly find the resultant answer. | 340 | 1,478 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2023-40 | longest | en | 0.906731 |
https://sofsource.com/algebra-2-chapter-4-resource-book/relations/1-best-selling-book-for.html | 1,590,509,349,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347391277.13/warc/CC-MAIN-20200526160400-20200526190400-00481.warc.gz | 540,517,715 | 11,886 | Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
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Mox
Registered: 18.05.2004
From: 3cities, Poland
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Registered: 08.03.2007
From: egypt,alexandria
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Majnatto
Registered: 17.10.2003
From: Ontario
Posted: Monday 01st of Jan 09:16 I would advise using Algebrator. It not only helps you with your math problems, but also gives all the necessary steps in detail so that you can improve the understanding of the subject. | 652 | 2,487 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.897679 |
https://metanumbers.com/14681 | 1,675,912,655,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764501066.53/warc/CC-MAIN-20230209014102-20230209044102-00517.warc.gz | 396,954,552 | 7,536 | 14681 (number)
14,681 (fourteen thousand six hundred eighty-one) is an odd five-digits composite number following 14680 and preceding 14682. In scientific notation, it is written as 1.4681 × 104. The sum of its digits is 20. It has a total of 2 prime factors and 4 positive divisors. There are 14,352 positive integers (up to 14681) that are relatively prime to 14681.
Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 20
• Digital Root 2
Name
Short name 14 thousand 681 fourteen thousand six hundred eighty-one
Notation
Scientific notation 1.4681 × 104 14.681 × 103
Prime Factorization of 14681
Prime Factorization 53 × 277
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 14681 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 14,681 is 53 × 277. Since it has a total of 2 prime factors, 14,681 is a composite number.
Divisors of 14681
1, 53, 277, 14681
4 divisors
Even divisors 0 4 4 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 15012 Sum of all the positive divisors of n s(n) 331 Sum of the proper positive divisors of n A(n) 3753 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 121.165 Returns the nth root of the product of n divisors H(n) 3.9118 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 14,681 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 14,681) is 15,012, the average is 3,753.
Other Arithmetic Functions (n = 14681)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 14352 Total number of positive integers not greater than n that are coprime to n λ(n) 3588 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1724 Total number of primes less than or equal to n r2(n) 16 The number of ways n can be represented as the sum of 2 squares
There are 14,352 positive integers (less than 14,681) that are coprime with 14,681. And there are approximately 1,724 prime numbers less than or equal to 14,681.
Divisibility of 14681
m n mod m 2 3 4 5 6 7 8 9 1 2 1 1 5 2 1 2
14,681 is not divisible by any number less than or equal to 9.
Classification of 14681
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
Other numbers
• LucasCarmichael
Base conversion (14681)
Base System Value
2 Binary 11100101011001
3 Ternary 202010202
4 Quaternary 3211121
5 Quinary 432211
6 Senary 151545
8 Octal 34531
10 Decimal 14681
12 Duodecimal 85b5
20 Vigesimal 1ge1
36 Base36 bbt
Basic calculations (n = 14681)
Multiplication
n×y
n×2 29362 44043 58724 73405
Division
n÷y
n÷2 7340.5 4893.67 3670.25 2936.2
Exponentiation
ny
n2 215531761 3164221783241 46453939999761121 681990293136493017401
Nth Root
y√n
2√n 121.165 24.486 11.0075 6.8132
14681 as geometric shapes
Circle
Diameter 29362 92243.4 6.77113e+08
Sphere
Volume 1.32543e+13 2.70845e+09 92243.4
Square
Length = n
Perimeter 58724 2.15532e+08 20762.1
Cube
Length = n
Surface area 1.29319e+09 3.16422e+12 25428.2
Equilateral Triangle
Length = n
Perimeter 44043 9.3328e+07 12714.1
Triangular Pyramid
Length = n
Surface area 3.73312e+08 3.72907e+11 11987
Cryptographic Hash Functions
md5 628f7dc50810e974c046a6b5e89246fc 0bb4f098e2b308fdfe5c5f7351d9cf54638f37a9 41bef2ec14f0edd1df07db4ac5554a008a73373d5b7e677232f736d29d37034a 05b12dc538ca75085a153a17accde157bc02cb41ce2eeed81bd7896fd1c885e7d1254e001afed79e545b75169a95423d5da88422d6637701cbac76d585ab4612 c97868867c2d706b4d3038c28816d17c82f4d50a | 1,422 | 4,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2023-06 | latest | en | 0.815719 |
https://www.sofolympiadtrainer.com/forum/imo/0/488 | 1,618,657,258,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038119532.50/warc/CC-MAIN-20210417102129-20210417132129-00236.warc.gz | 1,087,106,832 | 34,396 | # User Forum
Subject :IMO Class : Class 3
## Ans 1:
Class : Class 3
a
Subject :IMO Class : Class 5
## Ans 1:
Class : Class 8
the answer is c not b because 1*140=140,2*70=140,5*28=140,10*14=140
## Ans 2:
Class : Class 5
Subject :IMO Class : Class 3
Class : Class 3
d
Class : Class 3
## Ans 3:
Class : Class 4
Subject :IMO Class : Class 2
## Ans 1:
Class : Class 2
Subject :IMO Class : Class 6
Class : Class 8
b is correct
Class : Class 6
ans is b
Class : Class 7
Class : Class 7
b
Class : Class 6
## Ans 6:
Class : Class 6
some are saying a, some are saying b, some are saying c and some are saying d. what is the correct answer??????
## Ans 7:
Class : Class 6
sorry, a is the correct one
Class : Class 6
## Ans 9:
Class : Class 6
I am confused what is the answer because some are saying a , some are saying b ,some are saying c and some are saying d .
Class : Class 9
## Ans 11:
Class : Class 7
According to me, A is the correct answer.
Class : Class 6
Class : Class 8
b is correct
Class : Class 6
Class : Class 8
its b
## Ans 16:
Class : Class 6
according to me , its B as if two numbers are divisible by a number , their sum and difference is not divisible by the same number . For example : 2
Class : Class 7
a is correct
## Ans 18:
Class : Class 9
According to me ,it is C
## Ans 19:
Class : Class 6
it should be A because if we take 39 and 26 as multiples and 13 as factor then 39-26=13. which is divisible by 13
Class : Class 6
## Ans 21:
Class : Class 6
its b because say if you take 6,and you say 2 and 3 are divisible by it. then they are saying that their sum and diff is also divisible but 2 3 is 5 which is not divisible by 6
Class : Class 6
## Ans 23:
Class : Class 6
Subject :IMO Class : Class 4
Class : Class 5
Class : Class 6
a
## Ans 3:
Class : Class 5
The option is A
Subject :IMO Class : Class 6
## Ans 1:
Class : Class 5
5 million
Subject :IMO Class : Class 6
## Ans 1:
Class : Class 6
Subject :IMO Class : Class 7
## Ans 1:
Class : Class 10
the angles are 80 degree and 100 degree. Method- 4:5=4x and 5x , 4x+5x=180 , 9x=180, x=180/9, x=20, therefore, 4:5= 4*20and 5*20 4:5=80 and 100
## Ans 2:
Class : Class 7
80,100
Subject :IMO Class : Class 5
Class : Class 5
IT IS 8 ONLY
## Ans 2:
Class : Class 7
The answer is obviously 4.There cannot be 8 at all costs
## Ans 3:
Class : Class 8
So report the error man
## Ans 4:
Class : Class 8
it cannot be 8 at all costs
Class : Class 5
B
## Ans 6:
Class : Class 6
No, it is 8. You can see 4 of them easily. The other four are hidden between the sharp edges of the figure. Tilt it 2 find out.
## Ans 7:
Class : Class 4
there cannot be 8
## Ans 8:
Class : Class 7
Very good Nilarka Pahari and Khushie. Friends , the correct answer is B only. Just try to see . Tilt your head to your right and try to see the cuts at the end of the ribbons . You will find the error of your own. Did you find that? I know that only 5 to 37 percent of you can nfind the eight angles. Be very accurate and intelligent. I also did the same mistake while giving the test . Please look carefully!
## Ans 9:
Class : Class 8
It cannot be 8 at all costs There is an error in this question
Class : Class 5
Class : Class 7
## Ans 12:
Class : Class 4
No ,the answer is 8 because those small angles are also right angle
## Ans 13:
Class : Class 4
Look closely. The answer is 8. If you look at the cuts in the ribbon which looks like a V at the top, you can see the right angle by tilting your device. Same follows with the other cuts
Class : Class 7
Class : Class 5
Class : Class 6
Class : Class 5
Class : Class 5
D.8
## Ans 19:
Class : Class 7
Thank you very much friends if you all co-operate me, Nilarka and Khushie!
Class : Class 5
How ??
Class : Class 7
4
Class : Class 5
Class : Class 6
Class : Class 6
## Ans 25:
Class : Class 5
4/8 votei vote 4 they have done wrong with no solution
## Ans 26:
Class : Class 6
answer is 4. how can it be 8
Class : Class 5
It should be 4 | 1,315 | 4,044 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | longest | en | 0.92048 |
https://www.coursehero.com/file/51606/Chapter14/ | 1,547,885,712,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583662863.53/warc/CC-MAIN-20190119074836-20190119100836-00449.warc.gz | 769,362,224 | 165,847 | # Chapter14 - 14.1 Solve The frequency generated by a guitar...
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14.1. Solve: The frequency generated by a guitar string is 440 Hz. The period is the inverse of the frequency, hence T f = = = × = 1 1 2 27 10 2 27 3 440 Hz s ms . .
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14.2. Solve: Your pulse or heart beat is 75 beats per minute. The frequency of your heart’s oscillations is thus f = = = 75 beats 60 s beats s Hz 1 25 1 25 . . The period is the inverse of the frequency, hence T f = = = 1 1 0 80 1.25 Hz s .
14.3. Model: The air-track glider oscillating on a spring is in simple harmonic motion. Solve: The glider completes 10 oscillations in 33 s, and it oscillates between the 10 cm mark and the 60 cm mark. (a) T = = = 33 s 10 oscillations s oscillation s 3 3 3 3 . . (b) f T = = = 1 1 0 303 3.3 s Hz . (c) ω π π = = ( ) = 2 2 1 904 f 0.303 Hz rad s . (d) The oscillation from one side to the other is equal to 60 cm – 10 cm = 50 cm = 0.50 m. Thus, the amplitude is A = ( ) = 1 2 0 25 0.50 m m. . (e) The maximum speed is v A T A max 1.904 rad s 0.25 m m s = = = ( ) ( ) = ω π 2 0 476 .
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14.4. Model: The air-track glider attached to a spring is in simple harmonic motion. Visualize: The position of the glider can be represented as x ( t ) = A cos ω t . Solve: The glider is pulled to the right and released from rest at t = 0 s . It then oscillates with a period T = 2 0 . s and a maximum speed v max cm s m s = = 40 0 40 . . (a) v A T A v max max and s rad s m s rad s m cm = = = = = = = = ω ω π π π ω π 2 2 2 0 0 40 0 127 12 7 . . . . (b) The glider’s position at t = 0.25 s is x 0.25 s 0.127 m rad s 0.25 s m cm = ( ) ( ) ( ) [ ] = = cos . . π 0 090 9 0
14.5. Model: The oscillation is the result of simple harmonic motion. Visualize: Please refer to Figure Ex14.5. Solve: (a) The amplitude A = 10 cm. (b) The time to complete one cycle is the period, hence T = 2 0 . s and f T = = = 1 1 2 0 0 50 . . s Hz (c) The position of an object undergoing simple harmonic motion is x t A t ( ) = + ( ) cos ω φ 0 . At s, cm, t x = = 0 5 0 thus 5 cm cm s 5 cm 10 cm rad or 60 = ( ) ( ) + [ ] = = = = ° 10 0 1 2 1 2 3 0 0 0 1 cos cos cos ω φ φ φ π
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14.6. Model: The oscillation is the result of simple harmonic motion. Visualize: Please refer to Figure Ex14.6. Solve: (a) The amplitude A = 20 cm. (b) The period T = 4.0 s, thus f T = = = 1 1 0 25 4.0 s Hz . (c) The position of an object undergoing simple harmonic motion is x t A t ( ) = + ( ) cos ω φ 0 . At t x = = − 0 10 0 s, cm. Thus, = ( ) = = = ± = ± ° 10 1 2 2 3 120 0 0 1 1 cm 20 cm 10 cm 20 cm rad cos cos cos φ φ π Because the object is moving to the right at t = 0 s, it is in the lower half of the circular motion diagram and thus must have a phase constant between π and 2 π radians. Therefore, φ π 0 2 3 120 = − = − ° rad .
14.7. Visualize: The phase constant 2 3 π has a plus sign, which implies that the object undergoing simple harmonic motion is in the second quadrant of the circular motion diagram. That is, the object is moving to the left. Solve: The position of the object is x t A t A ft t ( ) = + ( ) = + ( ) = ( ) ( ) + [ ] cos cos cos ω φ π φ π π 0 0 2 3 2 4 4.0 cm rad s rad The amplitude is A = 4 cm and the period is T f = = 1 0 50 . s. A phase constant φ π 0 2 3 1 = = ° rad 20 (second quadrant) means that x starts at 1 2 A and is moving to the left (getting more negative).
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https://mcqslearn.com/electronics/advance-electromagnetic-theory/quiz/quiz.php?page=12 | 1,718,734,420,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861773.80/warc/CC-MAIN-20240618171806-20240618201806-00851.warc.gz | 334,145,259 | 16,740 | Engineering Courses
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# Bohr's Model Multiple Choice Questions (MCQ) PDF - 12
Books:
Apps:
The Bohr's Model MCQ with Answers PDF, Bohr's Model Multiple Choice Questions (MCQs) PDF Download e-Book Ch. 2-12 to prepare Electromagnetic Theory Practice Tests. Learn Electrical Properties of Matter Test PDF, Bohr's Model Multiple Choice Questions (MCQ Quiz) for online undergraduate engineering schools. The Bohr's Model MCQ App Download: Free certification app for metamaterials basics, dielectric constant of dielectric materials, dielectrics basics, differential form of maxwell equations, bohr's model test prep to study online schools courses.
The MCQ Quiz: Electrons can neither absorbs nor radiates energy when electron's; "Bohr's Model" App (Play Store & App Store) with answers: Energy level is zero; Energy level is maintained; Energy level is varying; Energy level is absorbed; for online undergraduate engineering schools. Practice Electrical Properties of Matter Questions and Answers, Apple Book to download free sample for online high school college acceptance.
## Bohr's Model Questions and Answers : Quiz 12
MCQ 56:
Electrons can neither absorbs nor radiates energy when electron's
1. energy level is maintained
2. energy level is zero
3. energy level is varying
4. energy level is absorbed
MCQ 57:
Divergence of the curl of any vector field is always
1. 0
2. 1
3. infinite
4. can't possible
MCQ 58:
Dielectrics are basically
1. insulators
2. semiconductors
3. superconductors
4. conductors
MCQ 59:
Dieelectric constant of air is
1. 1.0006
2. 1.03
3. 2.1
4. 2.2
MCQ 60:
Metamaterials literally stand for
1. materials beyond conduction ones
2. non referential material
3. materials beyond artificial ones
4. materials beyond natural ones | 458 | 1,853 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.770932 |
http://metamath.tirix.org/mpeascii/anim2d | 1,719,080,526,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862410.56/warc/CC-MAIN-20240622175245-20240622205245-00398.warc.gz | 17,471,260 | 1,588 | # Metamath Proof Explorer
## Theorem anim2d
Description: Add a conjunct to left of antecedent and consequent in a deduction. (Contributed by NM, 14-May-1993)
Ref Expression
Hypothesis anim1d.1
`|- ( ph -> ( ps -> ch ) )`
Assertion anim2d
`|- ( ph -> ( ( th /\ ps ) -> ( th /\ ch ) ) )`
### Proof
Step Hyp Ref Expression
1 anim1d.1
` |- ( ph -> ( ps -> ch ) )`
2 idd
` |- ( ph -> ( th -> th ) )`
3 2 1 anim12d
` |- ( ph -> ( ( th /\ ps ) -> ( th /\ ch ) ) )` | 171 | 465 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-26 | latest | en | 0.736827 |
http://www.discuss.wmie.uz.zgora.pl/php/discuss3.php?ip=&url=plik&nIdA=16147&sTyp=HTML&nIdSesji=-1 | 1,537,781,238,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160337.78/warc/CC-MAIN-20180924090455-20180924110855-00170.warc.gz | 304,264,045 | 3,037 | ## TOTAL EDGE IRREGULARITY STRENGTH OF TREES
Jaroslav Ivanco and Stanislav Jendrol'
Institute of Mathematics
P.J. Safarik University
Jesenna 5, SK-041 54 Košice, Slovak Republic
e-mail: ivanco@science.upjs.sk
e-mail: jendrol@Košice.upjs.sk
## Abstract
A total edge-irregular k-labelling ξ:V(G)∪ E(G)→{1,2,...,k} of a graph G is a labelling of vertices and edges of G in such a way that for any different edges e and f their weights wt(e) and wt(f) are distinct. The weight wt(e) of an edge e = xy is the sum of the labels of vertices x and y and the label of the edge e. The minimum k for which a graph G has a total edge-irregular k-labelling is called the total edge irregularity strength of G, tes(G). In this paper we prove that for every tree T of maximum degree Δ on p vertices
tes(T) = max {⎡(p+1)/3⎤, ⎡(Δ+1)/2⎤}.
Keywords: graph labelling, tree, irregularity strength, total labellings, total edge irregularity strength.
2000 Mathmatics Subject Classification: 05C78, 05C05.
## References
[1] M. Aigner and E. Triesch, Irregular assignment of trees and forests, SIAM J. Discrete Math. 3 (1990) 439-449, doi: 10.1137/0403038. [2] D. Amar and O. Togni, Irregularity strength of trees, Discrete Math. 190 (1998) 15-38, doi: 10.1016/S0012-365X(98)00112-5. [3] M. Bača, S. Jendrol' and M. Miller, On total edge irregular labelling of trees, (submitted). [4] M. Bača, S. Jendrol', M. Miller and J. Ryan, On irregular total labellings, Discrete Math. 307 (2007) 1378?1388, doi: 10.1016/j.disc.2005.11.075. [5] T. Bohman and D. Kravitz, On the irregularity strength of trees, J. Graph Theory 45 (2004) 241-254, doi: 10.1002/jgt.10158. [6] L.A. Cammack, R.H. Schelp and G.C. Schrag, Irregularity strength of full d-ary trees, Congr. Numer. 81 (1991) 113-119. [7] G. Chartrand, M.S. Jacobson, J. Lehel, O.R. Oellermann, S. Ruiz and F. Saba, Irregular networks, Congr. Numer. 64 (1988) 187-192. [8] A. Frieze, R.J. Gould, M. Karoński and F. Pfender, On graph irregularity strength, J. Graph Theory 41 (2002) 120-137, doi: 10.1002/jgt.10056. [9] J.A. Gallian, Graph labeling, The Electronic Jounal of Combinatorics, Dynamic Survey DS6 (October 19, 2003). [10] J. Lehel, Facts and quests on degree irregular assignment, in: Graph Theory, Combin. Appl. vol. 2, Y. Alavi, G. Chartrand, O.R. Oellermann and A.J. Schwenk, eds., (John Wiley and Sons, Inc., 1991) 765-782. [11] T. Nierhoff, A tight bound on the irregularity strength of graphs, SIAM J. Discrete Math. 13 (2000) 313-323, doi: 10.1137/S0895480196314291. [12] W. D. Wallis, Magic Graphs (Birkhäuser Boston, 2001), doi: 10.1007/978-1-4612-0123-6. | 941 | 2,608 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-39 | latest | en | 0.611454 |
http://www.ehow.co.uk/how_6902083_calculate-gross-investment.html | 1,493,271,275,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121869.65/warc/CC-MAIN-20170423031201-00325-ip-10-145-167-34.ec2.internal.warc.gz | 500,512,061 | 16,625 | # How to calculate gross investment
Written by carter mcbride
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Gross investment is the amount a company has invested in an asset or business without factoring in depreciation. Factoring in depreciation creates net investment. For example, a company buys a car for £3,250 that has depreciated by £1,950 after three years. In year three, the gross investment is £3,250 and the net investment is £1,300. This is important for tracking how much was actually used as an expenditure on the investment. Businesses also use this calculation for business formulas such as cash return on gross investment.
Skill level:
Easy
## Instructions
1. 1
Find the asset on the company's balance sheet. For example, the company has property valued at £325,000 on the balance sheet.
2. 2
Find the accumulated depreciation on the company's balance sheet. In the example, the property has £130,000 of accumulated depreciation.
3. 3
Add the accumulated depreciation to the company's book value of the asset to find the gross investment in the asset. In the example, £325,000 plus £130,000 equals a gross investment of £455,000.
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• Most recent | 301 | 1,285 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | longest | en | 0.924064 |
https://www.unitconverters.net/length/angstrom-to-decimeter.htm | 1,716,721,007,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058876.64/warc/CC-MAIN-20240526104835-20240526134835-00816.warc.gz | 913,148,953 | 4,046 | Home / Length Conversion / Convert Angstrom to Decimeter
# Convert Angstrom to Decimeter
Please provide values below to convert angstrom [A] to decimeter [dm], or vice versa.
From: angstrom To: decimeter
### Angstrom to Decimeter Conversion Table
Angstrom [A]Decimeter [dm]
0.01 A1.0E-11 dm
0.1 A1.0E-10 dm
1 A1.0E-9 dm
2 A2.0E-9 dm
3 A3.0E-9 dm
5 A5.0E-9 dm
10 A1.0E-8 dm
20 A2.0E-8 dm
50 A5.0E-8 dm
100 A1.0E-7 dm
1000 A1.0E-6 dm
### How to Convert Angstrom to Decimeter
1 A = 1.0E-9 dm
1 dm = 1000000000 A
Example: convert 15 A to dm:
15 A = 15 × 1.0E-9 dm = 1.5E-8 dm | 249 | 580 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-22 | latest | en | 0.445637 |
https://4geeks.com/lesson/k-nearest-neighbors | 1,726,036,284,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651344.44/warc/CC-MAIN-20240911052223-20240911082223-00530.warc.gz | 63,621,899 | 46,515 | Self-paced
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# K-nearest neighbors
K-nearest neighbors (KNN)
## K-nearest neighbors (KNN)
The K-nearest neighbors model, better known by its acronym KNN is an algorithm used for classification and regression tasks. In KNN, a data point is classified or predicted based on the most classes or values of the K nearest data points in the feature space.
For example, if we wanted to predict how much money a potential customer spends in our business, we could do it based on the 5 most similar customers to him and average their likes to make the prediction.
### Structure
The model is built according to well-defined steps, which are as follows:
1. Selection of the value of K: A value is chosen for K, which represents the number of nearest data points to be considered for classifying or predicting the new data point. A small value may lead to a noisier model sensitive to outliers, while a large value may smooth the decision boundaries.
2. Distance measurement: A metric is used to calculate the distance between the data point to be classified or predicted and the other data points in the training set.
3. Identification of the K nearest neighbors: The K nearest data points are selected (depending on the selected metric).
4. Prediction: If it is a classification problem, the new point is classified into the most frequent class among the K nearest neighbors. If it is a regression problem, the target value for the new point is calculated as the mean or median of the values of the K nearest neighbors.
Furthermore, the model does not involve a training phase per se, as the entire training set is stored in memory to perform nearest-neighbor classifications or predictions.
It is important to note that the performance of this model can be highly dependent on the value of K and the choice of distance metric. In addition, it can be computationally expensive for large data sets, since it must compute the distance to all training points for each prediction:
This distance orders the points surrounding the point to be predicted, so that depending on the value of K the closest points can be chosen:
One of the most common questions in this type of models deals with what optimal value of K we should choose. This number cannot be calculated a priori and is approximated in the hyperparameter optimization phase. As can be seen in the case of the figure, its value can bias a particular prediction towards the opposite or another one with slight changes.
#### Distance metrics
Distance metrics are functions used to measure the proximity or similarity between two data points in a KNN model. There are a large number of proposals, but the best known are the following:
• Euclidean: measures the straight-line distance between two points. Suitable for numerical data.
• Manhattan: Measures the distance as the difference of the Cartesian coordinates of the two points. Suitable for numerical data as well.
• Minkowski: It is an intermediate point between the two previous ones.
• Chebyshev: Also known as the maximum distance between the difference of heights (Y-axis) or widths (X-axis).
• Cosine: Used to measure the similarity between two vectors.
• Hamming: Used for categorical or binary data. It measures the difference between two character strings of equal length.
### Model hyperparameterization
We can easily build a KNN model in Python using the scikit-learn library and the KNeighborsClassifier and KNeighborsRegressor functions. Some of its most important hyperparameters, and the first ones we should focus on are:
• n_neighbors: This is the K value we mentioned earlier. It represents the number of nearest data points to be considered when classifying or predicting a new data point. It is the most important hyperparameter in KNN and directly affects the shape of the decision boundaries of the model. A small value can lead to a model more sensitive to noise and outliers, while a large value can simplify the model.
• metric: Function for calculating the distance between the data points and the new point. The choice of metric can affect the way the model interprets the proximity between points and, thus, the resulting classification or prediction.
• algorithm: Different implementations of the KNN model, which will be more or less effective depending on the characteristics and complexity of the data set. | 973 | 4,912 | {"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} | 3.265625 | 3 | CC-MAIN-2024-38 | latest | en | 0.941171 |
https://kidsworksheetfun.com/5th-grade-mixed-numbers-worksheets/ | 1,713,273,404,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817095.3/warc/CC-MAIN-20240416124708-20240416154708-00085.warc.gz | 299,340,476 | 24,716 | # 5th Grade Mixed Numbers Worksheets
These worksheets are pdf files. Adding mixed numbers with like denominators.
Adding Mixed Fractions Worksheet Education Com Adding Mixed Fractions Mixed Fractions Fractions Worksheets
### Parenting worksheets adding mixed numbers 2.
5th grade mixed numbers worksheets. The fractional part of the mixed number has the same denominator as the addend fraction. Here is a collection of our printable worksheets for topic mixed numbers of chapter fraction concepts in section fractions. Click on the images to view download or print them.
Assess fifth grade fraction concepts including equivalent fractions adding with unlike denominators improper fractions and mixed numbers. Keep learning interesting as your child converts improper fractions to mixed ones with a colorful worksheet and a simple example. This math worksheet gives your child practice adding mixed numbers with different denominators.
Adding mixed numbers and fractions like denominators below are six versions of our grade 5 math worksheet on adding mixed numbers to fractions. Worksheets math grade 5 fractions. Math adding mixed numbers 2.
Grade 5 number operations. A brief description of the worksheets is on each of the worksheet widgets. Fifth grade math worksheets.
Below are six versions of our grade 5 math worksheet on adding mixed numbers where the fractional part of the numbers have like denominators. Add subtract adding mixed numbers. Worksheets math grade 5 fractions.
Add subtract mixed numbers. 5th grade mixed numbers and improper fractions printable worksheets. These worksheets are pdf files.
Convert Improper Fractions 5th Grade Fraction Worksheets Fractions Improper Fractions Math Fractions
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Convert Improper Fractions To Mixed Fractions 1 Gif 790 1 022 Pixels Fractions Worksheets Improper Fractions Fractions
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5th Grade Fraction Review Mixed Numbers And Improper Fractions By Nastaran In 2020 Fractions Worksheets Free Fraction Worksheets Fractions
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Mixed And Improper Fractions Worksheet Education Com Fractions Worksheets Fifth Grade Math Improper Fractions
Worksheets Word Lists And Activities Greatschools 5th Grade Math Math Worksheets Fifth Grade Math | 553 | 2,987 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-18 | latest | en | 0.807069 |
https://voiceofwave.com/what-is-meant-by-spherical-wave | 1,642,515,135,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300849.28/warc/CC-MAIN-20220118122602-20220118152602-00669.warc.gz | 679,723,682 | 11,048 | # What is meant by spherical wave?
1
Date created: Sun, Feb 21, 2021 3:11 AM
Date updated: Mon, Jan 17, 2022 6:36 PM
Content
## Top best answers to the question «What is meant by spherical wave»
[′sfir·ə·kəl ′wāv] (physics) A wave whose equiphase surfaces form a family of concentric spheres; the direction of travel is always perpendicular to the surfaces of the spheres.
FAQ
Those who are looking for an answer to the question «What is meant by spherical wave?» often ask the following questions:
### 👋 What is spherical wave example?
One common example of a spherical wave is a sound wave. When an object oscillates or vibrates in the presence of medium, a sound wave is produced and this wave propagates outward in all possible directions. As the wave travels outward, it carries energy.
### 👋 What is a transverse spherical wave?
• Propagation of a transverse spherical wave in a 2d grid (empirical model) A transverse wave is a moving wave that consists of oscillations occurring perpendicular (right angled) to the direction of energy transfer (or the propagation of the wave).
### 👋 Is sound a spherical wave?
Spherical Wave (Sound) This is the wave produced by an elementary point source… This wave propagates isotropically - that is, the wave amplitude is spherically symmetric.
We've handpicked 23 related questions for you, similar to «What is meant by spherical wave?» so you can surely find the answer!
What is meant by wave packets?
Wave packets is a term used to rectify the wave particle duality or light and matter. A wave packet is simply a group of coinciding waves that is used to represent a particle.
What is meant by wave period?
• We can define wave period as the measure of time it takes for a wave cycle to complete or time taken by a wave to complete one oscillation. Formula to calculate wave period using frequency. Wave period is the reciprocal of frequency.
What kind of light sources produce spherical wave front?
• When the source of light is a point source the wavefront formed will be spherical wavefront.
• Point source means the source of light is so small that it is considered as point…
• For example: - Ripples in water are in the form of concentric circles which are spherical wavefronts.
How is a circular wave different from a spherical wave?
• The inverse square law. A plane wave of a single frequency in theory will propagate forever with no change or loss. This is not the case with a circular or spherical wave, however.
How do you get a spherical wave surface?
• The particles around the source will receive this energy and begin to oscillate. Thus, waves will be originating from the source. These waves will be travelling in all directions and in time t, this wave will travel a distance of ct from the source. Thus, we get a spherical wave surface.
Which is an example of a spherical wave?
• For example, a sound speaker mounted on a post above the ground may produce sound waves that move away from the source as a spherical wave. Sound waves are discussed in more detail in the next chapter, but in general, the farther you are from the speaker, the less intense the sound you hear.
What is meant by a longitudinal wave?
Longitudinal wave, wave consisting of a periodic disturbance or vibration that takes place in the same direction as the advance of the wave… Sound moving through air also compresses and rarefies the gas in the direction of travel of the sound wave as they vibrate back and forth.
What is meant by a rogue wave?
• In oceanography, rogue waves are more precisely defined as waves whose height is more than twice the significant wave height (Hs or SWH), which is itself defined as the mean of the largest third of waves in a wave record.
What is meant by a transverse wave?
Transverse wave, motion in which all points on a wave oscillate along paths at right angles to the direction of the wave's advance. Surface ripples on water, seismic S (secondary) waves, and electromagnetic (e.g., radio and light) waves are examples of transverse waves.
What is meant by an electromagnetic wave?
Definition: Electromagnetic waves or EM waves are waves that are created as a result of vibrations between an electric field and a magnetic field… They are hence known as 'electromagnetic' waves. The electric field and magnetic field of an electromagnetic wave are perpendicular (at right angles) to each other.
What is meant by electromagnetic wave propagation?
In simple words, electromagnetic waves are oscillations produced due to crossing over of an electric and a magnetic field. The direction of the propagation of such waves is perpendicular to the direction of the force of either of these fields as seen in the above figure.
What is meant by full wave rectifier?
• Full Wave Rectifier. Definition: Full wave rectifier is the semiconductor devices which convert complete cycle of AC into pulsating DC. Unlike half wave rectifiers which uses only half wave of the input AC cycle, full wave rectifiers utilize full wave.
What is meant by overlapping wave functions?
Pretty much the same thing as when ANY type wave activity overlaps. If a fluid wave from one source meets up with wave in the same fluid, but from a different source, you have overlapping waves. The result can be interference, both destructive and constructive. The same thing happens when you have EM waves (ie, light) coming from two different sources. This can happen if the EM radiation has a wavelength of one meter (ie, radio waves) or one angstrom (x-rays), or anything in between. Erwin Schroendinger hypothesized the usefulness (existence?) of a wave function, which can be used to accurately predict the behavior of sub-atomic particles. It has been found that, when predicting such behavior, one can assume that this wave function also acts like a fluid wave or a EM wave. You can accurately predict maxima and minima of the probability of an electron, over time, reaching a specific point in space given two possible paths for that electron. Simply assume that the two possible paths of the electron are like sources of this wave function, then assume the two wave functions overlap (just like other waves), and then do the math -- and, at the end, you have an accurate prediction of the probability that an electron will reach somewhere. Scientists still debate just what this MEANS -- is this wave function something REAL, or just a mathematical trick that just happens to work? No matter what the case, assuming an overlapping wave function is like other overlapping waves, allows us to predict what will happen.
What is meant by simple harmonic wave?
simple harmonic motion, in physics, repetitive movement back and forth through an equilibrium, or central, position, so that the maximum displacement on one side of this position is equal to the maximum displacement on the other side. The time interval of each complete vibration is the same.
What is meant by sine wave ups?
• The sine wave waveform means the UPS is going to be most compatible with the broadest range of electronics , and it is line interactive with Automatic Voltage Regulation, which gives you better ...
What is meant by third wave countries?
Countries that got democracy post 1974 are called third wave countries. e.g. nepal
Are there any non-spherical wave solutions to the wave equation?
• Such waves exist only in cases of space with odd dimensions. For physical examples of non-spherical wave solutions to the 3D wave equation that do possess angular dependence, see dipole radiation .
How is the energy of a spherical wave distributed?
• The intensity, or energy per unit of length along the circumference of the circle, will therefore decrease in an inverse relationship with the growing radius of the circle, or distance from the source of the wave. In the same way, as a spherical wave front expands, its energy is distributed over a larger and larger surface area.
When does spherical spreading occur in a sound wave?
• Spherical spreading describes the decrease in level when a sound wave propagates away from a source uniformly in all directions. This situation occurs for a sound source at mid-depth in the ocean, for example. One can picture the crests and troughs of the sound waves as spheres centered on the source location.
What is the difference between a diverging and concave spherical wave?
• converging spherical wave must be the inverse of the diverging one, the sign of the wavevector k must change (better change the sign of k rather than the sign of as you did above, this will mean time reversal).
What is meant by amplitude of a wave?
amplitude, in physics, the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position… Waves are generated by vibrating sources, their amplitude being proportional to the amplitude of the source.
What is meant by phase of a wave?
Wave phase is the offset of a wave from a given point. When two waves cross paths, they either cancel each other out or compliment each other, depending on their phase. These effects are called constructive and destructive… The word phase is used to describe a specific location within a given cycle of a periodic wave.
What is meant by speed of a wave?
• Speed of a wave can be thought of as, the speed at which a disturbance at a certain point can cause a disturbance at a nearby point. For example, light is disturbance of electric and magnetic fields, if we look at it with the wave picture. | 1,971 | 9,502 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2022-05 | latest | en | 0.917859 |
http://scripting.molab.eu/tutorials/air-traffic-control-tower-tutorial-creating-structure/ | 1,726,122,824,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651422.16/warc/CC-MAIN-20240912043139-20240912073139-00034.warc.gz | 23,929,294 | 13,027 | # Air Traffic Control Tower Tutorial – Creating Structure
Author:
Categories: Tutorials
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Tutorial created by: Lauren Bailey
Introduction
This past semester I worked on utilizing Grasshopper in studio and applying all the new skills I had learned from the Grasshopper course. This tutorial contains five sections, each representing a different part of the air traffic control tower I had worked on. Grasshopper allowed me to work out the structural principles and problems I had been facing with my design while allowing me the flexibility to adjust dimensions and geometries as my design progressed. First we will create the tower structure, then move on to creating the enclosure for that. The second and third sections will cover how to create floor and roof plates for a unique circular geometry and how to structure it, and finally we will create the control cab using the principles discussed as well as a seemingly complex roof. The last section is focused on finishing up and an evaluation on this series of scripts as a whole.
Part 1: Tower Structure
This section sets the basework for what the control cab would sit on in my design. I began by creating four columns that would be the primary structure for my tower’s shaft. The tower would be a little over 60 meters tall and so I began by setting four coordinates (four separate but similar scripts) and extruding cylinders from those points.
From there I added another cylinder extrusion from those points to create a thicker base for my column that would support the structure all the way up to the cab. I did this four times, as shown below.
It is important to group your scripts so they stay nice and neat and you do not forget what elements they modify. To do this, you select a script or portion of the script by holding down the left mouse button and dragging across the components, creating a rectangle. Right click outside of the area and select “Group” from the drop-down menu. You can also choose the color and name your groups.
Next, the structure needed some sort of a shell to create the form of the tower shaft. To do this I set a base point (based off of my columns) and entered in the X and Y coordinates. From there, I attached a series to the Z sector of my component. This determines the number of floors and heights between them. As required by code, my tower must have floors on the way up to the control cab, so the XY plane is added to create floor plates based on the series.
Now you should see a series of ascending floor plates. But they need some sort of geometry, not just points and rough grid outlines. Next add the polygon component and link the plane to it. This script is beginning to allow a lot of flexibility. For example, the polygon component allows changes in the type of polygon and how rounded the corners are (Rf = filet). This was essential for my design because sharp edges and corners are effected by wind more than smooth, rounded edges and corners. I also wanted my tower shaft to either increase in size as it ascended, or decrease in size. Because of this, a domain was created (bottom portion of the script). The script shown below and its components allow me to not only manipulate the geometry of the tower, but also the dimensions. Some examples are shown below.
Finally, the actual shaft shell had to be created. Based off of the outlines shown, I was able to loft all of the polygons (loft component). But once I did that, I realized that my coordinates did not align with my columns and the shaft shell needed to be rotated. This was added before the loft command so everything could be rotated all at once, including the floor plates. Finally, color was added to the lofted surface. To clean it up before baking, I went through and right clicked the components and turned the preview off in order to clean up the look and get rid of some points that were still necessary for calculations, but did not necessarily need to be seen.
The final script for the shaft shell:
Part 2: Creating Base Plates
Now that the columns and tower shaft had been created it was time to create the control cab and employee break area. My design dictated that these structures be two large cylinders that needed thick floor and roof plates to withstand the forces at that elevation. I began by working on the bottom volume by creating the floor plate. Once again, I started with a point and used sliders to make sure it was at the right elevation and offset from the tower. Then I added the ellipse component and created panels with the values of the x and y axes for the plate.
From there I created a second ellipse, slightly smaller and slightly lower than the original. This determines the geometry and thickness of the plate. I then lofted the two ellipses and also “capped” the geometry by creating surfaces for each of the curves, as shown below.
This is all nice, but I need to create the roof plate for my employee building now. Following the same steps as previously shown I added a second plate to my design.
This is another time that organization and grouping is crucial. Note that the two curves that I wish to attach my structure to are located near one another (the top ellipse of my bottom plate and bottom ellipse of my top plate).
Part 3: Cab Structure
The cab we just created is not yet complete! It must be structured. This is not a separate script, but rather a unification of the parts we created in the previous section.
I will then divide the curves I had previously highlighted both into 10 sections/points that represent the amount of columns needed. The second image shows a better view of what the divide polyline component looks like. I used a slider because throughout the semester the values changed and I needed increased flexibility in order to sculpt my roofs and their supporting structural systems. The final value I needed was 10.
Now we need to unite the two plates and create the columns. I chose to use the line tool and wrap cylinders around to form the needed geometry.
Both the divide polygon components are attached separately to the line command. I named this small section “Primary” because diagonal supports are needed as well. We still use the same points from both the divide polygons components, except this time we need to add a data component and shift list in order to connect the pairs of points in our desired formation.
Here is a final look at the script for this section. The lower employee break area is now structured!
Part 4: The Control Cab and Roof
Now that the employee area has been created, the control cab where the air traffic controllers work must be added. This form will be placed surrounding the thinner portions of my columns (the thicker support the floor, the thinner only support the roof). and will only need columns, not a truss-like structure due to the increased visibility required for this area. Thankfully it is quite similar to the previous section, only this will focus on the unique roof structure, created to increase efficiency and stability for my design.
First we create a pair of ellipses, just like the previous section. The bottom section is shown above and you can go ahead and loft and create surfaces from the curves to complete the floor plate.
Next the roof, the only major difference here is that a third ellipse is required for the top section. The roof we are aiming to create is thicker below, has a bit of a lip for rainwater control, and then gets significantly smaller near the top.
Go ahead and loft all three surfaces. Note that if you try to do it with one loft component it turns out looking quite strange and not like a roof at all. After some trial-and-error I discovered that in order to be successful, two lofts were required, one for ellipse 1 and 2, and the other for ellipse 2 and 3. Surfaces were added to curves 1 and 3.
Finally we add the columns by dividing the curves and connecting them first with lines, to help visualize, and then cylinders.
The final script for this section should look something like this:
Part 5: Finishing Up
After I had completed the structure and tweaked the sliders on many of my components to get the geometries and dimensions the way I wanted them, I then baked my creation. Baking allows the script to come to life and become an object/objects in Rhino. I simply selected the entirety of my script, then right clicked and selected the “Bake” option form the drop-down menu. The final script and product is shown below:
This series of scripts proved to be quite useful and quite the learning experience over the past few months. It was first created before a meeting with a structural engineer as part of my design studio coursework, then expanded and manipulated over the course of a month until the design was technically correct and aesthetically interesting. However, the design has a lot of repeated components. I must admit I created this script on my own based off our coursework and a few tutorials, I did not look too deeply how to increase efficiency or link the different scripts better. Linking them in some way would definitely increase cohesiveness and unity of changes, I feel as though my “mess-up” and needing to rotate my tower could have been avoided. Keeping the script neat and tidy was a challenge as well, but overall I am quite happy with the end result and the solutions it helped me explore and create.
ATCT Grasshopper File: Lauren_Bailey Air Traffic Control Tower Script | 1,929 | 9,498 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-38 | latest | en | 0.946279 |
http://www.algebra.com/algebra/homework/Expressions-with-variables/Expressions-with-variables.faq.question.550555.html | 1,368,883,420,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696382398/warc/CC-MAIN-20130516092622-00007-ip-10-60-113-184.ec2.internal.warc.gz | 309,602,958 | 4,679 | # SOLUTION: the lenght of the hypotenuse of a right triangle is 34 inches and the length of one of its legs is 16inches what is the length in inches of the other legf the right triangle
Algebra -> Algebra -> Expressions-with-variables -> SOLUTION: the lenght of the hypotenuse of a right triangle is 34 inches and the length of one of its legs is 16inches what is the length in inches of the other legf the right triangle Log On
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Algebra: Expressions involving variables, substitution Solvers Lessons Answers archive Quiz In Depth
Question 550555: the lenght of the hypotenuse of a right triangle is 34 inches and the length of one of its legs is 16inches what is the length in inches of the other legf the right triangleAnswer by mananth(12269) (Show Source): You can put this solution on YOUR website!Pythagoras theorem states that Hypotenuse ^2= leg1^2+leg2^2 Hypotenuse = 34 leg1 = 16 leg2^2= hypotenuse ^2 - leg1^2 Leg2^2= 34 ^2 - 16 ^2 Leg2^2= 1156 - 256 Leg2^2= 900 take the square root leg2= 30 inches m.ananth@hotmail.ca | 328 | 1,225 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2013-20 | latest | en | 0.828026 |
http://www.jiskha.com/display.cgi?id=1276905634 | 1,498,151,432,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319636.73/warc/CC-MAIN-20170622161445-20170622181445-00065.warc.gz | 557,967,785 | 4,075 | # Algebra
posted by on .
Need to get these checked--
33. What is the area of a square with the dimension of 1 + sqrt of 2?
9.64
34. A matrix with 5 columns and six rows added to another matrix with 5 columns and 6 rows would result in a matrix with:
a. 12 columns and 10 rows
b. 5 columns and 6 rows
c. 10 columns and 12 rows
d. 6 columns and 5 rows
Not sure..
43. If on the first day someone gave you \$5, and then doubled that amount each day, how much money would you have total, on the sixth day?
\$160
Thanks
-MC
• Algebra - ,
(1+sqrt 2)(1+sqrt 2) = 1+2sqrt 2 + 2
3 + 2 sqrt 2
You just add elements, no change in number of rows or columns
5*2^0 day 1
5*2^1 day 2
5*2^2 day 3
5*2^5 = 5*32 = 160 day 6 | 248 | 717 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2017-26 | latest | en | 0.822588 |
https://muammar-kc.me/educational/how-to-do-dcf-valuation.php | 1,582,387,296,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145708.59/warc/CC-MAIN-20200222150029-20200222180029-00536.warc.gz | 490,457,793 | 5,022 | ### How to do dcf valuation
The concept of DCF valuation is based on the principle that the value of a business or asset is inherently based on its ability to generate cash flows for the. A discounted cash flow model (DCF model) is a type of financial model that values a At some point, you must make some high level assumptions about cash. Discounted cash flow (DCF) is a valuation method used to estimate the In order to conduct a DCF analysis, an investor must make estimates.
## discounted cash flow valuation excel
The first step in the DCF model process is to build a forecast of the three financial statements, based on assumptions about how the business will perform in the. How to perform a DCF valuation. There are certain steps in performing a DCF valuation. These are: Calculate the WACC. The weighted. As such, discounted cash flow valuation analysis tries to calculate the value of a company IMPORTANT – Do have a look at this step by step guide to Financial .
Valuation using discounted cash flows is a method for determining the current value of a Discounted Cash Flow valuation was used in industry as early as the s or s, widely discussed in financial economics in the s, and. 1. Basics of Discounted Cash Flow. Valuation. Aswath Damodaran The dividend discount model is a specialized case of equity valuation, you need to do?. How to do Discounted Cash Flow (DCF) Analysis. The discounted . Here's a streamlined input model I use for stock analysis, called StockDelver: Discounted .
Additionally, DCF does not take into account any market-related valuation information, such as the valuations of comparable companies, as a “sanity check” on. Learn how to do DCF valuations on companies financial statements; Learn how to find the per share intrinsic value; Learn DCF Valuation techniques. But how do you define the valuation of a startup; a company that, per definition, might not have any How to perform a DCF valuation you say?.
## discounted cash flow calculator
The discounted cash flow model is one common way to value an entire company, and, by extension, its shares of stock. See examples and more. grasp of a large part of what investment bankers and corporate financiers do in terms of valuing something. Discounted Cash Flow (D.C. Investors in privately-held companies do not have such a readily available value for their ownership interests. How are values of privately-held businesses. You will understand discounted cash flow (DCF) valuation and how it compares to other the seemingly complex decisions that finance professionals make. So as we do in our seminars, we are going to present a simplified In finance, discounted cash flow (DCF) analysis is a method of valuing a project, company. Discounted cash flow (DCF) valuation views the intrinsic value of a security as ( or expects another investor to do so), dividends may be changed substantially;. A discounted cash flow (DCF) model is a financial model used to value companies What does the company do and where does it operate?. This calculator uses future earnings to find the fair value of stock shares. So, in order to apply the Discounted Cash Flow method, one has to calculate those free cash flows. For this, we make a multi-year business plan and a financial. This article explains how to calculate your company's valuation by using the most The DCF analyses future free cash flows (FCF) and discounts them. It helps I made an example in Excel for you to make it more hands on. | 721 | 3,482 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-10 | latest | en | 0.928568 |
https://www.allinterview.com/interview-questions/80-216/civil-engineering.html | 1,544,630,791,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824059.7/warc/CC-MAIN-20181212155747-20181212181247-00618.warc.gz | 798,860,692 | 8,971 | Civil Engineering Interview Questions
pls tell me about method of cutting length of ties column
1295
pls tell me , what is the water content
861
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4712
where we use one way slab and two way slab?
SRM,
10811
You have casted a column above your footing, and you found out it is off its location by 5cm. What would you Do?
5133
hello; sir. i would like to be an assistant executive engineer in irrigation department. is it really necessary to take coaching? or can i prepare myself by studying any study materials available in market? if so; what are the books i have to purchase. please reply soon.
NCC,
1112
how to read the drawings to find the quantity of steel in a storm water manhole or any other structure?
1266
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6404
how many type of billing
L&T,
9186
comparision difference between GSB and WBM subbase in road construction
CPWD,
3249
how much concrete solid/hollow block are manufacture per 50 kg cement bag in differnent sizes like 400x200x200mm, 400x200x150mm, 400x200x100mm. And need proper ratio as per sale quantity. pls send my mail mathi7599@gmail.com searching last two years not yet get pls find and enclosed pls. need practical analysis.
1198
For M25 concrete ,what is the maximum compressive strength at 28 days ?
9551
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3963
What is the procedure of Waterproofing in toilet sunken area ?
ACIL,
9172
How much percentage of overlap allowed in distribution steel as per IS Code.
5113
Un-Answered Questions { Civil Engineering }
How to find out slope footing concrete
1597
plz drive the formula d2/162.162 for the calculation of weight of steel bars????
27075
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1086
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279
what is the steel difference between np-2 and np-3 pipes and weight also
943
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I just got a first job in Maharashtra and m a fresher. My task is estimate the cost of excavation of earth with J.C.B. can anyone tell me how to do this??
518
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1097
3. A vane 12.5 cm long, 8.8 cm dia was pressed into a soft clay at the bottom of borehole. Torque was applied and the value at failure was 55 N-m. find the shear strength of the clay on a horizontal plane.
818
what"s the sort of beams?
979
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HOW MUCH STEEL REQUIRED FOR SLAB OF AREA 23'3"*18.3"? FOR SLAB HOW MUCH DIAMETER BAR WILL USE?
836 | 787 | 3,204 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-51 | latest | en | 0.915377 |
https://theses.gla.ac.uk/76428/ | 1,726,873,692,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701425385.95/warc/CC-MAIN-20240920222945-20240921012945-00401.warc.gz | 540,876,097 | 8,498 | Finite Group Theory: Odd Primes and CC-Subgroups
McLean, Joseph Francis (1985) Finite Group Theory: Odd Primes and CC-Subgroups. MSc(R) thesis, University of Glasgow.
Full text available as:
Abstract
The majority of this thesis comprises a survey of existing knowledge. Let G be a finite group and suppose that M is a subgroup of G such that (i)1<M<G, and (ii)for all x∈M#,CG(x)≤M. Such a subgroup M is called a CC-subgroup the concept to which this thesis is predominantly devoted. Following a brief introduction. Chapter 11 consists of a survey of the known results on the odd prime structure of finite groups. This survey is split into three sections as follows. The first gives an account of the development of a unified theory for characterising groups with CC-subgroups of order divisible by three. Section II introduces the twin ideas of closure and homogeneity, concluding with a theorem which has Important applications later in Chapter V. Section ill consists of a straightforward listing of the remaining odd prime structure results. Chapter III is the theoretical base of the thesis, contributing all the major results which are required before proceeding. Section II of this chapter is itself an integral part of the survey, being a systematic exposition of basic CC-subgroup theory. Chapter IV is a discussion on the various techniques and proofs Involved In Chapter II, giving a readable yet rigorous explanation of the theory. Chapter V highlights more recent, and more general, results involving CC-subgroups, giving detailed proofs, and sets the scene for the final chapter. Chapter VI consists of two sections. Section I is given over entirely to the statement and proof of a single theorem which completely classifies groups containing CC-subgroups, a simple corollary of which Initiates Section II, an outline of the search for CC-subgroups of the finite simple groups. This section, and the thesis, ends with four tables that give as complete a list as possible of the Information currently available on the CC-subgroups of the simple groups.
Item Type: Thesis (MSc(R)) Masters Adviser: Neil K Dickson Theoretical mathematics 1985 Enlighten Team glathesis:1985-76428 Copyright of this thesis is held by the author. 19 Nov 2019 14:31 19 Nov 2019 14:31 https://theses.gla.ac.uk/id/eprint/76428 | 500 | 2,318 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.932457 |
https://oeis.org/A252696 | 1,653,261,999,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662550298.31/warc/CC-MAIN-20220522220714-20220523010714-00252.warc.gz | 496,897,894 | 4,447 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A252696 Number of strings of length n over a 3-letter alphabet that do not begin with a nontrivial palindrome. 9
0, 3, 6, 12, 30, 78, 222, 636, 1878, 5556, 16590, 49548, 148422, 444630, 1333254, 3997884, 11991774, 35969766, 107903742, 323694636, 971067318, 2913152406, 8739407670, 26218074588, 78654075342, 235961781396, 707884899558, 2123653365420, 6370958763006 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS A nontrivial palindrome is of length at least 2. 3 divides a(n) for all n. lim n -> infinity a(n)/3^n ~ 0.278489919882115 is the probability that a random, infinite string over a 3-letter alphabet does not begin with a palindrome. This sequence gives the number of walks on K_3 with loops that do not begin with a palindromic sequence. LINKS Peter Kagey, Table of n, a(n) for n = 0..1000 Daniel Gabric, Jeffrey Shallit, Borders, Palindrome Prefixes, and Square Prefixes, arXiv:1906.03689 [cs.DM], 2019. FORMULA a(n) = 3^n - A248122(n) for n > 0. a(2n) = k*a(2n-1) - a(n) for n >= 1; a(2n+1) = k*a(2n) - a(n+1) for n >= 1. - Jeffrey Shallit, Jun 09 2019 EXAMPLE For n = 3, the first 10 of the a(3) = 12 solutions are (in lexicographic order) 011, 012, 021, 022, 100, 102, 120, 122, 200, 201. MATHEMATICA b[0] = 0; b[1] = 0; b[n_] := b[n] = 3*b[n-1] + 3^Ceiling[n/2] - b[Ceiling[n/2]]; a[n_] := 3^n - b[n]; a[0] = 0; Table[a[n], {n, 0, 28}] (* Jean-François Alcover, Jan 19 2015 *) PROG (Ruby) seq = [1, 0]; (2..N).each { |i| seq << 3 * seq[i-1] + 3**((i+1)/2) - seq[(i+1)/2] }; seq = seq.each_with_index.collect { |a, i| 3**i - a } CROSSREFS A248122 gives the number of strings of length n over a 3 letter alphabet that DO begin with a palindrome. Analogous sequences for k-letter alphabets: A252697 (k=4), A252698 (k=5), A252699 (k=6), A252700 (k=7), A252701 (k=8), A252702 (k=9), A252703 (k=10). Sequence in context: A245774 A049941 A219634 * A288147 A026079 A066710 Adjacent sequences: A252693 A252694 A252695 * A252697 A252698 A252699 KEYWORD easy,nonn,walk AUTHOR Peter Kagey, Dec 20 2014 STATUS approved
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Last modified May 22 17:42 EDT 2022. Contains 353957 sequences. (Running on oeis4.) | 931 | 2,532 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2022-21 | latest | en | 0.532056 |
https://it.mathworks.com/matlabcentral/cody/problems/2235-back-to-basics-mean-of-corner-elements-of-a-matrix/solutions/2075367 | 1,586,097,854,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371604800.52/warc/CC-MAIN-20200405115129-20200405145629-00417.warc.gz | 509,281,020 | 15,843 | Cody
# Problem 2235. Back to basics - mean of corner elements of a matrix
Solution 2075367
Submitted on 31 Dec 2019 by Carlo Venettacci
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = [1 2 3; 4 5 6; 7 8 9] y_correct = 5; assert(isequal(MeanCorner(x),y_correct))
x = 1 2 3 4 5 6 7 8 9
2 Pass
x = magic(4) y_correct = 8.5; assert(isequal(MeanCorner(x),y_correct))
x = 16 2 3 13 5 11 10 8 9 7 6 12 4 14 15 1
3 Pass
x = eye(4) y_correct = 0.5; assert(isequal(MeanCorner(x),y_correct))
x = 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1
4 Pass
x = repmat(5, 4,4); y_correct = 5; assert(isequal(MeanCorner(x),y_correct))
5 Pass
x = [1:10]; y_correct = 5.5; assert(isequal(MeanCorner(x),y_correct)) | 336 | 818 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-16 | latest | en | 0.529402 |
https://aakashdigitalsrv1.meritnation.com/ask-answer/question/if-a-and-b-are-square-matrix-of-order-3-such-that-a-1-and-b/determinants/10541637 | 1,695,933,805,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510454.60/warc/CC-MAIN-20230928194838-20230928224838-00825.warc.gz | 86,976,940 | 8,006 | If A and B are square matrix of order 3 such that |A|= -1 and |B|= 3, then find the values of |3AB|
Dear student,
Regards
• 48
What are you looking for? | 50 | 155 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.893386 |
https://www.doomworld.com/profile/102-fredrik/?status=610775&type=status | 1,548,324,938,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584520525.90/warc/CC-MAIN-20190124100934-20190124122934-00187.warc.gz | 755,181,590 | 12,355 | Search In
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# Fredrik
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1. ## Fredrik
1 + 2 + 3 + 4 + ... = -1/12
1. Show previous comments 11 more
2. Fredrik said:
This may not seem different from the standard false proofs that 1=2 (typically using an obscured division by zero).
There was a guy that proved that God existed using math. A buddy of mine has a mathemagician friend that proved that he didn't because he divided by zero.
Psyonisis said:
You know I understand 0.999... = 1 just fine, but this is really pushing ridiculousness.
I remember there was a long discussion about why that's true and the thread apparently got out of hand before I could pop in and explain it:
If there's an infinitely repeating single digit decimal number, place it over 9. 0.999 . . . = 1 because 9/9 =1.
Another example is 0.333 . . . = 1/3 because 3/9 = 1/3.
I was surprised nobody else said it.
3. Peizo, anyone who remembers that thread will likely get stomach pains just thinking of it. It was so divided that I'd say at least 40% of people were wrong about it.
As much as I'd like to toss in my opinion on 0.repeating9, it's not worth it and would just derail a thread.
4. Fredrik said:
(is nature playing tricks on us?)
More like math is mental stuff.
× | 370 | 1,422 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | latest | en | 0.969233 |
https://www.univerkov.com/in-triangle-abc-angle-c-is-90-degrees-ac-10-sin-a-0-8-find-the-height-ch/ | 1,719,319,238,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865972.21/warc/CC-MAIN-20240625104040-20240625134040-00881.warc.gz | 918,599,849 | 6,165 | # In triangle ABC, angle C is 90 degrees, AC = 10, sin A = 0.8. Find the height CH.
If in triangle ABC from angle C to draw the height CH, then the resulting triangle AНС will be rectangular (angle H is 90 degrees).
It is necessary to consider the AНС triangle. The AC = 10 side lies opposite the right angle and is the hypotenuse. By condition sin A = 0.8.
It is known that sine is a trigonometric function of the angle in a right triangle, equal to the ratio of the leg of the opposite angle to the hypotenuse.
Or sin A = CH / AC,
whence CH = sin A * AC.
CH = 0.8 * 10 = 8. | 167 | 581 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-26 | latest | en | 0.824796 |
https://math.stackexchange.com/questions/2141558/irreducible-spectrum-implies-zero-divisors-are-nilpotent/2141644 | 1,563,675,954,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526818.17/warc/CC-MAIN-20190721020230-20190721042230-00533.warc.gz | 471,218,504 | 37,969 | # Irreducible Spectrum implies Zero-Divisors are Nilpotent
Let $A$ be a commutative unital ring, and let $\mathrm{Spec}(A)$ be it's set of prime ideals endowed with the Zariski Topology.
In this topological space, my text had me prove a couple facts, specifically:
1. $\{X_a \colon a \in A\}$ forms a basis for the topology, where $X_a = \{ x \in \mathrm{Spec}(A) \colon a \notin x\}$
2. $X_a \cap X_b = X_{ab}$
3. $X_a = \emptyset \iff a$ is nilpotent in $A$
I have a conjecture that $\mathrm{Spec}(A)$ irreducible implies that every zero-divisor is nilpotent. The definition I have of irreducible is
$\mathrm{Spec}(A)$ is irreducible if every two non-empty open sets have a non-empty intersection.
I thought I successfully proved my claim, but after review I think I actually proved a weaker claim (technically a stronger claim, depending on how you look at it). Here is my attempted proof.
Proof: Suppose $z \in A$, $z \neq 0$, and $zy = 0$, for some $y \neq 0$. Proceeding by way of contrapositive suppose $z$ is not nilpotent. Then by fact 3 above we have $X_z \neq \emptyset$, but $X_z \cap X_y = X_{zy} = X_0 = \emptyset$. Thus $\mathrm{Spec}(A)$ is reducible.
It appears to be an obvious flaw that I do not make an assertion about whether or not $y$ is nilpotent. Therefore the claim I actually proved is
If $\mathrm{Spec}(A)$ is irreducible and $xy = 0$, for non-zero $x$ and $y$, then one of $x$ or $y$ is nilpotent.
Upon inspection, this actually made sense because the text had me prove the characterization that $\mathrm{Spec}(A)$ is irreducible if and only if the nilradical is a prime ideal. With this characterization, if the nilradical is prime and $xy = 0$, since $0$ is in the nilradical then $xy$ is in the nilradical, and if the nilradical is prime then $x$ is in or $y$ is in. Which is consistent with what I proved, that atleast $\textbf{one}$ of $x$ or $y$ is nilpotent.
Obviously it is no such luck that one could prove if $xy = 0$ and $x$ is not nilpotent, then $y$ is not nilpotent, since we are not in an integral domain (otherwise my original claim would be vacuously true).
$\textbf{HOWEVER:}$ I still believe my conjecture to be true. I think either 1) I am overlooking something very obvious, a quick fix, or 2) In order to prove my original claim in full generality I might need more machinery than I have at my disposal right now. Some thoughts I had to try and get my claim proven were to 1. Consider when and why the nilradical is a prime ideal, 2. Consider the fact that the set of zero divisors is a union of prime ideals, 3. quit wasting time thinking about this and move forward in my text and maybe when I finish the book I will either realize why this claim is not even important and thus proving it is not a good use of time, or the proof will be obvious.
Any help, thoughts, opinions appreciated.
Edit: I should also mention that the reason I still believe my conjecture is true is not because I have some intuition (although I did check some specific examples), it is because I googled "when are zero divisors nilpotent" and found an answer to this question Under what conditions is a zero divisor element $a$ in commutative ring $R$ nilpotent? that says the nilradical being prime as a sufficient condition for zero divisors being nilpotent. As mentioned the nilradical being prime is iff the spectrum is irreducible, thus my claim should hold.
Also can someone give me a counter example for why the converse does not hold? That is an example where the zero divisors and nilpotents coincide but the nilradical is not prime? There is probably a simple $\mathbb{Z}_n$ case that is apparent using elementary number theory but I don't have anything.
• To help avert any future confusion: I went ahead and removed the spurious claim about $Nil(R)$ being prime from my solution (the one referred to in this post.) – rschwieb Feb 13 '17 at 14:48
## 1 Answer
Counterexample: Consider $A=k[x,y]/(x^2,xy)$. Then $Nil(A)=(x)$ is prime, so $\textrm{Spec}(A)$ is irreducible. But $y$ is a zero-divisor which is not nilpotent.
The key to this example is that $A$ has an embedded prime $(x,y)$. The nilradical of any ring is the intersection of the minimal primes, so we have $\textrm{Spec}(A)$ is irreducible iff $A$ has only a single minimal prime. But $A$ still may have embedded primes, and then any element in an embedded prime but not in the minimal prime will be a zero-divisor but not nilpotent.
The answer you're citing in the thread you linked to seems to be incorrect. If you go back to the thread that person linked to, you'll see that the answer they cite actually requires a stronger condition: there is a unique minimal prime containing all the zero-divisors (not the nilpotents).
• Yes, the linked thread is incorrect and you are right: +1 – Georges Elencwajg Feb 13 '17 at 0:39
• Great example, thanks – Prince M Feb 13 '17 at 6:12
• I have no idea what I was thinking. Looking at the timestamps, it looks like I wrote the two solutions two weeks apart, so perhaps I misremembered recent history in the second one (the one being cited here.) Having a prime nilradical obviously is only sufficient to make "half" the zero divisors nilpotent. – rschwieb Feb 13 '17 at 12:26
• @PrinceM You really should have read Manny's solution (which was parallel to the one of mine you read) which already included this example! His posts are very good. – rschwieb Feb 13 '17 at 14:49 | 1,447 | 5,438 | {"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} | 3.140625 | 3 | CC-MAIN-2019-30 | longest | en | 0.890106 |
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