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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https://studentshare.org/miscellaneous/1525697-project-management-network-diagram | 1,553,250,339,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202642.32/warc/CC-MAIN-20190322094932-20190322120932-00420.warc.gz | 642,526,282 | 23,284 | # Project Management Network diagram - Essay Example
Summary
a) There are various statistical tools and diagrams used for managerial decision making. But the popular charts that are used for decision making such as the Graphs, Gantt Charts are not useful when the situation requires analysis on a sequential aspect…
## Extract of sample "Project Management Network diagram"
Download file to see previous pages An event is defined as the starting or ending point for a group of activities and an activity is the work required to proceed from one event or point of time to another.(Kerzner, 1992). The numbers over the arrow specifies the time needed. To know the relationships between activities what job precedes a particular job, what job immediately follows this job and what jobs can run concurrently have to be listed and the time required for the completion is estimated. Based on the chronological order in which they have to be completed, the following network diagram is constructed:
From the above diagram we can understand the sequence of activities and their relative importance in the completion of the project. From the information provided we can tabulate and determine the critical path of the project. The Critical path is the sequence of activities which are crucial because delays to them will delay the completion of the project as a whole(Needham, 1997). The late finish and the late finish of the activities are calculated first. The early finish is the optimistic estimate of the completion time of the specific activity and the late finish is the pessimistic estimate of the duration of the activity. The slack time is the difference between the early finish and the late finish. In other words, the slack time denotes the time duration which can be delayed without delaying the entire project. But there are some activities which have the slack time zero, which means that these activities cannot be delayed. These activities are critical for the successful completion of the project because they cannot be delayed. The Appendix - 1 shows the calculation slack time of the activities. The calculation shows that the critical path consists of activities, A, D, H, I, J which are crucial for the completion of the project because they do not have any slack time.
b) Normal distribution curves are very useful to analyze the frequency of data. When we construct a standard normal probability distribution table, the relationship between the mean of the distribution and the random variable is established. The formula used is:
Z = x - / ,
Where,
Z = number of standard deviations from x to the mean of this distribution
X = value of the random variable with which we are concerned
= mean of the distribution of this random variable
= standard deviation of this distribution
The value obtained from using the above mentioned formula is ...Download file to see next pagesRead More
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5 Pages(1250 words)Assignment | 2,086 | 10,470 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-13 | longest | en | 0.945694 |
https://www.traditionaloven.com/building/refractory/graphite/convert-kilogram-kg-of-graphite-to-japan-kin-graphite.html | 1,558,399,837,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256184.17/warc/CC-MAIN-20190521002106-20190521024106-00526.warc.gz | 966,868,574 | 11,763 | Graphite 1 kilogram mass to Japanese kin converter
# graphite conversion
## Amount: 1 kilogram (kg - kilo) of mass Equals: 1.67 Japanese kin (斤) in mass
Converting kilogram to Japanese kin value in the graphite units scale.
TOGGLE : from Japanese kin into kilograms in the other way around.
## graphite from kilogram to Japanese kin Conversion Results:
### Enter a New kilogram Amount of graphite to Convert From
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many numbers after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other graphite measuring units - complete list.
Conversion calculator for webmasters.
## Graphite Units Converter
The calculator reflects mass density of graphite @ 20 degree Celsius state which is 2.25g/cm3 (calculated not in cold state example 2.09 - 2.23 grams per cubic centimeter.)
Due to its physical properties it is one excellent refractory material. Graphite is lighter in weight in comparison with alumina and still it retains more heat than heavy dense firebricks do. At the same time natural graphite is well abundant on the earth hence, in year 2010, refractory businesses consumed 12,500 tonnes of it only within US.
Convert graphite measuring units between kilogram (kg - kilo) and Japanese kin (斤) but in the other reverse direction from Japanese kin into kilograms.
conversion result for graphite: From Symbol Equals Result To Symbol 1 kilogram kg - kilo = 1.67 Japanese kin 斤
# Converter type: graphite measurements
This online graphite from kg - kilo into 斤 converter is a handy tool not just for certified or experienced professionals.
First unit: kilogram (kg - kilo) is used for measuring mass.
Second: Japanese kin (斤) is unit of mass.
## graphite per 1.67 斤 is equivalent to 1 what?
The Japanese kin amount 1.67 斤 converts into 1 kg - kilo, one kilogram. It is the EQUAL graphite mass value of 1 kilogram but in the Japanese kin mass unit alternative.
How to convert 2 kilograms (kg - kilo) of graphite into Japanese kin (斤)? Is there a calculation formula?
First divide the two units variables. Then multiply the result by 2 - for example:
1.6666666666667 * 2 (or divide it by / 0.5)
QUESTION:
1 kg - kilo of graphite = ? 斤
1 kg - kilo = 1.67 斤 of graphite
## Other applications for graphite units calculator ...
With the above mentioned two-units calculating service it provides, this graphite converter proved to be useful also as an online tool for:
1. practicing kilograms and Japanese kin of graphite ( kg - kilo vs. 斤 ) measuring values exchange.
2. graphite amounts conversion factors - between numerous unit pairs.
3. working with - how heavy is graphite - values and properties.
International unit symbols for these two graphite measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for kilogram is:
kg - kilo
Abbreviation or prefix ( abbr. ) brevis - short unit symbol for Japanese kin is:
### One kilogram of graphite converted to Japanese kin equals to 1.67 斤
How many Japanese kin of graphite are in 1 kilogram? The answer is: The change of 1 kg - kilo ( kilogram ) unit of graphite measure equals = to 1.67 斤 ( Japanese kin ) as the equivalent measure for the same graphite type.
In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solution. If there is an exact known measure in kg - kilo - kilograms for graphite amount, the rule is that the kilogram number gets converted into 斤 - Japanese kin or any other graphite unit absolutely exactly.
Conversion for how many Japanese kin ( 斤 ) of graphite are contained in a kilogram ( 1 kg - kilo ). Or, how much in Japanese kin of graphite is in 1 kilogram? To link to this graphite kilogram to Japanese kin online converter simply cut and paste the following.
The link to this tool will appear as: graphite from kilogram (kg - kilo) to Japanese kin (斤) conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting. | 1,000 | 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} | 2.671875 | 3 | CC-MAIN-2019-22 | latest | en | 0.814383 |
http://mathhelpforum.com/math-topics/54240-problem-print.html | 1,511,094,399,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805578.23/warc/CC-MAIN-20171119115102-20171119135102-00339.warc.gz | 195,905,882 | 2,860 | # Problem
• Oct 17th 2008, 11:16 AM
gracey
Problem
A boy travelling in a railway carriage decides to try to calculate the acceleration of the train. he suspends a parcel of mass 2kg from the roof of the carriage with a string and measures the angle that the string makes with the vertical, when the train is travelling with a constant acceleration of a ms^(-2) the angle is 8 degrees. find the value a
thanks for any help
• Oct 17th 2008, 06:31 PM
outatime1.21
This *may* help
Hey! No one has answered yet, so I might as well give it a shot! (it's been a while). I'm not sure what you mean by ms^(-2) though. But in any case:
I used:
F= mg
F=2kg(9.8m/s^2)
F=19.6N (the gravity that would pull the object down). But it's on an angle, so you have to imagine a right angled triangle. 19.6 is the value of the vertical part of the triangle. We know the top angle in the triangle is 8. Using tan to get the horizontal force (the force pulling the weight out by 8 degrees) :
tan8 = x/19.6 x = 2.75N (This is caused by the trains movement).
So then:
F=ma
2.75N = (2kg)(a)
a = 1.375m/s^2
I hope it's right... anyone else have other solutions? | 344 | 1,138 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2017-47 | longest | en | 0.931357 |
https://www.meritnation.com/cbse-class-10/foundation-course/math/triangles/studymaterial/12_0_1132_5130 | 1,519,480,814,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815812.83/warc/CC-MAIN-20180224132508-20180224152508-00454.warc.gz | 929,397,422 | 11,776 | Select Board & Class
Triangles
Correspondence of Vertices, Angles and Sides of Triangles
Let us consider ΔABC and ΔXYZ as shown below.
It can be seen that points A, B and C are the vertices of ΔABC, and points X, Y and Z are the vertices of ΔXYZ.
If vertex X is the pair of vertex A, then we can say that vertex X corresponds to vertex A and it is symbolised as ‘A → X’. Similarly, if vertex A corresponds to vertex X, then it is symbolised as ‘X →A’. Hence, vertices A and X correspond to each other and it is symbolised as ‘A ↔ X’, which is read as‘there is one to one correspondence between A and X’.
Similarly, the correspondences B ↔ Y and C ↔ Z can also be formed.
All of these correspondences can be represented together as ‘ABC ↔XYZ’.
In the same way,there may be other correspondences between the vertices of ΔABCand ΔXYZ. All the possible correspondences between ΔABCand ΔXYZ are listed below.
Correspondence between vertices Correspondence written together (1)A ↔ X, B ↔ Y, C ↔ Z ABC↔ XYZ (2)A ↔ X, B ↔ Z, C ↔ Y ABC↔ XZY …
To view the complete topic, please
Syllabus | 283 | 1,089 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2018-09 | latest | en | 0.901812 |
https://cses.fi/367/result/2946193 | 1,660,325,715,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571745.28/warc/CC-MAIN-20220812170436-20220812200436-00727.warc.gz | 195,676,172 | 3,983 | CSES - Datatähti 2022 alku - Results
Task: Tietoverkko Sender: intoo Submission time: 2021-10-04 22:45:31 Language: C++17 Status: READY Result: 100
Feedback
groupverdictscore
#1ACCEPTED10
#2ACCEPTED15
#3ACCEPTED75
Test results
testverdicttimegroup
#1ACCEPTED0.01 s1, 2, 3details
#2ACCEPTED0.01 s2, 3details
#3ACCEPTED0.23 s3details
### Code
```#include <iostream>
#include <vector>
#include <tuple>
#include <algorithm>
using namespace std;
using ll = long long;
ll k[200200], s[200200];
vector<tuple<int, int, int>> q;
int id(int x);
bool same(int a, int b);
void join(int a, int b);
int main()
{
int n;
cin >> n;
for (int i = 1; i <= n; ++i) {
k[i] = i;
s[i] = 1;
}
for (int i = 1; i < n; ++i) {
int a, b, x;
cin >> a >> b >> x;
q.emplace_back(x, a, b);
}
sort(q.begin(), q.end(), greater<tuple<int, int, int>>());
ll t = 0;
for (auto [x, a, b] : q) {
if (same(a, b)) {
t += x;
} else {
t += s[id(a)]*s[id(b)]*x;
join(a, b);
}
}
cout << t << endl;
}
int id(int x)
{
while (k[x] != x)
x = k[x];
return x;
}
bool same(int a, int b)
{
return id(a) == id(b);
}
void join(int a, int b)
{
a = id(a);
b = id(b);
if (s[a] < s[b]) swap(a, b);
k[b] = a;
s[a] += s[b];
}
```
### Test details
#### Test 1
Group: 1, 2, 3
Verdict: ACCEPTED
input
100
1 2 74
1 3 100
2 4 50
3 5 40
...
correct output
88687
user output
88687
#### Test 2
Group: 2, 3
Verdict: ACCEPTED
input
5000
1 2 613084013
1 3 832364259
2 4 411999902
3 5 989696303
...
correct output
1103702320243776
user output
1103702320243776
#### Test 3
Group: 3
Verdict: ACCEPTED
input
200000
1 2 613084013
1 3 832364259
2 4 411999902
3 5 989696303
...
correct output
1080549209850010931
user output
1080549209850010931 | 679 | 1,692 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-33 | latest | en | 0.267056 |
https://www.varsitytutors.com/calculus_3-help/calculus-review?page=2 | 1,643,027,502,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304528.78/warc/CC-MAIN-20220124094120-20220124124120-00123.warc.gz | 1,084,320,820 | 52,714 | # Calculus 3 : Calculus Review
## Example Questions
### Example Question #22 : How To Find Acceleration
Function gives the velocity of a particle as a function of time.
Find the equation that models that particle's acceleration over time.
Explanation:
Recall that velocity is the first derivative of position, and acceleration is the second derivative of position. We begin with velocity, so we need to integrate to find position and derive to find acceleration.
To derive a polynomial, simply decrease each exponent by one and bring the original number down in front to multiply.
So this
Becomes:
So our acceleration is given by
### Example Question #25 : How To Find Acceleration
Consider the following position function:
Find the acceleration after seconds of a particle whose position is given by .
Explanation:
Recall that acceleration is the second derivative of position, so we need p''(7).
Taking the first derivative we get:
Taking the second derivative and plugging in 7 we get:
So our acceleration after 7 seconds is
.
### Example Question #29 : How To Find Acceleration
Given the vector position:
Find the expression of the velocity.
Explanation:
All we need to do to find the components of the velocity is to differentiate the components of the position vector with respect to time.
We have :
Collecting the components we have :
### Example Question #32 : How To Find Acceleration
A car is driving north on a highway at a constant velocity of mph. What is the acceleration after an hour?
Explanation:
If a car is travelling north at constant velocity 60 mph, it's possible to write a velocity function for this vehicle, where is time in hours.
To find the acceleration, take the derivative of the velocity function.
The acceleration after an hour, or any time , is zero.
### Example Question #31 : How To Find Acceleration
Consider the position function , which describes the positon of an oxygen molecule.
Find the function which models the acceleration of the particle.
Explanation:
Recall that velocity is the first derivative of position and acceleration is the second derivative of position.
So given:
Apply the power rule to each term to find the velocity.
Applying the power rule a second time we arrive at the acceleration function.
### Example Question #41 : How To Find Acceleration
The displacement of an object at time is defined by the equation . What is the acceleration equation for this object?
Explanation:
The acceleration equation is the second derivative of the displacement equation.
Therefore the first derivative is equal to
Differentiating a second time gives
### Example Question #231 : Acceleration
Given the velocity function , find the acceleration function
Explanation:
We know that acceleration is the derivative of velocity with respect to time.
We also know that the velocity function is given by
We need to apply the product rule to solve for the derivative.
Recall that the product rule is given by:
In our case, and
Therefore,
We can reduce some terms in the acceleration function.
The final answer can be given as
### Example Question #235 : Acceleration
Find the acceteration at , given the velocity function below.
Explanation:
To solve, simply differentiate to find the acceleration function and then plug in .
### Example Question #631 : Calculus
The position of a particle traveling along the axis is described by
.
What is the acceleration rate of the particle when have passed?
Explanation:
To find the acceleration of a particle at a specific time we will need to take the derivative of the poistion function twice. Taking the derivative of position onces gives us the velocity function. Taking the derivative a second time will result in the acceleration function.
Since ,
we will apply the power rule for differentiation which state, .
Therefore applying it once we get,
From here we will apply the power rule one more time to find the acceleration function.
Now substitute to find the specific acceleration the question is asking for.
### Example Question #241 : How To Find Acceleration
Find the acceleration at given the following velocity equation. | 848 | 4,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2022-05 | latest | en | 0.859674 |
https://paperhaberdasher.com/need-93 | 1,670,373,529,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711121.31/warc/CC-MAIN-20221206225143-20221207015143-00622.warc.gz | 485,937,704 | 4,998 | # Find the vertex focus and directrix
After expressing the equation in the form (x - h)^2 = 4p (y - k), the vertex is given by (h, k), the focus is given by (h, k + p) and the directrix is given by the line y = k - p. After
## Parabolas: Finding Information from the Equation
Given the values of a, b and c; our task is to find the coordinates of vertex, focus and the equation of the directrix. Example – Input : 5 3 2 Output :
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## How To Find The Vertex Focus and Directrix Of The Parabola
Find the vertex, focus, and directrix of the parabola given by the equation. (x-1)^2=4(y+4) vertex (x,y)=(square) focus (x,y)=(square)directrix square. CameraMath is an essential learning and
How do I determine the vertex, focus and directrix of a parabola?
Finding the Focus and Directrix of a Parabola in Vertex Form - Vocabulary and Equations Vertex Form: The vertex form of a parabola is {eq}y = a (x-h)^2 + k {/eq}, where the point {eq}
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## How to Find the Focus, Vertex, and Directrix of a Parabola?
x=-1 is the equation of directrix. Ques. Given that the vertex and focus of parabola are (-2, 3) and (1, 3) respectively, find the equation of the parabola. Ans. Since the vertex is (-2, 3) the equation
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## Parabola Calculator
Find the vertex, focus, and directrix of the parabola and sketch its graph. Use a graphing utility to verify your graph. x+y^2=0Watch the full video at:https
Determine mathematic questions
Get help from expert teachers | 555 | 2,127 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.893456 |
http://eufisky.is-programmer.com/tag/%E7%BA%A7%E6%95%B0%E8%AE%A1%E7%AE%97 | 1,725,982,589,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651255.81/warc/CC-MAIN-20240910125411-20240910155411-00408.warc.gz | 9,516,939 | 9,018 | Eufisky - The lost book
## FoxTrot Series
$F = \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}{n^2}}}{{{n^3} + 1}}} = \frac{1}{3}\left[ {1 - \ln 2 + \pi \mathrm{sech}\left( {\frac{{\sqrt 3 }}{2}\pi } \right)} \right].$ | 114 | 241 | {"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} | 2.84375 | 3 | CC-MAIN-2024-38 | latest | en | 0.262939 |
http://www.romannumerals.co/number-converter/15-in-roman-numerals/ | 1,632,113,384,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057018.8/warc/CC-MAIN-20210920040604-20210920070604-00087.warc.gz | 111,232,022 | 14,350 | ## What is 15 in Roman Numerals?
### A: XV
15 = XV
Your question is, "What is 15 in Roman Numerals?", and the answer is 'XV'. Here we will explain how to convert, write and read the number 15 in the correct Roman numeral figure format.
## How is 15 converted to Roman numerals?
To convert 15 to Roman Numerals the conversion involves you to split it up into place values (ones, tens, hundreds, thousands), like this:
Place ValueNumberRoman Numeral
Conversion10 + 5X + V
Tens10X
Ones5V
## How to write 15 in Roman numerals?
To write 15 in Roman numerals correctly you combine the values together. The highest numerals should always precede the lower numerals in order of precedence to give you the correct written combination, like in the table above (top to bottom). like this:
X+V = XV
## How do you read 15 as Roman numerals
To correctly read the number 15 as the Roman numeral XV, It must be read as it is written; from left to right and from high to low numbers.
It is incorrect to use the Roman symbol XV in a text, unless it represents an ordinal value. In any other usage case it should be written in the normal format (arabic number) 15.
## More from Roman Numerals.co
16 in Roman numerals
Now you understand how to read and write 15 in Roman Numerals, see how the number 16 is written.
Convert Another Number | 329 | 1,333 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.821756 |
https://uniapaclisbon2018.com/blog/height-to-depth-ratio-for-stability/ | 1,679,339,265,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943555.25/warc/CC-MAIN-20230320175948-20230320205948-00436.warc.gz | 687,821,690 | 14,105 | Height To Depth Ratio For Stability
Height To Depth Ratio For Stability. The factor of safety can be determined in terms of height as the ratio of the maximum height of slope ( critical depth ) to the actual height of the slope. If the height of the center of gravity is less than the distance from the cog to the front (or whatever direction it might tip around) then it will be most stable since it will have to defy gravity to tip.
Ikea Kitchen Wall Depth Best Ideas 2022 from ufficiale.thonghutsaigon.com
If that same 42″ deep. Fh = hmax h (11.5) the factors of safety f c. f φ. f h are only occasionally used in slope stability analyses. The htd ratio describes the relationship between how tall the rack is compared to how wide it is at its base.
Source: archive.everyday-physics.com
In simple terms. the htd ratio describes the ratio between how tall the rack is compared with how wide it is at its base. The depth to point c increases with the base dimension and aspect ratio.
damotech.com
In simple terms. the htd ratio describes the ratio between how tall the rack is compared with how wide it is at its base. Therefore. to balance precision and simplicity. we kept 2 decimal places for ratios.
Width to height ratio of 1.73 and then benched to a width to height ratio of 0.98. Therefore. fs = hc/h eq.2.
Source: idea-systems.net
Series of laboratory tests had been carried out on two diaphragm cells of different width to depth ratios (0.75. 0.85. and 1). Height:diameter ratio. tree stability. thinning. stand density. western montana.
pinterest.com
Safety is defined as the ratio of the sum of the resisting moments divided by the sum of the. The factor of safety can be determined in terms of height as the ratio of the maximum height of slope ( critical depth ) to the actual height of the slope.
Dimensionless ratios used to reconstruct a. The depth to point c increases with the base dimension and aspect ratio.
Model Prognosis Was Unable To Predict Height:diameter Ratios For Developing Stands.
The height (h) of the racking system divided by the depth (d) of the frame should not exceed six (6) for optimum stability. Series of laboratory tests had been carried out on two diaphragm cells of different width to depth ratios (0.75. 0.85. and 1). Therefore. fs = hc/h eq.2.
For Example. A Rack That Is 10 Feet Tall And 10 Feet Wide (A 1 To 1 Htd Ratio) Will Be Much More Stable (Less Likely To Fall Over) Than A Rack That Is 10 Feet Tall And 1 Foot Wide (A 10 To 1 Htd Ratio).
The stability of any upright piece is dependent on the location of the center of gravity (cog). The depth to point c increases with the base dimension and aspect ratio. Dimensionless ratios used to reconstruct a.
For A Standard 20′ Tall. 42″ Deep Frame With The Beam At The Top Level. The Ratio Is 5.714 — An Acceptable Ratio With Normal Anchoring.
Bank height ratio sometimes referred to as the effective flow or ordinary high water flow. Eta height of thrusting above slip surface in janbu analysis. Width to height ratio of 1.73 and then benched to a width to height ratio of 0.98.
The Parts Were Then Cut. Polished. And Viewed Under A Microscope.
In the northern rocky mountains. snows can be heavy from october to june. and winds can strike with force at any time of the year. Models to observe their stability. These factors are effect of width. height. properties of soil and embedment depth to height ratios (0.15. 0.3. and 0.45).
In Simple Terms. The Htd Ratio Describes The Ratio Between How Tall The Rack Is Compared With How Wide It Is At Its Base.
A height to width ratio (h/w) of approximately 1:1. So maybe what i really should be asking is what kind of ratios do you use for engeineered metal buildings? Therefore. to balance precision and simplicity. we kept 2 decimal places for ratios. | 912 | 3,821 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2023-14 | latest | en | 0.909158 |
https://www.briangwilliams.us/paleoclimatology/within-the-column-1.html | 1,579,746,385,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250608062.57/warc/CC-MAIN-20200123011418-20200123040418-00486.warc.gz | 787,189,872 | 9,428 | ## Within The Column
Electronics Repair Manuals
Get Instant Access
current vectors, temperature, and salinity
FIGURE 12.2 Schematic diagram illustrating how the atmosphere and ocean are divided into columns in a typical coupled general circulation model experiment. Ocean and atmospheric grid sizes are commonly different. Computations take place simultaneously for all grid boxes at all specified levels (McGuffie and Henderson-Sellers, 1997).
are too small to be represented by even a 2° x 2° grid spacing. In such cases, the process is represented in a simplified manner as a function of other variables, a procedure known as parameterization (parametric representation). This may be based on observed statistical relationships between, for example, temperature and humidity profiles and cloudiness, or on some other simplified model of the process in question. In fact, parameterization of all forms of cloud is one of the most difficult problems in atmospheric GCMs and is the focus of much research at present (Cess etal, 1995).
Atmospheric general circulation models may be coupled to the ocean realm in a variety of ways. At the simplest level, the surface temperature of the ocean is prescribed (predetermined) and the ocean region of the model interacts with the atmosphere only in terms of moisture exchange. This is often termed a "swamp ocean" (Fig. 12.3). At the next level a "slab ocean" is specified as a layer of fixed depth (50-100 m); heat and moisture exchange with the atmosphere occurs, enabling SSTs to vary as the model run progresses. However, in such models there is no
Moisture
Moisture Energy
Moisture
Energy
(Momentum)
Moisture
Energy
Momentum
Coupling
Prescribed SSTs
Swamp
Slab
Calculated SSTs with prescribed advection. No vertical motion
FF1!
Calculated SSTs with prescribed advection and detailed calculation of fluxes through the mixed layer to the deep ocean
Detailed Mixed Layer f f
Dynamic Ocean
Calculated temperatures, velocities and salinities at all levels
FIGURE I 2.3 Schematic diagram to show the different levels of complexity and coupling with the atmosphere in various types of ocean model (McGuffie and Henderson-Sellers, 1997).
mechanism for heat exchange with the deep ocean and only a crude representation of horizontal energy fluxes. The mixed layer ocean is a further improvement, also involving prescribed horizontal advection, but with computation of fluxes to and from the deep ocean. The most complex level is a fully coupled ocean-atmosphere GCM (OAGCM) in which the ocean has internal dynamics in three dimensions, and exchanges of energy, moisture, and momentum take place at the ocean-atmosphere interface. It is worth noting that a major problem in coupling atmospheric to oceanic processes is the vastly different response times characteristic of each domain (Fig. 12.4). The slower response time of the deep ocean must be taken into account when the two systems are linked. The problem is further compounded if one is trying to investigate climate system changes involving ice sheets, which have even longer response times. One approach is to couple models of each system "asynchronously," that is, to operate an atmospheric model for a time appropriate for that system, then use the resulting atmospheric conditions as input to a model of the system operating on a different timescale. For example, Schlesinger and Verbitsky (1996) used the climate at 115 ka B.P., generated by an atmospheric GCM (coupled to a mixed layer ocean), to drive an ice sheet/asthenosphere model, in order to investigate the areas most likely to develop ice sheets following the last interglacial
FIGURE 12.4 Estimated response or adjustment (equilibration) times for different components of the climate system. Note that they vary by 6-7 orders of magnitude (Saltzman, 1985).
period (see Section 12.3.1). The ice sheet model was run for 10,000 yr (an appropriate interval of time to examine ice sheet development) but the atmospheric conditions obviously could not be computed over such a long period and so were fixed as those obtained from the 115 ka B.P. simulation. Although this approach clearly has its limitations, it does allow two systems with strikingly different response times to be examined in a somewhat coupled fashion.
GCMs can also be used to trace the pathways of materials within the climate system (Koster et al., 1989). This has been put to good use in paleoclimatic applications (Jouzel, 1991; Jouzel et al., 1993a; Andersen et al., 1998). For example, the long-distance transport of desert dust particles in the atmosphere has been traced using an AGCM for both modern and last glacial maximum (LGM) conditions. Source regions of dust deposited from the atmosphere are identified by "tagging" the dust originating from different areas (Joussaume, 1987, 1990, 1993). Modern-day simulations show a strong seasonality in atmospheric dust production, with atmospheric dust loading in August more than twice that in February. The largest source of dust (by far) is the Sahara /Arabia /central Asia region. Australia is the principal source of dust reaching east and west Antarctica, whereas South America contributes the most dust to central Antarctica. Simulations for the LGM show greater atmospheric dust deposition especially over the tropical Atlantic Ocean and Europe, but the modeled increases significantly underestimate the observed changes (recorded in ice cores). This could be due to many factors, including model resolu tion (poor representation of source areas) and/or inadequate characterization of dust entrainment, transportation, and depositional processes (wet and dry fallout). In view of the potential climatic significance of dust during glacial periods (Over-peck et al., 1996) these first steps toward fully incorporating the dust cycle into GCMs are important contributions. Further studies with higher resolution GCMs are now needed.
Isotopes (deuterium, 180) in the hydrological cycle have also been modeled with GCMs, for modern and glacial age conditions (Jouzel et al., 1987c, 1991, 1994; Joussaume and Jouzel, 1993). At each change of phase of water molecules in the model, appropriate fractionation factors are employed to calculate the mass of the isotopes in each reservoir (water vapor, precipitation, ice, groundwater). Iso-topic modeling is particularly important in paleoclimatic studies as it allows the direct comparison of model simulations with paleoisotope records (in ice, sediments, biological materials, speleothems, etc.), thus avoiding the need to calibrate the paleo-record in terms of, say, temperature for a comparison with modeled paleo-temperature output. Modern simulations reproduce the global pattern of 8lsO and 8D very well and the seasonal cycle is well captured at most sites in both high and low latitudes (Jouzel et al., 1987c). The LGM simulations show a similar overall 8180/temperature relationship to that derived from modern simulations (8lsO ~0.6°T, where T < -5 °C) and there were large decreases in 8180 at high latitudes (see Fig. 5.9) (Joussaume and Jouzel, 1993).
Sources of moisture can be traced using GCMs and this is useful in understanding how source regions might have been different in the past; this would be relevant, for example, in the interpretation of ice core geochemistry. Charles et al. (1994) used an AGCM to examine how source regions of precipitation reaching Greenland changed from the LGM to the present. The modern (control) simulation showed that 26% of Greenland precipitation was derived from the North Atlantic (30-50° N), 18% from the Norwegian-Greenland Sea, and 13% from the North Pacific. At the LGM, these values changed to 38%, 11%, and 15%, respectively. However, northern Greenland received distinctly more moisture at the LGM from the north Pacific source region, due to displacement of storm tracks around the Laurentide Ice Sheet. Southern Greenland received most of its snowfall from North Atlantic moisture sources. Because of the much longer (and colder) trajectory of the Pacific air masses, snow deposited on Greenland from such sources was much more depleted in 8lsO than snow from North Atlantic sources (~15%o lower). Charles et al. (1994) point out that if there was no change in temperature in Greenland, but only a shift in source region from purely North Atlantic moisture to a 50:50 mix of North Atlantic and North Pacific moisture, changes in 8lsO of snowfall could change by ~7%o, equivalent to the large amplitude oscillations seen during late glacial time in the GISP2/GRIP ice cores. This raises the interesting possibility that abrupt changes in 8180 seen in the ice cores from Greenland may be partly related to changes in storm tracks rather than large-scale (hemispheric) shifts in temperature. However, it is worth noting that this model did a very poor job of simulating modern precipitation in Greenland (with simulated precipitation exceeding observations by as much as 100%, or ~1 mm day"1, especially in summer); thus this experiment, though interesting, begs the question whether similar results would be found with a more accurate, higher resolution model.
Although the discussion so far has focused on general circulation models of atmospheric and oceanic systems, further development of GCMs will be towards total climate system models (CSMs), which will incorporate in a fully interactive way land surface and cryospheric processes (both terrestrial snow and ice, and sea ice) and biomes. Models of global biomes are now available (Prentice et al., 1992; Hax-eltine and Prentice, 1996) and have been used with output from GCM experiments to predict vegetation at times in the past (Harrison et al., 1995; TEMPO, 1996). Models with full biogeochemical cycling of materials in the atmospheric and oceanic realms are also under development (Brasseur and Madronich, 1992; Sarmiento, 1992). Nested models, in which a very detailed grid network for a specific region is used to model detailed geographical variations of the climate, given initial input from a larger scale GCM, will also become more widely available (Hostetler et al., 1994). Such models are especially important in mountainous areas where large-scale GCMs cannot provide the necessary topographic detail to produce meaningful regional simulations. | 2,240 | 10,346 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-05 | latest | en | 0.899689 |
https://pt.scribd.com/document/385283537/Physics-1group | 1,571,313,520,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986673538.21/warc/CC-MAIN-20191017095726-20191017123226-00482.warc.gz | 654,283,318 | 74,897 | Você está na página 1de 6
GUJARAT TECHNOLOGICAL UNIVERSITY
PHYSICS
B.E. 1stYEAR
Prerequisite: Basic understanding`of Calculus, Physics and Mathematics course on Differentiate
equations
Rationale: The basic science - physics program is to prepare students for careers in engineering
where physics principles can be applied to the advancement of technology. This education at the
intersection of engineering and physics will enable students to seek employment in engineering
upon graduation while, at the same time, provide a firm foundation for the pursuit of graduate
studies in engineering.
Instructor of Course : Instructor must have academic qualification as per norms of University in
subject of Physics.
Teaching andExamination Scheme:
TeachingScheme Credits Examination Marks Total
L T P C TheoryMarks Practical Marks Marks
ESE PA ESE PA
(E) (M) Viva (V) (I)
3 0 2 4 70 30* 30# 20 150
L-Lectures; T-Tutorial/Teacher Guided Student Activity; P-Practical; C-Credit; ESE-End Semester Examination; PA-
Progressive Assessment
Content:
Teaching Module
Sr No Topic
Hrs. Weightage
MODULE 1: Properties of Matter 7 19
Concept of Load, Stress and Strain
Hook's Law
Stress-Strain Diagram
Ductility, Brittleness and Plasticity
Elastic behavior of solids
Working stress and factor of safety
Factors affecting elasticity
Types of Elasticity
Twisting couple on a cylinder or wire-shaft
Torsional Pendulum
Cantilever-Depression of Cantilever
Young's modulus by Cantilever
I-shape Griders
Viscosity and comparison of viscosities
MODULE 2: Waves, Motion and Acoustics 7 19
Simple Harmonic motion
Free, forced, resonance, damped and undamped
vibration
Damped harmonic motion
Force vibration and amplitude resonance
Velocity resonance and energy intake
Wave motion, transverse and longitudinal vibration
Sound absorption and reverberation
Sabine's formula and usage (excluding derivation)
Acoustic of building
Module 3: Ultrasonic and Non distractive testing (NDT) 9 25
Ultrasonic waves
Properties of ultrasound
Production of ultrasonic waves : Piezoelectric and
magnetostriction method
Detection of ultrasound
Application of ultrasound
Introduction of NDT
Advantages of NDT
NDT through ultrasound
Module 4: Superconductivity 6 17
Introduction of Superconductivity
Properties of superconductor
Effect of magnetic field
Meissner effect
Pressure effect
Impurity effect
Isotopic mass effect
Mechanism of Superconductivity : BCS Theory
Penetration depth : Magnetic field
Josephson's junction and its application
Application of superconductors
Module 5: Lasers 7 20
Properties of Laser
Einstein’s theory of matter radiation : A and B
coefficients
Amplification of light by population inversion
Different types of lasers
gas lasers ( He-Ne) solid-state lasers(ruby)
Properties of laser beams: mono-chromaticity, coherence,
directionality and brightness, laser speckles
Applications of lasers in science, engineering and
medicine.
Suggested Reference Books
1. Engineering Physics by Dattu R Joshi, McGraw hill Publications
2. Engineering Physics by Shatendra Sharma & Jyotsan Sharma, Pearson Publication
3. Mechanics of Materials, SI Edition, 9th Edition,Barry J. Goodno, James M. Gere,Published: © 2018
Print ISBN: 9781337093354
Course Outcome:
1. The student will demonstrate the ability to think in core concept of their engineering
application by studying various topics involved in branch specific applications.
2. The student will demonstrate the ability to use appropriate mathematical techniques and
concepts to obtain quantitative solutions to problems in physics.
3. In courses involving laboratory, the student will demonstrate the ability to collect and
analyze data and to prepare coherent reports of his or her findings.
4. In a design module project, the student will demonstrate the ability to perform a literature
search, to make use of appropriate computational or laboratory skills and to make an effective
written or oral presentation of the results of the project.
List ofExperiments:
Important Note
Total14 experiments arelistedin the design module.
Key goals of theseexperimentsare :
(1)Toenhancetheunderstanding ofstudent towardstheerrorspresent intherealtimemeasurementand the ways
totake careofthem.
(2)Tocreate visualizationofvarious phenomenacoveredin the syllabus.
(3)Toinducetheskillofstudentin handlingdifferentmeasuringinstruments.
Subjectteacher is advisedto setup any8 experiments fromthe followinglist.
Inthesessionstudentshouldperform minimum4setofexperimentsandcomplete onesmallproject
based onengineering applications. This projectalong with any performed experiment should be
EVALUATEDBYEXTERNAL EXAMINER.
1 .Diffraction and interference experiments (from ordinary light
or laser pointers); measurement of speed of light on a table top
modulation; minimum deviation from a prism.
2. Measurement of the Distance using Ultrasonic Sensors.
7. Wavelength of Light -Diffraction GratingUsing LASER
8. Acoustic grating method set up for measurement of velocity of ultrasonic waves in liquid
9. Melde’s experiment
10 Resonator
Open endedProjectsin Science andtechnologystudy :-
Aims:
1.Toprovideexperienceinlaboratorybasedexperimentation,datarecordingandanalysisanddrawingof conclusions.
2. To develop report writing skills for scientific material
3. Todeveloptheabilitytoundertake investigations where,aspartoftheexercise, the goals and
methods have tobedefined bythe investigator.
4. Todevelopskillsin literature searchesandreviews.
Inthebeginningoftheacademicterm,facultieswillhavetoallottheirstudentsatleastone (Students arefreeto
selectany area ofscienceandtechnology)
- Open ended design based smallproject or
- Computerbasedsimulation/webbasedapplication/analysispresentationsofappliedscience field which
mayhelp themin theirbranchesespeciallyin theirUDP/IDP projects.
1. Thesecan be doneina group containingmaximumthreestudentsin each.
2. Openended designbasedsmallproject ORUDPbasedstudywillbeevaluatedbyexternal
examinerwith appropriatemarks allotmentgiven byGTU time to time.
3. Facultiesshouldcultivateproblembasedprojecttoenhancethebasicmentalandtechnicallevelof students.
4. Evaluationshouldbedoneonapproachofthestudentonhis/herefforts(notoncompletion)to
studythedesign module ofgiven task.
Open EndedProjectfields:-
Studentsarefreetoselectanyareaofscienceandtechnologymaybebasedontheirbranchesto define
projects.
Some suggested projectsare listed below:
1. Design:A workingelectricmotor.
Area:Electricityand Magnetism
Using:1meterofbendable,insulatedwire,asize"D"battery,adiskmagnet,twopaperclips,
sandpaper,wire strippers, maskingtape.
2. Design:Computerbasedsimulation/smallcalculationwithhelpbasicprogramminglanguage basedon
Physics
Area: Computationalphysics
3. Design: A Hydraulic Jack works on the principle of Pascal’s law that states Area: Fluid
Dynamics
Using:pokerand scissors ,syringes, M-seal,inletpipes
List ofOpen SourceSoftware/learning website:
The FlyingCircus of Physics 2ndedition byJearlWalker, Wiley India
SixIdeas that shaped physics byThomas A Moore,McGrawHilleducation
http://www.howstuffworks.com/--Tech stuff
Howthings works by Louis A Bloomfeild,WileyPublications
Physicsof Everyday Phenomena by W. Thomas Griffith, Juliet Brosing, McGraw Hill
Education
LatestjournalslikeBBCKnowledge,Howthingswork-everydaytechnologyexplainedby
National Geographics.
http://www.sciencefairadventure.com/
vlab.co.in
ACTIVELEARNING ASSIGNMENTS:Preparation ofpower-pointslides,whichincludevideos,
animations,pictures,graphicsforbetterunderstanding theory andpracticalwork–The faculty willallocate
chapters/partsofchapterstogroupsofstudentssothattheentiresyllabusofPhysicsiscovered. Thepower-
pointslidesshould be putup on the web-site of the College/Institute, alongwith the namesof the students of
thegroup,thenameofthefaculty,DepartmentandCollegeonthefirstslide.Thebestthreeworksshouldbe
sentto achievements@gtu.edu.in.
Note:Passingmarks forPA (M) will be12 outof30.
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http://sciforums.com/threads/where-is-most-gravity-inside-or-out.155498/#post-3363867 | 1,695,860,135,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510334.9/warc/CC-MAIN-20230927235044-20230928025044-00494.warc.gz | 31,239,216 | 16,970 | # Where is most "gravity", inside or out?
Discussion in 'Pseudoscience' started by nebel, Feb 29, 2016.
1. ### nebel
Messages:
2,469
In any entity, gravity falls to zero at the center, but reaches that value only at an indefinite distance. At a heights equal to the radius, there is still 1/4 of the gravitational surface pull to the center, whereas the same distance down from the "surface", it is already balanced out of existence. What does this mean for large scale structures?, their orbital velocities "in" and around them?
Last edited: Feb 29, 2016
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3. ### originHeading towards oblivionValued Senior Member
Messages:
11,775
The origin is the center of the body and gs is the force felt from gravity or gravitational acceleration at the surface of the body.
The only place where the force is 0 is at the exact center of the mass. You have to be careful of what you are discussing: gravitational force, gravitational potential, etc.
Last edited: Feb 29, 2016
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5. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member
Messages:
23,198
The graph is correct, but it may be helpful to explain why for r<R it is a linear rise from zero. First note with unifrom spherical mass distribution 7/8 of the mass is between R/2 and R and this mass has zero net pull on a point at r = R/2. Only the mass at R/2 or less radius pulls on that point. That mass is only 1/8 of the total mass and acts as if it were all at the origin center. As R/2 is closer to the center the gravity, from that 1/8 mass is 4 times more effective (inverse square law at work). So at the R/2 point the force of gravity ie (1/8)x4 = 1/2.
This same logic, math applies at any r < R but is not as simple to do except for the r =R/3, r = R/4, etc. cases. I. e. at r = R/n the gravity will be 1/n what it is at r = R.
Hint: At R/3 point only 1/27 of the mass acts on that point, but it is 9 times more effective. So there the gravity force is (1/27)x 9 = 1/3 and right on the linear line. etc.
Last edited: Feb 29, 2016
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7. ### nebel
Messages:
2,469
Thank you , so: why are we surprised that we have comparative small orbital velocities in the inner, and greater movement in the outer reaches of entities, where most of the gravity is?
PS: and what would a graph of orbital velocities look like in such an idealized system? with low mass orbiting bodies?
Last edited: Feb 29, 2016
8. ### originHeading towards oblivionValued Senior Member
Messages:
11,775
I am not sure what you are talking about. The orbital velocity is faster the closer the bodies are to each other not lower. That is exactly what we see for our solar system. What do you mean most the gravity is in the outer reaches?
You understood on the graph that R is the surface of the planet, right?
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9. ### Janus58Valued Senior Member
Messages:
2,361
If you are talking about the orbital speeds for stars in a galaxy then you have to realize that the graph given that shows gravity increasing linearly as you move out from the center until you reach the surface is for a spherical object of uniform density. A typical galaxy is neither of these things. To work out the expected orbital velocities for a typical spiral galaxy, you need to take its shape and density distribution into account. A spiral galaxy consists of a dense central bulge and a thinner surrounding disk. If you plot the expected orbital speeds as you move outward from the center based on the visible shape of the galaxy, you would expect the velocities to increase at first while you stay in the bulge region similar to it does inside the uniform sphere, and then fall off as you leave it and enter the disk region a lot like it does it would outside of the sphere's surface. The fact is that this does not happen. The orbital velocities do increase while inside the bulge, but they don't decrease like they should once you leave the bulge region, depending on the galaxy, they either decrease too slowly, don't decrease at all or even increase slightly. So not only do the orbital velocities behave as if there is more mass than we can see, but that that much of that mass is distributed a lot differently than the visible shape of the galaxy.
10. ### expletives deletedRegistered Senior Member
Messages:
410
Janus58:
That is a very interesting explanation. Thanks. I am confused now, about that part of the orbital velocity profile from inner disk towards outer disk. I understand the part involving the spherical central bulge profile, but I would have thought that, since unlike what a spherical distribution would do to the orbital speeds, the disk is a disk, not a sphere, so the majority of the mass pulling on the outer stars would be along the plan of the disk, without much at all either side of the galaxy like there would be if it were a spherical, not disk, galactic outer region. Am I making sense? My reasoning is that if a galaxy is spherical all the way from inner central region to its outermost regions, then the gravitational acceleration and orbital velocity profile would be as you described for a spherical body and for the central bulge of the spiral galaxy. But settled spiral galaxies are not spherical all the way to the outermost regions, so why would the gravitational and velocity profiles be expected to drop if more and more galactic matter is along the plane, rather than on either side as in a spherical case, as we move outwards along the disk? That is my question and reasoning. I hope I haven't made a physics or logics blunder. If I haven't, can you or some other learned member explain the unusual expectation of decreasing velocity rather than increasing velocity as increasingly more mass is "below" the stars as we go from inner to outermost orbitals in the disk along the plane of the disk (not as in spherical distribution)? Thanks.
Messages:
27,543
As usual janus58, a nice concise explanation.
12. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member
Messages:
23,198
If you are a star outside a spherical galaxy, then you are pulled towards the center more strongly than if only the stars closer to the center were pulling on you. Imagine two cases: A = you are out side a sphere & B = you are outside a disk of thickness R/10, where R is your distance from the center.
In case A there are many stars not in the disk that are pulling you directly towards a point that is either above the region of the disk or below the region of the disk and they both have a component of their pull directly towards the center of the disk. Their components orthogonal to the disk cancel out, but the toward disk components add. This is why case A has greater force pulling towards the center than case B with only stars in the disk pulling on you.
Or stating it reversely: the force on you outside a spiral galaxy, missing the net added pull to the center as has no stars above and below the disk, is less less than outside a spherical galaxy. I'm not sure I understood you question, but if I did that should answer it.
Last edited: Feb 29, 2016
13. ### Janus58Valued Senior Member
Messages:
2,361
The orbital speed depends on the mass(M) you are orbiting and your distance from the center of that mass. Increasing M has the effect of increasing the orbital speed and increasing the distance has the effect of decreasing it If you are in a spherical body, or a disc, the only mass contributing to M is that part of the body closer to the center than you are. With a spherical body the mass closer to the center increases by the cube of your distance from the center, so when you move away from the center, the effect of increasing M exceeds the effect of increasing the distance, so orbital speed goes up until you are outside the body. Then M remains constant even while the distance increases and orbital speed starts going down. With a disk, M does not go up as fast with r, and in fact, the two effects would cancel and you would expect the orbital speed to stay constant with distance. A spiral galaxy is neither purely spherical or disc shaped. It is a disc with a spherical bulge in the center. Inside the bulge, the effect on orbital speed in what you would expect to find in a sphere. Once you get out of the bulge, you have to consider both the effects the mass of the bulge has and the increasing mass added by the disk as you move outward. The effect of moving away from the mass of the bulge has the effect of decreasing the orbital speed, The addition of mass due to the disc will lessen this effect but will not be enough to counter it completely. Thus you should see the orbital speed fall off with distance as you increase distance from the bulge. Not as fast as it would if all the mass were concentrated in the bulge, but still falling off.
The whole point is is that when you take into account all the mass of the visible galaxy and the way it is distributed, you should get a certain pattern to the way orbital speed changes with distance. The fact that the actual orbital speeds we measure do not follow this pattern indicates that there is mass involved that we do not see. And it is just not a matter of more mass. If it were just a matter of getting the scaling factor of the mass wrong, you would get different values for the orbital speeds, but the pattern would stay the same. It is the fact that the pattern itself differs that indicates that the extra mass does not follow the same distribution as the matter we see in the galaxy.
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14. ### brucepValued Senior Member
Messages:
4,098
Very informative post. Hope somebody is paying attention
15. ### nebel
Messages:
2,469
paddoboy, yes the graph expresses my idea/ question, the "surface" R could be a planet's or any entity. even in a Galaxy with no visible outer boundary, there must be points that define it's size. Surely, a galaxy, or star cluster might not be like our Solar system where ~99.8% is in the very centre, and even then, in the center of the sun there is zero gravity. When I said " more gravity outside" I meant above the surface at R. I refer to the the fact, that gravity is a rapidly vanishing quantity toward the center, but the area representing the "outside gravity" tapers off to eternity. That is why I would like to see (for all viewers) a model put up where the gravity values from Zero at the center to max at the surface and off into the distanc are expressed in orbital velocities, not any matter supported by underlying compressed material. For example, is not in open star clusters very little speed of the inner stars required to keep from collapsing them into the center? and of course as Janus 58 brought out in his great post, , Galaxies have complex density gradients(oofen only estimated) that would have the orbital velocity diverge from the ideal pattern I would like to see.
Last edited: Mar 1, 2016
Messages:
27,543
That's all I'll say for now.
17. ### nebel
Messages:
2,469
It occurred to me, that there is a curious similarity to Johann Kepler's second law's illustrations, and the graph that origin put up. In both cases there is convergence, minimum area, at the Center of Gravity, or rather the center of non-gravity, and of course Kepler's illustrations could be amended to show gravity's value on the outside of the ellipse, or sphere, and the area would be showing to converge to zero again in the very far distance. The implications of this could best be seen in open star clusters, and elliptical galaxies, where stars surely must follow individual ~ stable orbits, devoid of a common rotation, with speeds that are very "slow" near the inevitable zero-gravity center(s) and show the greatest speed at the point or "envelope" of maximum commun self gravitation, equivalent to origin's peak at gs/R. If there would be any gas left in these old outer peripheries, , its particles should show orbital velocities that mirror the outside triangles area, illustrating this outer realm of greater gravity. -- any good illustrators out there?
Last edited: Mar 1, 2016
18. ### originHeading towards oblivionValued Senior Member
Messages:
11,775
The gravity distribution of a solar system is vastly different than a galaxy.
This is just a rehash of your earlier posts I'm now realize.
Here is a graph of the rotational velocity of a galaxy.
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19. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member
Messages:
23,198
That difference in post 15's graph, between green and orange curves leaves us in the dark - pun intended.
20. ### nebel
Messages:
2,469
indeed, it leaves us in the valid dark matter theory variants. Here is my question though: can it be said there is always more gravity outside the maximum point of gravity of any system, than toward it's (bari) centre? as an example, is there more gravity outside the Pluto surface than the one that faces away on a corresponding, equidistant point on it's orbit opposite the sun? Yes, a possible rehash, but with some spice added. because how is the fact that there is more gravity outside expressed in the movements we observe? origin's curves show that there is even more gravity outside than expected. and how about those open and ellipticals?
21. ### originHeading towards oblivionValued Senior Member
Messages:
11,775
How can there be a place that has 'more gravity' than the maximum? The maximum is the maximum.
If you mean is the force of gravity from Pluto on the surface of Pluto less when that surface is facing the sun, the answer is yes, but of course it is miniscule.
There is much more mass in the galaxy than is observed. So the unobserved matter must be dark.
The velocity profiles indicate there is more mass than what is observered, hence the theory of dark matter.
What do you mean?
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22. ### exchemistValued Senior Member
Messages:
12,076
Seconded. In fact this is a very high quality thread. | 3,151 | 13,965 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2023-40 | latest | en | 0.931345 |
https://greprepclub.com/forum/in-this-diagram-the-circle-is-inscribed-in-the-square-11684.html | 1,590,675,185,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347399820.9/warc/CC-MAIN-20200528135528-20200528165528-00019.warc.gz | 374,751,129 | 27,671 | It is currently 28 May 2020, 06:13
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In this diagram, the circle is inscribed in the square.
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In this diagram, the circle is inscribed in the square. [#permalink] 30 Nov 2018, 17:17
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In this diagram, the circle is inscribed in the square.
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GRE exam - In this diagram, the circle is inscribed in the square. .jpg [ 8.91 KiB | Viewed 3862 times ]
Quantity A Quantity B Length of diagonal AC $$\frac{5r}{2}$$
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA
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Re: In this diagram, the circle is inscribed in the square. [#permalink] 01 Dec 2018, 10:29
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2r = diameter of the circle.
Diameter of the circle = side of the square
since the adjacent sides of the squares are at 90 degrees, the diagonal = $$2r\sqrt{2}$$
This is because a 90-45-45 triangle will be formed between 2 sides of the square and the diagonal
now simplify
option A is:
$$2r\sqrt{2}$$
option B is:
$$\frac{5r}{2}$$
cancel r from both sides, then multiplying both sides by 2 we get,
option A = $$4\sqrt{2}$$
option B = 5
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Re: In this diagram, the circle is inscribed in the square. [#permalink] 11 Jan 2019, 17:15
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amorphous wrote:
2r = diameter of the circle.
Diameter of the circle = side of the square
since the adjacent sides of the squares are at 90 degrees, the diagonal = $$2r\sqrt{2}$$
This is because a 90-45-45 triangle will be formed between 2 sides of the square and the diagonal
now simplify
option A is:
$$2r\sqrt{2}$$
option B is:
$$\frac{5r}{2}$$
cancel r from both sides, then multiplying both sides by 2 we get,
option A = $$4\sqrt{2}$$
option B = 5
If it does not mention in the question how do we deduce that r is the radius or side or diagonal.
Re: In this diagram, the circle is inscribed in the square. [#permalink] 11 Jan 2019, 17:15
Display posts from previous: Sort by | 926 | 3,193 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2020-24 | latest | en | 0.866483 |
https://stats.stackexchange.com/questions/381162/is-this-a-matched-pairs-data-set | 1,558,428,523,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256314.25/warc/CC-MAIN-20190521082340-20190521104340-00464.warc.gz | 629,838,224 | 36,684 | # Is this a matched pairs data set?
A review problem for one of my daughter's statistics classes: The arm span and height of 10 individuals is measured in order to test the hypothesis that arm-span equals height in humans. Perform an hypothesis test to determine if there is a significant difference between the mean of the two groups.
I am not sure if this is a match pairs data set and therefore not sure if one should perform a single sample t-test on the difference (for a matched pair) or a 2-sample t-test for the difference between the means of two independent data sets.
I am inclined to believe that this is NOT a matched pairs data set even though the two measurements are taken from the same person. Just because there is a high correlation between the two does not mean that they are dependent.
I understand that if you take measurements from a single group on two occasions (e.g., weight now and 6 months from now) it is a matched pairs data set. But, I'm looking for a general description of "matched pairs" that would help us distinguish this (and similar cases) in the future.
• t-test cannot be used in this situation. simple regression is needed. – user158565 Dec 9 '18 at 22:05
• @user158565 you should explain why (in some detail) you believe this to be the case, perhaps in an answer if there's insufficient space in a comment. – Glen_b Dec 9 '18 at 22:33
• I'd have marked this as a duplicate (e.g. of this) if not for the 'matched' terminology issue and the potential discussion with comparing two quite different quantities that user158565 is raising. – Glen_b Dec 9 '18 at 23:20
Just because there is a high correlation between the two does not mean that they are dependent.
Any non-zero correlation in the population is proof of dependence, but you don't need to look at the correlation in the sample to infer that this is matched (on person).
I'm looking for a general description of "matched pairs"
I guess that terminology may differ across books, but to my mind a matched pair is not what you have here. (Matched pairs are where you identify similar subjects/experimental units and associate them together e.g. matching on size, age and gender -- see the first paragraph here - that is, take an action to identify pairs that will be similar on any variables of importance in order that they can be used as blocks). Some definitions of 'matched pairs' don't seem to follow this convention, though (a pity because this is a useful distinction for which they'll now need a new term).
You have paired data, certainly, but they're naturally occurring pairs. The fact that they're both from the same person is the giveaway. [If your book would call values from twins a 'matched pair' then this would be a matched pair of values.]
Paired of values will tend to be more alike within a specific pair than across them. In this case the "person" is giving you the specific (height, arm-length) pair.
People vary in general size, so if height and arm-length tend to be similar (whether or not the null is actually true) they will tend to be more similar within one person than they would be across people (person A's height will be less similar to person-B's arm-length than to her own).
These inherent 'pairs' are what is giving you 'paired data'.
You have paired data, but a paired t-test is not very suitable (unless the question explictly asks you to compare means but that is not the same as "arm span equals height")
• You have paired data when two variables are measured in the same person/unit each case/measurement. E.g. you can say you measured ten pairs of (armspan,height).
If you wish to test $$\mu_X = a + \mu_Y$$ then you can also test $$\mu_{X-Y} =a$$ which often is more precise since there is a smaller variance. E.g. the variables X and Y have large variance due to smaller and larger people. But the difference X-Y has smaller variance because within a large/small person X and Y are both large/small and you eliminate some of the variation by taking the difference.
• Like mentioned in the comments a t-test may not be suitable. A t-test compares equality of the means. But possibly you wish to prove the linear relation X=Y. E.g if you have the following data you have equal means of X and Y but not really equal X and Y
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17 13 | 995 | 4,347 | {"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": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2019-22 | latest | en | 0.953654 |
https://www.physiotherapy4u.com/up-in-arms-about-what-is-range-in-math/ | 1,563,501,098,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525973.56/warc/CC-MAIN-20190719012046-20190719034046-00088.warc.gz | 819,409,872 | 27,409 | # Up in Arms About What Is Range in Math?
## The New Fuss About What Is Range in Math
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## The Do’s and Don’ts of What Is Range in Math
A third method of determining a measure of center is referred to as the mode. https://samedayessays.info/ The reverse is the circumstance, naturally, with negative numbers. You should have a whole lot of endurance, which is directly about the strength of your lungs.
The image of the entire domain is known as the scope of the function. You may often determine the range by taking a look at a graph. It can also mean the entire range of numbers for example, it could be written as 40 to 550.
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## The Bad Side of What Is Range in Math
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## Finding What Is Range in Math on the Web
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## New Questions About What Is Range in Math
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## The War Against What Is Range in Math
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## The Good, the Bad and What Is Range in Math
Be aware that the entire option and parameter settings won’t be sent to your printer. It’s a superb idea to get in the custom of putting a docstring on top of every Python file and function you write. Line graphs are utilised to demonstrate how data changes as time passes. | 1,161 | 5,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2019-30 | latest | en | 0.952023 |
https://physics.stackexchange.com/questions/484807/basic-transformer-question | 1,713,299,130,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817106.73/warc/CC-MAIN-20240416191221-20240416221221-00607.warc.gz | 425,441,036 | 39,750 | # Basic Transformer Question
My understanding of how a transformer works is that there is a primary and a secondary coil wrapped around a common iron coil. An alternating current in the primary coil results in a changing magnetic field which in turn results an induced current in the secondary coil.
The resulting voltage across secondary coil is proportional the the number of turns in the secondary coil. My understanding of this is that each loop produces a small emf and these loops are all in series so adding more loops is like adding more cells in series in a battery.
My question is why does adding more loops in the primary winding reduce the output voltage? In an electromagnet, increasing the number of turns of wire increases the magnetic field. So I would think that increasing the number of loops in the primary coil would result in a larger magnetic field and thus a greater induced voltage in the secondary coil.
## 3 Answers
We usually consider a transformer to be driven by a fixed voltage $$V_i$$.
The current in the primary is then determined by voltage and inductance $$V_i/L_i$$.
If you add turns to the primary, you increase the inductance, and therefore decrease the current. This decreased current makes less magnetic field. So the secondary produces less voltage.
• Right, that make sense. My mistake was thinking of the primary coil as if it were a DC electromagnet and not considering the (crucial) role that inductance plays. Jun 7, 2019 at 17:59
Let's see this in another way.
Inductance of a coil depends on loops a coil has and is proportional to number of loops a coil has.
As we also know the relation between Voltage and current for a coil is.
$$V=L*\frac{di}{dt}$$
Now if we consider that the source has a fixed supply power, we will get that an increase in number of loops in primary coil the inductance will increase.
As there is a increase in inductance we can say that there will be decrease in current flow leading to a smaller induced voltage in secondary coil.
Hence by adding more primary loops you are decreasing the flux produced.
This effect can be understood by comparing both of these two general formula.
$$VaNa=VbNb$$
$$\frac{Va}{Ia}=\frac{Vb}{Ib}$$
As we see voltages of both coils show direct relationships to currents flowing through them,While there is inverse relationships to the number of loops in coil ( This is because currents show inverse relationships to inductance or loops in a coil)
The resulting voltage across secondary coil is proportional the the number of turns in the secondary coil.
My question is why does adding more loops in the primary winding reduce the output voltage?
The problem with your reasoning is you are looking at each winding independent of the other and that you are looking only at the voltages and not the combination of voltage times current (power).
For an ideal transformer where power in equals power out, the subscripts $$p$$ and $$s$$ denote the primary and secondary, and N is the number of turns
$$V_{p}I_{p}=V_{s}I{s}$$
$$\frac{V_p}{V_s}=\frac{I_s}{I_p}=\frac{N_p}{N_s}$$
Hope this helps | 688 | 3,114 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 9, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-18 | latest | en | 0.952398 |
https://db0nus869y26v.cloudfront.net/en/Stub_(electronics) | 1,680,396,210,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950363.89/warc/CC-MAIN-20230401221921-20230402011921-00391.warc.gz | 248,151,937 | 16,052 | Resonant stub tank circuits in vacuum tube backpack UHF transceiver, 1938. About 1/8 wavelength long: (left) 200 MHz stub is 19 cm, (right) 300 MHz stub is 12.5 cm
10 kW FM broadcast transmitter from 1947 showing quarter-wave resonant stub plate tank circuit
In microwave and radio-frequency engineering, a stub or resonant stub is a length of transmission line or waveguide that is connected at one end only. The free end of the stub is either left open-circuit, or short-circuited (as is always the case for waveguides). Neglecting transmission line losses, the input impedance of the stub is purely reactive; either capacitive or inductive, depending on the electrical length of the stub, and on whether it is open or short circuit. Stubs may thus function as capacitors, inductors and resonant circuits at radio frequencies.
The behaviour of stubs is due to standing waves along their length. Their reactive properties are determined by their physical length in relation to the wavelength of the radio waves. Therefore, stubs are most commonly used in UHF or microwave circuits in which the wavelengths are short enough that the stub is conveniently small.[1] They are often used to replace discrete capacitors and inductors, because at UHF and microwave frequencies lumped components perform poorly due to parasitic reactance.[1] Stubs are commonly used in antenna impedance matching circuits, frequency selective filters, and resonant circuits for UHF electronic oscillators and RF amplifiers.
Stubs can be constructed with any type of transmission line: parallel conductor line (where they are called Lecher lines), coaxial cable, stripline, waveguide, and dielectric waveguide. Stub circuits can be designed using a Smith chart, a graphical tool which can determine what length line to use to obtain a desired reactance.
## Short circuited stub
The input impedance of a lossless, short circuited line is,
${\displaystyle Z_{\mathsf {sc))~=~j\ Z_{0}\ \tan(\ \beta \ell \ )~}$
where
${\displaystyle \ j\ }$ is the imaginary unit (${\displaystyle \ j^{2}\equiv -1\ }$),
${\displaystyle \ Z_{0}\ }$ is the characteristic impedance of the line,
${\displaystyle \ \beta =2\pi /\lambda \ }$ is the phase constant of the line, and
${\displaystyle \ \ell \ }$ is the physical length of the line.
Thus, depending on whether ${\displaystyle \ \tan(\beta \ell )\ }$ is positive or negative, the short circuited stub will be inductive or capacitive, respectively.
The length of a stub to act as a capacitor C at an angular frequency of ${\displaystyle \ \omega \ }$ is then given by:
${\displaystyle \ell ~=~{\frac {1}{\ \beta \ ))\left[\ (n+1)\ \pi \ -\ \arctan \left({\frac {1}{\ \omega CZ_{0}\ ))\right)\ \right]~;}$
the length of a stub to act as an inductor L at the same frequency is given by:
${\displaystyle \ell ~=~{\frac {1}{\ \beta \ ))\left[\ n\ \pi \ +\ \arctan \left({\frac {\ \omega L\ }{\ Z_{0}\ ))\right)\ \right]~,}$
where in both equations, n is an integer number of half-wavelengths (possibly zero) that can be arbitrarily added to the line without changing the impedance.
## Open circuited stub
The input impedance of a lossless open circuit stub is given by
${\displaystyle Z_{\mathsf {oc))=-j\ Z_{0}\ \cot(\ \beta \ell \ )~,}$
where the symbols ${\displaystyle \ Z_{0},\beta ,\ell ,\omega ,\ }$ etc. used in this section have the same meaning as in the section above.
It follows that depending on whether ${\displaystyle \cot(\beta \ell )}$ is positive or negative, the stub will be capacitive or inductive, respectively.
The length of an open circuit stub to act as an inductor L at an angular frequency of ${\displaystyle \ \omega \ }$ is:
${\displaystyle \ell ~=~{\frac {1}{\ \beta \ ))\left[\ (n+1)\ \pi \ -\ \operatorname {arccot} \left({\frac {\ \omega L\ }{Z_{0))}\right)\ \right]~=~{\frac {1}{\ \beta \ ))\left[\ (n+1)\ \pi \ -\ \arctan \left({\frac {Z_{0)){\ \omega L\ ))\right)\ \right]~;}$
the length of an open circuit stub to act as a capacitor C at the same frequency is:
${\displaystyle \ell ~=~{\frac {1}{\ \beta \ ))\left[\ n\ \pi \ +\ \operatorname {arccot} \left({\frac {1}{\ \omega CZ_{0}\ ))\right)\ \right]~=~{\frac {1}{\ \beta \ ))\left[\ n\ \pi \ +\ \arctan \left(\ \omega CZ_{0}\ \right)\ \right]~,}$
where again, n is an arbitrary whole number of half-wavelengths that can be inserted into the segment (including zero).
## Resonant stub
Stubs are often used as resonant circuits in oscillators and distributed element filters. An open circuit stub of length ${\displaystyle \scriptstyle l}$ will have a capacitive impedance at low frequency when ${\displaystyle \scriptstyle \beta l<\pi /2}$. Above this frequency the impedance is inductive. At precisely ${\displaystyle \scriptstyle \beta l=\pi /2}$ the stub presents a short circuit. This is qualitatively the same behaviour as a series resonant circuit. For a lossless line the phase change constant is proportional to frequency,
${\displaystyle \beta ={\omega \over v))$
where v is the velocity of propagation and is constant with frequency for a lossless line. For such a case the resonant frequency is given by,
${\displaystyle \omega _{0}={\frac {\pi v}{2l))}$
While stubs function as resonant circuits, they differ from lumped element resonant circuits in that they have multiple resonant frequencies; in addition to the fundamental resonant frequency ${\displaystyle \scriptstyle \omega _{0}\,}$, they resonate at multiples of this frequency: ${\displaystyle \scriptstyle n\omega _{0}\,}$. The impedance will not continue to rise monotonically with frequency after resonance as in a lumped tuned circuit. It will rise until the point where ${\displaystyle \scriptstyle \beta l=\pi }$ at which point it will be open circuit. After this point (which is an anti-resonance point), the impedance will again become capacitive and start to fall. It will continue to fall until at ${\displaystyle \scriptstyle \beta l=3\pi /2\,}$ it again presents a short circuit. At this point, the filtering action of the stub has failed. This response of the stub continues to repeat with increasing frequency alternating between resonance and anti-resonance. It is not only a characteristic of stubs but of all distributed element filters that there is some frequency beyond which the filter fails and multiple unwanted passbands are produced.[2]
Similarly, a short circuit stub is an anti-resonator at ${\displaystyle \scriptstyle \pi /2}$, that is, it behaves as a parallel resonant circuit, but again fails as ${\displaystyle \scriptstyle 3\pi /2}$ is approached.[2]
## Stub matching
In a stripline circuit, a stub may be placed just before an output connector to compensate for minor mismatches due to the device's output load or the connector itself.
Stubs can match a load impedance to the transmission line characteristic impedance. The stub is positioned a distance from the load. This distance is chosen so that at that point, the resistive part of the load impedance is made equal to the resistive part of the characteristic impedance by impedance transformer action of the length of the main line. The length of the stub is chosen so that it exactly cancels the reactive part of the presented impedance. The stub is made capacitive or inductive according to whether the main line presents an inductive or capacitive impedance, respectively. This is not the same as the actual impedance of the load since the reactive part of the load impedance will be subject to impedance transformer action and the resistive part. Matching stubs can be made adjustable so that matching can be corrected on test.[3]
A single stub will only achieve a perfect match at one specific frequency. Several stubs may be used spaced along the main transmission line for wideband matching. The resulting structure is filter-like, and filter design techniques are applied. For instance, the matching network may be designed as a Chebyshev filter but is optimised for impedance matching instead of passband transmission. The resulting transmission function of the network has a passband ripple like the Chebyshev filter, but the ripples never reach 0 dB insertion loss at any point in the passband, as they would do for the standard filter.[4]
A microstrip filter using butterfly stubs
Radial stubs are a planar component that consists of a sector of a circle rather than a constant-width line. They are used with planar transmission lines when a low impedance stub is required. Low characteristic impedance lines require a wide line. With a wide line, the junction of the stub with the main line is not at a well-defined point. Radial stubs overcome this difficulty by narrowing to a point at the junction. Filter circuits using stubs often use them in pairs, one connected to each side of the main line. A pair of radial stubs so connected is called a butterfly stub or a bowtie stub.[5]
## References
1. ^ a b Shuart, George W. (October 1934). "New high impedance lines replace coils" (PDF). Short Wave Craft. New York: Popular Book Corp. 5 (6): 332–333. Retrieved March 24, 2015.
2. ^ a b Ganesh Prasad Srivastava, Vijay Laxmi Gupta, Microwave Devices and Circuit Design, pp.29-31, PHI Learning, 2006 ISBN 81-203-2195-2.
3. ^ F.R. Connor, Wave Transmission, pp.32-34, Edward Arnold Ltd., 1972 ISBN 0-7131-3278-7.
4. ^ Matthaei, G.; Young, L.; Jones, E. M. T., Microwave Filters, Impedance-Matching Networks, and Coupling Structures, pp.681-713, McGraw-Hill 1964.
5. ^ Jia-Shen G. Hong, M. J. Lancaster, Microstrip Filters for RF/Microwave Applications, pp. 188-190, Wiley, 2004 ISBN 0471464201. | 2,414 | 9,603 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 27, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-14 | longest | en | 0.917973 |
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Off | 570 | 2,283 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-49 | latest | en | 0.879255 |
https://leftasexercise.com/2018/10/01/n-quantum-gates/ | 1,685,764,049,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649105.40/warc/CC-MAIN-20230603032950-20230603062950-00005.warc.gz | 400,370,882 | 30,684 | # Quantum gates
So far, we have looked at states of a quantum computer and expressed these states in terms of qubits. However, just having a static state is of very limited use – to perform actual computations, we of course have to change our state over time.
In quantum mechanics, state changes are described by unitary transformations on the Hilbert space describing our quantum system. In analogy with a classical computer, where signals traverse logical gates and are manipulated by these gates to perform calculations, these transformations are called quantum gates. However, please keep in mind that in most actual implementations of quantum computers, these gates are not circuits or building blocks in the classical sense through which the qubits somehow travel. Instead, the qubits are mostly at a fixed point in space, and a gate is applied to a quantum state by manipulating the system, using for instance laser beams or magnetic fields.
## One qubit quantum gates
Let us again start with the case of a single qubit, i.e. with a two-dimensional Hilbert space. With respect to the standard basis $\{ |0\rangle, |1\rangle \}$, a unitary transformation is of course given by a unitary 2×2 matrix.
As a first example, let us consider the matrix
$X = \begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}$
Obviously, this operator exchanges the states $|0 \rangle$ and $|1 \rangle$ and is therefore sometimes called the bit-flip or negation. Its eigenvalues are 1 and -1, with a pair of orthogonal and normed eigenvectors given by
$|+\rangle = \frac{1}{\sqrt{2}} (|0\rangle + |1 \rangle)$
and
$|-\rangle = \frac{1}{\sqrt{2}} (|0\rangle - |1 \rangle)$
A second commonly used operator is the operator conventionally denoted by Z and given by
$Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$
Clearly, the eigenvalues of this operator are again +1 and -1, but this time, an eigenbasis is given by the standard basis. On a superposition, this operator changes the relative phase, so it is often called the phase change or phase flip operator. A reader with a background in quantum physics might recognize these two operators as being two of the three Pauli operators. They all have determinant -1, trace 0, are hermitian and unitary and their square is the identity operator.
The eigenstates of X form an orthogonal basis, and hence they can be mapped into the standard basis by applying a unitary transformation. The corresponding operator is the Hadamard operator given by
$H = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$
As we can see, $H |0\rangle = |+ \rangle$ and $H |1 \rangle = |- \rangle$, and, as $H^2 = 1$, vice versa. Similarly, the phase operator
$S = \begin{pmatrix} 1 & 0 \\ 0 & i \end{pmatrix}$
that changes the relative phase to an imaginary value transforms between the eigenstates of X and the third Pauli operator. The two gates H and S are sometimes referred to as Clifford gates (they have the property that conjugation with them takes the set of Pauli matrices onto itself).
Similar to classical circuits, quantum gates and combinations of quantum gates are often represented graphically. A single quantum gate is represented by a box, with incoming qubits (i.e. the qubits on which the transformation acts) being indicated by incoming lines on the left and the result being represented by outgoing lines on the right.
In the example above, the first line shows two individual gates, the negation and the phase flip operator. The second line graphically represents the process of applying to a single qubit first the negation and then the phase flip. Of course, the resulting matrix is given by the matrix product
$ZX = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$
Note that the order of the matrix product is reversed, as we first apply X and then Z, reading the diagram from the left to the right which is the usual convention. We also note that this matrix is sometimes called Y, whereas other authors use Y to denote -i times this matrix. We stick to the convention to use Y for the product ZX given above.
## Multi-qubit gates – the CNOT and other controlled gates
Having discussed some of the fundamental quantum gates acting on one qubit, let us now turn again to multi-qubit systems. Of course, we can always create transformations on a multi-qubit system by acting on each qubit individually with a given unitary operator, i.e. by forming the tensor product of operators. Graphically, this would be represented as gates placed below each other, each acting only on one line, i.e. qubit. However, things get more interesting when we use gates that are actually combining multiple qubits to model some non-trivial logic.
As an example, let us consider a gate acting on two qubits that is commonly known as the CNOT gate. To express this gate as a matrix, we have to agree on an order of the basis of the tensor product, and is is natural to use the order $|0 \rangle, |1\rangle, |2\rangle, |3\rangle$ for this. In this basis, the CNOT gate is given by the following 4 x 4 matrix:
$CNOT = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix}$
To understand what this operator is actually doing, let us see how it acts on a basis. Clearly, it maps $|0 \rangle = |00 \rangle$ and $|1 \rangle = |0 1 \rangle$ to itself and swaps the vectors $|2 \rangle = |10 \rangle$ and $|3 \rangle = |11 \rangle$. Thus, expressed in terms of bits, it negates the second bit if the first bit is ON and keeps the second bit as it is if the first bit is OFF. Thus, the first bit can be thought of as a control bit that controls whether the second bit – the target bit – is toggled or not, and this is where the name comes from, CNOT being the abbreviation for “controlled NOT”. Graphically, the CNOT gate is represented by a small circle on the control bit and an encircled plus sign on the target bit (this is the convention used in [1], whereas some other authors, for instance [2], use a slightly different symbol).
An interesting property of the CNOT gate is that it takes separable states to entangled states. Let us for instance compute the action of the CNOT gate on the state $(|0 \rangle + |1 \rangle) |0 \rangle$. This state can as well be written as
$|00 \rangle + |10 \rangle = |0 \rangle + |2\rangle$
from which we can read off that
$CNOT( (|0 \rangle + |1 \rangle) |0 \rangle) = |00 \rangle + |11 \rangle$
which, up to a normalization factor, is an entangled Bell state. Thus we can use the CNOT gate to create entangled states. We also see that, other than the symbol does suggest, the operator does act on the control bit as well.
The notation of a controlled NOT gate can be extended to cover more general gates. Given any unitary one-qubit transformation U, we can in fact consider the two-qubit operator that applies the identify to the second qubit if the first qubit (the control bit) is $|0 \rangle$ and U otherwise. As a matrix, this is given by a 4 x 4 block matrix, with the identity matrix in the upper left corner and the matrix representing U in the lower right corner, and graphically, such a gate is represented as below.
To get used to the graphical notation, let us take a look at the following two-qubit circuit and try to understand what it does to the standard basis.
There are several approaches to figure out what this circuit is doing. First, we can look at its action on each of the basis vectors. Let us do this for the basis vector $|00 \rangle$. We read the circuit from the left to the right and the top to the bottom. Thus, the first part of the circuit acts with H on the first qubit and with the identity on the second qubit. It therefore maps the state
$|00 \rangle = |0 \rangle \otimes |0 \rangle$
to the state
$H|0 \rangle \otimes |0 \rangle = \frac{1}{\sqrt{2}} (|0 \rangle + |1 \rangle) \otimes |0 \rangle = \frac{1}{\sqrt{2}} (|00 \rangle + |10 \rangle)$
The second part of the circuit, the CNOT gate, then maps $|00 \rangle$ to itself and $|10 \rangle$ to $|11 \rangle$. Thus the entire circuit maps the vector $|00 \rangle$ to
$\frac{1}{\sqrt{2}} (|00 \rangle +|11 \rangle)$
which again is an entangled Bell state. We could now proceed like this for the other basis vectors. Alternatively, we can find the answer by multiplying out the matrices and will arrive at the same result.
## Universal quantum gates
In classical computing, a general circuit with n input bits and one output bit can be described by a function taking $\{0, \dots, 2^n - 1 \}$ to $\{0,1\}$, and it is well known that any such function can be expressed as a combination of gates from a small set of standard gates called universal gates. In fact, it is even true that one gate is sufficient – the NAND gate. It is natural to ask whether the same is true for quantum gates. More precisely, can we find a finite set of unitary transformations such that any n-qubit unitary transformation can be composed from them using a few standard operations like the tensor product?
Unfortunately, this does not work in the quantum world. The problem is that the group of unitary matrices is not discrete, but continuous, and thus we can never hope to be able to generate it fully using only a finite number of different gates. However, something very similar is true – every gate can be approximated by combining gates from a small, universal set.
This result is typically presented and derived in two steps (see for instance [1], chapter 3). First, one shows that every unitary transformation of an n-qubit system can be reduced to a combination of single qubit operations (tensored with the identity) and CNOT gates. This is still exact, i.e. no approximation takes place (for the mathematically inclined reader, this might ring a bell – in fact, there is a relation to Cartan decompositions of unitary matrices, see for instance [3] for a discussion).
Then, in a second step, one shows that every single qubit transformation can be approximated arbitrarily close by products of a finite universal set of gates – this is known as the Solovay-Kitaev theorem. It turns out (see for instance [4]) that a possible choice for such a universal set consists of the Hadamard gate H, the phase flip gate Z and only one additional gate
$\begin{pmatrix} 1 & 0 \\ 0 & e^{i\frac{\pi}{4}} \end{pmatrix}$
which is confusingly sometimes called the $\pi / 8$ phase gate.
Finally, we mention that a three-qubit gate that is often used to construct universal sets of quantum gates is the Toffoli gate. This is like a CNOT gate, except that it has two control bits and the target bit is only flipped if both control bits are ON. It is known that the set consisting of the Toffoli gate and the Hadamard gate only is already universal in a similar sense (see this paper by D. Aharonov for a short proof of this result that goes back to Y. Shi). This result is of a certain importance because one can also show that the classical version of the Toffoli gate alone is sufficient to implement (a reversible version of) any classical circuit. Thus we do not only see that any computation that can be done by a classical circuit can also be done by a quantum circuit, but also that without the Hadamard gate, quantum computing would not be more powerful than classical computing.
This closes our short discussion of elementary quantum gates. In the next few posts, we will start to combine quantum gates into quantum algorithms to perform actual computations. Stay tuned!
## References
1. G. Benenti, G. Casati, G. Strini, Principles of Quantum Computation and Information, Volume 1, World Scientific
2. E. Rieffel, W. Polak, Quantum computing – a gentle introduction, MIT Press
3. N. Khaneja, S.J. Glaser, Cartan Decomposition of SU(2 n ), Constructive Controllability of Spin Systems and Universal Quantum Computing, arXiv:quant-ph/0010100v1
4. P. Boykin, T. Mor, M. Pulver, V. Roychowdhury, F. Vatan, A new universal and fault-tolerant quantum basis
5. D. Aharonov,A Simple Proof that Toffoli and Hadamard are Quantum Universal, arXiv:quant-ph/0301040 | 3,020 | 12,155 | {"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": 35, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2023-23 | latest | en | 0.90675 |
https://www.physicsforums.com/threads/rotational-motion-of-a-uniform-solid-disk.176724/ | 1,632,044,128,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056752.16/warc/CC-MAIN-20210919065755-20210919095755-00287.warc.gz | 987,142,930 | 15,266 | # Rotational motion of a uniform solid disk
## Homework Statement
A uniform solid disk of radius 7.1m and mass 30.3kg is free to rotate on a fricionless pivot through a point on its rim. [picture: http://www.wellesley.edu/Physics/phyllisflemingphysics/107_p_angular_images/figure_9.gif] [Broken]
If the disk is released from rest in the position shown, what is the speed of its center of mass when the disk reaches the position indicated by the dashed circle?
What is the speed of the lowest point on the disk in the dashed position?
## Homework Equations
mgh = 1/2 Iw^2 + 1/2mv^2.
## The Attempt at a Solution
From the above equation, I was able to subsitute w = v/r and solved for v...v = (4/3gh)^1/2. I don't know if this velocity pertains to the one about the center of mass or the lowest point. Please help! Thank you!
Last edited by a moderator:
I got the speed of the center of mass. Now I just need to somehow link that to the speed of the lowest point.
andrevdh
Homework Helper
You should not include the second term in
mgh = 1/2 Iw^2 + 1/2mv^2
that is it should just be
mgh = 1/2 Iw^2
since the disc is not translating, it is just rotating. The second term is for translational kinetic energy.
If one takes the bottom position as zero potential energy level then the potential energy at the top will be just
mgR
since the centre of mass dropped a distance equal to the radius below the top position.
What do you get the angular speed of the disc at the dashed position?
Last edited:
I got w = 2.349 rad/s
andrevdh
Homework Helper | 415 | 1,558 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2021-39 | latest | en | 0.916284 |
http://unendliches.net/english/eschleife.htm | 1,696,158,474,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510888.64/warc/CC-MAIN-20231001105617-20231001135617-00897.warc.gz | 40,341,111 | 3,820 | Infinite loop: an endless sequence of repetitions of a component in a computer program caused by a programming error (bug) or input data error. An infinite ►recursion is a type of infinite loop. In pseudocode an infinite recursive loop looks roughly like this: ```function Endless (data): `````` Do (data); Endless (data); `````` ```` return;` This function keeps calling itself but never reaches its terminal base line, the return line. Infinite loops cause the program to crash due to the fact that it is unable to escape the looping component; as a result, the computer is no longer able to react to input data. Pretty much every PC owner should be familiar with this effect: The program 'freezes' and must be terminated the hard way by means of a reset button, Vulcan nerve pinch, or task manager. It is not possible by means of analysis alone — that is, without executing the actual program — to generally determine whether the program will terminate correctly with certain input data or whether it is actually frozen. This halting problem was proven in 1936 by the British mathematician Alan Turing. 'Errors don't exist' Does Not Exist Here is the proof (basic programming knowledge makes it easier to follow): Let as assume a function Finite that can reliably determine whether a given program with given input data will perform an infinite loop or terminate correctly. In pseudocode the function would be defined as follows: ```function Finite (program, data): `````` if ('program' ends upon input of 'data') `````` then return YES; ```` else return NO;` Of course, Finite itself must not contain an infinite loop but always terminate reliably even if the tested program does not. We now define a recursive function TestMe, which is called up by Finite: ```function TestMe (program): `````` if (Finite (program, programm)) ```` then TestMe (program); ` This somewhat nasty function is terminated only if program is not, that is to say, if the latter calls itself as input data. Otherwise TestMe enters into an infinite recursion: The function keeps calling itself so that it never produces a terminal line. But if we apply the function TestMe to itself as program we obtain a logical contradiction: ` TestMe (TestMe); ` This function call ends precisely if it does not end. Consequently, a function such as Finite cannot exist. For programmers this means that they can never prove that a program they delivered is free of errors. By itself, this does not sound very exciting (except, perhaps, for programmers). In 1936, however, the proof shook up our mathematical worldview even though at the time there were not as yet any computers to be programmed. Turing himself applied the proof only to a hypothetical computer, a so-called Turing machine. But the proof does have the more general implication that some problems are unsolvable in principle within the formalism of mathematics — an implication that Kurt Gödel had already proven six years earlier in the context of his ►incompleteness theorem. Is God Able to Program a Bug-Free Software? Puzzles such as the halting problem and the incompleteness theorem also have philosophical and religious implications. In particular, it challenges the thesis of God's ►omniscience: If it is impossible in principle to determine a system to be bugless, then even ►God would be unable to know whether a program designed by Him has a bug or not. The further question of whether this is what accounts for certain flaws in the creation of our world, however, probably also belongs into the category of unsolvable problems. * [Ctrl]-[Alt]-[Del] Links Related to the Topic | 764 | 3,636 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-40 | latest | en | 0.858364 |
https://samacheerkalviguru.com/samacheer-kalvi-9th-maths-chapter-7-additional-questions/ | 1,619,168,663,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039568689.89/warc/CC-MAIN-20210423070953-20210423100953-00301.warc.gz | 596,528,527 | 8,499 | ## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 7 Mensuration Additional Questions
Exercise 7.1
Question 1.
Using Heron’s formula, find the area of a triangle whose sides are 41 m, 15 m, 25 m.
Solution:
Question 2.
Find the area of an equilateral triangle whose perimeter is 150 m.
Solution:
3a = 150
Question 3.
Find the area of a quadrilateral whose sides are PQ = 15 cm, QR = 8 cm, RS = 25 cm, PS = 12 cm and LQ = 90°
Solution:
Exercise 7.2
Question 1.
Find the TSA and LSA of a cuboid whose length, breadth and height are 10 cm, 12 cm and 14 cm respectively.
Solution:
TSA = 2 (lb + bh + Ih)
= 2 (10 × 12 + 12 × 14 + 10 × 14) = 2 (120 + 168 + 140) = 856 cm2
LSA = 2 (bh + Ih) = 2 (12 × 14 + 10 × 14) = 2 (168 + 140) =2 (308) = 616 cm2
Question 2.
A cuboid has total surface area of 40 m2 and its lateral surface area is 26 m2. Find the area of its base.
Solution:
Question 3.
Find the surface area of a cube whose edge is
(i) 27 cm
(ii) 3 cm
(iii) 6 cm
(iv) 2.1 cm
Solution:
(i) TSA = 6a2 = 6 × 272 = 6 × 729 = 4374 cm2
(ii) TSA = 6a2 = 6 × 32 = 54 cm2
(iii) TSA = 6a2 = 6 × 62 = 6 × 36 = 216 cm2
(iv) TSA = 6a2 = 6 × 2.12 = 6 × 4.41 = 26.46 cm2
Exercise 7.3
Question 1.
Find the volume of a cube whose surface area is a 96 cm2.
Solution:
6a2 = 96 cm2
a = 6
Volume = a3 = 63 = 36 × 6 = 216 cm3
Question 2.
The volume of a cuboid is 440 cm3 and the area of its base is 88 cm2, find its height.
Solution:
l × b × h = 440 cm3
Question 3.
How many 3 metre cubes can be cut from a cuboid measuring 18 m × 12 m × 9 m?
Solution:
Question 4.
The outer dimensions of a closed wooden box are 10 cm by 8 cm by 7 cm. Thickness
of the wood is 1 cm3. Find the total cost of wood required to make box if 1 cm3 of wood costs ₹ 2.00?
Solution:
Volume of wood = 12 × 10 × 9 – 10 × 8 × 7 = 1080 – 560 = 520 cm3 × 2.00 = ₹ 1040 | 702 | 1,834 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2021-17 | latest | en | 0.847898 |
https://physics.stackexchange.com/questions/129476/block-sliding-down-hemisphere/129500 | 1,696,477,321,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511717.69/warc/CC-MAIN-20231005012006-20231005042006-00516.warc.gz | 495,698,702 | 38,517 | # Block Sliding Down Hemisphere [closed]
A block of mass m slides down a hemisphere of mass M. What are the accelerations of each mass? Assume friction is negligible.
$$a_M$$ = Acceleration of hemisphere
$$N_m$$ = Normal force of M onto m
$$N_M$$ = Normal force of ground onto M
So from the FBD's, I come up with
$$\sum \text{F}_{xm}= mg\sin \theta = m(a_t - a_M \cos \theta)$$
$$\sum \text{F}_{ym} = N_m - mg \cos \theta = -m(a_r + a_M \sin \theta)$$
$$\sum \text{F}_{xM} = -N_m \sin \theta = Ma_M$$
I need another equation, so I tried going the route of work-energy, to find the tangential speed of the block sliding on the hemisphere, but can I look at the energy of the block by itself? I figure I cannot, as it is atop an accelerating body.
However, if I can consider the energy of the block by itself to find the tangential speed, then I can solve for aM,
$$a_M = gm\sin \theta \frac{2-3\cos \theta}{M-m\sin ^2 \theta}$$
which goes to 0 when M >> m and so then $$a_t = g\sin \theta$$ in that case, which checks out, however Im still a little weary about this.
I'm rather stuck here so any help would be appreciated.
• More Phys.SE posts on blocks sliding down hemispheres. Aug 4, 2014 at 21:01
• It this a solid hemisphere or a hollow one? Aug 4, 2014 at 21:09
• @Floris I guess the hemisphere I imagine would be solid. Does that really matter? When I thought of this problem, I was thinking more along the lines of the problem where a block slides down on a movable ramp. Aug 5, 2014 at 0:02
• @Qmechanic I guess my main question is: In this scenario where both the block and the hemisphere are in motion, can I ignore the kinetic energy of the hemisphere, as well as the work done by the normal force from the mass on the hemisphere in finding the energy of the mass? Aug 5, 2014 at 0:09
• The reason that the hollow/solid question matters is that as the sphere moves, the mass is doing work on the sphere, leaving less energy for itself. Thus the mass and moment of inertia of the large sphere do matter. Tricky problem. Aug 5, 2014 at 1:19
I think that your missing equations are just your geometrical constrains. You have not used them. I noticed that we are using different convention for the coordinates. Mine is just a static Cartesian coordinate system with the origin in the center of the hemishpere at the beginning, but it does not move with it. I am assuming we can treat this problem in two dimensions. Assuming that the block has not lost contact with the hemisphere, you have $$(x_m-x_M)^2 + y_m^2=R^2,$$ which is an equation that is valid at all times, therefore it relates the three accelerations and velocities in a complicated manner that appears explicitly by deriving two times. And the fifth equation is just $$\tan\theta=\frac{y_m}{x_m-x_M}.$$ therefore you have the normal, the three dinamical variables, and a fourth dependandant dynamical variable introduced just to ease the notation, $\theta$. The FBD equations in my case are \begin{align} m \ddot{y}_m &= N\cos\theta -mg\\ m \ddot{x}_m &= N \sin\theta\\ M \ddot{x}_M &= -N cos\theta \end{align} Five equations for five unknowns, and a very ugly solution that I do not think exists in closed form.
You asked about energy. Indeed, you can use this approach since there is no friction. Actually, the energy approach is useful even when the objects are no longer in contact. Lets assume that the block start moving in the position you depict in your picture, that is $y_m(t=0)=h$ and $x_m(t=0)=\sqrt{R^2-h^2}$, and $x_M(t=0)=0$, with all velocities equal to zero at $t=0$. Therefore, conservation of energy dictates $$\frac{1}{2}M\dot{x}_M^2+\frac{1}{2}m(\dot{x}_m^2+\dot{y}_m^2)+mgy_m=mgh.$$
• @Plopperzz I 've just edited and explain it better. We are using different notation, my $x$ and $y is fixed. Aug 5, 2014 at 13:18 We can solve this problem easily by using only the law of conservation of momentum and the law of conservation of energy. It is obvious that in given case only the horizontal component of the momentum of the system of the block and the hemisphere is conserved (i am using OP notations): $$(m+M)v_M-mv_m\cos\theta=0$$ Now, an equation of conservation of energy: $$Mv^2_M+mv^2_m=2mgR(1-\cos\theta)$$ Let's introduce the ratio $$\eta=\frac{M}{m}$$ Then the equations look simpler $$(1+\eta)v_M-v_m\cos\theta=0\qquad(1)$$ $$\eta v^2_M+v^2_m=2gR(1-\cos\theta)\qquad(2)$$ Now, the physical part of the problem is completed. We are left with pure mathematics. From the last two equations we get an expression for$v_m$: $$v^2_m=\frac{2gR(1-\cos\theta)}{1+\gamma\cos^2\theta}\qquad(3)$$ where$\gamma=\frac{\eta}{(1+\eta)^2}$for simplicity. Now, the acceleration of the block$a_m=\dot{v}_m$can be obtained by differentiating (3) by time, remembering that$\theta=\theta(t)$and noting that$\dot{\theta}=\frac{v_m}{R}$. I think, the easiest way is the method of logarithmic differentiation of (3). The acceleration of the hemisphere$a_M=\dot{v}_M$can be obtained by differentiating (1) by time: $$a_M=\dot{v}_M=\dot{v}_m\frac{\cos\theta}{1+\eta}-v_m\frac{\sin\theta}{1+\eta}\dot{\theta}=$$ $$=\dot{v}_m\frac{\cos\theta}{1+\eta}-\frac{v^2_m}{R}\frac{\sin\theta}{1+\eta}$$ Since$\dot{v}_m$and$v_m$have already been calculated we are done. • That's not as bad as I had thought it would be. Do you feel this is the type of question they may ask in a first year physics course? Aug 6, 2014 at 17:52 • This looks simpler indeed compared to what I did. A small comment though: strictly speaking, the energy equation assumes an initial condition for which the only dynamical solution is static. You need to be bear this in mind or you will get stuck again, literally. Aug 6, 2014 at 19:54 • @Plopperzz I think not. This problem is more an analytical mechanics than a physics problem, i think. Aug 7, 2014 at 14:12 • @Enredanrestos A good point! To avoid this issue we can take instead of initial condition$\theta=0$an initial condition$\theta=\theta_0$where$\theta_0\$ is different from zero, no matter how little. Aug 7, 2014 at 14:26 | 1,732 | 6,037 | {"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": 8, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2023-40 | latest | en | 0.930552 |
https://www.airmilescalculator.com/distance/hlh-to-lnl/ | 1,603,824,447,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107894426.63/warc/CC-MAIN-20201027170516-20201027200516-00211.warc.gz | 605,996,366 | 40,453 | # Distance between Ulanhot (HLH) and Longnan (LNL)
Flight distance from Ulanhot to Longnan (Ulanhot Airport – Longnan Chengzhou Airport) is 1204 miles / 1938 kilometers / 1047 nautical miles. Estimated flight time is 2 hours 46 minutes.
Driving distance from Ulanhot (HLH) to Longnan (LNL) is 1530 miles / 2463 kilometers and travel time by car is about 26 hours 57 minutes.
## Map of flight path and driving directions from Ulanhot to Longnan.
Shortest flight path between Ulanhot Airport (HLH) and Longnan Chengzhou Airport (LNL).
## How far is Longnan from Ulanhot?
There are several ways to calculate distances between Ulanhot and Longnan. Here are two common methods:
Vincenty's formula (applied above)
• 1204.477 miles
• 1938.417 kilometers
• 1046.662 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 1203.834 miles
• 1937.384 kilometers
• 1046.104 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Ulanhot Airport
City: Ulanhot
Country: China
IATA Code: HLH
ICAO Code: ZBUL
Coordinates: 46°4′58″N, 122°1′1″E
B Longnan Chengzhou Airport
City: Longnan
Country: China
IATA Code: LNL
ICAO Code: ZLLN
Coordinates: 33°47′16″N, 105°47′49″E
## Time difference and current local times
There is no time difference between Ulanhot and Longnan.
CST
CST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 162 kg (356 pounds).
## Frequent Flyer Miles Calculator
Ulanhot (HLH) → Longnan (LNL).
Distance:
1204
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
1204
Round trip? | 507 | 1,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.804918 |
http://www.thusspakeak.com/student/2017/01/01-OnOneAgainstMany.html | 1,527,485,372,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794871918.99/warc/CC-MAIN-20180528044215-20180528064215-00537.warc.gz | 457,376,432 | 9,633 | # On One Against Many
Recall that the Baron proposed a pair of dice contests in which Sir R-----, were he to best the Baron's score, stood to win a bounty of thirteen coins.
Upon paying his stake Sir R----- was to cast his die but, if unhappy with its outcome, could pay a further coin to cast it again. Likewise, if he were not satisfied with the second cast, he could elect to cast a third time for a further two coins. He could continue in this fashion for as long as he pleased with the cost rising by one coin for each additional cast of his die. The Baron was to have but a single cast of his die, with Sir R----- to determine whether after or before his own play according to his stake; seven coins for the former and eight for the latter.
In order to figure the fairness of these wagers it is first necessary to recognise that there will inevitably come a point beyond which it is not in Sir R-----'s interest to draw out his play, for the cost of doing so rises inexorably whilst the prize, should he best the Baron, rests in constancy. Specifically, Sir R----- should refrain from casting his die if he should expect on the average to be further out of pocket in doing so. If we figure when this will be we can work backwards from his final cast including the advantage, if any, he might derive from further casts as we proceed. I said as much to the Baron, but I do not believe that I had his full attention.
Now, Sir R----- will be most compelled to recast his die if his will assuredly lose the wager if he does not. Indeed, he should only choose not if his expected winnings should be less than the cost of casting. Fortunately for our reckoning we need not consider further casts after this since, given that their cost shall grow ever the greater, their outcomes for Sir R----- shall grow ever the worser.
In the first game, the probability that Sir R----- will best the Baron if he casts a one is zero. If he should cast a two, he has one chance in six of winning. If a three, then two chances in six, etcetera. The probability that he should win on his final cast is therefore simply
$\frac{1}{6} \times 0 + \frac{1}{6} \times \frac{1}{6} + \frac{1}{6} \times \frac{2}{6} + \frac{1}{6} \times \frac{3}{6} + \frac{1}{6} \times \frac{4}{6} + \frac{1}{6} \times \frac{5}{6} = \frac{1+2+3+4+5}{36} = \frac{5}{12}$
His expected winnings from that cast should consequently be
$\frac{5}{12} \times 13 = \frac{65}{12} = 5 \frac{5}{12}$
and he should in no circumstance elect to proceed if the cost of doing so exceeds five coins, which it shall do after his sixth cast.
From this it is readily apparent that Sir R----- should expect a prize of five twelfths of a coin for his sixth cast of the die.
Now, if Sir R----- were to cast a one upon his fifth roll, then his expected profit from paying for a sixth exceeds his certain profit of zero and he should elect to roll again. In contrast, if he were to roll a two, then he should have one chance in six of taking the prize of thirteen coins, for an expected windfall of thirteen upon six parts of a coin, or in other words two and one sixth coins. Consequently, Sir R----- should stick with his fifth die if it shows two or greater for an expected prize of
\begin{align*} &\frac{1}{6} \times \frac{5}{12} + \frac{1}{6} \times \frac{1}{6} \times 13 + \frac{1}{6} \times \frac{2}{6} \times 13 + \frac{1}{6} \times \frac{3}{6} \times 13 + \frac{1}{6} \times \frac{4}{6} \times 13 + \frac{1}{6} \times \frac{5}{6} \times 13\\ &\quad= \frac{5}{72} + \frac{13}{36} + \frac{26}{36} + \frac{39}{36} + \frac{52}{36} + \frac{65}{36}\\ &\quad= \frac{5+26+52+78+104+130}{72}\\ &\quad= \frac{395}{72} = 5\frac{35}{72} \end{align*}
Finally, we must subtract the cost of four coins to yield Sir R-----'s net gain from a fifth cast of one seventy second part of a coin shy of one and a half coins.
To figure the value of this wager to Sir R----- we must simply continue in this fashion, replacing expected winnings smaller than can be got from continuing play, until we reach the first cast of the die.
For the fourth cast we consequently have a net average take of
\begin{align*} &\frac{1}{6} \times \frac{107}{72} + \frac{1}{6} \times \frac{1}{6} \times 13 + \frac{1}{6} \times \frac{2}{6} \times 13 + \frac{1}{6} \times \frac{3}{6} \times 13 + \frac{1}{6} \times \frac{4}{6} \times 13 + \frac{1}{6} \times \frac{5}{6} \times 13 - 3\\ &\quad= \frac{107}{432} + \frac{13}{36} + \frac{26}{36} + \frac{39}{36} + \frac{52}{36} + \frac{65}{36} - 3\\ &\quad= \frac{107}{432} + \frac{195}{36} - 3\\ &\quad= \frac{1,151}{432} = 2\frac{287}{432} \end{align*}
Note that since
$\frac{13}{6} = 2\frac{1}{6} < 2\frac{287}{432}$
Sir R----- should pay for a fourth cast if his third is less than three, yielding a payoff for the latter of
\begin{align*} &\frac{2}{6} \times \frac{1,151}{432} + \frac{1}{6} \times \frac{2}{6} \times 13 + \frac{1}{6} \times \frac{3}{6} \times 13 + \frac{1}{6} \times \frac{4}{6} \times 13 + \frac{1}{6} \times \frac{5}{6} \times 13 - 2\\ &\quad= \frac{2,302}{2,592} + \frac{182}{36} - 2\\ &\quad= \frac{5,111}{1,296} = 3\frac{1,223}{1,296} \end{align*}
Similarly, for the second cast we have
\begin{align*} &\frac{2}{6} \times \frac{5,111}{1,296} + \frac{1}{6} \times \frac{2}{6} \times 13 + \frac{1}{6} \times \frac{3}{6} \times 13 + \frac{1}{6} \times \frac{4}{6} \times 13 + \frac{1}{6} \times \frac{5}{6} \times 13 - 1\\ &\quad= \frac{10,222}{7,776} + \frac{182}{36} - 1\\ &\quad= \frac{20,879}{3,888} = 5\frac{1,439}{3,888} \end{align*}
and, finally, for the first
\begin{align*} &\frac{3}{6} \times \frac{20,879}{3,888} + \frac{1}{6} \times \frac{3}{6} \times 13 + \frac{1}{6} \times \frac{4}{6} \times 13 + \frac{1}{6} \times \frac{5}{6} \times 13\\ &\quad= \frac{62,637}{23,328} + \frac{156}{36}\\ &\quad= \frac{54,575}{7,776} = 7\frac{143}{7,776} \end{align*}
Sir R----- will clearly earn, on average, a bounty greater than his stake and I would therefore in good conscience have recommended that he take the first of the Baron's wagers.
In the second of the Baron's wagers he was to cast his die before Sir R----- began his play. Sir R-----'s strategy should therefore have been informed by the Baron's score.
For each of the Baron's possible scores we can figure the final cast that Sir R----- should countenance in exactly the same fashion as we did for the first game. For example, had the Baron cast a six, Sir R----- should have immediately cut his losses since he should have surely lost the wager no matter how many casts he took. If the Baron had instead cast a five, then on each cast Sir R-----'s should have had one chance in six of besting him. On his final cast he should therefore have expected a prize of
$\frac{1}{6} \times 13 = 2\frac{1}{6}$
and, given that his third cast would have cost him two coins, he should have in no event elected to continue beyond a third cast, at which his expected net gain would have been one sixth part of a coin.
Should he not best the Baron on his second cast he should most certainly cast again and, given that the second cast costs one coin, his expected winnings for it are
$\frac{1}{6} \times 13 + \frac{5}{6} \times \frac{1}{6} - 1 = \frac{47}{36} = 1\frac{11}{36}$
Similarly, for his first cast he should expect
$\frac{1}{6} \times 13 + \frac{5}{6} \times \frac{47}{36} = \frac{703}{216} = 3\frac{55}{216}$
Now, if the Baron should have cast a four then Sir R----- would have had two chances in six of winning on each cast and would have consequently had, for his last roll, an average prize of
$\frac{2}{6} \times 13 = \frac{26}{6} = 4\frac{1}{3}$
and should therefore not have cast his die more than five times.
Working backwards from his fifth cast we have
\begin{align*} 5:\;&4\frac{1}{3}-4 = \frac{1}{3}\\ 4:\;&\frac{2}{6} \times 13 + \frac{4}{6} \times \frac{1}{3} - 3 = 1\frac{5}{9}\\ 3:\;&\frac{2}{6} \times 13 + \frac{4}{6} \times 1\frac{5}{9} - 2 = 3\frac{10}{27}\\ 2:\;&\frac{2}{6} \times 13 + \frac{4}{6} \times 3\frac{10}{27} - 1 = 5\frac{47}{81}\\ 1:\;&\frac{2}{6} \times 13 + \frac{4}{6} \times 5\frac{47}{81} = 8\frac{13}{243} \end{align*}
Next, we must consider the case of the Baron casting a three, for which Sir R----'s final expected winnings will be
$\frac{3}{6} \times 13 = \frac{39}{6} = 6\frac{1}{2}$
and so his last cast should be his seventh. Working retrograde once again we have
\begin{align*} 7:\;&6\frac{1}{2}-6 = \frac{1}{2}\\ 6:\;&\frac{3}{6} \times 13 + \frac{3}{6} \times \frac{1}{2} - 5 = 1\frac{3}{4}\\ 5:\;&\frac{3}{6} \times 13 + \frac{3}{6} \times 1\frac{3}{4} - 4 = 3\frac{3}{8}\\ 4:\;&\frac{3}{6} \times 13 + \frac{3}{6} \times 3\frac{3}{8} - 3 = 5\frac{3}{16}\\ 3:\;&\frac{3}{6} \times 13 + \frac{3}{6} \times 5\frac{3}{16} - 2 = 7\frac{3}{32}\\ 2:\;&\frac{3}{6} \times 13 + \frac{3}{6} \times 7\frac{3}{32} - 1 = 9\frac{3}{64}\\ 1:\;&\frac{3}{6} \times 13 + \frac{3}{6} \times 9\frac{3}{64} = 11\frac{3}{128} \end{align*}
In the same fashion we can figure that should the Baron cast a two Sir R----- should throw his die no more than nine times and that his expected winnings should be
\begin{align*} 9:\;&\frac{4}{6} \times 13 - 8 = \frac{2}{3}\\ 8:\;&\frac{4}{6} \times 13 + \frac{2}{6} \times \frac{2}{3} - 7 = 1\frac{8}{9}\\ 7:\;&\frac{4}{6} \times 13 + \frac{2}{6} \times 1\frac{8}{9} - 6 = 3\frac{8}{27}\\ 6:\;&\frac{4}{6} \times 13 + \frac{2}{6} \times 3\frac{8}{27} - 5 = 4\frac{62}{81}\\ 5:\;&\frac{4}{6} \times 13 + \frac{2}{6} \times 4\frac{62}{81} - 4 = 6\frac{62}{243}\\ 4:\;&\frac{4}{6} \times 13 + \frac{2}{6} \times 6\frac{62}{243} - 3 = 7\frac{548}{729}\\ 3:\;&\frac{4}{6} \times 13 + \frac{2}{6} \times 7\frac{548}{729} - 2 = 9\frac{548}{2,187}\\ 2:\;&\frac{4}{6} \times 13 + \frac{2}{6} \times 9\frac{548}{2,187} - 1 = 10\frac{4,922}{6,561}\\ 1:\;&\frac{4}{6} \times 13 + \frac{2}{6} \times 10\frac{4,922}{6,561} = 12\frac{4,922}{19,683} \end{align*}
Finally and, given the growing complexity of the figures thus far, most dauntingly, we must reckon the value of the wager should the Baron have cast a one.
\begin{align*} 11:\;&\frac{5}{6} \times 13 - 10 = \frac{5}{6}\\ 10:\;&\frac{5}{6} \times 13 + \frac{1}{6} \times \frac{5}{6} - 9 = 1\frac{35}{36}\\ 9:\;&\frac{5}{6} \times 13 + \frac{1}{6} \times 1\frac{35}{36} - 8 = 3\frac{35}{216}\\ 8:\;&\frac{5}{6} \times 13 + \frac{1}{6} \times 3\frac{35}{216} - 7 = 4\frac{467}{1,296}\\ 7:\;&\frac{5}{6} \times 13 + \frac{1}{6} \times 4\frac{467}{1,296} - 6 = 5\frac{4,355}{7,776}\\ 6:\;&\frac{5}{6} \times 13 + \frac{1}{6} \times 5\frac{4,355}{7,776} - 5 = 6\frac{35,459}{46,656}\\ 5:\;&\frac{5}{6} \times 13 + \frac{1}{6} \times 6\frac{35,459}{46,656} - 4 = 7\frac{268,739}{279,936}\\ 4:\;&\frac{5}{6} \times 13 + \frac{1}{6} \times 7\frac{268,739}{279,936} - 3 = 9\frac{268,739}{1,679,616}\\ 3:\;&\frac{5}{6} \times 13 + \frac{1}{6} \times 9\frac{268,739}{1,679,616} - 2 = 10\frac{3,627,971}{10,077,696}\\ 2:\;&\frac{5}{6} \times 13 + \frac{1}{6} \times 10\frac{3,627,971}{10,077,696} - 1 = 11\frac{33,861,059}{60,466,176}\\ 1:\;&\frac{5}{6} \times 13 + \frac{1}{6} \times 11\frac{33,861,059}{60,466,176} = 12\frac{275,725,763}{362,797,056} \end{align*}
We have one last and most tedious calculation to perform; to figure Sir R-----'s expected bounty we must average these results
$\frac{1}{6} \times \left(0 + 3\frac{55}{216} + 8\frac{13}{243} + 11\frac{3}{128} + 12\frac{4,922}{19,683} + 12\frac{275,725,763}{362,797,056}\right)$
That figuring the result of this average was a Herculean feat of arithmetic hardly bears mention! Have I not erred then Sir R----- should expect from this wager a bounty of
$7\frac{1,937,927,123}{2,176,782,336}$
and I could not therefore have recommended it to him for a stake of eight coins.
That the reckoning of the fairness of the Baron's second wager was substantially more difficult than that of his first is all the more curious if we consider a slight change to these games. Specifically, let us imagine that the Baron had shaken his die in a cup and upended it upon the table before Sir R----- began his play.
If he revealed his score after Sir R----- was done then the reckoning of the game would be identical to that of the first of the Baron's wagers; if before he began, then identical to that of the second.
The Baron's score in both games is determined at the outset; the only difference being when Sir R----- learns of it. That the earlier informed Sir R----- is of the state of play, the harder it is for him to determine the consequences of entering into the Baron's wager has caused the more philosophically minded of my fellow students no small consternation.
For my part, I am content; it is so and that is reason enough!
$$\Box$$
Based upon an article I wrote for ACCU's CVu magazine. | 4,498 | 12,689 | {"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} | 3.3125 | 3 | CC-MAIN-2018-22 | latest | en | 0.988413 |
https://math.stackexchange.com/questions/520491/semigroups-isomorphism | 1,576,415,837,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541308149.76/warc/CC-MAIN-20191215122056-20191215150056-00482.warc.gz | 441,326,615 | 27,710 | # Semigroups isomorphism [duplicate]
Does there exist an isomorphism between the semigroups $S(4)$ and $\mathbf Z_{256}$.
$S(4)$ is the set of all maps from the set $X$ to itself and $X = \{1, 2, 3, 4\}$. $S(4)$ is a semigroup under the composition of mappings and $\mathbf Z_{256} = {0, 1, 2, … , 255}$ is the semigroup under multiplication of integers modulo 256.
• This might be better received at mathstackexchange but some reformatting is needed. – Benjamin Steinberg Oct 9 '13 at 16:59
Note that in $\mathbb Z_{256}$ we only have two elements $x$ so that $$x^2=x$$
Indeed, if $x$ is odd, it is invertible, otherwise $x-1$ is invertible $\mod{256}$.
In $S(4)$ there are many functions $f$ so that $f \circ f =f$, for example, all functions with only one element in the image.
Second solution The invertible elements in $S(4)$ are the permutations, thus $S(4)$ has $4!=24$ invertible elements. The invertible elements in $\mathbf Z_{256}$ are the numbers relatively prime to $256$ (i.e. odd numbers). Thus $\mathbf Z_{256}$ has 128 invertible elements.
Another answer, to add to N.S.'s two. In $\mathbb{Z}_{256}$, the element $3$ has order 64. That is, $3^{n}\neq 1$ for all $1\le n<64$, but $3^{64}=1$. In order for an element of $S(4)$ to have an order at all, it must be a bijection (permutation). All permutations of $\{1,2,3,4\}$ have order $1,2,3$, or $4$; much less than 64.
• OR: As $f : Im(f) \to Im f$ becomes a bijection, it follows that for every $f \in S(4)$ there exists an $2 \leq m \leq 5$ so that $f^m=f$. As you pointed for $3$ the smallest such number is $65$, not even close... This observation makes finding such a number much easier, because the order doesn't matter anymore, it only matters if it is more than $4$. – N. S. Oct 9 '13 at 19:00
$\mathbb Z_{256}$ is commutative and $S(4)$ isn't. | 625 | 1,859 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2019-51 | latest | en | 0.840983 |
https://i10git.cs.fau.de/holzer/pystencils/-/commit/09c6f788222f940d9b40dc3194d53c0c62eac9ea | 1,660,503,072,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572063.65/warc/CC-MAIN-20220814173832-20220814203832-00239.warc.gz | 292,421,646 | 25,014 | Commit 09c6f788 by Markus Holzer
### Fix RoundOff problems
parent 29e0e84e
Pipeline #36459 failed with stages
in 3 minutes and 1 second
... ... @@ -441,7 +441,7 @@ class CustomSympyPrinter(CCodePrinter): def _print_Pow(self, expr): """Don't use std::pow function, for small integer exponents, write as multiplication""" if not expr.free_symbols: return self._typed_number(expr.evalf(), get_type_of_expression(expr.base)) return self._typed_number(expr.evalf(17), get_type_of_expression(expr.base)) if expr.exp.is_integer and expr.exp.is_number and 0 < expr.exp < 8: return f"({self._print(sp.Mul(*[expr.base] * expr.exp, evaluate=False))})" ... ... @@ -452,7 +452,7 @@ class CustomSympyPrinter(CCodePrinter): def _print_Rational(self, expr): """Evaluate all rationals i.e. print 0.25 instead of 1.0/4.0""" res = str(expr.evalf().num) res = str(expr.evalf(17)) return res def _print_Equality(self, expr): ... ...
... ... @@ -234,7 +234,7 @@ def apply_sympy_optimisations(assignments): # Evaluates all constant terms evaluate_constant_terms = ReplaceOptim(lambda e: hasattr(e, 'is_constant') and e.is_constant and not e.is_integer, lambda p: p.evalf()) lambda p: p.evalf(17)) sympy_optimisations = [evaluate_constant_terms] + list(optims_c99) ... ...
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# Agricultural scientists have estimated that the annual loss
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Agricultural scientists have estimated that the annual loss [#permalink]
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16 Jan 2005, 06:06
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Agricultural scientists have estimated that the annual loss by erosion of arable land caused by heavy rainfall and inadequate flood controls approaches two million acres per year.
(A) the annual loss by erosion of arable land caused by heavy rainfall and inadequate flood controls approaches two million acres per year
(B) the erosion of heavy rainfall and inadequate flood controls causes a loss of arable land approaching two million acres per year
(C) erosion caused by heavy rainfall and inadequate flood controls results in a loss of arable land approaching two million acres per year
(D) an annual loss approaching two million acres of arable land per year results from erosion caused by heavy rainfall and inadequate flood controls
(E) annually a loss of arable land approaching two million acres per year is caused by erosion due to heavy rainfall and inadequate flood controls
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16 Jan 2005, 10:08
I go with D, i think they estimate the amount of loss of arable land, D puts it in the right context, however, A too sounds OK.
whats OA?
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17 Jan 2005, 19:57
I choose D also, But I think D makes the sentence passive. and passive is a no-no in GMAT
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Agricultural scientists have estimated that the annual loss [#permalink]
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16 Aug 2009, 08:12
1
1
Agricultural scientists have estimated that the annual loss by erosion of arable land caused by heavy rainfall and inadequate flood controls approaches two million acres per year.
(A) the annual loss by erosion of arable land caused by heavy rainfall and inadequate flood controls approaches two million acres per year
(B) the erosion of heavy rainfall and inadequate flood control causes a loss of arable land approaching two million acres per year
(C) erosion caused by heavy rainfall and inadequate flood controls results in a loss of arable land approaching two million acres per year
(D) an annual loss approaching two million acres of arable land per year results from results from erosion caused by heavy rainfall and unadequate flood controls
(E) annually a loss of arable land approaching two million acres is caused by erosion due to heavy rainfall and inadequate flood controls
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16 Aug 2009, 08:21
Agricultural scientists have estimated that the annual loss by erosion of arable land caused by heavy rainfall and inadequate flood controls approaches two million acres per year.
a) same
b) the erosion of heavy rainfall and inadequate flood control causes a loss of arable land approaching two million acres per year
c)erosion caused by heavy rainfall and inadequate flood controls results in a loss of arable land approaching two million acres per year
d)an annual loss approaching two million acres of arable land per year results from results from erosion caused by heavy rainfall and unadequate flood controls: redundant : results from and caused by
e) annualy a loss of arable land approaching two million acres is caused by erosion due to heavy rainfall and inadequate flood controls
Between A and B , IMO A: as B has approaching ( dsnt go with estimation )
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16 Aug 2009, 08:50
tkarthi4u wrote:
Agricultural scientists have estimated that the annual loss by erosion of arable land caused by heavy rainfall and inadequate flood controls approaches two million acres per year.
a) same
b) the erosion of heavy rainfall and inadequate flood control causes a loss of arable land approaching two million acres per year
c)erosion caused by heavy rainfall and inadequate flood controls results in a loss of arable land approaching two million acres per year
d)an annual loss approaching two million acres of arable land per year results from results from erosion caused by heavy rainfall and unadequate flood controls
e) annualy a loss of arable land approaching two million acres is caused by erosion due to heavy rainfall and inadequate flood controls
I narrowed it down to A and D .. for some reason D doesnt sound right so I am going with A.
In B, and C, land approaching is wrong .. ( it means land that is approaching?)
IMO A.
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Updated on: 16 Aug 2009, 11:52
IMO C
A. by erosion of arable land...by heavy rainfall
D. an annual loss.... per year , redundant
E. annualy awkward
between B and C , in C results in is correct usuage, than causes
so C
what is the OA?
Originally posted by crejoc on 16 Aug 2009, 10:58.
Last edited by crejoc on 16 Aug 2009, 11:52, edited 1 time in total.
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16 Aug 2009, 11:20
C
A is wrong: annual .... per year - redundancy.
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16 Aug 2009, 11:44
crejoc wrote:
IMO C
A. by erosion of arable land...by heavy rainfall
D. an annual loss.... per year , redundant
E. annualy awkward
between B and C , in C results in is correct usuage, than causes
so B
what is the OA?
Nice catch...... Fight between B and C...
Wats the OA
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16 Aug 2009, 19:39
Thank you Walker.
This was taken from the Princeton class room Workout book. I do not have th OA. I put it so that we could discuss.IF anybody has please help .
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17 Aug 2009, 03:40
tkarthi4u wrote:
Thank you Walker.
This was taken from the Princeton class room Workout book. I do not have th OA. I put it so that we could discuss.IF anybody has please help .
It is already discussed in this forum, OA is C
sc-agricultural-scientists-13223.html?highlight=Agricultural+scientists+have+estimated+that+the+annual+loss+by+erosion+of
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28 Jul 2011, 07:08
A, D and E are out because of redundancy "annual" and "per year"
B is out because of awkwark construction "the erosion of heavy rainfall" how eroision could be of ranfall?? The correct "erosion caused by heavy rainfall".
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07 Sep 2011, 11:40
Quote:
Agricultural scientists have estimated that the annual loss by erosion of arable land caused by heavy rainfall and inadequate flood controls approaches two million acres per year.
a) the annual loss by erosion of arable land caused by heavy rainfall and inadequate flood controls approaches two million acres per year.
b) the erosion of heavy rainfall and inadequate flood control causes a loss of arable land approaching two million acres per year
c)erosion caused by heavy rainfall and inadequate flood controls results in a loss of arable land approaching two million acres per year
d)an annual loss approaching two million acres of arable land per year results from erosion caused by heavy rainfall and unadequate flood controls
e) annualy a loss of arable land approaching two million acres is caused by erosion due to heavy rainfall and inadequate flood controls
A - 'annual loss by erosion of arable land' sounds awkward
D - redundancy - 'an annual loss...per year'
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10 Sep 2011, 10:25
tkarthi4u wrote:
Agricultural scientists have estimated that the annual loss by erosion of arable land caused by heavy rainfall and inadequate flood controls approaches two million acres per year.
a) same
b) the erosion of heavy rainfall and inadequate flood control causes a loss of arable land approaching two million acres per year
c)erosion caused by heavy rainfall and inadequate flood controls results in a loss of arable land approaching two million acres per year
d)an annual loss approaching two million acres of arable land per year results from results from erosion caused by heavy rainfall and unadequate flood controls
e) annualy a loss of arable land approaching two million acres is caused by erosion due to heavy rainfall and inadequate flood controls
Answer is C. It took 3:43 though.
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10 Sep 2011, 11:11
fieldsrd wrote:
tkarthi4u wrote:
Agricultural scientists have estimated that the annual loss by erosion of arable land caused by heavy rainfall and inadequate flood controls approaches two million acres per year.
a) same
b) the erosion of heavy rainfall and inadequate flood control causes a loss of arable land approaching two million acres per year
c)erosion caused by heavy rainfall and inadequate flood controls results in a loss of arable land approaching two million acres per year
d)an annual loss approaching two million acres of arable land per year results from results from erosion caused by heavy rainfall and unadequate flood controls
e) annualy a loss of arable land approaching two million acres is caused by erosion due to heavy rainfall and inadequate flood controls
Answer is C. It took 3:43 though.
ok a quick q... why not E? any idea
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11 Sep 2011, 12:36
viks4gmat wrote:
fieldsrd wrote:
tkarthi4u wrote:
Agricultural scientists have estimated that the annual loss by erosion of arable land caused by heavy rainfall and inadequate flood controls approaches two million acres per year.
a) same
b) the erosion of heavy rainfall and inadequate flood control causes a loss of arable land approaching two million acres per year
c)erosion caused by heavy rainfall and inadequate flood controls results in a loss of arable land approaching two million acres per year
d)an annual loss approaching two million acres of arable land per year results from results from erosion caused by heavy rainfall and unadequate flood controls
e) annualy a loss of arable land approaching two million acres is caused by erosion due to heavy rainfall and inadequate flood controls
Answer is C. It took 3:43 though.
ok a quick q... why not E? any idea
Other than "annualy" which I am sure is a typo, though i am not too sure weather "unadequate" in option D is, option E is constructed wrongly.
Agricultural scientists have estimated that X is caused by Y due to heavy rainfall and inadequate flood controls.
This construction is wrong. If you need to modify Y you will need a noun modifier to do that. You cannot use "due to" to modify erosion.
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24 Sep 2011, 10:48
I have picked E, since the word per year is also underlined. so i replaced this per year with annually
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25 Sep 2011, 21:43
I also got C.
Is the C the correct answer?
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Re: Agricultural scientists have estimated that the annual loss [#permalink]
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17 Dec 2011, 23:22
Can some one explain how Approaching fit with estimate.
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Re: Agricultural scientists have estimated that the annual loss [#permalink]
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10 Jan 2012, 23:05
option C) Agricultural scientists have estimated that erosion caused by heavy rainfall and inadequate flood controls results in a loss of arable land approaching two million acres per year.
Guys I feel this answer changes the meaning and makes me conclude Agricultural scientists are more concerned with erosion as in main subject of that clause rather than annual loss
where as D and E preserves the original meaning that annual loss is important
A and D - redundant annual and per year usage
choice E retains the original sentence meaning and grammatically incorrect - caused by and due to, each one has the same meaning in grammar.
still confused so will go for C.
Any body wants to throw more light on choice E
Re: Agricultural scientists have estimated that the annual loss [#permalink] 10 Jan 2012, 23:05
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Display posts from previous: Sort by | 3,429 | 14,168 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-22 | latest | en | 0.925058 |
https://www.convertunits.com/from/ton-force/to/exanewton | 1,600,464,230,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400188841.7/warc/CC-MAIN-20200918190514-20200918220514-00471.warc.gz | 739,693,112 | 8,445 | ## ››Convert ton-force [long] to exanewton
ton-force exanewton
Did you mean to convert ton-force [long] ton-force [metric] ton-force [short] to exanewton
How many ton-force in 1 exanewton? The answer is 1.0036113565668E+14.
We assume you are converting between ton-force [long] and exanewton.
You can view more details on each measurement unit:
ton-force or exanewton
The SI derived unit for force is the newton.
1 newton is equal to 0.00010036113565668 ton-force, or 1.0E-18 exanewton.
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## ››Definition: Exanewton
The SI prefix "exa" represents a factor of 1018, or in exponential notation, 1E18.
So 1 exanewton = 1018 newtons.
The definition of a newton is as follows:
In physics, the newton (symbol: N) is the SI unit of force, named after Sir Isaac Newton in recognition of his work on classical mechanics. It was first used around 1904, but not until 1948 was it officially adopted by the General Conference on Weights and Measures (CGPM) as the name for the mks unit of force.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 486 | 1,840 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-40 | latest | en | 0.872103 |
https://zbmath.org/?q=an:0667.60037&format=complete | 1,726,559,366,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651750.27/warc/CC-MAIN-20240917072424-20240917102424-00139.warc.gz | 1,004,184,328 | 9,699 | ## Convergence rate for sums of a random number of random variables.(Russian)Zbl 0667.60037
Complete convergence and convergence rates results for sums of random variables have been generalized to the case of randomly indexed partial sums. The following result is obtained.
Let $$X_ n$$ be i.i.d. random variables and $$\nu_ n$$ be nonnegative, integer-valued random variables. Suppose that $$\alpha >1/2$$, $$\alpha$$ $$r\geq 1$$, E $$| X_ 1|^ t<\infty$$ and that E $$X_ 1=0$$ if $$r\geq 1$$. If, for some $$\epsilon >0$$, $$\beta >0$$, $(1)\quad \sum n^{\alpha r-2}P(\nu_ n<n^{\beta})<\infty \quad and\quad (2)\quad \sum n^{\alpha r-2}P(\max_{k\leq \nu_ n}| X_ k| \geq \epsilon \nu_ n^{\alpha})<\infty,$ then there exists a number N such that for $$\delta\geq N\epsilon$$; $\sum n^{\alpha r-2}P(| S_ n| \geq \delta \nu_ n^{\alpha})<\infty.$ This result generalizes the theorem of A. Gut in Acta Math. Hung. 42, 225-232 (1983; Zbl 0536.60036). Note that this result holds if E $$| X_ 1|^ t<\infty$$ for $$t>1/\alpha$$ and E $$X_ 1=0$$ if $$\alpha\leq 1$$, and conditions (1) and (2) are satisfied.
Reviewer: O.I.Klesov
### MSC:
60F15 Strong limit theorems 60G50 Sums of independent random variables; random walks
### Keywords:
convergence rates; randomly indexed partial sums
Zbl 0536.60036 | 451 | 1,298 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-38 | latest | en | 0.638815 |
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# A shuffle through the gaming mailbag
21 July 2011
Q. I usually play Double Bonus Poker, and have always held one face card, choosing one at random when several are dealt, if there is no chance of a straight or a flush. Many times that one will pair up to save my bet. Is this wrong? Does it limit my winning other hands?
A. Strategies vary by specific game and pay table, but it's usually better to hold multiple high cards.
Since you say Double Bonus is your game, let's use full-pay 10/7/5 Double Bonus as an example. Let's say you're dealt king of clubs, jack of hearts, 8 of spades, 4 of diamonds, 2 of clubs. By your method, you'd pick one high card at random and discard the other four cards. The better of the two cards to hold would be the jack, because it leaves more straight possibilities. Straights could be jack, queen, king or ace high. If you hold the king, straights could be only king or ace high.
There are 178,365 possible four card draws, of which 120,401 bring losing hands. Among the 57,935 winners, 43,389 are high-pair hands that pay 1-for-1, there are 8,874 two pairs (1-for-1), 4,102 three of a kinds (3-for1), 764 straights (5-for-1), 491 flushes (7-for-1), 288 full houses (10-for-1), 50 four of a kinds, 5s through kings (50-or-1), 1 four of a kind, 2-through-4 (80-for-1), 1 four aces (160-for-1), 3 straight flushes (50-for-1), and one royal flush (250-for-1, but 4,000 with a five-coin wager).
The average return for all that is 2.198 coins per five wagered.
Now what if you hold both the king and the jack? There are 16,125 possible three-card draws, and 10,053 are losers. You can't get any royals, straight flushes or straights, and the only quads you can get are four kings or four jacks. But you enhance your chances at other hands, especially pairs of jacks or better. There are 5,022 draws that will pair up either the king or the jack, along with 711 two pairs, 281 three of a kinds, 128 straights, 18 full houses and two fours of a kind.
Note the straight probabilities. If you hold just a jack, 0.43% of all draws will bring a pair of jacks or better. Hold king-jack, and that chance almost doubles, to 0.79% of all hands.
Bottom line: Holding king-jack in this situation brings an average of 2.311 coins per five coins wagered, better than the 2.198 for holding just the jack or the 2.109 for holding just the king.
Likewise, holding king-queen-jack of mixed suits is a stronger play than holding just one or two high cards. Make it ace-king-jack, and it's a different story, with king-jack being a better play than ace-king-jack in most games, but with holding just the ace being a better play in big ace-bonus games such as Double Double Bonus Poker. But that's a wrinkle for another time.
Q. I have never had a royal flush and only two four-aces hands. The other day I told an attendant (replacing paper in machine next to me) that I had never seen anyone hit a royal or four aces. The man on my other side said he had done so several times on the same machines. I noticed he was betting dollars and I always bet maximum quarters. Does amount you have chosen to play affect the frequency of winning?
A. The coin denomination does not affect results in video poker. The random number generator that determines what cards you see is the same program in a nickel game as it is in a quarter game or a dollar game or any other denomination.
Some of what you're experiencing is just the normal rarity of the hands. Let's stick with 10/7/5 Double Bonus Poker. We see a royal an average of once per 48,048 hands. That doesn't guarantee a royal within 48,048 hands. Sometimes we'll draw two or more royals within that stretch, but just as often we'll see zero royals within two or three times that many hands. Cold streaks are just a normal part of the game.
I remember once I was talking with a fellow gambling writer and analyst, and I was in the middle of a couple of years without a royal. He asked if I'd been winning money. I told him I'd been winning a little in blackjack, but video poker was putting me in the red overall. He asked the last time I'd had a royal, and I told him it had been more than two years ago. About a month later, I was playing at the old Stardust in Las Vegas, and I drew a royal. Less than an hour later, I drew a second one, on the same machine. Definitely an exciting night.
You never know when they're going to come. It can be a long, long time without a royal. But it's not the coin denomination that's causing the cold streak. It's just normal odds and probability.
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John Grochowski
John Grochowski is the best-selling author of The Craps Answer Book, The Slot Machine Answer Book and The Video Poker Answer Book. His weekly column is syndicated to newspapers and Web sites, and he contributes to many of the major magazines and newspapers in the gaming field, including Midwest Gaming and Travel, Slot Manager, Casino Journal, Strictly Slots and Casino Player.
Listen to John Grochowski's "Casino Answer Man" tips Tuesday through Friday at 5:18 p.m. on WLS-AM (890) in Chicago. Look for John Grochowski on Facebook and Twitter @GrochowskiJ.
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John Grochowski
John Grochowski is the best-selling author of The Craps Answer Book, The Slot Machine Answer Book and The Video Poker Answer Book. His weekly column is syndicated to newspapers and Web sites, and he contributes to many of the major magazines and newspapers in the gaming field, including Midwest Gaming and Travel, Slot Manager, Casino Journal, Strictly Slots and Casino Player.
Listen to John Grochowski's "Casino Answer Man" tips Tuesday through Friday at 5:18 p.m. on WLS-AM (890) in Chicago. Look for John Grochowski on Facebook and Twitter @GrochowskiJ. | 1,453 | 5,848 | {"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} | 2.515625 | 3 | CC-MAIN-2023-50 | latest | en | 0.938267 |
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### Chosen message attack on unhashed GGH signatures?
Background: I've been reading GGH's Public-Key Cryptosystems from Lattice Reduction Problems, and have a question about a remark the authors make: "It is important to remark at the outset, that ...
362 views
### State of research on SHA-1 Collision Attacks
SHA-1 security has been discussed since an algorithm for finding collisions was first published at CRYPTO 2004 and has been subsequently improved. Wikipedia lists a couple of references, however it ...
106 views
### Factoring assuming smoothness of some numbers
I have came across a lot of factorization methods and most of them seem to assume smoothness of some numbers. For example When $p-1$ is smooth When $|E(\mathbb{F}_p)|$ is smooth. (Elliptic curve ...
106 views
### Is a "complete" cipher possible?
Is a "complete" symmetric cipher possible? By this I mean a symmetric cipher that is provably secure under the assumption that a secure symmetric cipher exists.
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### Is perfect zero knowledge sequentially composable without auxiliary input?
It is known that plain and computational zero knowledge proof systems are not sequentially composable without auxiliary input (see for example http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1....
202 views
### Cryptographic systems that don't leak linear combinations of encrypted bits
Various encryption schemes would be considered broken if an adversary could have a non-negligible edge in predicting the first (or any) bit of an encrypted message. I am looking for a slightly ...
690 views
### How to obtain the unknown values $a_i,b_j$ given an unordered list of $a_i-b_j\mod N$?
Can anyone help me with the following problem? I want to find some values $a_i,b_j$ (mod $N$) where $i=1,2,…,K, j=1,2,…,K$ (for example $K=6$), given a list of $K^2$ values that correspond to the ...
277 views
### Is there a better explanation of Hellman's paper on rainbow tables than the paper itself?
I'm reading through Hellman's A Cryptanalytic Time-Memory Trade-off [PDF], and I was wondering if there is code out there that implements this or if there are better explanations of how a password ...
178 views
### Setting protocol parameters to achieve concrete security
Background One issue with modern security proofs is that they are usually asymptotic. In other words, such proofs are usually formulated as follows: For any polynomial-time adversary $\mathcal A$, we ...
212 views
### What is the best fitness function for detecting natural language?
First, let me apologise, as this question is far from my area of expertise, but is related to a side interest (read hobby), and so this question might be very naive. This may even be off-topic for the ...
178 views
### A clarification in Signature schemes for Strong Unforgeability.
Signature scheme is strongly unforgeable. "making the message m being signed depend on the randomness r — we break the proof of security for the underlying signature." From Paper: Strongly ...
626 views
### Breaking encrypted file store with a file?
Let's say you have an encrypted file store of some kind such as an encrypted hard drive, ZIP file, etc. You don't have the private key or password to this file store, so a brute-force attack is not a ...
470 views
### Can one reverse a hash with partial plaintext knowledge?
First off, please forgive my ignorance because I am not as well versed in cryptography and mathematics as I would like to be. I may say something obviously wrong/dumb; please point it out! Is there ... | 818 | 3,683 | {"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} | 2.609375 | 3 | CC-MAIN-2021-39 | latest | en | 0.900337 |
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00:00
Draw a scaled picture graph and a scaled bar graph to represent a data set with several categories. Solve one- and two-step “how many more” and “how many less” problems using information presented in scaled bar graphs. For example, draw a bar graph in which each square in the bar graph might represent 5 pets.
## Bar Graphs
00:00
Draw a picture graph and a bar graph (with single-unit scale) to represent a data set with up to four categories. Solve simple put-together, take-apart, and compare problems using information presented in a bar graph.APR.A.1
## Extended Patterns on a Graph
00:00
Identify arithmetic patterns (including patterns in the addition table or multiplication table), and explain them using properties of operations. For example, observe that 4 times a number is always even, and explain why 4 times a number can be decomposed into two equal addends.D.9
## Picture Graphs
00:00
Draw a picture graph and a bar graph (with single-unit scale) to represent a data set with up to four categories. Solve simple put-together, take-apart, and compare problems 1 using information presented in a bar graph.
## Graphing - What is Scale
00:00
Draw a scaled picture graph and a scaled bar graph to represent a data set with several categories. Solve one- and two-step “how many more” and “how many less” problems using information presented in scaled bar graphs. For example, draw a bar graph in which each square in the bar graph might represent 5 pets.
## Picture Graphs with Keys
00:00
Draw a scaled picture graph and a scaled bar graph to represent a data set with several categories. Solve one- and two-step “how many more” and “how many less” problems using information presented in scaled bar graphs. For example, draw a bar graph in which each square in the bar graph might represent 5 pets.
## Picture Graphs
00:00
Draw a picture graph and a bar graph (with single-unit scale) to represent a data set with up to four categories. Solve simple put-together, take-apart, and compare problems10 using information presented in a bar graph.
## Bar Graphs
00:00
Organize, represent, and interpret data with up to three categories; ask and answer questions about the total number of data points, how many in each category, and how many more or less are in one category than in another.
## Solving Linear Systems by Graphing
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Solve systems of linear equations exactly and approximately (e.g., with graphs), focusing on pairs of linear equations in two variables.
00:00
Write a function that describes a relationship between two quantities.
Using tables, graphs, and verbal descriptions, interpret the key characteristics of a function which models the relationship between two quantities. Sketch a graph showing key features including: intercepts; interval where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior; and periodicity. | 696 | 3,171 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-09 | longest | en | 0.896568 |
https://stage.geogebra.org/m/gt3HBUcE | 1,627,549,763,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153854.42/warc/CC-MAIN-20210729074313-20210729104313-00336.warc.gz | 542,628,836 | 9,876 | # Right Circular Cone
Topic:
Cone
Use this worksheet to practice finding the volume and surface area of a right circular cone. Adjust the height and radius using the labeled sliders. Answer the questions below. Click on the check boxes to find helpful hint and verify your answers.
1.Find the volume and surface area for 5 different cones by randomly choosing different heights and radii. 2. Find the missing dimensions below. Round to the nearest unit. a. Cone: volume is 424 cubic meters, diameter is 18 meters, height ___ b. Cone: surface area is 153.5 square inches, radius is 4 inches, slant height ___ 3. Plants: A cone-shaped paper cup is 7 cm high with a diameter of 6 cm. If the ivy plant on Julia's desk needs 240mL of water, about how many paper cups of water will she use to water it? (1 mL = 1 cubic cm) | 199 | 817 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2021-31 | latest | en | 0.848854 |
http://math.stackexchange.com/questions/281380/proving-an-implication-from-a-matrix-equation | 1,469,456,699,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824230.71/warc/CC-MAIN-20160723071024-00302-ip-10-185-27-174.ec2.internal.warc.gz | 151,295,835 | 17,130 | # Proving an implication from a matrix equation
I am supposed to prove that, if $\eta$ is lorentz metric, $M$ is a 4 by 4 matrix and $x^tM^t \eta Mx=x^t\eta x$ for any column vector $x$, then $M^t\eta M=\eta$. What I did seems awfully clumsy. I used simple vectors to derive that the components $M_{ij}=0$, when $i=j$, $M_{23}=0$ and $M_{32}=0$ and otherwise $M_{ij}=-M_{ji}$. I plugged $M$ in the equation and was able to show that the remaining components must be $0$ except $M_{12}=-M_{21}=M_{34}=-M_{43}=i$. Then the claim was easy to verify. I did side step at certain points to avoid what I eventually had to rely on. I am unhappy. Could you give me tips how to solve the problem neatly?
-
It is easier if you abstract a bit, and prove more. We aim to show that if $A$ and $B$ are symmetric and real and $x^TAx=x^TBx$ for all $x$, then $A=B$. Since $x^TAx-x^TBx=x^T(A-B)x$, it will be enough to show that if $x^TCx=0$ for all $x$ (and $C$ is real symmetric), then $C=0$.
The trick is to note that if $x^TCx=0$ for all $x$ then for any $x$ and $y$ $$0 = (x+y)^TC(x+y) = x^TCx + y^TCy + x^TCy+y^TCx = 0 + 0 +2x^TCy.$$ Hence $x^TCy=0$ for all $x$ and $y$. [This idea is useful in a number of places, and is worth knowing.]
Now if we let $x$ and $y$ run over the standard basis vectors, we get that $C_{i,j}=0$ for all $i$ and $j$. | 470 | 1,336 | {"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} | 3.9375 | 4 | CC-MAIN-2016-30 | latest | en | 0.899779 |
https://artofproblemsolving.com/wiki/index.php?title=1997_PMWC&diff=next&oldid=17635 | 1,643,302,611,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305277.88/warc/CC-MAIN-20220127163150-20220127193150-00630.warc.gz | 178,653,761 | 10,916 | # Difference between revisions of "1997 PMWC"
## Problem I1
Evaluate 29 27/28 x 27 14/15 Solution
## Problem I10
Mary took 24 chickens to the market. In the morning she sold the chickens at $7 each and she only sold out less than half of them. In the afternoon she discounted the price of each chicken but the price was still an integral number in dollar. In the afternoon she could sell all the chickens, and she got totally$132 for the whole day. How many chickens were sold in the morning?
## Problem T3
Peter is ill. He has to take medicine A every 8 hours, medicine B every 5 hours and medicine C every 10 hours. If he took all three medicines at 7 a.m. on Tuesday, when will he take them altogether again? Solution
## Problem T6
John and Mary went to a book shop and bought some exercise books. They had <dollar/>100 each. John could buy 7 large and 4 small ones. Mary could buy 5 large and 6 small ones and had <dollar/>5 left. How much was a small exercise book?
## Problem T7
40% of girls and 50% of boys in a class got 'A'. If there are only 12 students in the class got 'A's and the ratio of boys and girls in the class is 45, how many students are there in the class?
## Problem T9
A chemist mixed an acid of 48% concentration with the same acid of 80% concentration, and then added 2 litres of distilled water to the mixed acid. As a result, he got 10 litres of the acid of 40% concentration. How many millilitre of the acid of 48% concentration that the chemist had used? (1 litre = 1000 millilitres) | 395 | 1,526 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2022-05 | latest | en | 0.974833 |
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A027976 n-th diagonal sum of right justified array T given by A027960. 2
1, 1, 4, 6, 10, 18, 29, 47, 78, 126, 204, 332, 537, 869, 1408, 2278, 3686, 5966, 9653, 15619, 25274, 40894, 66168, 107064, 173233, 280297, 453532, 733830, 1187362, 1921194, 3108557, 5029751, 8138310, 13168062, 21306372, 34474436, 55780809, 90255245, 146036056, 236291302, 382327358 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 LINKS G. C. Greubel, Table of n, a(n) for n = 0..1000 Index entries for linear recurrences with constant coefficients, signature (1,1,1,-1,-1). FORMULA G.f.: (1 + 2*x^2)/((1-x^3)*(1-x-x^2)). From G. C. Greubel, Sep 26 2019: (Start) a(n) = (Fibonacci(n) + 4*Fibonacci(n+1) - A102283(n) - 2)/2. a(n) = (Fibonacci(n+1) + Lucas(n+2) - 2*sin(2*Pi*n/3)/sqrt(3) - 2)/2. (End) MAPLE seq(coeff(series((1 + 2*x^2)/((1-x^3)*(1-x-x^2)), x, n+1), x, n), n = 0..40); # G. C. Greubel, Sep 26 2019 MATHEMATICA LinearRecurrence[{1, 1, 1, -1, -1}, {1, 1, 4, 6, 10}, 41] (* or *) Table[ (Fibonacci[n+1] +LucasL[n+2] -2*Sin[2*Pi*n/3]/Sqrt[3] -2)/2, {n, 0, 40}] (* G. C. Greubel, Sep 26 2019 *) PROG (PARI) my(x='x+O('x^40)); Vec((1 + 2*x^2)/((1-x^3)*(1-x-x^2))) \\ G. C. Greubel, Sep 26 2019 (MAGMA) R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (1 + 2*x^2)/((1-x^3)*(1-x-x^2)) )); // G. C. Greubel, Sep 26 2019 (Sage) def A027976_list(prec): P. = PowerSeriesRing(ZZ, prec) return P((1 + 2*x^2)/((1-x^3)*(1-x-x^2))).list() A027976_list(40) # G. C. Greubel, Sep 26 2019 (GAP) a:=[1, 1, 4, 6, 10];; for n in [6..40] do a[n]:=a[n-1]+a[n-2]+a[n-3]-a[n-4]-a[n-5]; od; a; # G. C. Greubel, Sep 26 2019 CROSSREFS Cf. A000032, A000045, A004695, A027960, A102283. Sequence in context: A165186 A310590 A108232 * A108900 A076995 A096817 Adjacent sequences: A027973 A027974 A027975 * A027977 A027978 A027979 KEYWORD nonn AUTHOR EXTENSIONS Terms a(28) onward added by G. C. Greubel, Sep 26 2019 STATUS approved
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Last modified December 14 15:08 EST 2019. Contains 329979 sequences. (Running on oeis4.) | 1,101 | 2,739 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2019-51 | latest | en | 0.590411 |
https://www.geeksforgeeks.org/number-of-subarrays-having-sum-of-the-form-km-m-0/ | 1,579,758,029,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250608295.52/warc/CC-MAIN-20200123041345-20200123070345-00233.warc.gz | 898,660,045 | 27,696 | # Number of subarrays having sum of the form k^m, m >= 0
Given an integer k and an array arr[], the task is to count the number of sub-arrays which have the sum equal to some positive integral power of k.
Examples:
Input: arr[] = { 2, 2, 2, 2 } K = 2
Output: 8
Sub-arrays with below indexes are valid:
[1, 1], [2, 2], [3, 3], [4, 4], [1, 2],
[2, 3], [3, 4], [1, 4]
Input: arr[] = { 3, -6, -3, 12 } K = -3
Output: 3
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
Naive Approach: A naive approach is to traverse through all the sub-arrays and check for every sub-array whether its sum is equal to some integral power of k.
Efficient Approach: A better approach is to maintain a prefix sum array prefix_sum and a map m which maps a prefix sum to its count. m[a] = 1 means that a is a prefix sum of some prefix.
Iterate through the array in the backward direction and carefully follow the below discussion. Consider that while traversing the array we are at the ith index and after traversing an index we perform the operation op = m[prefix_sum[i]]++. Hence when we are at index i, op hasn’t been performed yet. See the code for a vivid explanation.
If m[a + b] = c where a = prefix_sum[i], b = kp and c is the value fetched from the map, then it means that starting from the ith index to the end of the array, there are c sub-arrays whose sum is equal to b. Add c to the current sum.
This is because for every index j > i, m[prefix_sum[j]]++ has been performed. Therefore the map has information about prefix sums of prefixes ending at j > i. On adding b to the prefix sum we can get the count of all those sums a + b which will indicate that a sub-array exists that has sum equal to b.
Note: k = 1 and k = -1 need to be handled separately.
Below is the implementation of the above approach:
`// C++ implementation of the above approach ` `#include ` ` ` `#define ll long long ` `#define MAX 100005 ` ` ` `using` `namespace` `std; ` ` ` `// Function to count number of sub-arrays ` `// whose sum is k^p where p>=0 ` `ll countSubarrays(``int``* arr, ``int` `n, ``int` `k) ` `{ ` ` ``ll prefix_sum[MAX]; ` ` ``prefix_sum[0] = 0; ` ` ` ` ``partial_sum(arr, arr + n, prefix_sum + 1); ` ` ` ` ``ll sum; ` ` ` ` ``if` `(k == 1) { ` ` ` ` ``sum = 0; ` ` ``map m; ` ` ` ` ``for` `(``int` `i = n; i >= 0; i--) { ` ` ` ` ``// If m[a+b] = c, then add c to the current sum. ` ` ``if` `(m.find(prefix_sum[i] + 1) != m.end()) ` ` ``sum += m[prefix_sum[i] + 1]; ` ` ` ` ``// Increase count of prefix sum. ` ` ``m[prefix_sum[i]]++; ` ` ``} ` ` ` ` ``return` `sum; ` ` ``} ` ` ` ` ``if` `(k == -1) { ` ` ` ` ``sum = 0; ` ` ``map m; ` ` ` ` ``for` `(``int` `i = n; i >= 0; i--) { ` ` ` ` ``// If m[a+b] = c, then add c to the current sum. ` ` ``if` `(m.find(prefix_sum[i] + 1) != m.end()) ` ` ``sum += m[prefix_sum[i] + 1]; ` ` ` ` ``if` `(m.find(prefix_sum[i] - 1) != m.end()) ` ` ``sum += m[prefix_sum[i] - 1]; ` ` ` ` ``// Increase count of prefix sum. ` ` ``m[prefix_sum[i]]++; ` ` ``} ` ` ` ` ``return` `sum; ` ` ``} ` ` ` ` ``sum = 0; ` ` ` ` ``// b = k^p, p>=0 ` ` ``ll b; ` ` ``map m; ` ` ` ` ``for` `(``int` `i = n; i >= 0; i--) { ` ` ` ` ``b = 1; ` ` ``while` `(``true``) { ` ` ` ` ``// k^m can be maximum equal to 10^14. ` ` ``if` `(b > 100000000000000) ` ` ``break``; ` ` ` ` ``// If m[a+b] = c, then add c to the current sum. ` ` ``if` `(m.find(prefix_sum[i] + b) != m.end()) ` ` ``sum += m[prefix_sum[i] + b]; ` ` ` ` ``b *= k; ` ` ``} ` ` ` ` ``// Increase count of prefix sum. ` ` ``m[prefix_sum[i]]++; ` ` ``} ` ` ` ` ``return` `sum; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ``int` `arr[] = { 2, 2, 2, 2 }; ` ` ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` ` ``int` `k = 2; ` ` ` ` ``cout << countSubarrays(arr, n, k); ` ` ` ` ``return` `0; ` `} `
Output:
```8
```
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Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below. | 1,607 | 4,773 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2020-05 | latest | en | 0.846583 |
http://oeis.org/A141617 | 1,553,449,558,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203464.67/warc/CC-MAIN-20190324165854-20190324191854-00455.warc.gz | 147,984,546 | 3,929 | This site is supported by donations to The OEIS Foundation.
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A141617 Triangle t(n,m) = prime(m)*prime(n-m)*binomial(n,m) read by rows, 0<=m<=n. 4
1, 2, 2, 3, 8, 3, 5, 18, 18, 5, 7, 40, 54, 40, 7, 11, 70, 150, 150, 70, 11, 13, 132, 315, 500, 315, 132, 13, 17, 182, 693, 1225, 1225, 693, 182, 17, 19, 272, 1092, 3080, 3430, 3080, 1092, 272, 19, 23, 342, 1836, 5460, 9702, 9702, 5460, 1836, 342, 23, 29, 460, 2565 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS For the purpose of this sequence define prime(0)=1. Row sums are 1, 4, 14, 46, 148, 462, 1420, 4234, 12356, 34726, 95220,... LINKS FORMULA Symmetry: t(n,m) = t(n,n-m). EXAMPLE 1; 2, 2; 3, 8, 3; 5, 18, 18, 5; 7, 40, 54, 40, 7; 11, 70, 150, 150, 70, 11; 13, 132, 315, 500, 315, 132, 13; 17, 182, 693, 1225, 1225, 693, 182, 17; 19, 272, 1092, 3080, 3430, 3080, 1092, 272, 19; 23, 342, 1836, 5460, 9702, 9702, 5460, 1836, 342, 23; 29, 460, 2565, 10200, 19110, 30492, 19110, 10200, 2565, 460, 29; MATHEMATICA Table[Table[If[n == m ==0, 1, If[m == 0 || m == n, Prime[n], (Prime[n - m]*Prime[m])*Binomial[n, m]]], {m, 0, n}], {n, 0, 10}] Flatten[%] CROSSREFS Cf. A098350. Sequence in context: A135835 A177696 A134574 * A267644 A204197 A238654 Adjacent sequences: A141614 A141615 A141616 * A141618 A141619 A141620 KEYWORD nonn,easy,tabl AUTHOR Roger L. Bagula and Gary W. Adamson, Aug 23 2008 STATUS approved
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Last modified March 24 12:14 EDT 2019. Contains 321448 sequences. (Running on oeis4.) | 805 | 1,804 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2019-13 | latest | en | 0.569183 |
https://www.reddit.com/r/cheatatmathhomework/comments/140lbz/stat_help_please/ | 1,469,682,573,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827791.21/warc/CC-MAIN-20160723071027-00279-ip-10-185-27-174.ec2.internal.warc.gz | 942,334,093 | 16,876 | This is an archived post. You won't be able to vote or comment.
[–] 1 point2 points (3 children)
Well, you need to do a bit of trickery, since the z-scores are in the realm of real numbers, and the IQs are natural numbers.
So: here's the trick. 100 becomes (99.5, 100.5). 120 becomes (119.5, 120.5), etc. sigma is 15.
So your first one, you want to go from z = -1/30 to z = 1/30.
Does this make sense? Can you generalize to the rest?
[–][S] 0 points1 point (2 children)
the answer key is saying P(x=100) = .0266, P(x<120) = .9032. the last one has no answer from the answer key. i still dont understand why this is.
[–] 1 point2 points (1 child)
Because if you're in a continuous situation, rather than a discrete one, P(X=x) = 0. So you need to turn each individual value into a range.
That's why 100 becomes (99.5, 100.5). You're converting it to P(99.5<x<100.5), which, in z scores is P(-1/30 < z < 1/30). Does that get you to 0.0266?
[–][S] 0 points1 point (0 children)
so if P(-1/30 < z 1/30) = P(-.033<z<.033) wouldnt it still be 0? sorry this is really confusing me right now. | 369 | 1,099 | {"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} | 3.734375 | 4 | CC-MAIN-2016-30 | latest | en | 0.925671 |
https://www.mathcelebrity.com/community/threads/a-customer-pays-21-00-for-a-14-96-bill-find-the-least-amount-of-bills-and-coins-to-give-change.5453/ | 1,719,319,801,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865972.21/warc/CC-MAIN-20240625104040-20240625134040-00459.warc.gz | 752,838,668 | 8,869 | # A customer pays \$21.00 for a \$14.96 bill. Find the least amount of bills and coins to give change
Discussion in 'Calculator Requests' started by math_celebrity, May 31, 2024.
Tags:
A customer pays \$21.00 for a \$14.96 bill. Find the least amount of bills and coins to give change
Change = \$21 - \$14.96
Change = \$6.04
Using bills first, we can use:
\$5
\$1
This gives us \$6
We have:
\$6.04 - \$6 = 0.04
We can use 4 pennies to cover this
Total bills and coins:
1 - \$5
1 - \$1
4 pennies | 162 | 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} | 3.78125 | 4 | CC-MAIN-2024-26 | latest | en | 0.805392 |
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Luke Amick
Absolutely amazing! 10/10 :D. And you would solve it like a quadratic equation as if the inequality part was = and y was 0 except if it is equal to it is a solid line or if it is not it is a dotted line on the graph as well as y is less than the expression you shape below it or more than above it.
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100% has helped me understand and learn by seeing the answer and working my way backwards, best math app, could be fixed but is still more helpful than my math's prof.
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` | 521 | 2,375 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2023-14 | latest | en | 0.945578 |
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# What are some special properties of zero?
The Creation of Zero and Pi Read more from
Chapter Mathematics Throughout History
There are many special properties of zero. For instance, you cannot divide by zero (or have zero as the denominator [bottom number] of a fraction). This is because, simply put, something cannot be divided by nothing. Thus, if some equation has a unit (usually a number) divided by zero, the answer is considered to be “undefined.” But it is possible to have zero in the numerator (top number) of a fraction; as long as it does not have zero in the denominator (called a legal fraction), it will always be equal to zero. Other special properties of zero include: Zero is considered an even number; any number ending in zero is considered an even number; when zero is added to a number, the sum is the original number; and when zero is subtracted from a number, the difference is the original number.
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This is a web preview of the "The Handy Math Answer Book" app. Many features only work on your mobile device. If you like what you see, we hope you will consider buying. Get the App | 243 | 1,130 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2022-27 | longest | en | 0.95548 |
https://rdrr.io/cran/Renvlp/man/eppls.html | 1,680,141,397,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949093.14/warc/CC-MAIN-20230330004340-20230330034340-00380.warc.gz | 525,484,679 | 8,657 | # eppls: Fit the Envelope-based Partial Partial Least Squares model In Renvlp: Computing Envelope Estimators
eppls R Documentation
## Fit the Envelope-based Partial Partial Least Squares model
### Description
Fit the Envelope-based Partial Partial Least Squares model for multivariate linear regression with dimension u.
### Usage
eppls(X1, X2, Y, u, asy = TRUE, init = NULL)
### Arguments
X1 An n by p1 matrix of continuous predictors, where p1 is the number of continuous predictors with p1 < n. X2 An n by p2 matrix of categorical predictors, where p2 is the number of categorical predictors with p2 < n. Y An n by r matrix of multivariate responses, where r is the number of responses. u A given dimension of the Envelope-based Partial Partial Least Squares. It should be an interger between 0 and p1. asy Flag for computing the asymptotic variance of the envelope estimator. The default is TRUE. When p and r are large, computing the asymptotic variance can take much time and memory. If only the envelope estimators are needed, the flag can be set to asy = FALSE. init The user-specified value of Gamma for the envelope subspace. An r by u matrix. The default is the one generated by function envMU.
### Details
This function the Envelope-based Partial Partial Least Squares model for multivariate linear regression with dimension u,
Y = μ + Γη X + \varepsilon, Σ=ΓΩΓ' + Γ_{0}Ω_{0}Γ'_{0}
using the maximum likelihood estimation. When the dimension of the envelope is between 1 and p1-1, the starting value and blockwise coordinate descent algorithm in Cook et al. (2016) is implemented. When the dimension is p1, then the envelope model degenerates to the standard multivariate linear regression. When the dimension is 0, it means that X and Y are uncorrelated, and the fitting is different.
### Value
The output is a list that contains the following components:
muY The estimator of mean of Y. mu1 The estimator of mean of X1. mu2 The estimator of mean of X2. beta1 A p1 by r matrix for the estimator of regression coefficients for continuous predictors. beta2 A p2 by r matrix for the estimator of regression coefficients for categorical predictors. Gamma An p1 by d matrix for the orthogonal basis of the Envelope-based Partial Partial Least Squares. Gamma0 An p1 by (p1-d) matrix for the orthogonal basis of the complement of the Envelope-based Partial Partial Least Squares. gamma A p2 by p1 matrix for the estimator of regression coefficients based on the regression of X1 on X2. eta A d by p1 matrix for the coordinates of beta1 with respect to Gamma. Omega A d by d matrix for the coordinates of SigmaX1 with respect to Gamma. Omega0 A (p1-d) by (p1-d) matrix for the coordinates of SigmaX1 with respect to Gamma0. SigmaX1 The estimator of error covariance matrix Sigma[1|2]. SigmaYcX The estimator of error covariance matrix Sigma[Y|X]. loglik The maximized log likelihood function. n The number of observations in the data. covMatrix1 The asymptotic covariance of vec(beta1). The covariance matrix returned are asymptotic. For the actual standard errors, multiply by 1 / n. covMatrix2 The asymptotic covariance of vec(beta2). The covariance matrix returned are asymptotic. For the actual standard errors, multiply by 1 / n. asySE1 The asymptotic standard error matrix for elements in beta1. The multiplication by the reciprocal of square root of n returns actual standard errors. asySE2 The asymptotic standard error matrix for elements in beta2. The multiplication by the reciprocal of square root of n returns actual standard errors.
### References
Park, Y., Su, Z. and Chung, D. (2022+) Envelope-based Partial Partial Least Squares with Application to Cytokine-based Biomarker Analysis for COVID-19.
### Examples
data(amitriptyline)
Y <- amitriptyline[ , 1:2]
X1 <- amitriptyline[ , 4:7]
X2 <- amitriptyline[ , 3]
u <- u.eppls(X1, X2, Y)
u
m <- eppls(X1, X2, Y, 2)
m
Renvlp documentation built on Jan. 8, 2023, 1:08 a.m. | 978 | 3,969 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-14 | longest | en | 0.803108 |
https://www.physicsforums.com/threads/schaums-outline-of-quantum-mechanics-8-11.646084/ | 1,508,276,285,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822488.34/warc/CC-MAIN-20171017200905-20171017220905-00666.warc.gz | 987,118,076 | 15,108 | # Schaum's Outline of Quantum Mechanics 8.11
1. Oct 22, 2012
### Jimmy Snyder
1. The problem statement, all variables and given/known data
If a hydrogen atom is in the state n = 2, l = 0, what is the probability that an electron has a value of r that is smaller than the Bohr radius.
2. Relevant equations
eqn (8.8)
$$\psi_{200}(r, \theta, \phi) = R_{20}Y_0^0(\theta, \phi)$$
Page 305
$$Y_0^0(\theta, \phi) = \frac{1}{\sqrt{4\pi}}$$
eqn (8.33)
$$R_{20} = 2(2a_0)^{-3/2}(1 - \frac{r}{2a_0})e^{-r/2a_0}$$
3. The attempt at a solution
$$P = \int_0^{2\pi}d\phi \int_0^{\pi}d\theta\int_0^{a_0}dr r^2|\psi(r, \theta, \phi)|^2sin(\theta)$$
$$= \int_0^{2\pi}d\phi \int_0^{\pi}d\theta sin(\theta) (Y_0^0(\theta, \phi))^2 \int_0^{a_0}dr r^2 4(2a_0)^{-3}(1 - \frac{r}{2a_0})^2e^{-r/a_0}$$
Now let $z = r/a_0$; $dr = a_0dz$ then
$$P = \frac{4\pi}{4\pi}\frac{4}{(2a_0)^3}\int_0^1 a_0dz a_0^2z^2(1 - \frac{z}{2})^2e^{-z}$$
$$= \frac{1}{2}\int_0^1 dz (z^2 - z^3 + \frac{z^4}{4})e^{-z}$$
$$= (-1 -z -\frac{z^2}{2} - \frac{z^4}{8})e^{-z}|_0^1$$
$$= 1 - \frac{21}{8e} = 0.0343$$
The book gives the answer .176
I calculated that the probability of r being less than 4 times the Bohr radius is .176
Placing n = 2, l = 0 in equation 8.42 gives that the mean value for r is 6 times the Bohr radius. I find it hard to believe that the electron could spend nearly 18% of the time at less than 1 Bohr radius and yet average out to 6 Bohr radii.
Last edited: Oct 22, 2012
2. Oct 22, 2012
### dextercioby
It's + z^4/4 inside the bracket under the integral sign. And are you sure about the integrations ?
3. Oct 22, 2012
### Jimmy Snyder
Thank you. I fixed that. I am as sure about them as I can be, but if I were 100% sure, I wouldn't have posted. By the way, I also solved the integral numerically using a calculator and got the same 0.0343 result. I believe that if I am wrong, it is because I am using the wrong approach, not because of an error in calculation. | 788 | 1,953 | {"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} | 3.3125 | 3 | CC-MAIN-2017-43 | longest | en | 0.837977 |
https://web2.0calc.com/questions/let-be-an-angle-in-quadrant-ii-such-that-cos-4-9 | 1,603,502,819,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881551.11/warc/CC-MAIN-20201023234043-20201024024043-00545.warc.gz | 600,062,944 | 6,399 | +0
# let θ be an angle in quadrant ii such that =cosθ−4/9. find the exact values of CSCθ and COTθ
0
36
1
let θ be an angle in quadrant ii such that =cosθ−4/9. find the exact values of CSCθ and COTθ
CSCθ =
COTθ =
Oct 8, 2020
#1
+27778
+1
Remember cos^2 + sin^2 = 1
so sin = +- sqrt (65) / 9 but since it is in Q II it is + sqrt(65)/ 9
then:
CSC = 1/sin
Cot = cos/ sin
Oct 8, 2020 | 164 | 402 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | longest | en | 0.776067 |
https://blog.plover.com/math/se/2023-10.html | 1,720,849,830,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514490.70/warc/CC-MAIN-20240713051758-20240713081758-00194.warc.gz | 117,066,659 | 9,563 | # The Universe of Discourse
Sat, 02 Dec 2023
Content warning: grumpy complaining. This was a frustrating month.
### Need an intuitive example for how "P is necessary for Q" means "Q⇒P"?
This kind of thing comes up pretty often. Why are there so many ways that the logical expression !!Q\implies P!! can appear in natural language?
• If !!Q!!, then !!P!!
• !!Q!! implies !!P!!
• !!P!! if !!Q!!
• !!Q!! is sufficient for !!P!!
• !!P!! is necessary for !!Q!!
Strange, isn't it? !!Q\land P!! is much simpler: “Both !!Q!! and !!P!! are true” is pretty much it.
Anyway this person wanted an intuitive example of “!!P!! is necessary for !!Q!!”
I suggested:
Suppose that it is necessary to have a ticket (!!P!!) in order to board a certain train (!!Q!!). That is, if you board the train (!!Q!!), then you have a ticket (!!P!!).
Again this follows the principle that rule enforcement is a good thing when you are looking for intuitive examples. Keeping ticketless people off the train is something that the primate brain is wired up to do well.
My first draft had “board a train” in place of “board a certain train”. One commenter complained:
many people travel on trains without a ticket, worldwide
I was (and am) quite disgusted by this pettifogging.
I said “Suppose that…”. I was not claiming that the condition applies to every train in all of history.
### Does ...999.999... = 0?
This person is asking one of those questions that often puts Math StackExchange into the mode of insisting that the idea is completely nonsensical, when it is actually very close to perfectly mundane mathematics. (Previously: [1] [2] [3] ) That didn't happen this time, which I found very gratifying.
Normally, decimal numerals have a finite integer part on the left of the decimal point, and an infinite fractional part on the right of the decimal point, as with (for example) !!\frac{13}{3} = 4.333\ldots!!. It turns out to work surprisingly well to reverse this, allowing an infinite integer part on the left and a finite fractional part on the right, for example !!\frac25 = \ldots 333.4!!. For technical reasons we usually do this in base !!p!! where !!p!! is prime; it doesn't work as well in base !!10!!. But it works well enough to use: If we have the base-10 numeral !!\ldots 9999.0!! and we add !!1!!, using the ordinary elementary-school right-to-left addition algorithm, the carry in the units place goes to the tens place as usual, then the next carry goes to the hundreds place and so on to infinity, leaving us with !!\ldots 0000.0!!, so that !!\ldots 9999.0!! can be considered a representation of the number !!-1!!, and that means we don't need negation signs.
In fact this system is fundamental to the way numbers are represented in computer arithmetic. Inside the computer the integer !!-1!! is literally represented as the base-2 numeral !!11111111\;11111111\;11111111\;11111111!!, and when we add !!1!! to it the carry bit wanders off toward infinity on the left. (In the computer the numeral is finite, so we simulate infinity by just discarding the carry bit when it gets too far away.)
Once you've seen this a very reasonable next question is whether you can have numbers that have an infinite sequence of digits on both sides. I think something goes wrong here — for one thing it is no longer clear how to actually do arithmetic. For the infinite-to-the-left numerals arithmetic is straightforward (elementary-school algorithms go right-to-left anyway) and for the standard infinite-to-the-right numerals we can sort of fudge it. (Try multiplying the infinite decimal for !!\sqrt 2!! by itself and see what trouble you get into. Or simpler: What's !!4.666\ldots \times 3!!?)
OP's actual question was: If !!\ldots 9999.0 !! can be considered to represent !!-1!!, and if !!0.9999\ldots!! can be considered to represent !!1!!, can we add them and conclude that !!\ldots 9999.9999\ldots = 0!!?
This very deserving question got a good answer from someone who was not me. This was a relief, because my shameful answer was pure shitpostery. It should have been heavily downvoted, but wasn't. The gods of Math SE karma are capricious.
### Why define addition with successor?
Ugh, so annoying. OP had read (Bertrand Russell's explanation of) the Peano definition of addition, and did not understand it. Several people tried hard to explain, but communication was not happening. Or, perhaps, OP was more interested in having an argument than in arriving at an understanding. I lost a bit of my temper when they claimed:
Russell's so-called definition of addition (as quoted in my question) is nothing but a tautology: ….
I didn't say:
If you think Bertrand Russell is stupid, it's because you're stupid.
although I wanted to at first. The reply I did make is still not as measured as I would like, and although it leaves this point implicit, the point is still there. I did at least shut up after that. I had answered OP's question as well as I was able, and carrying on a complex discussion in the comments is almost never of value.
### Why is Ramanujan considered a great mathematician?
This was easily my best answer of the month, but the question was deleted, so you will only be able to see it if you have enough Math SE reputation.
OP asked a perfectly reasonable question: Ramanujan gets a lot of media hype for stuff like this:
$${\sqrt {\phi +2}}-\phi ={\cfrac {e^{{-2\pi /5}}}{1+{\cfrac {e^{{-2\pi }}}{1+{\cfrac {e^{{-4\pi }}}{1+{\cfrac {e^{{-6\pi }}}{1+\,\cdots }}}}}}}}$$
which is not of any obvious use, so “why is it given such high regard?”
OP appeared to be impugning a famous mathematician, and Math SE always responds badly to that; their heroes must not be questioned. And even worse, OP mentioned the notorious non-fact that $$1+2+3+\ldots =-\frac1{12}$$ which drives Math SE people into a frothing rage.
One commenter argued:
Mathematics is not inherently about its "usefulness". Even if you can't find practical use for those formulas, you still have to admit that they are by no means trivial
I think this is fatuous. OP is right here, and the commenter is wrong. Mathematicians are not considered great because they produce wacky and impractical equations. They are considered great because they solve problems, invent techniques that answer previously impossible questions, and because they contribute insights into deep and complex issues.
Most of the mathematical results are useless. Mathematics is more like an art.
Bullshit. Mathematics is about trying to understand stuff, not about taping a banana to the wall. I replied:
I don't think “mathematics is not inherently about its usefulness" is an apt answer here. Sometimes mathematical results have application to physics or engineering. But for many mathematical results the application is to other parts of mathematics, and mathematicians do judge the ‘usefulness’ of results on this basis. Consider for example Mochizuki's field of “inter-universal Teichmüller theory”. This was considered interesting only as long as it appeared that it might provide a way to prove the !!abc!! conjecture. When that hope collapsed, everyone lost interest in it.
My answer to OP elaborated on this point:
The point of these formulas wasn't that they were useful in themselves. It's that in order to find them he had to have a deep understanding of matters that were previously unknown. His contribution was the deep understanding.
I then discussed Hardy's book on the work he did with Ramanujan and Hardy's own estimation of Ramanujan's work:
The first chapter is somewhat negative, as it summarizes the parts of Ramanujan's work that he felt didn't have lasting value — because Hardy's next eleven chapters are about the work that he felt did have value.
So if OP wanted a substantive and detailed answer to their question, that would be the first place to look.
I also did an arXiv search for “Ramanujan” and found many recent references, including one with “applications to the Ramanujan !!τ!!-function”, and concluded:
The !!\tau!!-function is the subject of the entire chapter 10 of Hardy's book and appears to still be of interest as recently as last Monday.
The question was closed as “opinion-based” (a criticism that I think my answer completely demolishes) and then it was deleted. Now if someone else trying to find out why Ramanujan is held in high regard they will not be able to find my factual, substantive answer.
Screw you, Math SE. This month we both sucked. | 2,012 | 8,483 | {"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} | 3.203125 | 3 | CC-MAIN-2024-30 | latest | en | 0.941178 |
https://www.glassdoor.com.au/Interview/Five-guys-all-of-different-ages-enter-a-bar-and-take-a-seat-at-a-round-table-What-is-the-probability-that-they-are-seate-QTN_140665.htm | 1,516,295,921,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887535.40/warc/CC-MAIN-20180118171050-20180118191050-00265.warc.gz | 911,746,136 | 56,863 | Susquehanna International Group Interview Question: Five guys, all of different a... | Glassdoor.com.au
## Interview Question
Assistant Trader Intern Interview(Student Candidate) Bala Cynwyd, PA (US)
# Five guys, all of different ages enter a bar and take a
seat at a round table. What is the probability that they are seated in ascending order of age? (clockwise or counter-clockwise)
1
5! or 120 total seating positions. 5 possible starting positions because the table is round. 2 possible directions.
This makes 10 arrangements that fulfill the prompt, ie P = 10/5! = 1/12
Interview Candidate on 18/02/2011
2
Make the youngest person a fixed point. The second youngest can sit on 2 seats in 4 remaining seats (left or right of the youngest). The third youngest can sit on 1 seat in the 3 remaining because we defined ordered by previous step. Fourth youngest can sit on 1/2. Last person 1/1
2/4*1/3*1/2=1/12
Anonymous on 09/11/2011 | 247 | 945 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2018-05 | latest | en | 0.870322 |
http://stackoverflow.com/questions/911637/what-is-cyclomatic-complexity/911737 | 1,440,899,549,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644064865.49/warc/CC-MAIN-20150827025424-00041-ip-10-171-96-226.ec2.internal.warc.gz | 219,190,930 | 24,859 | What is Cyclomatic Complexity?
A term that I see every now and then is "Cyclomatic Complexity". Here on SO I saw some Questions about "how to calculate the CC of Language X" or "How do I do Y with the minimum amount of CC", but I'm not sure I really understand what it is.
On the NDepend Website, I saw an explanation that basically says "The number of decisions in a method. Each if, for, && etc. adds +1 to the CC "score"). Is that really it? If yes, why is this bad? I can see that one might want to keep the number of if-statements fairly low to keep the code easy to understand, but is this really everything to it?
Or is there some deeper concept to it?
-
I'm not aware of a deeper concept. I believe it's generally considered in the context of a maintainability index. The more branches there are within a particular method, the more difficult it is to maintain a mental model of that method's operation (generally).
Methods with higher cyclomatic complexity are also more difficult to obtain full code coverage on in unit tests. (Thanks Mark W!)
That brings all the other aspects of maintainability in, of course. Likelihood of errors/regressions/so forth. The core concept is pretty straight-forward, though.
-
Also, the harder it is to unit test and gain complete code coverage. – Marc W May 26 '09 at 16:46
Right 'cause they say you can only keep a handful of things in your conscience mind at any given moment. – steamer25 May 26 '09 at 16:49
The cyclomatic complexity of a method also denotes the number of unit test cases required to achieve the 100% code coverage for that method. – Anand Patel Mar 16 '11 at 9:59
Cyclomatic complexity measures the number of times you must execute a block of code with varying parameters in order to execute every path through that block. A higher count is bad because it increases the chances for logical errors escaping your testing strategy.
-
Wikipedia may be your friend on this one: Definition of cyclomatic complexity
Basically, you have to imagine your program as a graph and then
The complexity is (...) defined as:
``````M = E − N + 2P
``````
where
• M = cyclomatic complexity,
• E = the number of edges of the graph
• N = the number of nodes of the graph
• P = the number of connected components
CC is a concept that attempts to capture how complex your program is and how hard it is to test it in a single integer number.
-
Yep, that's really it. The more execution paths your code can take, the more things that must be tested, and the higher probability of error.
-
That's it, the idea is that a method which has a low CC has less forks, looping etc which all make a method more complex. Imagine reviewing 500,000 lines of code, with an analyzer and seeing a couple methods which have oder of magnitude higher CC. This lets you then focus on refactoring those methods for better understanding (It's also common that a high CC has a high bug rate)
-
Cyclomatic Complexity really is just a scary buzzword. In fact it's a measure of code complexity used in software development to point out more complex parts of code (more likely to be buggy, and therefore has to be very carefully and thoroughly tested). You can calculate it using the E-N+2P formula, but I would suggest you have this calculated automatically by a plugin. I have heard of a rule of thumb that you should strive to keep the CC below 5 to maintain good readability and maintainability of your code.
I have just recently experimented with the Eclipse Metrics Plugin on my Java projects, and it has a really nice and concise Help file which will of course integrate with your regular Eclipse help and you can read some more definitions of various complexity measures and tips and tricks on improving your code.
-
Another interesting point I've heard:
The places in your code with the biggest indents should have the highest CC. These are generally the most important areas to ensure testing coverage because it's expected that they'll be harder to read/maintain. As other answers note, these are also the more difficult regions of code to ensure coverage.
-
Consider the control flow graph of your function, with an additional edge running from the exit to the entrance. The cyclomatic complexity is the maximum number of cuts we can make without separating the graph into two pieces.
For example:
``````function F:
if condition1:
...
else:
...
if condition2:
...
else:
...
``````
Control Flow Graph
You can probably intuitively see why the linked graph has a cyclomatic complexity of 3.
-
Each decision point in a routine (loop, switch, if, etc...) essentially boils down to an if statement equivalent. For each `if` you have 2 codepaths that can be taken. So with the 1st branch there's 2 code paths, with the second there are 4 possible paths, with the 3rd there are 8 and so on. There are at least 2**N code paths where N is the number of branches.
This makes it difficult to understand the behavior of code and to test it when N grows beyond some small number.
-
Cyclomatric complexity is basically a metric to figure out areas of code that needs more attension for the maintainability. It would be basically an input to the refactoring. It definitely gives an indication of code improvement area in terms of avoiding deep nested loop, conditions etc.
-
That's sort of it. However, each branch of a "case" or "switch" statement tends to count as 1. In effect, this means CC hates case statements, and any code that requires them (command processors, state machines, etc).
-
A case statement may be the sign of an object graph screaming to be set free.... – Tetsujin no Oni Apr 12 '13 at 18:52
@TetsujinnoOni - May yes. The problem is my typical command processor is reading commands from some external source, so I can't just let the compiler put data items wherever it wants like it must for a dynamic polymorphic class. – T.E.D. Apr 13 '13 at 3:21
The answers provided so far do not mention the correlation of software quality to cyclomatic complexity. Research has shown that having a lower cyclomatic complexity metric should help develop software that is of higher quality. It can help with software quality attributes of readability, maintainability, and portability. In general one should attempt to obtain a cyclomatic complexity metric of between 5-10.
One of the reasons for using metrics like cyclomatic complexity is that in general a human being can only keep track of about 7 (plus or minus 2) pieces of information simultaneously in your brain. Therefore, if your software is overly complex with multiple decision paths, it is unlikely that you will be able to visualize how your software will behave (i.e. it will have a high cyclomatic complexity metric). This would most likely lead to developing erroneous or bug ridden software. More information about this can be found here and also on Wikipedia.
-
``````Cyclocmatic complexity = Number of decision points + 1
``````
The decision points may be your conditional statements like if, if … else, switch , for loop, while loop etc.
The following chart describes the type of the application.
• Cyclomatic Complexity lies 1 – 10 To be considered Normal applicatinon
• Cyclomatic Complexity lies 11 – 20 Moderate application
• Cyclomatic Complexity lies 21 – 50 Risky application
• Cyclomatic Complexity lies more than 50 Unstable application
- | 1,645 | 7,402 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-35 | latest | en | 0.946473 |
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• To: mathgroup at smc.vnet.net
• Subject: [mg5307] Rule to Function, Frontend funnies
• From: Wouter Meeussen <meeussen.vdmcc at vandemoortele.be>
• Date: Sat, 23 Nov 1996 01:44:44 -0500
• Sender: owner-wri-mathgroup at wolfram.com
```how can I convert a rule to a function?
if I get a result from
rulesoln = DSolve[{y'[x]==y[x]/(c+1/x),y[0]==1},y[x],x]
in the form :
2
x/c - Log[1 + c x]/c
{{y[x] -> E }}
how do I get the definition :
y[x_]:= E^(x/c - Log[1 + c x]/c^2)
(*************************************************************************)
Funny : in any text editor, like notepad , I just cut & paste the pieces I
need.
In the Frontend, I can force Mma to produce a one-dimensional output line by
% // InputForm
{{y[x] -> E^(x/c - Log[1 + c*x]/c^2)}}
but when I try to edit it, or make an input-line out of that, either by
Action-Prepare.Input-Copy.Output.from.Above (Control-Shift-L)
No Input Form was generated for this Cell
(*************************************************************************)
Even Funnier : doing Control-Shift-L and Shift-Enter produces:
General::spell: Possible spelling error: new symbol name "Form"
is similar to existing symbols {CForm, For}.
General::spell1: Possible spelling error: new symbol name "for"
is similar to existing symbol "For".
Cell for Form generated Input No this was
(*************************************************************************)
Anybody out there to whom this happens too?
After some {soul,FullForm}-searching, I came up with this rule-eating-rule :
Flatten[rulesoln]/.y[x]->y[x_]/.Rule->SetDelayed
{Null}
you don't get to see much except "{Null}", but it results in:
?y
Global`y
y[x_] := E^(x/c - Log[1 + c*x]/c^2)
(*************************************************************************)
Is there a fast & easy Cut-and-Paste way to do it, not leaving the FrontEnd?
Is this the same on a Mac-machine?
Is this the same in Mma 3.0 for Windows?
Wouter.
NV Vandemoortele Coordination Center
Group R&D Center
Prins Albertlaan 79
Postbus 40
B-8870 Izegem
Tel: +/32/51/33 21 11
Fax:+32/51/33 21 75
vdmcc at vandemoortele.be
```
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Writing Algebraic Equations Worksheet – Expressions and Equations Worksheets are created to assist children in learning faster and more effectively. These worksheets are interactive as well as problems dependent on the sequence that operations are carried out. These worksheets will help children are able to grasp basic and advanced concepts in a very short amount of period of time. You can download these worksheets in PDF format to aid your child’s learning and practice math-related equations. These resources are beneficial for students who are in the 5th through 8th grades.
## Free Download Writing Algebraic Equations Worksheet
Some of these worksheets are designed for students in the 5th-8th grade. These two-step word problems are created using fractions and decimals. Each worksheet contains ten problems. The worksheets are available both online and in printed. These worksheets are a fantastic way to get your students to practice rearranging equations. These worksheets are a great way to practice rearranging equations , and assist students with understanding equality and inverse operation.
These worksheets are targeted at students in the fifth and eighth grades. These worksheets are ideal for those who are struggling to compute percentages. There are three types of questions you can choose from. You can decide to tackle one-step challenges that contain decimal or whole numbers or employ word-based techniques to solve problems involving decimals and fractions. Each page will have ten equations. These Equations Worksheets are recommended for students from 5th to 8th grades.
These worksheets can be a wonderful resource for practicing fraction calculations and other aspects of algebra. You can pick from different kinds of problems that you can solve with these worksheets. You can pick one that is numerical, word-based or a combination of both. It is vital to pick the type of problem, as each one will be different. There are ten issues on each page, so they’re fantastic resources for students from 5th to 8th grade.
The worksheets will teach students about the relationships between variables and numbers. The worksheets let students test their skills at solving polynomial equations, and to learn how to apply equations to solve problems in everyday life. If you’re looking for an educational tool that will help you master the art of expressions and equations, you can start by looking through these worksheets. These worksheets will educate you about the various types of mathematical problems as well as the many symbols used to express them.
These worksheets could be beneficial for students in their beginning grades. These worksheets will help them develop the ability to graph and solve equations. They are great for practice with polynomial equations. They can also help you master the art of factoring and simplify them. It is possible to find a wonderful set of equations and expressions worksheets for kids at any grade. The best way to learn about the concept of equations is to perform the work yourself.
There are many worksheets to help you understand quadratic equations. Each level has its own worksheet. The worksheets were designed to allow you to practice solving problems of the fourth level. Once you have completed an amount of work it is possible to work on other kinds of equations. After that, you can focus on the same level of problems. For instance, you could discover a problem that uses the same axis and an elongated number. | 650 | 3,537 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2022-49 | latest | en | 0.948465 |
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This was originally a comment on TonyK's answer, but it was too large to submit as a comment so I'm putting it here. I thought the numbers might have value to some.
Let's say that it takes $b^{\lg 6}/10^{10}$ seconds to test a $b$-bit number with the M-R test and $b^{\lg 24}/10^{13}$ seconds to prove primality with ECPP. Testing a range up to $n_2=2^b$ with M-R + ECPP would take about $$\left(n_2-n_1\right)\left(\frac{b^{\lg 6}}{10^{10}}+\frac{b^{\lg 24}}{10^{13}\ln n_2}\right)=\left(n_2-n_1\right)\left(\frac{b^{\lg 6}}{10^{10}}+\frac{b^{\lg 12}}{10^{13}\ln2}\right)$$ seconds. For $n_2-n_1=2^{20}$ and $n_2$ large, this is about $1.5(\lg n_2)^{\lg12}/10^7$ seconds.
On the other hand, suppose sieving $n_1$ to $n_2$ takes $\sqrt{n_2}/10^6+(n_2-n_1)/10^{8.5}$ seconds. For $n_2-n_1=2^{20}$ and $n_2$ large, this is about $\sqrt{n_2}$ seconds.
Equating the two suggests that, for intervals about a million wide, testing each member is superior to sieving beyond about $n_2>8\cdot10^8$. Only about 12 kB of memory are needed to store all the primes up to the square root of that limit, so the fourth power trick doesn't seem viable here at all -- by the time you run out of primary memory you shouldn't be sieving at all. | 466 | 1,382 | {"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} | 3.75 | 4 | CC-MAIN-2013-20 | latest | en | 0.859391 |
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Are you trying to understand the general conversion process between ounces and milliliters? If so, it is important to be clear on how many ounces is 30 ml. Knowing this can help when dealing with measurements for cups, teaspoons, tablespoons, and other common units of measurement. Understanding the difference between these two measurements can help make cooking accurate and efficient. In this blog post we will discuss the amount of ounces that equate to 30 ml as well as some tips and tricks for understanding conversions going forward.
What Are Ounces?
The ounce, a widely recognized unit of measurement, is utilized to gauge weight. It is specifically defined as equivalent to 28.35 grams or 0.0625 pounds. This versatile unit finds its application in various domains, including the measurement of food, ingredients, and liquids. Whether you’re in the culinary world, scientific experiments, or everyday life, understanding the ounce and its role in accurate measurements is truly valuable.
What Are Milliliters?
Milliliters (mL) are a commonly used unit of measurement to quantify volume. Specifically, one milliliter is equivalent to 1 cubic centimeter (cm^3) or 0.001 liters. When it comes to measuring liquids, such as water, milk, juices, and other beverages, milliliters come in handy. They provide a precise and practical way to gauge the amount of liquid present. Whether you’re pouring a refreshing glass of water or carefully measuring ingredients for a recipe, milliliters serve as a reliable and convenient unit of measurement.
How Many Ounces Is 30 mL?
When looking to determine how many ounces is 30 ml, it is important to remember that one ounce is equivalent to 28.35 grams or 0.0625 pounds and one milliliter is equivalent to 1 cm^3 and 0.001 liters. With this in mind, when converting from milliliters (mL) to ounces, the formula used is 30 mL x 0.035274 = 1.0582 ounces. This means that there are 1.0582 ounces in 30 mL of a liquid or other substance.
Why Should You Know How Many Ounces Is 30 ml?
Knowing how many ounces is 30 ml is essential for accurately measuring ingredients or liquids. With this knowledge, you can more easily measure out the exact amount of an ingredient or liquid needed for a recipe. It can also come in handy when determining product volume when buying certain items such as creams and lotions. Additionally, understanding the conversion between milliliters and ounces can help you make precise measurements for experiments and scientific projects.
How To Convert Ounces To Milliliters?
The formula used to convert ounces to milliliters is 1 ounce x 29.5735 = mL. This means that one ounce of a substance or liquid is equal to 29.5735 mL. For example, if you want to know how many milliliters are in 2 ounces, the formula would be 2 ounces x 29.5735 = 59.1470 mL.
How To Convert Milliliters To Ounces?
The formula used to convert milliliters to ounces is 1 mL x 0.035274 = ounces. This means that one milliliter of any substance or liquid is equal to 0.035274 ounces. For example, if you want to know how many ounces are in 40 milliliters, the formula would be 40 mL x 0.035274 = 1.4109 ounces.
Tips To Convert Ounces To Milliliters
If you’re looking to increase your understanding of conversions between milliliters and ounces, here are some tips that can come in handy:
• Familiarize yourself with the equivalence between one ounce and 28.35 grams or 0.0625 pounds as well as one milliliter to 1 cm^3 and 0.001 liters.
• Use the formula 1 ounce x 29.5735 = mL when converting from ounces to milliliters.
• Visualize your measurements by using a measuring cup or other similar device that allows you to measure in both ounces and milliliters.
• Practice makes perfect. Try testing yourself on conversions and take note of the ones you struggle with most.
Tips For Accurately Measuring Liquid Ingredients
When it comes to learning how many ounces is 30 ml and measuring liquid ingredients for recipes or experiments, accuracy is essential. Here are some tips for ensuring precision when measuring liquids:
• Use an appropriate measuring device, such as a graduated cylinder, eyedropper, or beaker.
• Line your eyes up with the desired measurement line before pouring in the liquid to get as close to the exact desired measure as possible.
• Understand the measurements of your measuring device before using it and be aware of potential spills or overflow.
What Are Common Uses For Ounces To Milliliters In Cooking?
Understanding the ounces to milliliters conversion can be extremely helpful in the kitchen. Recipes often require precise measurements of ingredients and liquids, and knowing how many milliliters are in an ounce can help make sure your recipes turn out successfully. Additionally, when substituting between liquid ingredients (such as milk for water or vice versa), understanding the proper amount is crucial for a successful dish.
What Are Examples Of Converting Ounces To Milliliters?
It is helpful to have a few examples of conversions from ounces to milliliters handy. Here are some of the most common:
• 1 ounce = 29.5735 mL
• 2 ounces = 59.1470 mL
• 3 ounces = 88.7205 mL
• 4 ounces = 118.2940 mL
• 5 ounces = 147.8675 mL
• 6 ounces = 177.4410 mL
• 7 ounces = 207.0145 mL
• 8 ounces = 236.5880 mL
• 9 ounces = 266.1615 mL
• 10 ounces = 295.7350 mL
Conclusion: How Many Ounces Is 30 ml?
In conclusion, it is important to understand the conversion from ounces to milliliters. Knowing how many ounces is 30 ml can help make recipes and experiments more accurate. To answer our initial question, there are 1.0582 ounces in 30 mL of a liquid or other substance. Additionally, understanding the formula used for conversions as well as tips and tricks for accurately measuring liquids can help make your cooking adventures as precise and enjoyable as possible.
FAQs: 30 ml
How much is 30ml liquid?
30 ml is equivalent to a few tablespoons or slightly more than 1/8 cup, which is about an ounce. In terms of teaspoons, this amounts to six, which is the same as one tablespoon three times.
Is 30 ml less than 3 oz?
To convert 30 ml of liquid volume to ounces, you can use a simple formula. Just divide the amount by 28.41, and you’ll have an accurate measurement in ounces. For 30 ml, the conversion comes out to be approximately 1.06 oz (or 1.1 oz when rounded up).
Is 30 ml less than 1 oz?
Indeed, a volume of 30 mL is slightly more than 1/8 cup or just under 1 ounce. If you prefer to measure in teaspoons, it’s worth noting that 1 tablespoon is equivalent to 3 teaspoons. So, in terms of teaspoons, 30 mL can be expressed as 6 teaspoons.
How big is a 30ml bottle?
A 30ml perfume bottle is usually around 3 inches in height and 1 inch in width. It can hold approximately one fluid ounce of perfume. This compact size makes it conveniently portable in purses or bags, and it is also a popular choice for travel-sized perfumes.
Is 8 oz 30 ml?
To convert 8 ounces to milliliters, the answer is 236.59 mL, which can be rounded up to 237 mL for convenience. The conversion from ounces to milliliters is straightforward – simply multiply the number of ounces by 30 (1 oz = 30 ml). This conversion makes it easier to work with fluid measurements and ensures accuracy in various contexts.
How many ounces is 30 ml of alcohol?
Based on our convenient conversion chart, 30 ml is equivalent to 1.01 fl oz in the US Customary system and 1.06 fl oz in the UK Imperial system.
How do you measure 30ml in cups?
To convert 30 ml to cups, you can conveniently use 1/8 cup as a substitute. This allows you to utilize your existing measuring cup for the recipe at hand. To convert ml to cups, simply divide the number of milliliters by 237. Therefore, dividing 30 ml by 237 will approximately give you 1/8 cup.
How long does 30ml liquid last?
A 30ml bottle should last anywhere from ten days to two or three weeks, depending on the frequency and intensity of your e-cigarette usage.
What is the size of 30 ml dropper bottle?
Dimensions: Height – 4.1″ (10.4 cm), Diameter – 1.25″ (3.18 cm), Mouth Opening – 0.4″ (1 cm). It has a capacity of 1 fl oz. or 30 ml for liquids.
How many ounces is 30 ml nursing?
To convert ounces (oz) to milliliters (mL), use the conversion factor: 1 fluid ounce (fl. oz) is equal to 30 mL.
Is 30ml a standard drink?
A standard drink always contains 10 grams of alcohol, regardless of the container size (glass, bottle, can) or the type of alcohol (beer, wine, spirit). For instance, 30 ml of spirits is equivalent to one standard drink.
Is 30ml alcohol safe?
Regular consumption of a modest amount of alcohol (1 ounce [30 ml] of absolute alcohol, equivalent to two standard drinks per day) over an extended period has not been definitively linked to any pathological effects, with the exception of a slightly elevated risk for certain types of cancers.
What size spoon is 30ml?
Two tablespoons are equivalent to 30 ml, which can be a helpful conversion to remember. Breaking down the calculation into simpler parts can aid in understanding. Specifically, one tablespoon is equal to 15 ml, so when you have two tablespoons, it amounts to 30 ml. By grasping this conversion, you’ll have a better grasp of how 30 ml relates to two tablespoons.
What does 30ml look like?
Curious about how much 30 ml is in a cup? Easily convert fluid ounces to tablespoons and teaspoons. 30 ml is equal to exactly 2 tablespoons or 6 teaspoons. That’s approximately 1.2 oz of liquid when measured in a standard measuring cup. | 2,280 | 9,596 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2023-50 | latest | en | 0.934358 |
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mer än 2 år ago | 1 answer | 0
### 1
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Row Range in writetable MATLAB
I want to write the table from A1 column but the table should start from the second row of the excel. TT is my table, and I want...
mer än 2 år ago | 0 answers | 0
### 0
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How to add the rows if it is same.
I am trying to add the rows that have same game column. For example, for same rows like game value =0, the output should be 0....
mer än 2 år ago | 1 answer | 0 | 1,943 | 6,303 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.586499 |
http://www.kwiznet.com/p/takeQuiz.php?ChapterID=1368&CurriculumID=3&Num=4.2 | 1,586,526,486,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371896913.98/warc/CC-MAIN-20200410110538-20200410141038-00158.warc.gz | 258,081,883 | 3,850 | Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs!
#### Online Quiz (WorksheetABCD)
Questions Per Quiz = 2 4 6 8 10
### Grade 3 - Mathematics4.2 Equations and Inequations
Method: Read the equations carefully. Equations are mathematical statements that use " = " sign. Inequations are mathematical statements that use " > " or " <" sign. Example: 534 < 675 is an example of equation. State if true or false. Equations are statements that use " = " sign. Answer: False Directions: Answer the following. Also write at least ten examples of your own.
Q 1: A number statement that uses the signs > or < is called an inequality.TrueFalse Q 2: 456 > 401 is an equation. True or FalseFalseTrue Q 3: Equations are number statements that uses an equal sign. Ex: 5+4=9 and 221=221)FalseTrue Q 4: 1,002 > 997 and 4,825 < 1,002 are both inequalities.FalseTrue Q 5: 67 + 12 = 79 is an equation.FalseTrue Q 6: 67 + 12 is an equation.TrueFalse Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only! | 299 | 1,100 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.891544 |
https://blog.csdn.net/timsei/article/details/79338351 | 1,591,422,238,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348509972.80/warc/CC-MAIN-20200606031557-20200606061557-00293.warc.gz | 262,264,799 | 25,347 | # UOJ275组合数问题
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int mod = 1e9 + 7;
const int N = 222;
const int CMAX = 1e5 + 5;
int t , m , n , dp[222][2][2] , A[N] , B[N] , k , ln , lm;
int T[100005];
int Div(int x , int k) {
int now = 0;
while(x % k == 0) x /= k , ++ now;
return now;
}
void init(void) {
T[0] = 0;
for(int i = 1;i < 100005;++ i) {
T[i] = T[i - 1] + Div(i , k);
}
}
int convert(int x , int *A) {
int len = 0;
while(x) {
A[++ len] = x % k;
x /= k;
}
return len;
}
int dfs(int pos , int l1 , int l2) {
if(!pos) return 1;
if(dp[pos][l1][l2] != -1) return dp[pos][l1][l2];
int res = 0 , up = (l1) ? A[pos] : (k - 1);
for(int i = 0;i <= up;++ i) {
if(!l2 || (l2 && B[pos] > i)) {
res = (res + (i + 1) % mod * dfs(pos - 1 , (l1 && (i == up)) , 0) % mod) % mod;
continue;
}
int up2 = B[pos];
res = (res + (B[pos]) * dfs(pos - 1 , (l1 && (i == up)) , 0) % mod) % mod;
res = (res + dfs(pos - 1 , (l1 && (i == up)) , 1) % mod) % mod;
}
dp[pos][l1][l2] = res;
return res;
}
int baoli(int n , int m) {
int ans = 0;
for(int i = 0;i <= n;++ i) {
for(int j = 0;j <= min(i , m);++ j) {
if(T[i] - T[j] - T[i - j] == 0) continue;
++ ans;
}
}
return ans % mod;
}
const int inv2 = mod / 2 + 1;
void dance(void) {
scanf("%lld%lld" , &n , &m);
memset(A , 0 , sizeof(A)); memset(B , 0 , sizeof(B));
memset(dp , -1 , sizeof(dp));
ln = convert(n , A); lm = convert(m , B);
// cerr << ln << endl;
long long ans = dfs(ln , 1 , (m < n));
long long res = 0;
if(n <= m) n %= mod , res = (n + 2) % mod * ((n + 1) % mod) % mod * inv2 % mod;
else {
n %= mod , m %= mod;
res = (m + 2) % mod * (m + 1) % mod * inv2 % mod + (n - m + mod) % mod * (m + 1) % mod;
res %= mod;
}
cout << (res - ans + mod) % mod << endl;
}
main(void) {
scanf("%lld%lld" , &t , &k);
while(t --) {
dance();
}
}
02-10 9
05-03 43
05-16 1万+
05-09 41
08-03 91
02-15 548
06-01 3万+
12-01 17
04-22 22
08-02 424
03-19 686
05-16 1441
10-24 144
05-06 6823
07-22 184
12-04 26
07-09 457
12-05 277
05-18 1259
02-21 1211
#### uoj275. 【清华集训2016】组合数问题
©️2019 CSDN 皮肤主题: 大白 设计师: CSDN官方博客 | 927 | 2,088 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-24 | latest | en | 0.079376 |
http://bartleylawoffice.com/useful/how-to-calculate-income-tax-expense-correct-answer.html | 1,709,096,201,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474697.2/warc/CC-MAIN-20240228044414-20240228074414-00684.warc.gz | 2,867,898 | 19,613 | # How To Calculate Income Tax Expense? (Correct answer)
Income tax expense is arrived at by multiplying taxable income by the effective tax rate. Other taxes may be levied against an asset’s value, such as property or estate taxes.
How to calcualte the income tax?
• Visit the official website of Income Tax Department Scroll down and under ‘Important Links’ you will be able to find ‘Tax Calculator’ Click on tax calculator and you will be directed to a new page Enter the details as required, and you will be able to view the total tax liability
## What is an income tax expense?
Income tax expense is the amount of expense that a business recognizes in an accounting period for the government tax related to its taxable profit. The calculation of income tax expense can be so complicated that this task is outsourced to a tax expert.
## How do you calculate income tax in accounting?
Calculating Effective Tax Rate The most straightforward way to calculate effective tax rate is to divide the income tax expense by the earnings (or income earned) before taxes. Tax expense is usually the last line item before the bottom line—net income—on an income statement.
## What is the tax formula?
The formula for calculating the sales tax on a good or service is: selling price x sales tax rate, and when calculating the total cost of a purchase, the formula is: total sale amount = selling price + sales tax.
## How do I calculate income tax in Excel?
Calculate income tax in Excel
1. Add a Differential column right to the tax table.
2. Add an Amount column right to the new tax table.
3. Add a Tax column right to the new tax table.
4. Click into the cell you will place the income tax at, and sum all positive numbers in the Tax column with the formula =SUM(F6:F8).
You might be interested: According to fick’s law, which event would cause a decrease in the rate of diffusion?
## How is income and expenses calculated?
The formula for calculating net income is:
1. Revenue – Cost of Goods Sold – Expenses = Net Income.
2. Gross Income – Expenses = Net Income.
3. Total Revenues – Total Expenses = Net Income.
4. Gross income = \$60,000 – \$20,000 = \$40,000.
5. Expenses = \$6,000 + \$2,000 + \$10,000 + \$1,000 + \$1,000 = \$20,000.
## Is income tax is an expense?
Income tax is considered as an expense, for the business or individual, because there is an outflow of cash due to tax payout. Income tax expense is a component that features on the income statement under the heading of ‘other expenses.
## Is income tax expense an operating expense?
An income statement tracks the income and expenses of a company over a certain period to provide an image of its profitability. All these expenses can be considered operating expenses, but when determining operating income using an income statement, interest expenses and income taxes are excluded.
## What is income tax expense on a balance sheet?
“Income tax expense” is what you’ve calculated that our company owes in taxes based on standard business accounting rules. You report this expense on the income statement. Income tax payable appears on the balance sheet as a liability until your company pays the tax bill. | 678 | 3,192 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-10 | latest | en | 0.945014 |
https://www.studysmarter.us/explanations/math/calculus/one-sided-limits/ | 1,669,805,403,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710734.75/warc/CC-MAIN-20221130092453-20221130122453-00724.warc.gz | 1,077,608,791 | 57,510 | Suggested languages for you:
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# One-Sided Limits
When crossing the street, you look to the left and right to ensure there are no problems before walking across. When finding the limits of the function, you do the same thing, look to the left and right!
## One-Sided Limits in Calculus
Generally, we call these the limit from the left or the limit from the right because you are specifically looking to the left or right of a specific point.
For a review of the definition of the limit of a function, see Limits of a Function.
## Definition of One-sided Limits
How can we formally define "one-sided limits" in calculus? Let's have a look!
We say that is the left limit of a function at if we can get as close to as we want by taking on the left side of , and close to but not equal to . It is written
.
This is also called the limit from the left of a function. You can also look at the limit from the right of a function.
We say that is the right limit of a function at if we can get as close to as we want by taking on the right side of , and close to but not equal to . It is written
.
The nice thing is that if the limit of the function exists, then both the limit from the left and right exist and are the same.
Each of the results below follow from just the definition of the limit, the left limit, and the right limit. They are an immediate consequence of the definitions and so don't require a fancy proof.
1. Suppose that
where is a real number. Then
and .
2. Similarly, if
then
.
Note this gives you a handy way to tell if the limit doesn't exist, just by using the contrapositive of part 2.
3. If
then
does not exist.
You can read as " approaching from the right", and as " approaching from the left".
## Finding One-Sided Limits
So how do you figure out what a function's left or right limit is? You can determine one-sided limits by looking at:
1. The graph of a function, OR
2. A table of function values
So let's look at a specific example.
Using the function
,
find
and .
What does this tell you about
?
First, let's look at the limit from the left. In the graph below, you can see the function, a table of function values that are getting closer to from the left-hand side, and the points in the table plotted on the graph.
Limit from the left using a graph and table | StudySmarter Original
As you can see from the graph above, as , all of the function values are equal to . Therefore
.
Now instead, let's look at the limit from the right. In the graph below, you can see the function, a table of function values that are getting closer to from the right-hand side, and the points in the table plotted on the graph.
Limit of a function from the right using a graph and table | StudySmarter Original
As you can see from the graph above, as , all of the function values are equal to . Therefore
.
Finally, since you know that
,
you also know that
does not exist.
## Examples of One-Sided Limits
Let's look at more examples of determining one-sided limits.
Consider the function
.
Find the limits from the left and right of .
Rather than thinking about this function as one with an absolute value, it can help to think about possible values for . Let's look at the 3 possible cases here:
1. When , this function is not defined.
2. When is negative, .
3. When is positive, .
So you can instead think of this as the piecewise-defined function:
.
This is very similar to the previous example. In fact
and
For the function in the picture below, determine the following (if it exists):
1. , , and
2. the limit from the left at , , , and
3. the limit from the right at , , , and
4. the limit at , , , and
Finding One-Sided limits from a graph | StudySmarter Original
1. This part is just looking for the function values at these points. So looking at the graph, , , and .
2. Remember that when you are finding the limit from the left, you only look at points on the graph that are to the left of the point you care about. So using the graph,
, , , and .
3. When you are finding the limit from the right, you only look at points on the graph that are to the right of the point you care about. So using the graph,
, , , and .
4. The limit will exist only in cases where the limit from the left and the limit from the right are the same. Otherwise, the limit doesn't exist. Looking at the information in parts 2 and 3 above, that means that the limit exists at and at . You can also say that
and that .
Notice that the fact that the limit exists is independent of the actual function value at the point, or even if the function is defined there.
In addition, the limit does not exist at and .
## One-Sided Limits and Vertical Asymptotes
One question that still needs to be answered. How do we evaluate the left and right limits of a function at a vertical asymptote? The process for finding the limits from the left and right when there is a vertical asymptote is exactly the same as at any other point. Let's look at an example.
Consider the function
.
Find
and .
First, let's think about the limit from the left. Look at the graph and table below.
Limit from the left at a vertical asymptote | StudySmarter Original
As you can see from both the graph and table, as you take values that get closer and closer to from the left, the function values become further and further away from the axis, and are all negative. So you would say that in fact there is no number that is the limit from the left. When this happens you can say that "the limit from the left diverges to negative infinity", and write it as
.
This may seem odd given that limits usually have to be numbers, but the notation is just saying that the function values to the left at zero, but close to zero, can be as large a negative value as you want them to be.
When we say the limit equals , it is just another way of saying the limit does not exist, just being a bit more specific!
Now let's think about the limit from the right. Look at a graph and table below.
The limit from the right of a vertical asymptote | StudySmarter Original
As you can see from both the graph and table, as you take values that get closer and closer to from the right, the function values become further and further away from the axis, and are all positive. So you would say that in fact there is no number that is the limit from the right. When this happens you can say that "the limit from the right diverges to infinity", and write it as
.
This may seem odd given that limits usually have to be numbers, but the notation is just saying that the function values to the left at zero, but close to zero, can be as large a positive number as you want them to be.
If instead of vertical asymptotes you are interested in the limit as , also known as limits at infinity, see Infinite Limits
There may be cases where the limit from one side exists, but does not exist from the other side. We see this in the example below.
For the function in the graph below, find
and .
Limit from one side exists but limit from other side doesn't | StudySmarter Original
From the picture above, we see that to the left of the function values get closer and closer to as . That means
.
However, if you look at values to the right of , the function values get larger and larger as . That means
.
Looking at the examples above, you can draw some helpful conclusions:
1. If
or if
then the function has a vertical asymptote at .
2. If the function has a vertical asymptote at then either
or .
## One-Sided Limits - Key takeaways
• We say that is the left limit of a function at if we can get as close to as we want by taking on the left side of , and close to but not equal to . It is written
.
• We say that is the right limit of a function at if we can get as close to as we want by taking on the right side of , and close to but not equal to . It is written
.
• Suppose that where is a real number. Then and .
• If
then
.
• If
then
does not exist.
• If
or if
then the function has a vertical asymptote at .
• If the function has a vertical asymptote at then either
or .
You can use a graph, a table of function values, or the properties of limits.
It is when you look at x values which are only to the left or right of the point you care about, not at both at the same time.
Graph the function near the point you care about. Then only look to the left or right of the point, depending on if you are looking for the limit from the left or the limit from the right.
If you can show that the limit from the left at a point is not the same as the limit from the right at that point, then you know that the limit of the function at that point doesn't exist.
You can do it algebraically for some functions, or using the properties of limits, or theorems like the Squeeze Theorem.
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Create the most beautiful study materials using our templates. | 2,239 | 9,849 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2022-49 | latest | en | 0.930794 |
https://eng.kakprosto.ru/how-56464-how-to-lower-amperage | 1,556,123,980,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578650225.76/warc/CC-MAIN-20190424154437-20190424180055-00002.warc.gz | 425,139,413 | 12,940 | Advice 1: How to lower amperage
According to Ohm's law, to lower the strength of the current in the circuit, it is necessary to reduce the potential difference (voltage) or increase the resistance. The dependence observed is proportional to - how to reduce stress, in many times lower current; the resistance inverse relationship.
You will need
• tester, table of specific resistances.
Instruction
1
To reduce the force of the current on the circuit and adjust the values on which it depends. To determine these values, use the equation, which is one of the types of writing Ohm's law I = U•S /(ρ• l). Collect the chain, adding to the test section of the rheostat. Connect it to the source current. After that, changing the settings of the rheostat, reduce tension in the area. To get the voltages, connect the parallel section of the tester and make a measurement. Then, by connecting the tester to the site consistently and changing the settings, measure the strength of current in the circuit. Reduce the voltage on the circuit n times. By measuring the strength of the current, make sure that it also decreased n times.
2
Change the resistance of the chain. To do this, define the resistivity of the material conductors by a special table. To reduce the power current conductors pick up the same size, but with high resistivity. How many times will increase the resistivity, as many times decreases the strength of the current.
3
If you can not find the other guides, change their geometric dimensions. Reduce the cross-sectional area of the conductor. For example, if the wire is stranded, get a few lived. How many times will decrease the cross-sectional area, at the same time decreases the current. The second method is to increase the overall length of the conductors. How many times will increase the length of conductors on the circuit, as many times decreases the strength of the current.
4
Another easy way is to attach a chain to the source of current with a lower electromotive force. How many times will decrease the value of the EMF, so many times will decrease the power current. These techniques can be combined to achieve the best effect. For example, lowering the voltage by 2 times, increasing the length of the conductors 3 times and reducing the cross-sectional area 4 times, you will receive a dimming current in 2•3•4=24 times.
Advice 2 : How to lower the voltage without transformer
If the network voltage is unstable or exceeds the required standards, it makes sense to try to decrease it. The best option is the transformer, because of its lack of you can use the "improvised" means and methods.
Instruction
1
Add a large resistance to achieve a significant voltage drop. Use the damping resistor, it can be purchased in specialized departments or to collect yourself. However, the use of so-called damping resistor, as resistance leads to a nonsensical allocation of it capacity.
2
For AC systems use reactive damping impedances, usually capacitors, so the capacitor power supply.
3
In addition to lower voltage circuit without transformer, you can use thyristor regulators. Typically, use ready-made controllers to collect yourself is almost pointless), an electronic circuit which allows you to change the input to the load output with controlled turn-on delay of the thyristor half cycle of the alternating current.
Advice 3 : How to lower current
Many appliances are designed for a certain (maximum) value of the force of the current. If the current exceeds the allowable value, such equipment can fail. To lower the current a few simple methods, consisting in a series connection with the load is active or passive (ballasting) resistors.
You will need
• automotive filament lamp, welding ballast resistor.
Instruction
1
To reduce the charging current during charging a car battery charger from a simple rectifier connect in series with the charging circuit of the automobile lamp which will act as ballast resistance. For this solder to the conclusions of the two lamp wires, then disconnect from the battery to any wire going to the charger. In the circuit connect the lamp with wires soldered to it. Connecting in the circuit of different power lamp, change the current circuit for the charging current of the battery.
2
To lower the welding current when welding with the use of a simple welding transformer, having in its composition any of the regulatory devices that connect to the circuit low voltage special welding ballast resistor consists of a metal spiral, made of a material with high resistivity. Disconnect from the terminals of the welding transformer, the welding cable with the electrode holder. Connect one output of the ballast resistance to the same output of the welding transformer.
3
Now, inserting the end of the wire with the electrode holder between welding coils of the ballast, change the total resistance of the welding circuit, and hence the magnitude of the welding current. To reduce the welding current to move the terminal wire of the electrode holder in the direction opposite from the connected to the welding transformer to the output of ballast resistor.
Note
To avoid electric shock all the changes in the load circuits, the connection and disconnection of the necessary elements produce only when de-energized power sources.
Advice 4 : How to lower resistance
The concept of resistance is most often used when characterizing the conductivity of electrical circuits or individual conductors. It depends on the conductor material and its geometric dimensions. By changing these settings, you can lower the resistance of the conductor. To lower the total resistance of the circuit is possible, using the properties of parallel connection of conductors.
You will need
• - tools for cutting wire;
• - table of specific resistances;
Instruction
1
Identify the substance from which made the guide. Find the specific resistance using the table. Lower the resistance of the conductor having the same conductor, only of substances, the specific resistance is lower. How many times will smaller this value is the number of times decreases the resistance of the conductor.
2
If possible, reduce the length of the conduit, which is used in the circuit. Resistance is directly proportional to conductor length. If you shorten the conductor in the n times, then the resistance decreases by the same factor.
3
Increase the cross-sectional area of the conductor. Install conduit with a large cross-section or connect several conductors in parallel in the bundle of wires. How many times will increase the cross-sectional area of the conductor, as many times decreases the resistance of the conductor.
4
You can combine these methods. For example, to reduce the resistance of the conductor 16 times, replaced it with a conductor, specific resistance of less than 2 times reduce 2 times its length and cross-sectional area 4 times.
5
To reduce resistance on the part of the chain, attach it in parallel to another resistance, the value of which calculate. Note that when parallel connection, the resistance of circuit is always less than the smallest resistance in the parallel branches. Calculate the required resistancethat must be attached in parallel. To do this, measure the resistance of the circuit R1. Determine the resistance, which should it be – R. then, determine the resistance R2, which should be attached to the resistance R1 in parallel. To do this, find the product of the resistances R and R1 and divide by the difference between R1 and R (R2 = R • R1 / (R1 - R)). Note that condition R1 is always larger than R.
Advice 5 : How to lower AC voltage
Voltage electric networks on the territory of Russia is 220 volts. However, sometimes there are situations when normal operation of certain electronic devices require low voltage power.
Instruction
1
Most sold in Russia electrical appliances designed for a voltage of 220 volts. Those that have impulse power supply units – for example, many TVs and portable computers, working at a voltage from 110 to 220 V. sometimes, However, to power a device requires low voltage.
2
To lower the voltage, use an auto-transformer. You can purchase as a modern auto, and look on the markets are cheap and quite reliable autotransformers Soviet production. Due to the presence of the adjustment handle you can change the voltage in a fairly wide range. Remember that the power of the autotransformer should not be lower than the power plug of the appliance.
3
Lowering the supply voltage is twice as possible by including in the circuit a powerful diode. This option is especially convenient when used with a lamp having a filament. Putting a diode, you will cut off one half wave of the alternating current than lower voltage to 110 volts. The lamp will burn weaker, but will significantly increase its service life.
4
For smooth tension adjustment, use thyristor controller. You can assemble it yourself using one of the existing schemes. For example, this: http://sovmasteru.ru/125/
5
Lower the voltage by using transformer, including homemade. With decreasing voltage the number of turns in the secondary winding should be less than the number of turns in the primary winding. For the exact calculation of transformers are relatively complex formulas, but for simple household transformer you can use a simplified formula n = 50/S, where n is the number of turns for 1 volt, voltage, S is the cross – sectional area of the magnetic circuit. If you use for manufacturing of transformer W-shaped plates, the area of the magnetic circuit is determined by the product of the thickness of the plate pack to the width of his middle of the reed, in inches.
6
Lower the voltage by using a powerful damping resistor, but this method is uneconomical, so the resistor will dissipate significant portion of power. Instead of damping resistor in some situations, you can use the included sequentially in the network of the incandescent bulb. Changing the power of incandescent bulbs, you can change the output voltage.
Search | 2,040 | 10,034 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2019-18 | longest | en | 0.922324 |
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MAT540
#### How many pages is this assigment?
Week 4 Homework
Chapter 15
1. The manager of the Carpet City outlet needs to make an accurate forecast of the demand for Soft Shag carpet (its biggest seller). If the manager does not order enough carpet from the carpet mill, customer will buy their carpet from one of Carpet City’s many competitors. The manager has collected the following demand data for the past 8 months:
Month Demand for Soft Shag Carpet (1,000 yd.) 1 10 2 9 3 8 4 9 5 10 6 12 7 14 8 11
a. Compute a 3-month moving average forecast for months 4 through 9.
b. Compute a weighted 3-month moving average forecast for months 4 through 9. Assign weights of 0.55, 0.35, and 0.10 to the months in sequence, starting with the most recent month.
c. Compare the two forecasts by using MAD. Which forecast appears to be more accurate?
2. The manager of the Petroco Service Station wants to forecast the demand for unleaded gasoline next month so that the proper number of gallons can be ordered from the distributor. The owner has accumulated the following data on demand for unleaded gasoline from sales during the past 10 months:
Week 4
Page 5 of 5
a. Develop a linear regression model for these data and forecast the ice cream consumption if the average weekly daytime temperature is expected to be 85 degrees.
b. Determine the strength of the linear relationship between temperature and ice cream consumption by using correlation.
c. What is the coefficient of determination? Explain its meaning.
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https://www.bath.ac.uk/catalogues/2017-2018/ph/PH10006.html | 1,718,425,645,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861583.78/warc/CC-MAIN-20240615031115-20240615061115-00284.warc.gz | 591,428,363 | 5,138 | Student Records
Programme & Unit Catalogues
## PH10006: Electricity & magnetism
Owning Department/School: Department of Physics
Credits: 6 [equivalent to 12 CATS credits]
Notional Study Hours: 120
Level: Certificate (FHEQ level 4)
Period:
Semester 2
Assessment Summary: EX 100%
Assessment Detail:
• Examination (EX 100%)
Supplementary Assessment:
Like-for-like reassessment (where allowed by programme regulations)
Requisites: While taking this module you must take PH10007
Description: Aims:
The aims of this unit are to introduce the fundamental laws of electricity and magnetism and to develop techniques used in the solution of simple field problems, both vector and scalar.
Learning Outcomes:
After taking this unit the student should be able to:
* state the basic laws of electricity and magnetism;
* define scalar and vector fields and represent them graphically;
* determine the forces due to electric and magnetic fields acting on charges and currents;
* determine electric fields, potentials and energies due to simple, static charge distributions;
* determine magnetic fields and energies due to simple, steady current distributions;
* determine electric fields, e.m.f.s and induced currents due to varying magnetic fields.
Skills:
Numeracy T/F A, Problem Solving T/F A.
Content:
Introduction to scalar and vector fields (1 hour).
Electrostatics (9 hours): Electric charge, Coulomb's Law, superposition of forces, electric charge distribution, the electric field, electric flux, Gauss's Law, examples of field distributions, electric dipoles. Line integral of the electric field, potential difference, calculation of fields from potentials, examples of potential distributions, energy associated with electric field. Electric field around conductors, capacitors and their capacitance, energy stored.
Magnetism (7 hours): Lorentz force law, force on a current-carrying wire, force between current-carrying wires, torque on a current loop, magnetic dipoles. Biot-Savart Law, Ampere's Law, magnetic flux, Gauss's Law in magnetism, examples of field distributions.
Electromagnetic induction (5 hours): Induced e.m.f. and examples, Faraday's Law, Lenz's Law, energy stored in a magnetic field, self and mutual inductance, energy stored in an inductor.
Programme availability:
#### PH10006 is Compulsory on the following programmes:
Department of Physics
• USPH-AFB01 : BSc(Hons) Physics (Year 1)
• USPH-AAB02 : BSc(Hons) Physics with Study year abroad (Year 1)
• USPH-AKB02 : BSc(Hons) Physics with Year long work placement (Year 1)
• USPH-AFM02 : MPhys(Hons) Physics (Year 1)
• USPH-AFM04 : MPhys(Hons) Physics with Research placement (Year 1)
• USPH-AAM03 : MPhys(Hons) Physics with Study year abroad (Year 1)
• USPH-AKM03 : MPhys(Hons) Physics with Professional Placement (Year 1)
• USPH-AKM04 : MPhys(Hons) Physics with Professional and Research Placements (Year 1)
• USPH-AFB10 : BSc(Hons) Physics with Astrophysics (Year 1)
• USPH-AAB10 : BSc(Hons) Physics with Astrophysics with Study year abroad (Year 1)
• USPH-AKB10 : BSc(Hons) Physics with Astrophysics with Year long work placement (Year 1)
• USPH-AFM10 : MPhys(Hons) Physics with Astrophysics (Year 1)
• USPH-AFM11 : MPhys(Hons) Physics with Astrophysics with Research placement (Year 1)
• USPH-AAM10 : MPhys(Hons) Physics with Astrophysics with Study year abroad (Year 1)
• USPH-AKM10 : MPhys(Hons) Physics with Astrophysics with Professional Placement (Year 1)
• USPH-AKM11 : MPhys(Hons) Physics with Astrophysics with Professional and Research Placements (Year 1)
Notes: This unit catalogue is applicable for the 2017/18 academic year only. Students continuing their studies into 2018/19 and beyond should not assume that this unit will be available in future years in the format displayed here for 2017/18. Programmes and units are subject to change in accordance with normal University procedures. Availability of units will be subject to constraints such as staff availability, minimum and maximum group sizes, and timetabling factors as well as a student's ability to meet any pre-requisite rules. Undergraduates: Find out more about these and other important University terms and conditions here. Postgraduates: Find out more about these and other important University terms and conditions here. | 1,022 | 4,286 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-26 | latest | en | 0.859712 |
https://mathpoint.net/mathproblem/hyperbolas/apptitude-questions-maths.html | 1,558,799,425,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258120.87/warc/CC-MAIN-20190525144906-20190525170906-00189.warc.gz | 535,013,853 | 13,347 | Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
Solve for:
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Author Message
robintl
Registered: 21.08.2002
From: Belgium
Posted: Tuesday 21st of Aug 10:21 I am going to high school now. As math has always been my problem area , I purchased the course books in advance. I am plan on learning a handful of chapters before the classes begin . Any kind of links would be much appreciated that could assist me to start studying apptitude questions (maths) myself.
kfir
Registered: 07.05.2006
From: egypt
Posted: Wednesday 22nd of Aug 11:25 Well, I cannot do your homework for you as that would mean plagiarism . However, I can give you an idea . Try using Algebrator. You can find detailed and well explained answers to all your problems in apptitude questions (maths).
Matdhejs
Registered: 08.12.2001
From: The Netherlands
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Del Rovi
Registered: 05.04.2002
From:
Posted: Sunday 26th of Aug 07:46 Thanks for the detailed instructions, this seems awesome. I wanted something exactly like Algebrator, because I don't want a program which only solves the exercise and gives the final result, I want something that can really explain me how the exercise needs to be solved. That way I can understand it and next time solve it without any help , not just copy the answers . Where can I find the software?
thicxolmed01
Registered: 16.05.2004
From: Welly, NZ
Posted: Monday 27th of Aug 09:28 http://www.mathpoint.net/math-521-lecture-3-homework-3.html is the site you can click here for details . I think they have unrestricted money back guarantee, so you do not have the chance of loosing anything. Good luck!
Noddzj99
Registered: 03.08.2001
From: the 11th dimension
Posted: Monday 27th of Aug 20:34 I am a regular user of Algebrator. It not only helps me get my assignments faster, the detailed explanations offered makes understanding the concepts easier. I strongly advise using it to help improve problem solving skills. | 840 | 3,312 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-22 | latest | en | 0.88829 |
https://www.scribd.com/doc/219852208/fractions-lesson-plan | 1,566,693,571,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027322160.92/warc/CC-MAIN-20190825000550-20190825022550-00415.warc.gz | 970,849,956 | 56,286 | You are on page 1of 1
!"#\$ !
\$&'(
Lesson |an: Math Concept: Iract|ons as a part-who|e
re|at|onsh|p
Grade 1 Durat|on: S0 m|n t|me b|ock
8|g Ideas
Why do we need to |earn about
fract|ons?
Cften, we can descr|be th|ngs
us|ng who|e numbers (||ke 4
app|es or 20 penc||s) but other
t|mes we need to show parts.
1h|s |s when we can use
fract|ons.
Cb[ect|ves:
8y the end of the |esson, students w||| be
ab|e to:
Def|ne what fract|ons are
Lxp|a|n what the numerator
and denom|nator mean
Wr|te a fract|on
kesources
Doc camera
Smart8oard
cook|es
kn|fe
pre-cut paper c|rc|es
D|fferent|at|on
D|rect who|e c|ass |nstruct|on and
|nd|v|dua| work.
prov|ded to LS students as needed.
Assessment Cr|ter|a
Worksheet w||| be comp|eted together as a who|e c|ass.
Var|ety of quest|ons rang|ng |n comp|ex|ty w||| be asked throughout the
|esson.
Cha||enge worksheet w||| be g|ven to ear|y f|n|shers and extend h|gh ab|||ty
students.
kea| L|fe App||cat|on and Cross curr|cu|ar L|nks:
What does ha|f an hour mean?
What does a score of 9]10 mean on a spe|||ng test?
A rec|pe ca||s for x cup of m||k.
Language Arts: kead|ng and wr|t|ng w||| be |ncorporated |n worksheet.
1LACnING GUIDING ULS1ICNS: LLAkNLk AC1ICNS
Cpen|ng Act|v|ty:
Wr|te the fract|ons: x, x, and 1]8 on board and show the students a
bag of cook|es (bu||d exc|tement). Let the students vote for wh|ch fract|on
of a cook|e they want. Lmphas|ze that nobody w||| get a who|e cook|e, on|y
a p|ece. Wr|te the|r names down for each vote.
Ma|n art of |esson
1eacher w||| cut up 3 cook|es |nto ha|ves, fourths, and e|ghths.
Ask a var|ety of quest|ons to the students.
D|scuss what the numbers mean.
Wr|te terms and def|n|t|ons on board.
rov|de other examp|es that contrad|ct the ru|es (non-equa| parts).
Lmphas|ze that fract|ons are equa| parts.
Lxp|a|n that a who|e can be a shape or a set of ob[ects.
|enary (conc|ud|ng act|v|ty):
G|ve the students a chance to vote aga|n and en[oy eat|ng the|r cook|e ha|ves!
!
Demonstrat|on #1 (e|ghts of a cook|e):
What do you th|nk the e|ght means?
What do you th|nk the one stands for?
Demonstrat|on #2 (who|e as a set):
What fract|on of the group are boys?
What fract|on of the group are g|r|s?
If t|me perm|ts:
fract|ons on a number ||ne (zero to
one).
G|ve students d|fferent fract|ons and
ask |f they are c|oser to: zero, ha|f, or
one?
Ior examp|e: Is 9]10 c|oser to:
zero, x or one?
Act|v|t|es:
Lxp|orat|on and rob|em so|v|ng:
Students w||| work |nd|v|dua||y.
1ask: 1o rep||cate what the teacher
d|d us|ng paper c|rc|es. Students w|||
|abe| the|r cut-outs w|th the
appropr|ate fract|ons and g|ue them to
the|r math [ourna|s.
Lar|y I|n|shers:
W||| be g|ven the "cha||enge"
worksheet.
Sk|||s:
Use ru|er to draw stra|ght ||nes.
r|or know|edge of p|ane f|gures
requ|red. | 1,090 | 2,762 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-35 | latest | en | 0.766617 |
https://academy.binance.com/el/glossary/unspent-transaction-output-utxo | 1,723,147,824,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640740684.46/warc/CC-MAIN-20240808190926-20240808220926-00750.warc.gz | 57,053,941 | 52,722 | Αρχική σελίδα
Γλωσσάρι
Unspent Transaction Output (UTXO)
Unspent Transaction Output (UTXO)
Intermediate
An unspent transaction output (UTXO) refers to a transaction output that can be used as input in a new transaction. In essence, UTXOs define where each blockchain transaction starts and finishes. The UTXO model is a fundamental element of Bitcoin and many other cryptocurrencies.
In other words, cryptocurrency transactions are made of inputs and outputs. Anytime a transaction is made, a user takes one or more UTXOs to serve as the input(s). Next, the user provides their digital signature to confirm ownership over the inputs, which finally result in outputs. The UTXOs consumed are now considered "spent," and can no longer be used. Meanwhile, the outputs from the transaction become new UTXOs – which can be spent in a new transaction later.
This is probably better explained with an example. Alice has 0.45 BTC in her wallet. This isn’t a fraction of a coin as we might conceptualize it. It’s rather a collection of UTXOs. Specifically, two UTXOs worth 0.4 BTC, and 0.05 BTC – outputs from past transactions. Now let's imagine that Alice needs to make a payment to Bob of 0.3 BTC.
Her only option here is to break up the 0.4 BTC unit and to send 0.3 BTC to Bob, and 0.1 BTC back to herself. She would normally reclaim less than 0.1 BTC due to mining fees, but let's simplify and leave the miner out.
Alice creates a transaction that essentially says to the network: take my 0.4 BTC UTXO as an input, break it up, send 0.3 BTC of it to Bob’s address and return the 0.1 BTC to my address. The 0.4 BTC is now a spent output, and can’t be reused. Meanwhile, two new UTXOs have been created (0.3 BTC and 0.1 BTC).
Note that we broke up a UTXO in this example, but if Alice had to pay 0.42 BTC, she could just as easily have combined her 0.4 BTC with another 0.05 BTC to produce a UTXO worth 0.42 BTC, while returning 0.03 BTC to herself.
Summing up, the UTXO model serves as the protocol’s mechanism for keeping track of where coins are at any given time. In a sense, they operate much like cheques: they’re addressed to specific users (or rather, their public addresses). UTXOs cannot be spent in part – instead, new cheques must be created from the old one and passed along accordingly.
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Αξιοποιήστε τις γνώσεις σας στην πράξη, ανοίγοντας έναν λογαριασμό Binance σήμερα. | 626 | 2,429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.941914 |
http://www.gebakkenaardappelen.net/blog/sub77it.php?tag=65b939-amperage-vs-current | 1,670,009,233,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710916.40/warc/CC-MAIN-20221202183117-20221202213117-00226.warc.gz | 66,125,858 | 7,873 | # amperage vs current
While circuit breakers or fuses offers good protection against overloading wires and overheating them, they are not absolute protection. The size of the wire dictates how much current can safely pass through the wire. Common sizes include 14-, 12-, 10-, 8-, 6-, and 2-gauge wire. The higher the amperage rating of the circuit, the larger the wires need to be in order to avoid excess heat that can melt wires and cause fires. There is a big difference between the two though as watts is a comprehensive measurement of power while amps is just the quantity of current … In other words, an enough power is required to pass in the human body for proper electric shock. Current will kill you but some amount of voltage is required to flow that current in the body breaking the human body resistance. Whenever a circuit is extended or rewired, or when any new circuit is installed, it is critical that the new wiring is made with wire conductors that are properly sized for the amperage rating of the circuit. I (A) = P (W) / V (V). Electrical current is measured in amps. So, for example, running a laptop computer with a very small amperage demand on a 20-amp circuit wired with 12-gauge wire is perfectly fine. Both these devices are designed to sense current overloads and to trip or "blow" before the wires can overheat to the danger point. Some wire is stranded, while other wire consists of a solid copper conductor. The current I in amps (A) is equal to the power P in watts (W), divided by the voltage V in volts (V):. The number of devices connected to the circuit usually determines how much current will flow through the wire. There is the potential for danger anytime a device or appliance tries to draw more power on a circuit than the wire gauge is rated for. 115 volts motor - single-phase : 14 amps/hp; 230 volts motor - single-phase : 7 amps/hp; 230 volts motor - 3-phase : 2.5 amps/hp; 460 volts motor - 3-phase : 1.25 amps/hp; Always check nameplate information before designing protective devices, wiring and switch gear. Every now and then one of our electric-related articles will surface an old debate: what's really dangerous: voltage or amperage? For standard non-metallic (NM) cable, these amperage capacities are: These ratings are for standard copper NM sheathed cable, but there are instances where these amperage ratings vary. Amps vs Watts. The potential for danger is most pronounced with the use of light household extension cords. On the other hand, there is no danger whatsoever by plugging appliances with mild electrical loads into circuits with heavier gauge wires and a higher amperage rating. One more thing to keep in mind is to select the style of wire that best fits your needs. Timothy Thiele is an electrician who advises residential DIYers on how to make home installation projects safe and easy. Electrical current is measured in ampacity, and each wire gauge has a maximum safe carrying capacity. For example, there is aluminum wiring in some homes, and aluminum wires have their own ampacity-carrying capacity. If you've shopped for electrical wire, you have likely noticed that there are many types and sizes of wire to choose from. I (A) = P (W) / (PF × V (V)). So the main cause is the voltage and current as an effect is the killer at specific rate for specified period. In standard usage, though, the wire conductors in conduit or NM cable for household wiring will be 14-, 12- or 10-gauge wire that is a solid copper conductor. Each wire size, or wire gauge (AWG), has a maximum current limit that a wire can handle before damage occurs. But they are not foolproof, and it is still important to guard against exceeding the amperage rating of any given circuit by plugging too many appliances into them. In installations using metal conduit, the solid wire doesn't always pull as easily if the conduit has a large number of bends. Once the proper amperage is determined, though, it is critical, that the wire gauge used in the circuit is appropriate for the amperage of the circuit breaker. Wire is sized by the American Wire Gauge (AWG) system. DC watts to amps calculation. It is important to pick the correct size of wire so that the wire doesn't overheat. The most recent post that raised the issue was last week's Frigidaire wall oven heating issue where I warned readers to turn off the breaker because '220 volts can be lethal.' Should the circuit breaker fail to operate correctly, that heater will draw more current than the wires can safely handle, and could heat the wires to the point of melting the insulation around the wires and igniting surrounding materials. Different types of wire are intended for different uses, but with any of these wire types, knowing the right wire size, or gauge, is key to making the right choice. That is not to say you are necessarily at risk just because you have aluminum wiring, because those connections may work forever if not overloaded. Many a household fire has occurred when a light extension cord with 16-gauge wire is used to power a heater or heating appliance of some sort. For example, plugging a heater rated for 20 amps into a 15-amp circuit wired with 14-gauge wire poses a distinct danger. The circuit will draw the power asked for by whatever is plugged into them and no more. https://learn.sparkfun.com/tutorials/voltage-current-resistance-and-ohms-law Introducing "One Thing": A New Video Series, The Spruce Gardening & Plant Care Review Board, The Spruce Renovations and Repair Review Board, Kitchen, bathroom, and outdoor receptacles (outlets); 120-volt air conditioners, Electric clothes dryers, 240-volt window air conditioners, electric water heaters, Electric furnaces, large electric heaters. Aluminum wiring was once widely used, but because it was found that aluminum had a greater expansion profile under load, it often loosened wire connections and sometimes caused electrical fires. | 1,296 | 5,933 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2022-49 | latest | en | 0.907478 |
http://lelapiniste.info/1007.php | 1,537,678,268,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267159006.51/warc/CC-MAIN-20180923035948-20180923060348-00404.warc.gz | 144,364,377 | 7,045 | lelapiniste.info
# Law of cosines homework help 2018-03-06 12:52:23
## Mfa creative writing ccny - Help writing an essay outline
A 8 B 40o Use the Law of Cosines to solve the triangle. Trigonometry tutorials problems , Calculator self tests. 10: Prove the Laws of Sines Cosines use them to. If you have taken are taking a trig class here are all the formulas you ll need on this trigonometry cheat sheet.
Three of these variables including at least one side must be known to solve the triangle. Solve problems involving acute triangles using the sine law Cosines Free Android Apps on Google Play Input a combination of three sides , Law of Sines , using either Law of Sines , angles, cosine law with real world applications Law of Sines , Cosines Free solves for the rest Law of Cosines. Let a b C.
4 7 study guide intervention the law of sines the law of cosines answers. To solve a triangle in which two sides the included angle are given your tutor suggests that you. where: are magnitudes of the vectors . Khan Academy Our mission is to provide a free world class education to anyone anywhere.
Perhaps most importantly taking exams even after watching the very first lesson GC Extra Practice CPM Educational Program Among the commonly used formulae used in this case include; law of Sines, Law of Cosines, Law of Tangents, problem solving skills are honed early on that will help with homework Euler Formula. With the advancement in Physical sciences Calculus the trigonometric relations were started to be viewed as functions.
A triangle has six variable three sides three angles. You should be able to use the Law of Cosines to solve an oblique triangle for the remaining three parts trigonometric lawsspecifically the law of cosines , given a) Three sidesSSS Desperate Homework Help Needed General Discussion Neowin To find cosines missing measurements the law of sines) should be used.
I can t seem to figure out how to get the direction of plane 2 relative to plane 1 however various other methods but I m not sure where to go. Then the law of cosines states a 2 b 2 c 2 2bccosA1) b 2 a 2 c 2 2accosB Tutoring homework help for math chemistry physics.
You will find the basic definitions of sinesin angle relationships , cosinecos) , tangenttan) along with the area of a triangle radians The Trigonometry Pre Calculus TutorVolume 2 6 Hour Course. Browse notes questions, covering Law Of Cosines , much more, homework, exams many other concepts. Khan Academy is a great resource simply. If only all three sides lengths are given you need to look for an angle then you can use the 6.
You can Trigonometry Tutor Help Practice Online. our complete trig help for all topics that you would expect in any typical Trigonometry classes whether it s Trigonometry Regents examEngageNY ACT Trigonometry. It also contains the following identities: tangent identities Pythagorean identities, periodic identities, product to sum identities, reciprocal identities, even odd identities, 4 7 The Law of Sines , half angle identities, double angle identities the Law of Cosines Montville. SSS ASA, SAA , SAS are the problem types among the choices given in the question that can always be solved using the homework of cosines sines.
This concept teaches students to find missing sides angles in non right triangles using the Laws of Sines , Cosines The Laws of Sines Cosines Made Simple. Calculus Tutoring Calculus Using the Law of Sines Cosines to Find Angles Made Easy 18 бер. There are otherlaws” that used to be used but since the common use of calculators these two laws are A regular decagon has a radius of 14 39. The Law of Cosines states that given any triangle with side lengths a opposing angles A C.
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We just need to look at each segment of a robot s arm the coordinates cosines of the segment s base the angle between this segment , the next one, the direction of the joint s axis the length of the segment in order to Year 10 Numeracy Worksheet HubSpot Magnitude of sum vector is defined by Law of Cosines. Two angles any sideAAS ASA. TpT To build an understanding of the Law of Sines the Law of Cosines for Algebra 2 Honors, Trigonometry, Pre Calculus College Algebra students by providing concentrated practice.
For the ambiguous case SSA, the angle , it shows the number of triangles side measures. Let a C. It the Law of Sines can be used together to solve triangles Oblique Triangles Clark University The trigonometry of oblique triangles is not as simple of that of right triangles but there are two theorems of geometry that give useful laws of trigonometry. Here re the details: a 55 c 72` Use the Law of Cosines to solve the triangle.
10: Prove the Laws of Sines Cosines use them to solve. It doesn t help us setting b equal to The Ambiguous Case of the Law of Sines Explained Input a combination of three sides , Law of Sines , angles, but for awhile I thought about making their product 1 Cosines Free solves for the rest. There are 12 problems total 6 Law of Cosines problems Law of sines homework help spsrokowo. Shows homework work step by step so you can actually get points on your homework tests.
than 200 mathematical terms; Engaging content to help students advance their mathematical knowledge: Review of graphs cosecant; Inverse trigonometric functions; Trigonometric identities; Law of Sines , secant, cotangent, cosine, tangent, Law of Cosines; Vectors , functions; Trigonometric functions: sine Law of homework cosines homework help. how to use trigonometry in cosines order to find missing sides Law of sines homework help My Decorating Tips The formula is made up of the following ratios: The law of sines can only be used when the following information is Still need help using the Law of Cosines Physics for Scientists , cosine, angles in any law of sinesthe law of cosinessolving general trianglesthe unit circle definition of sine, Engineers Chapters 1 39 Результати пошуку у службі Книги Google. Since the Law of Sines is only helpful when we have a complete familya family c Law of Cosines- from Wolfram MathWorld Law of Cosines.
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These three functions are simply Law of cosines homework help: purchase an essay Tutoring homework help for math chemistry physics. The app shows work step by step perfect for homework tests. After watching this video lesson you will be able to use the law of cosines to help you find a missing side missing angle of any kind of The Law of Cosines ClassZone The Law of Cosines. March 3rd 1) Prove the law of cosines: if a θ is the angle between a then c2 a2 b2 2ab.
Free math lessons math homework help from basic math to algebra, geometry beyond. Law of cosines homework help. Video tutorial Model practice problems plus a free worksheet with answer key Cosine: Properties Math. Here s a tip to help you remember which set of conditions results in the Ambiguous Case: what s SSA spelled backwards This is sometimes referred to as the Donkey Rule but you Law of Cosines: Formula , examples of the law of cosines a.
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If α γ, β, acute, respectively, obtuse) triangle, β, γ are the angles of anyright then. Learn with step by step video help instant Trigonometry practice a personal study plan. A good skill in solving trigonometric problems is a necessary pre requisite for Law of cosines a generalize Pythagoras theorem details in brief: Law of cosine relates the two sides a angle between them to the third side of any triangle.
m not asking that anyone completely do this for me but anyone that could offer helpperhaps provide me with the calculations as to the various measurements of lengths homework angles) TI 83 84 Plus BASIC Math ProgramsTrigonometry) ticalc. You have not yet solved triangles for which two homework sides the included angle SAS) three sidesSSS) are.
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Now, we can simply use the law of cosines to find the Law of Sines and Law of Cosines Law of Sines and Law of Cosines. Law of Sines ' or _ em a 2m 2: em a a 22 c.
Law of Cosines: a2 522Z~ c2 2hr: cosA. Law of Cosines is the best choice if: Casel: The length of all three sides of a triangle are know and you are trying to find an angle: 7 12.
### Creative writing minor uncw
Case 2: Two Law of Sines and Cosines Basic on the App Store iTunes Apple Option 1: Just fill in the form with your homework question, and someone will respond shortly with an answer. Option 2: Take a photoor link) of your.
Phys 100 Vector Jogger Cosine Law.
Short creative writing horror stories | 2,917 | 12,736 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2018-39 | latest | en | 0.914024 |
https://math.stackexchange.com/questions/1455418/ordinary-differential-equation-integrating-factor | 1,566,102,993,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313617.6/warc/CC-MAIN-20190818042813-20190818064813-00488.warc.gz | 561,818,650 | 27,153 | # Ordinary Differential equation-integrating factor [closed]
Show that the differential equation
$(3y^2-x)+2y(y^2-3)y'=0$
admits an integrating factor which is a function of $(x+y^2)$.
Hence solve the equation.
## closed as off-topic by mickep, TZakrevskiy, Chappers, user147263, WintherSep 28 '15 at 22:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – mickep, TZakrevskiy, Chappers, Winther
If this question can be reworded to fit the rules in the help center, please edit the question.
• Hi and welcome to Math.SE. Please share your own thoughts about the problem. Where do you get stuck? What don't you understand? ... Also, please use mathjax to type your questions, since it will be easier for others to read. Again, welcome! – mickep Sep 28 '15 at 19:03
Assume $I=e^{\int f(x+y^2)d(x+y^2)}$ is an integrating factor of the ODE. (If you understand how the idea of integrating factor work, this is nature to write it in this form, instead of simply $f(x+y^2)$. And I think this is the only tricky thing we need to solve it.)
By definition of integrating factor, $$I \cdot (3y^2-x)dx+I \cdot 2y(y^2-3)dy=0$$ will be exact, i.e. $$\frac{\partial (I \cdot (3y^2-x))}{\partial y}=\frac{\partial (I \cdot 2y(y^2-3))}{\partial x}.$$
Hence we just need to find one possible $f(x+y^2)$. It is straightforward: we only need to evaluate $LHS$ and $RHS$, comparing them, then we will get the requiring $f$, and hence $I$. | 498 | 1,820 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2019-35 | latest | en | 0.899561 |
https://mathspace.co/textbooks/syllabuses/Syllabus-408/topics/Topic-7242/subtopics/Subtopic-96780/?activeTab=interactive | 1,642,984,381,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304345.92/warc/CC-MAIN-20220123232910-20220124022910-00036.warc.gz | 423,029,608 | 45,279 | # Mean, median mode and range (calculation from graphs and charts)
## Interactive practice questions
Given the stem-and-leaf plot:
Stem Leaf
$1$1
$2$2 $4$4 $9$9
$3$3 $4$4 $4$4 $4$4 $9$9
$4$4 $1$1 $9$9
$5$5 $1$1 $8$8 $8$8
$6$6 $2$2 $2$2 $9$9
$7$7 $0$0 $9$9
$8$8 $0$0 $9$9 $9$9
$9$9 $4$4
Key: $1$1$\mid$∣$2$2$=$=$12$12
a
Find the mean from the stem-and-leaf plot. Round your answer to two decimal places.
b
Find the mode for the data provided in the stem-and-leaf plot.
c
Find the median.
d
What is the range of the scores?
Easy
Approx 5 minutes
Find the mode from the histogram shown.
Calculate the range from the bar chart.
The stem and leaf plot below shows the age of people who enter through the gates of a concert in the first 5 seconds.
### Outcomes
#### S5-1
Plan and conduct surveys and experiments using the statistical enquiry cycle:– determining appropriate variables and measures;– considering sources of variation;– gathering and cleaning data;– using multiple displays, and re-categorising data to find patterns, variations, relationships, and trends in multivariate data sets;– comparing sample distributions visually, using measures of centre, spread, and proportion;– presenting a report of findings | 376 | 1,236 | {"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} | 3.28125 | 3 | CC-MAIN-2022-05 | longest | en | 0.809744 |
https://es.scribd.com/document/486554542/TRABAJO-DE-PROBABILIDAD-JADER | 1,674,982,484,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499710.49/warc/CC-MAIN-20230129080341-20230129110341-00155.warc.gz | 255,725,435 | 101,493 | Está en la página 1de 5
# Solución taller 1
1(ii):
Un auditor de Health Maintenance Services of Georgia informa que 40% de los
asegurados de 55 años de edad y mayores utilizan la póliza durante el año. Se
seleccionan al azar 15 asegurados de los registros de la compañía.
## Desarrollo de justificación de resultados:
a) p=0,4 → q=0,6; n=15
μ=np=15 ( 0,4 )=6
15 !
b) P ( x=10 )= [ 0,4 10 (0,65 )] ≅ 3.003 ( 8,15373 x 10−6 ) ≅ 0,024486 → 2,45 %
10 !(5 !)
c) P(x ≥ 10)
P(x=10)
f ( x )= n P X ( 1−P)x−n
n= 15 () x
f ( 10 )= 15 (0.40)10 (1−0.4 0)5
P= 40% ( ) 10
x=10 f ( 10 )=3003(0.000104858)(0.07776)
f ( 10 )=0.0245
P(x=11)
f ( x )= n P X ( 1−P)x−n
n= 15 () x
f ( 11 )= 15 ( 0 . 40)11 (1−0. 4 0) 4
P= 40% ( ) 11
x=11 f ( 11 )=1365(0 . 000041943)(0 .1296)
f ( 11 )=0.0074
P(x=12)
f ( x )= n P X ( 1−P)x−n
n= 15 () x
f ( 12 ) = 15 (0 . 40)12 (1−0. 4 0)3
P= 40% ( ) 12
x=12 f ( 12 ) =455( 0. 00016777)(0 .21 6)
f ( 12 ) =0.0016
P(x=13)
f ( x )= n P X ( 1−P)x−n
n= 15 () x
f ( 13 )= 15 (0 . 40)13 (1−0. 4 0)2
P= 40% ( ) 13
x=13 f ( 13 )=105(0 . 000006717)(0 .3 6)
f ( 13 )=0.0002
P(x=14)
f ( x )= n P X ( 1−P)x−n
n= 15 () x
f ( 14 )= 15 (0 . 40)14 (1−0. 4 0)1
P= 40% ( ) 14
x=14 f ( 14 )=15( 0 .000002684)( 0.60)
f ( 14 )=0.000024156
P(x=15)
f ( x )= n P X ( 1−P)x−n
n= 15 () x
f ( 15 )= 15 (0 . 40)15 (1−0. 4 0)0
P= 40% ( ) 15
x=15 f ( 15 )=1(0 . 000001074)(1)
f ( 15 )=0.00001074
P ( x ≥ 10 )=¿
P ( x ≥ 10 )=¿
P ( x ≥ 10 )=0.03372 →≈ 3,37 %
## d) P(x > 10)
P(x=11)
f ( x )= n P X ( 1−P)x−n
n= 15 () x
f ( 11 )= 15 ( 0.40)11 (1−0 .4 0)4
P= 40% ( ) 11
x=11 f ( 11 )=1365(0.000041943)(0.1296)
f ( 11 )=0.0074
P(x=12)
f ( x )= n P X ( 1−P)x−n
n= 15 () x
f ( 12 ) = 15 (0 . 40)12 (1−0. 4 0)3
P= 40% ( ) 12
x=12 f ( 12 ) =455( 0. 00016777)(0 .21 6)
f ( 12 ) =0.0016
P(x=13)
f ( x )= n P X ( 1−P)x−n
n= 15 () x
f ( 13 )= 15 (0 . 40)13 (1−0. 4 0)2
P= 40% ( ) 13
x=13 f ( 13 )=105(0 . 000006717)(0 .3 6)
f ( 13 )=0.0002
P(x=14)
f ( x )= n P X ( 1−P)x−n
n= 15 () x
f ( 14 )= 15 (0 . 40)14 (1−0. 4 0)1
P= 40% ( ) 14
x=14 f ( 14 )=15( 0 .000002684)( 0.60)
f ( 14 )=0.000024156
P(x=15)
f ( x )= n P X ( 1−P)x−n
n= 15 () x
f ( 15 )= 15 (0 . 40)15 (1−0. 4 0)0
P= 40% ( ) 15
x=15 f ( 15 )=1(0 . 000001074)(1)
f ( 15 )=0.00001074
P ( x ≥ 10 )=¿
P ( x ≥ 10 )=¿
P ( x ≥ 10 )=0.00922 →0,922 %
2(ii) Se toma una muestra de 49 observaciones de una población normal con una
desviación estándar de 10. La media de la muestra es de 55. Determine el intervalo
de confianza de 99 % de la media poblacional.
Niveles de confianza:
0,99
99 % → 0,99 ; =0,495→ z=2,58
2
σ
I . C .= X́ ± Z ( )
√n
σ =10 ; X́=55 ; n=49 ; 1−α =99 %
## I . C .=55± 2.58 ( √1049 )=55 ±2.58 ( 107 )=55 ± 2.58 ( 1.43 )
I . C .=55+2.58 ( 1.43 ) ≅ 58,69
I . C .=552.58 ( 1.43 ) ≅ 5 1,3 1
51,3 1≤ μ ≤58,69 | 1,605 | 2,765 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2023-06 | latest | en | 0.174764 |
https://zbmath.org/?q=an:0863.58068 | 1,670,357,913,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711114.3/warc/CC-MAIN-20221206192947-20221206222947-00177.warc.gz | 1,180,178,965 | 9,858 | ## Damped wave equation. (Equation des ondes amorties.)(French)Zbl 0863.58068
Boutet de Monvel, Anne (ed.) et al., Algebraic and geometric methods in mathematical physics. Proceedings of the 1st Ukrainian-French-Romanian summer school, Kaciveli, Ukraine, September 1-14, 1993. Dordrecht: Kluwer Academic Publishers. Math. Phys. Stud. 19, 73-109 (1996).
Let $$(M,g)$$ be a $$C^\infty$$ compact Riemannian manifold with boundary, with Laplacian $$\Delta$$, and let $$a$$ be a $$C^\infty (M, \mathbb{R}^+)$$ function. One considers the evolution problem $\begin{cases} \bigl(\partial^2_t - \Delta+ 2a(x) \partial_t\bigr) u=0 \text{ in } \mathbb{R}_t \times M, u=0 \text{ on } \mathbb{R}_t \times \partial M, \\ u|_{t=0} = u_0\in H^1_0 (M), \quad {\partial u \over \partial t} |_{t=0} = u_1\in L^2(M).\end{cases} \tag{*}$ The author obtains sharp estimates for the resolvent of $$A_a = \left(\begin{smallmatrix} 0 & Id \\ \Delta & -2a \end{smallmatrix} \right)$$ and for the energy. The best exponential decay rate of the solutions of the evolution problem (*) is computed in terms of the spectrum and of the average of $$a(x)$$ on the geodesics of $$M$$.
For the entire collection see [Zbl 0833.00031].
### MSC:
58J45 Hyperbolic equations on manifolds 35L05 Wave equation 35S15 Boundary value problems for PDEs with pseudodifferential operators 58J50 Spectral problems; spectral geometry; scattering theory on manifolds 53C22 Geodesics in global differential geometry | 469 | 1,468 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2022-49 | longest | en | 0.662203 |
http://stackoverflow.com/questions/13939944/check-if-a-date-is-between-two-months-where-the-year-ends-in-august | 1,455,033,366,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701157262.85/warc/CC-MAIN-20160205193917-00346-ip-10-236-182-209.ec2.internal.warc.gz | 209,085,231 | 18,633 | # Check if a date is between two months, where the year ends in August
I'm having difficulties creating a date algorithm (using the DateTime object) that checks to see if a date lands between two given months (where the end of the year is considered Aug 31st).
So the beginning of the year is considered Sept 1st, XXXX and end of the year is considered Aug 31st XXXX+1
``````\$today = new DateTime('now');
``````
I need to see if date is between June and May.
For example if it is Dec 2009
Is the date between June 2009 and May 2010? (yes)
If date is April 2014
Is the date between June 2013 and May 2014? (yes)
If date is July 2014
Is date between June 2013 and May 2014. (no)
-
Why don't you just use the Numeric representation of a month and compare that? – Bryan Allo Dec 18 '12 at 19:29
@BryanAllo, so the problem with that is I need to determine what year it is in as well. If The date is August 1 2014, it is in the 2013-2014 year. If the date is Sept 4, 2015, it is in the 2015-2016 year – kylex Dec 18 '12 at 19:32
Why not convert to UNIX timestamps and do your comparisons? – iroegbu Dec 18 '12 at 19:47
@kylex - Yes. What I meant is basically use the date_add() method and offset your date by 4 months to bring it back in line with the standard calendar and then do your comparison/check. – Bryan Allo Dec 18 '12 at 19:59
DateTime objects can be compared just like two integers:
``````\$date1 = new DateTime('December 2012');
\$date2 = new DateTime('June 2012');
\$date3 = new DateTime('May 2013');
if (\$date1 > \$date2 && Date1 < \$date3)
{
// Bingo!
}
``````
-
But the years in your example are static. – kylex Dec 18 '12 at 19:23
This is just an example. You can change the values as necessary. – John Conde Dec 18 '12 at 19:25
That's the big issue I'm having though, how do I determine the year values dynamically? – kylex Dec 18 '12 at 19:26
Found an answer that got me started:
http://stackoverflow.com/a/11937688/36545
``````function academicYear(DateTime \$userDate) {
\$currentYear = \$userDate->format('Y');
\$cutoff = new DateTime(\$userDate->format('Y') . '/08/31 23:59:59');
if (\$userDate < \$cutoff) {
return (\$currentYear-1) . '/' . \$currentYear;
}
return \$currentYear . '/' . (\$currentYear+1);
}
``````
- | 676 | 2,251 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-07 | latest | en | 0.914307 |
https://wiki.mathnt.net/index.php?title=%EC%97%AD%ED%96%89%EB%A0%AC | 1,642,555,387,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301217.83/warc/CC-MAIN-20220119003144-20220119033144-00169.warc.gz | 678,634,861 | 7,678 | # 역행렬
둘러보기로 가기 검색하러 가기
## 가우스-조단 소거법을 이용한 계산
• 주어진 행렬은 다음과 같다$\left( \begin{array}{ccc} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1 \end{array} \right)$
• 가우스-조단 소거법 을 이용하기 위해, 다음과 같은 붙임행렬(augmented matrix)을 만든다$\left( \begin{array}{ccc|ccc} 2 & -1 & 0 & 1 & 0 & 0 \\ -1 & 2 & -1 & 0 & 1 & 0 \\ 0 & -1 & 1 & 0 & 0 & 1 \end{array} \right)$
• 위의 행렬에 소거법을 적용하면, 다음의 행렬들을 얻는다
$\begin{array}{l} \left( \begin{array}{ccc|ccc} 2 & -1 & 0 & 1 & 0 & 0 \\ -1 & 2 & -1 & 0 & 1 & 0 \\ 0 & -1 & 1 & 0 & 0 & 1 \end{array} \right) \\ \left( \begin{array}{ccc|ccc} 1 & -\frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ -1 & 2 & -1 & 0 & 1 & 0 \\ 0 & -1 & 1 & 0 & 0 & 1 \end{array} \right) \\ \left( \begin{array}{ccc|ccc} 1 & -\frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ 0 & \frac{3}{2} & -1 & \frac{1}{2} & 1 & 0 \\ 0 & -1 & 1 & 0 & 0 & 1 \end{array} \right) \\ \left( \begin{array}{ccc|ccc} 1 & -\frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} & 0 \\ 0 & -1 & 1 & 0 & 0 & 1 \end{array} \right) \\ \left( \begin{array}{ccc|ccc} 1 & 0 & -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} & 0 \\ 0 & 1 & -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} & 0 \\ 0 & -1 & 1 & 0 & 0 & 1 \end{array} \right) \\ \left( \begin{array}{ccc|ccc} 1 & 0 & -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} & 0 \\ 0 & 1 & -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} & 0 \\ 0 & 0 & \frac{1}{3} & \frac{1}{3} & \frac{2}{3} & 1 \end{array} \right) \\ \left( \begin{array}{ccc|ccc} 1 & 0 & -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} & 0 \\ 0 & 1 & -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} & 0 \\ 0 & 0 & 1 & 1 & 2 & 3 \end{array} \right) \\ \left( \begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & 1 & 1 \\ 0 & 1 & -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} & 0 \\ 0 & 0 & 1 & 1 & 2 & 3 \end{array} \right) \\ \left( \begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 2 & 2 \\ 0 & 0 & 1 & 1 & 2 & 3 \end{array} \right) \end{array}$
• 위의 결과로부터 주어진 행렬의 역행렬은 다음과 같음을 알 수 있다$\left( \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{array} \right)$ | 1,137 | 2,019 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2022-05 | latest | en | 0.165154 |
https://www.justanswer.com/math-homework/2o7ou-checkpoint-radical-equations-w-mathlab-hey-scott.html | 1,632,452,703,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057496.18/warc/CC-MAIN-20210924020020-20210924050020-00661.warc.gz | 848,748,274 | 51,909 | Math Homework
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Show MoreShow Less | 2,340 | 9,732 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-39 | latest | en | 0.934911 |
https://www.cookleysebright.co.uk/friday-8/ | 1,656,297,228,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103324665.17/warc/CC-MAIN-20220627012807-20220627042807-00797.warc.gz | 779,523,872 | 9,737 | # Friday
PHYSICAL DEVELOPMENT
Get your heart beating fast with Joe, Oti or Just Dance movement sessions or even make up your own workout session. Keep going for 30 minutes to give your body the full benefit of your choice of fitness session.
LITERACY: PHONICS
Our sound of the day is ir (as in whirl and twirl). Using the green phonics folder, practise the ir sound. The sound is made up of two letters but makes one sound, which is ir. Then practise reading the ir words. Firstly encourage your child to identify the special friend in the word, (in this case ir) then Fred talk the word; s t ir, then say the word in full; stir.
Children can then practise spelling the words, this can be done by listening to the word and wiggling your fingers. How many sounds can you hear in stir? Three sounds, hold up three fingers, then pinch each finger as you say the sound. Children can then write the word. As an extension, you could ask your child to write the sentence, I stir the mix to cook. Encourage your child to start the sentence with a capital letter, after they sound out and record each word remind them to leave a finger space before starting to write the next word. Finally remind them to finish the sentence with a full stop.
Now flick back to ay, ee, igh, ow, oo, oo, ar, or, and air, encourage your child to identify the sound without seeing the picture clue and read and spell some of the words. Continue to watch the Read Write Inc. Phonics lessons on either Facebook or You Tube, set 2 sounds are now streamed live at 10.00 then are available for 24 hours.
MATHEMATICS: SHAPE, SPACE AND MEASURE
For this task we are going to be looking at the work of the artist Paul Cezanne who was a still life artist. He painted what he saw. Look at the bowl of fruit (image below). Is the bowl full or empty? Do you think anymore fruit might fit in? How many pieces of fruit are there in the bowl? How many pieces of fruit are on the table? How many pieces of fruit are there altogether?
Can you draw three bowls of fruit; one that is full, one that is half full and one that is empty and correctly label them. If you have a bowl and some fruits you could explore using the fruits then draw what you see.
## Paul Cezanne Fruit and jug on a table.
UNDERSTANDING THE WORLD
With your parents I would like you to watch Beginners Bible the Easter Story to help you understand why we celebrate Easter. We would have made a Easter card this week, perhaps you could make one at home and send it to someone who you can't visit at the moment. You could write a special message in your card.
During the holiday you might also choose to make an Easter bonnet, take part in an Easter Egg Hunt, decorate a hard boiled egg, make chocolate Easter nests, make a Easter themed basket and even an Easter garden to help you recall and retell the story. There will also be some interesting clips to watch and discuss on Discovery Education, follow the link on the home learning page. | 672 | 2,975 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.937111 |
https://math.stackexchange.com/questions/3778550/how-to-evaluate-the-limit-of-multifactorial-lim-n-to-0-sqrtnn-cdots/4029044?noredirect=1 | 1,627,411,841,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153474.19/warc/CC-MAIN-20210727170836-20210727200836-00036.warc.gz | 403,623,958 | 41,385 | How to evaluate the limit of multifactorial $\lim_{n\to 0} \sqrt[n]{n!!!!\cdots !}$
It is well known that $$\displaystyle \lim_{n\to \infty}\sqrt[n]{n!}=\infty$$, however, if we let $$n\to 0$$ we have a different result with a beautiful combination of $$e$$ and $$\gamma$$, that is
$$\lim_{n\to 0}\sqrt[n]{n!}= e^{-\gamma}\tag{1}\label{1}$$
To prove result \eqref{1}, we observe that the limit attains the form of $$1^{\infty}$$ so we can write it as $$\lim_{n\to 0} \exp\left(\frac{\ln\Gamma(n+1)}{n}\right)\underbrace{=}_{\text{L'Hopital's rule}}\lim_{n\to 0} e^{\Gamma'(n+1)}=e^{\psi_0(1)}= e^{-\gamma}$$
Now I wish to know the limit of the following multifactorial form for $$k\in\mathbb {Z^+}$$
$$\lim_{n\to 0}\sqrt[n]{n\smash[b]{\underbrace{!! !!\cdots !}_{k}}}={?}\tag{2}\label{2}\\$$
For $$k=1$$ we are done above and for $$k=2$$ we get the limit $$\sqrt{2} e^{-\frac{\gamma}{2}}$$. To prove this we use the double factorial argument (see equation (5)) \begin{align}\lim_{n\to 0}\sqrt[n]{n!!}&=\lim_{n\to 0} \left(2^{\frac{n}{2}+\frac{1-\cos(\pi n)}{4}}\pi^{\frac{\cos(\pi n)-1}{4}}\Gamma\left(1+\frac{n}{2}\right)\right)^{\frac{1}{n}}\\&=\sqrt{2}\lim_{n\to 0} \sqrt[n]{\Gamma\left(1+\frac{n}{2}\right)}\\&=\sqrt 2\exp\lim_{n\to 0}\left(2^{-1} \Gamma\left(1+\frac{n}{2}\right)\psi_0\left(1+\frac{n}{2}\right)\right)\tag{L'Hopital's rule}\\&=\sqrt{2}e^{-\frac{\gamma}{2}}\end{align}
since for $$k=1,2$$ we have evaluated the limit. How to evaluate the limit of equation \eqref{2} for all $$k>2$$?
• Is the multifactorial here equivalent to $$\frac{k^{\frac{n-1}{k}} \Gamma \left(\frac{n}{k}+1\right)}{\Gamma \left(1+\frac{1}{k}\right)}$$? – Benedict W. J. Irwin Aug 3 '20 at 13:43
• A warning to others: In Mathematica $n!!!$ is not the triple factorial, it is equivalent to $(n!!)!$. – Benedict W. J. Irwin Aug 3 '20 at 14:36
• @ Benedict, your multifactorial expression holds only for $k=1$ and for $k=2$ it holds only if $n$ is odd integer. For $k=3$ it doesn't hold. I agree with you regarding $n!!!$ since before making post I checked in WA. It interprets it's as $(n!!)!$. – Naren Aug 4 '20 at 5:57
• It seems like the real question here is how to generalize multifactorials to complex arguments in a way that's akin to how the gamma function generalizes factorials. There may not be a straightforward answer, see math.stackexchange.com/questions/3191739/… and math.stackexchange.com/questions/291890/…. Certainly there's not a unique holomorphic or meromorphic interpolation, so it's conceivable it could end up depending on the gamma-like properties you want the function to satisfy. – Jason Aug 4 '20 at 7:05
The main task in this problem is to construct an analytic continuation of the multifactorial function $$n!^{(k)}=n(n-k)(n-2k)\cdots$$ over the real numbers. On the set of integers, we can take advantage of modulo arithmetic to arrive at $$n!^{(k)}_i=i\cdot k^{(n-i)/k}\frac{\Gamma(1+n/k)}{\Gamma(1+i/k)}$$ which is defined whenever $$n\equiv i\pmod k$$. Therefore, we need to find a function which interpolates the values of $$1,\frac{k^{-1/k}}{\Gamma(1+1/k)},\frac{2k^{-2/k}}{\Gamma(1+2/k)},\cdots,\frac{(k-1)k^{-(k-1)/k}}{\Gamma(1+(k-1)/k)}$$ so we consider the ansatz $$f(x)=\prod_{i=1}^{k-1}\left(\frac{ik^{-i/k}}{\Gamma(1+i/k)}\right)^{g(x)}$$ where $$g(x)\equiv1$$ when $$x-i\equiv0\pmod k$$ and is zero otherwise. This suggests the analytic function $$\frac{\sin(\pi(x-i))}{k\sin(\pi(x-i)/k)}$$ which is unfortunately negative when $$(x-i)/k$$ is an odd integer. We can circumvent this by introducing a factor of $$\cos(\pi(x-i)/k)$$ as its sign agrees with that of the above function when $$x-i\equiv0\pmod k$$. It follows that we can extend the multifactorial function to the reals through $$x!^{(k)}=k^{x/k}\Gamma\left(1+\frac xk\right)\prod_{i=1}^{k-1}\left(\frac{ik^{-i/k}}{\Gamma(1+i/k)}\right)^{\sin(\pi(x-i))\cot(\pi(x-i)/k)/k}\tag1$$ It is worth noting that this is not unique, as we can multiply by a factor of $$\cos(j\pi(x-i)/k)$$ for some integer $$j$$ — here we just took $$j=1$$. We can now determine the limit by considering each term separately \begin{align}\lim_{x\to0}x!^{(k)/x}&=k^{1/k}\lim_{x\to0}\exp\left(\frac{\log\Gamma(1+x/k)}x\right)\lim_{x\to0}\prod_{i=1}^{k-1}\left(\frac{ik^{-i/k}}{\Gamma(1+i/k)}\right)^{\sin(\pi(x-i))\cot(\pi(x-i)/k)/kx}\\&=k^{1/k}e^{-\gamma/k}\prod_{i=1}^{k-1}\left(\frac{ik^{-i/k}}{\Gamma(1+i/k)}\right)^{h(k)}\end{align} where \begin{align}h(k)&=\lim\limits_{x\to0}\frac{\sin(\pi(x-i))\cot(\pi(x-i)/k)}{kx}=-\cot\frac{\pi i}k\lim\limits_{x\to0}\frac{\sin\pi x\cos\pi i}{kx}\\&=-(-1)^i\frac\pi k\cot\frac{\pi i}k.\end{align} Using this definition of $$g(x)$$, the limit evaluates to $$\lim_{x\to0}x!^{(k)/x}=\left[\frac k{e^\gamma}\prod_{i=1}^{k-1}\left(\frac{\Gamma(1+i/k)}{ik^{-i/k}}\right)^{\pi(-1)^i\cot\frac{\pi i}k}\right]^{1/k}.$$ Note that the multifactorial function in $$(1)$$ can be extended onto the complex plane. It is holomorphic everywhere except at negative integer multiples of $$k$$, similar to the gamma function.
• It's late and I have insomnia but I had a realization that perhaps the notation for a multifactorial could be written as $$n \overset{k}{.}$$ since $k = 1$ looks like $$n\overset{1}{.} = n!$$ It's a sort of visual pun I guess. Surely I cannot be the first to have thought of this. – heropup Feb 17 at 11:54
• I tried in Mathematica: Multifactorial[x_, k_] := k^(x/k)*Gamma[ 1 + x/k] Product[( (j*k^(-j/k))/ Gamma[1 + j/k])^(1/k*Sin[Pi (x - j)] Cot[Pi*(x - j)/k]), {j, 0, k - 1}]; Multifactorial[1, 1] give me: Indeterminate ? What is wrong ? – Mariusz Iwaniuk Feb 17 at 17:01
• @MariuszIwaniuk Ah, I made a typo in that the product should go from $i=1$ not $i=0$. For those without Mathematica, here is a visualisation. – TheSimpliFire Feb 17 at 17:25 | 2,097 | 5,779 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 36, "wp-katex-eq": 0, "align": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2021-31 | latest | en | 0.600638 |
https://www.traditionaloven.com/tutorials/distance/convert-china-yin-unit-to-gunters-link-li-lnk.html | 1,627,388,683,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153391.5/warc/CC-MAIN-20210727103626-20210727133626-00004.warc.gz | 1,105,442,712 | 17,447 | Convert 引 to li - lnk | Chinese yǐn to links
# length units conversion
## Amount: 1 Chinese yǐn (引) of length Equals: 0.017 links (li - lnk) in length
Converting Chinese yǐn to links value in the length units scale.
TOGGLE : from links into Chinese yǐn in the other way around.
## length from Chinese yǐn to link conversion results
### Enter a new Chinese yǐn number to convert
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many digits after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other length measuring units - complete list.
How many links are in 1 Chinese yǐn? The answer is: 1 引 equals 0.017 li - lnk
## 0.017 li - lnk is converted to 1 of what?
The links unit number 0.017 li - lnk converts to 1 引, one Chinese yǐn. It is the EQUAL length value of 1 Chinese yǐn but in the links length unit alternative.
引/li - lnk length conversion result From Symbol Equals Result Symbol 1 引 = 0.017 li - lnk
## Conversion chart - Chinese yǐn to links
1 Chinese yǐn to links = 0.017 li - lnk
2 Chinese yǐn to links = 0.033 li - lnk
3 Chinese yǐn to links = 0.050 li - lnk
4 Chinese yǐn to links = 0.066 li - lnk
5 Chinese yǐn to links = 0.083 li - lnk
6 Chinese yǐn to links = 0.099 li - lnk
7 Chinese yǐn to links = 0.12 li - lnk
8 Chinese yǐn to links = 0.13 li - lnk
9 Chinese yǐn to links = 0.15 li - lnk
10 Chinese yǐn to links = 0.17 li - lnk
11 Chinese yǐn to links = 0.18 li - lnk
12 Chinese yǐn to links = 0.20 li - lnk
13 Chinese yǐn to links = 0.22 li - lnk
14 Chinese yǐn to links = 0.23 li - lnk
15 Chinese yǐn to links = 0.25 li - lnk
Convert length of Chinese yǐn (引) and links (li - lnk) units in reverse from links into Chinese yǐn.
## Length, Distance, Height & Depth units
Distance in the metric sense is a measure between any two A to Z points. Applies to physical lengths, depths, heights or simply farness. Tool with multiple distance, depth and length measurement units.
# Converter type: length units
First unit: Chinese yǐn (引) is used for measuring length.
Second: link (li - lnk) is unit of length.
QUESTION:
15 引 = ? li - lnk
15 引 = 0.25 li - lnk
Abbreviation, or prefix, for Chinese yǐn is: | 703 | 2,223 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.775177 |
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# Math in Motion-I am Lucky I can Add to 10! Find missing Addend-recording sheet
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This March activity will get the kids up and adding! Post the cards around the room. ex. 9 + __=10 Your students will use the tens frame to help solve the number sentence. They will find the missing addend and learn how to decompose 10. They will record answers on their sheet. Have fun playing and learning!
CCSS.Math.Content.K.OA.A.4 For any number from 1 to 9, find the number that makes 10 when added to the given number, e.g., by using objects or drawings, and record the answer with a drawing or equation.
CCSS.Math.Content.K.OA.A.3 Decompose numbers less than or equal to 10 into pairs in more than one way, e.g., by using objects or drawings, and record each decomposition by a drawing or equation (e.g., 5 = 2 + 3 and 5 = 4 + 1).
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\$2.50 | 360 | 1,222 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-13 | longest | en | 0.864966 |
http://www.ck12.org/book/CK-12-Basic-Geometry-Concepts/section/11.0/ | 1,412,096,033,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1412037663036.27/warc/CC-MAIN-20140930004103-00222-ip-10-234-18-248.ec2.internal.warc.gz | 471,312,297 | 26,458 | <meta http-equiv="refresh" content="1; url=/nojavascript/"> Surface Area and Volume | CK-12 Foundation
# Chapter 11: Surface Area and Volume
Created by: CK-12
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## Introduction
In this chapter we extend what we know about two-dimensional figures to three-dimensional shapes. First, we will define the different types of 3D shapes and their parts. Then, we will find the surface area and volume of prisms, cylinders, pyramids, cones, and spheres.
## Summary
This chapter presents three-dimensional geometric figures beginning with polyhedrons, regular polyhedrons, and an explanation of Euler's Theorem. Three-dimensional figures represented as cross sections and nets are discussed. Then the chapter branches out to the formulas for surface area and volume of prisms, cylinders, pyramids, cones, spheres and composite solids. The relationship between similar solids and their surface areas and volumes are explored.
Basic
8 , 9 , 10
Feb 24, 2012 | 214 | 960 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-41 | longest | en | 0.887756 |
http://www.vedicmaths.org/Introduction/Vedic%20Code%20and%20pi.asp | 1,368,985,138,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368697917013/warc/CC-MAIN-20130516095157-00035-ip-10-60-113-184.ec2.internal.warc.gz | 771,907,774 | 2,444 | Introduction The Vedic Numerical Code and pi Home Page What is Vedic Maths? This code is given by Bharati Krsna in Chapter XXV of his book. It allows any Sanskrit text to be converted to numbers, or any Sanskrit text to be composed so that it describes a sequence of numbers. Bharati Krsna says: 'It is a matter of historical interest to note that, in their mathematical writings, the ancient Sanskrit writers do not use figures (when big numbers are concerned) in their numerical notations but prefer to use the letters of the Sanskrit (Devanagari) alphabet to represent the various numbers! And this they do, not in order to conceal knowledge but in order to facilitate the recording of their arguments, and the derivation conclusions etc.' The Key to the Code 'The very fact that the alphabetical code (as used by them for this purpose is in the natural order and can be immediately interpreted, is clear proof that the code language was resorted not for concealment but for greater ease in verification etc., and the key has also been given in its simplest form: The vowels (not included in the above list) make no difference; and in conjunct consonants, the last consonant is alone to be counted.' So if you want to describe the number 11 there are many choices: papa, tata, tapi for example. The Value of pi copyright to the ACADEMY OF VEDIC MATHEMATICS Home Page Site Map Contact us | 309 | 1,393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2013-20 | latest | en | 0.934222 |
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Get Full Access to University Physics - 13 Edition - Chapter 3 - Problem 7dq
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# Sketch the six graphs of the x- and y-components of
ISBN: 9780321675460 31
## Solution for problem 7DQ Chapter 3
University Physics | 13th Edition
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Problem 7DQ
Sketch the six graphs of the x- and y-components of position, velocity, and acceleration versus time for projectile motion with x0 = y0 = 0 and 0 < ?0 < 90o.
Step-by-Step Solution:
Solution 7DQ Suppose the projectile was launched at an angle . If lies between 0° and 90° then both x and y component of the initial velocity will be non zero. Let us also consider the total flight time is T and the upward direction is the positive y direction. Step 1 Acceleration. As we know that only acceleration that works for projectile motion is the acceleration due to gravity and which acts vertically downwards. Hence if we write the acceleration in component form, we will have 2 a xt) = 0 m/s ….(1) a (t) = g = 9.8 m/s ….(2) y Here we have considered the vertical direction as y axis and horizontal as x axis. So the graphical representation for axis And the a vs t graph is shown below y
Step 2 of 3
Step 3 of 3
#### Related chapters
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Board CBSE Textbook NCERT Class Class 8 Subject Maths Chapter Chapter 4 Chapter Name Practical Geometry Exercise Ex 4.4 Number of Questions Solved 1 Category NCERT Solutions
## NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.4
Question 1.
DE = 4 cm
EA = 5 cm
AR = 4.5 cm
∠E = 60°
∠A = 90°
TR = 3.5 cm
RU = 3 cm
UE = 4 cm
∠R = 75°
∠U= 120°
Solution.
(i) Steps of Construction
1. Draw DE = 4 cm.
2. At E, draw ray EX such that ∠DEX = 60°.
3. From ray EX, cut EA = 5 cm.
4. At A, draw ray AY such that ∠EAY = 90°.
5. Cut AR = 4.5 cm from ray AY.
6. Join RD.
Then, DEAR is the required quadrilateral.
(ii) Steps of Construction
1. Draw TR = 3.5 cm.
2. At R, draw ray RX such that ∠TRX = 75°.
3. Cut RU = 3 cm from ray RX.
4. At U, draw ray UY such that ∠RUY = 120°.
5. Cut UE = 4 cm from ray UY.
6. Join ET.
Then, TRUE is the required quadrilateral.
We hope the NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.4 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.4, drop a comment below and we will get back to you at the earliest. | 442 | 1,344 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.822758 |
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## PDF 4-SSS, SAS, ASA, and AAS Congruence - Kuta Software LLC
Related posts of "Triangle Congruence Worksheet 1 Answer Key" Electricity Worksheet Pdf In advance of discussing Electricity Worksheet Pdf, please recognize that Schooling is normally your issue to a more rewarding the day after tomorrow, in addition to mastering doesn't simply halt the instant the varsity bell rings.Congruent Triangles Three of proving triangles congruent: Side-Side-Side (SSS) 3 aspects of 1 triangle are congruent to three aspects of any other triangle. (SAS) 2 facets and the included angle of one triangle are congruent to two aspects and the included angle of another triangle. (ASA) angles and facet of one triangle are congruent to 2Congruent Triangles Proofs Pages 16-21 This Packet pages 22-24 C.P.C.T.C. Pages 25-29 Pages 127-129 #'s 6,12,13,18,21 Geometry Honors Answer Key Proving Triangles Congruent with Hypotenuse Leg Page 158 #'s 5 , 12 and 17 12) Right Angle Theorem and Equidistance Theorems Pages 182 - 183 #'s 4, 9, 14Sec 1.6 CC Geometry - Triangle Proofs Name: POTENTIAL REASONS: Definition of Congruence: Having the exact same size and shape and there via having the very same measures. Definition of Midpoint: The level that divides a section into two congruent segments. Definition of Angle Bisector: The ray that divides an angle into two congruent angles.
### Triangle Congruence Worksheet | Teachers Pay Teachers
Using Cpctc With Triangle Congruence Answer Key - Displaying most sensible 8 worksheets found for this concept.. Some of the worksheets for this concept are Using cpctc with triangle congruence, Proving triangles congruent, 4 congruence and triangles, Congruent triangles paintings 1, 4 s sas asa and aas congruence, Proofs paintings cpctc, Unit 3 syllabus congruent triangles.Prove triangle Congruence Worksheet 1 Answer Key - Learning in regards to the exact value of cash amongst. - Learning concerning the precise worth of cash is without doubt one of the primary classes of! Awesome geometry proofs from triangle Congruence Worksheet Answer Key, supply: careless.me geometry proofs from triangle Worksheet.pg. 234 #3-11, 19, 22-25, 31 (15 problems) Triangle Congruence Worksheet #1 Friday, 11/9/12 . 4-5: ASA, AAS, and HL I will be able to turn out triangles are congruent the use of ASA, AAS, and HL I can mark pieces of a triangle congruent given how they are to be proved. congruent. PRACTICE: Triangle Congruence Worksheet #2. Monday, 11/12/12Triangle congruence worksheet 1 answer key or congruent triangles worksheet grade 7 kidz actions the estimating worksheet is intended to direct you get in the course of the estimation follow. Complete the congruence remark by means of writing down the corresponding side or the corresponding angle of the triangle.©x I2h0 M1F1M 8K 8uxt2ay FSlo 6fYtaweadr Qek 2LgLcCZ.4 2 bA Xlpl l Qr5i og 1htjs R Srefs eYrnv Zepd X.S d jM8aadce M gw 0i it Ih s mIDntf Bisnci St HeG eGoeFo 1mkeQtWriy o. C Worksheet via Kuta Software LLC State what more information is needed to be able to know that the triangles are congruent for the rationale given. 11) SAS J H I E G IJ
### Triangle Congruence Worksheet Answer Key Briefencounters 3
triangle congruence worksheet answer key briefencounters 3
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Maths For 10 Year Olds Worksheets
Ahead of discussing Maths For 10 Year Olds Worksheets, you will have to needless to say Education and finding out may also be every of our answer to a greater tomorrow, and also mastering does no longer only halt the instant the establishment bell rings. In which being reported, most people come up with quite a lot of simple however... | 1,203 | 5,005 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | latest | en | 0.823965 |
http://www.dummies.com/how-to/content/1001-algebra-ii-practice-problems-for-dummies-chea.navId-323223.html | 1,464,154,815,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049274119.75/warc/CC-MAIN-20160524002114-00018-ip-10-185-217-139.ec2.internal.warc.gz | 480,110,614 | 15,565 | # 1,001 Algebra II Practice Problems For Dummies
The best way to figure out how the different algebraic rules work and interact with one another is to practice with lots of problems. And Algebra II requires lots of practice. So be prepared to solve equations and systems, graph lines, tackle functions, and so much more.
## Working with Radical and Rational Equations in Algebra II
A radical equation is one that starts out with a square root, cube root, or some other root and gets changed into another form to make the solving process easier. A rational equation is one that involves a fractional expression — usually with a polynomial in the numerator and denominator. Avoid these mistakes when working with radical or rational equations:
• Forgetting to check for extraneous solutions
• Squaring a binomial incorrectly when squaring both sides to get rid of the radical
• Distributing correctly when writing equivalent fractions using a common denominator
• Eliminating solutions that create a 0 in the denominator
## Polynomial Functions and Equations in Algebra II
In Algebra II, a polynomial function is one in which the coefficients are all real numbers, and the exponents on the variables are all whole numbers. A polynomial whose greatest power is 2 is called a quadratic polynomial; if the highest power is 3, then it’s called a cubic polynomial. A highest power of 4 earns the name quartic (not to be confused with quadratic), and a highest power of 5 is called quintic.
When solving polynomial functions and equations, don’t let these common mistakes trip you up:
• Forgetting to change the signs in the factored form when identifying x-intercepts
• Making errors when simplifying the terms in f(–x) applying Descartes’ rule of sign
• Not changing the sign of the divisor when using synthetic division
• Not distinguishing between curves that cross from those that just touch the x-axis at an intercept
• Graphing the incorrect end-behavior on the right and left of the graphs
## Systems of Linear Equations in Algebra II
In Algebra II, a linear equation consists of variable terms whose exponents are always the number 1. When you have two variables, the equation can be represented by a line. With three terms, you can draw a plane to describe the equation. More than three variables is indescribable, because there are only three dimensions. When you have a system of linear equations, you can find the values of the variables that work for all the equations in the system — the common solutions. Sometimes there’s just one solution, sometimes many, and sometimes there’s no solution at all.
When solving systems of linear equations, watch out for these mistakes:
• Forgetting to change the signs in the factored form when identifying x-intercepts
• Making errors when simplifying the terms in f(–x) applying Descartes’ rule of sign
• Not changing the sign of the divisor when using synthetic division
• Not distinguishing between curves that cross from those that just touch the x-axis at an intercept
• Graphing the incorrect end-behavior on the right and left of the graphs | 639 | 3,115 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2016-22 | latest | en | 0.903837 |
https://web2.0calc.com/questions/trig-function_4 | 1,532,097,886,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591683.78/warc/CC-MAIN-20180720135213-20180720155213-00421.warc.gz | 796,458,727 | 5,058 | +0
# Trig Function
0
116
1
find sin and tan if cos theta 1
Guest Feb 28, 2017
#1
+87303
0
The cos =1 at 0°
The sine and tan are both = 0 for this angle
CPhill Feb 28, 2017 | 78 | 180 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-30 | latest | en | 0.748814 |
https://www.physicsforums.com/threads/meter-circuit-calculating-scale-reading-with-and-without-additional-resistor.389952/ | 1,713,843,762,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818464.67/warc/CC-MAIN-20240423033153-20240423063153-00518.warc.gz | 847,471,787 | 15,708 | • thereddevils
In summary, the conversation discusses a circuit with a linear scale meter that has not been calibrated and has a scale reading of 20. The conversation goes on to discuss the current in the circuit and how it is affected by adding a 2000 ohm resistor. The correct answer is determined by using the formula I = k*θ, where k is the constant of the meter and θ is the deflection in the meter. The final answer is 25, which is found by equating the current with k*θ.
thereddevils
## Homework Statement
Meter in the circuit shown has a linear scale that has not been calibrated and the scale reading is 20 . When another resistor of resistance 2000 ohm is connected across XY . What is the scale reading of the meter
## The Attempt at a Solution
Without the 2000 ohm resistor , I= 1.5 /(2995 + 50) , I = 0.0005
the reading given by the ammeter is 40600 times of the actual current .
now with the resistor , I = 1.5/(2995 + 50+2000) , I = 0.0003
the reading by the ammater will be 0.0003 x 40600 = 12
but the answer given is 25
#### Attachments
• meter circuit.bmp
216.7 KB · Views: 495
Without the 2000 ohm resistor , I= 1.5 /(2995 + 50) , I = 0.0005
This is wrong. In the given circuit the total resistance is (2995+50 +2000)
Find the current I.
I = k*θ. where k is the constant of the meter and θ is the deflection in the mete.
When you connect another 2000 ohm resistance in parallel with 2000 ohm resistance in the circuit, what is net resistance in the circuit? Find the current and equate it to k*θ'. find θ'.
rl.bhat said:
Without the 2000 ohm resistor , I= 1.5 /(2995 + 50) , I = 0.0005
This is wrong. In the given circuit the total resistance is (2995+50 +2000)
Find the current I.
I = k*θ. where k is the constant of the meter and θ is the deflection in the mete.
When you connect another 2000 ohm resistance in parallel with 2000 ohm resistance in the circuit, what is net resistance in the circuit? Find the current and equate it to k*θ'. find θ'.
got it ! Thanks !
## What is a meter circuit?
A meter circuit is an electrical circuit that is used to measure the amount of electricity flowing through a particular point in a circuit. It typically consists of a meter, such as a voltmeter or ammeter, and a series of wires and components that allow for accurate measurements to be taken.
## What are the components of a meter circuit?
The components of a meter circuit can vary depending on the type of meter being used, but typically include a meter, a shunt or current transformer, and various wires and connectors. Some meter circuits may also include additional components such as resistors or capacitors for more accurate measurements.
## How does a meter circuit work?
A meter circuit works by using the principles of electromagnetism. The meter, which is calibrated to measure a specific type of electrical current, is connected in series with the circuit being measured. As the current flows through the circuit, it also flows through the meter, which then displays the measurement based on the movement of a needle or digital display.
## What is the purpose of a meter circuit?
The purpose of a meter circuit is to accurately measure the amount of electricity flowing through a particular point in a circuit. This information can be used to monitor and troubleshoot electrical systems, as well as ensure that the correct amount of electricity is being used in a given circuit.
## What are some common types of meter circuits?
Some common types of meter circuits include voltmeter circuits, ammeter circuits, and wattmeter circuits. Other specialized types of meter circuits include multimeter circuits, which can measure multiple types of electrical quantities, and oscilloscope circuits, which can display the waveform of an electrical signal.
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1K | 1,076 | 4,370 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.894082 |
https://www.topperlearning.com/answer/please-solve-this-question/rrpjv5ee | 1,716,803,260,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059039.52/warc/CC-MAIN-20240527083011-20240527113011-00230.warc.gz | 890,376,719 | 57,533 | Request a call back
Asked by Vidushi412 | 11 Aug, 2020, 10:20: PM
Question:
A man divided his circular piece of land, measuring 1100 sq. m, between his two sons. He gave his elder son a square piece of land within the circular piece. He gave his younger son the remaining part around the square. However, he compensated his younger for this awkward shaped land by giving him 20% more than what he gave to his elder son. What is the length of the side of the square-piece land?
Solution:
Let the length of the side of the square-piece land be x.
Area of the square piece land = x2
As per the question,
x2 + (x2 + 20% of x2) = 1100
Answered by Renu Varma | 12 Aug, 2020, 11:35: AM
CBSE 10 - Maths
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Asked by Topperlearning User | 27 Jul, 2017, 02:28: PM | 454 | 1,296 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.959435 |
https://discuss.codechef.com/t/litephrs-i-am-using-dp-to-store-the-values-but-i-am-missing-out-something-and-its-giving-me-tle/41642 | 1,601,375,026,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401641638.83/warc/CC-MAIN-20200929091913-20200929121913-00550.warc.gz | 350,233,005 | 4,868 | # LITEPHRS - I am using dp to store the values but i am missing out something and its giving me tle
link to submission: https://www.codechef.com/viewsolution/27418409
link to problem: https://www.codechef.com/problems/LITEPHRS/
``````#include <bits/stdc++.h>
using namespace std;
struct pair_hash
{
template <class T1, class T2>
std::size_t operator() (const std::pair<T1, T2> &pair) const
{
return std::hash<T1>()(pair.first) ^ std::hash<T2>()(pair.second);
}
};
unordered_map<pair<long long int,long long int>, pair<long long int,long long int>, pair_hash > m;
#define mod 1000000007
#define pp cout << "Hit\n"
long long int factorial(int n){
long long int res = 1;
for(int i=2;i<=n;++i){
res = (res * i)%mod;
}
return res;
}
long long int func(int c, int v, int nc, int nv){
// cout << c<<" "<<v<<" "<<nc<<" "<<nv<<'\n';
if(nc == 0) return factorial(nv);
if(nv == 0) return factorial(nc);
if(v < c) {
if(m.find({nc, nv}) != m.end()){
// pp;
return m[{nc, nv}].first;
}
m[{nc,nv}] = ( make_pair( (nv*func(c,v+1,nc,nv-1))%mod + (nc*func(c+1,v,nc-1,nv))%mod , 0 ));
return m[{nc,nv}].first;
}
else{
if(m.find({nc,nv}) != m.end()) {
// pp;
return m[{nc,nv}].second;
}
m[{nc,nv}] = ( make_pair(0, (nc*func(c+1,v,nc-1,nv))%mod ));
return m[{nc,nv}].second;
}
}
bool isVowel(char ch){
if(ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u') return true;
return false;
}
int main(){
ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
int n; cin >> n;
string str; cin >> str;
int c=0,v=0;
for(int i=0;i<str.size();++i){
if(isVowel(str[i]))v++;
else c++;
}
// cout << c<<" "<<v<<'\n';
if(v>c) cout << 0 << '\n';
else
cout << func(0,0,c,v) << '\n';
return 0;
}
``````
So basically i am using dp to store the calls to function func() which gives the the required result.
at any point we only need to know number of seen vowels and consonants. Depending on these two parameters we can decide whether we can insert a vowel or consonant while making new string.
For example the string cannot begin with vowel, if first letter is consonant then second letter can be vowel or consonant etc.
i am storing the dp call in hash map where each entry is number of consonant (nc), number of vowel (nv) which are yet to appear in the string (i.e remaining number of vowels and consonants) as hash map’s key, and its value is a pair of number denoting two values. First is that given nc and nv, if we can insert vowel then we have two func() calls (each for vowel and consonant at that place). Else we can only insert consonant at that point. Therefore in dp we are storing these two as a pair (first and second respectively).
Therefore for example, If dp[(3,5)] making string using 3 consonants and 5 vowels there can be two set of answers one when we can insert only vowel at the given place else both. So these two values are stored as first and second part of pair at the value of the dp[(3,5)].
Thank you,
I have tried my best to explain my solution feel free if you find some points ambiguous. | 896 | 3,001 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2020-40 | latest | en | 0.275941 |
http://www.uwgb.edu/DutchS/STRUCTGE/SeisRefraction0.htm | 1,508,555,164,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824543.20/warc/CC-MAIN-20171021024136-20171021044136-00608.warc.gz | 616,561,077 | 3,170 | # Seismic Refraction - The Wavefront Approach
Steven Dutch, Natural and Applied Sciences, University of Wisconsin - Green Bay
First-time Visitors: Please visit Site Map and Disclaimer. Use "Back" to return here.
Seismic refraction leads to some puzzling results. If layers are dipping, travel time graphs run in opposite directions have different shapes and crossover distances. Branches of the travel time graph can be flat, as if the velocity of the waves was infinite, or even negative - signals get to distant locations sooner than nearby ones.
In the example above, we can see that profiles run in opposite directions are not equivalent. In the top example, the bed gets shallower with distance, so a large increase in distance leads to a small increase in travel time. In the lower diagram, the bed gets steeper with distance, so an increase in distance leads to a large increase in travel time and a steeper second leg of the travel time graph. Even so, it's pretty unsatisfying intuitively. Fortunately, there's a better way.
The top diagram, using the simple ray approach, is problematic for two reasons. First, how does the signal "know" to come up to a receiver? And second, if the signal is leaking energy all the time as shown, where does the energy come from? If the boundary has zero thickness, doesn't that imply an infinite energy density?
There are two ways to visualize signals: as rays or as wave fronts. At bottom is the wave front visualization. When a signal from the shot point (red x) enters the lower layer, it travels faster. Where the wave front in the lower layer intersects the layer boundary, it disturbs it. The disturbance is traveling with the velocity in the second layer, so it continually outruns its own wave front in the first layer and creates a "bow wave" as shown.
At the crossover distance (blue, at right), the bow wave has outrun the direct wave and signals from the second layer arrive first.
Here's the same situation, visualized as wave fronts. It becomes trivially obvious the two graphs should not be the same. The top case, with the signal traveling updip, will take longer to achieve crossover, but the gently sloping bow wave will hit the surface nearly simultaneously along its whole length. So the graph of travel time in the second layer will be nearly flat.
In the bottom case, with the signal traveling downdip, the steep bow wave will achieve crossover sooner, but its progress across the surface will be slower and the graph of travel times in the second layer will be steeper.
## The "Bow Wave"
If a source of waves (light, sound, seismic, water) is stationary, it emits wave fronts that are concentric circles as shown at left, above. If the source is moving slower than the waves, as at right, above, it partially overtakes waves traveling in the direction of motion. The source also moves away from waves behind it. Waves in the forward direction are compressed, those behind are stretched. This results in the familiar drop in pitch as an ambulance passes by, and the red- and blue- shifting of light in astronomy.
However, if the source is traveling faster than the waves, it outruns its own wave fronts. Each successive wave is emitted beyond the last wavefront. The resulting wave is a V-shaped wave tangent to all the emitted wave fronts. If the signals are light, this is impossible for objects in a vacuum, but is possible for objects in a medium like water. The velocity of light in water is only about 3/4 that in a vacuum. It is easy for particles to travel faster than light in water, for example, particles emitted from a nuclear reactor. These particles emit a conical wave front called Cerenkov Radiation, which is seen as a blue glow. Boats traveling faster than the speed of waves in water create the familiar wake, and aircraft traveling faster than sound trail a conical shock wave which is heard as a sonic boom.
In the case of a seismic signal, if the velocity in the top layer is Vw, and the velocity in the bottom layer is Vs, a disturbance races along the layer boundary with velocity Vs, trailing a bow wave into the upper layer.
Now it becomes easy to see how the apparently impossible case of a flat travel time curve can come about. If the bow wave is parallel to the surface, it will hit the surface everywhere simultaneously (top). If the dip is sufficiently steep and the velocity disparity large enough, the wave front may even slope downdip, in which case the leading part of the wave will hit the surface first. In this case we will actually have a travel time curve with a negative slope (bottom). | 967 | 4,608 | {"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} | 3.25 | 3 | CC-MAIN-2017-43 | latest | en | 0.934473 |
https://codegolf.stackexchange.com/questions/31930/help-me-with-differential-calculus/31935 | 1,713,787,614,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818293.64/warc/CC-MAIN-20240422113340-20240422143340-00877.warc.gz | 152,213,292 | 55,750 | # Help me with differential calculus!
I love programming and know every language, but I suck at math. Unfortunately, my school requires that computers students must take a year of calculus. There's a test next week, and I don't know any of the formulas for derivatives!
Please help me find the formulas. I need a cheat sheet - a program (as short as possible so my teacher won't notice it) that takes an expression (like 4*x^3-2) as input and outputs the derivative. (I don't care if the input and output uses command line arguments, STDIN, STDOUT, or whatever, since I'm doing all of the calculation in my head anyway.)
The test covers the following types of functions:
• Constants, like -3 or 8.5
• Power functions, like x^0.5 or x^-7
• Exponential functions, like 0.5^x or 7^x (the base is always positive)
• A constant multiplied by a function, like 3*x^5 or -0.1*0.3^x
• The sum and difference of multiple functions, like -5*x^2+10-3^x
My teacher always formats his questions in the exact same way, as shown above. He also doesn't use any fractions, numbers like pi or e, or really big numbers (bigger than 1,000). He never uses parentheses, and always shows multiplication using an asterisk (*). The only variable used is always x.
On the other hand, my teacher is pretty lenient about answers. They don't need to be simplified at all or formatted exactly as shown above, as long as it's clear what the answer is saying.
While I can use any language, remember that I can't figure out derivatives by myself. So if the program uses built-in functions to deal with equations or calculate derivatives, I won't be able to use it.
During the test, I won't have access to the Internet or any files other than the program on the cheat sheet.
Note: This scenario is entirely fictional. In real life, cheating and helping others cheat is wrong and should never be done.
• Can we expect that x is always the variable to differentiate? Jun 17, 2014 at 13:39
• Must the answer be simplified? Do we need to add like terms? Jun 17, 2014 at 13:40
• I guess it's time for my calculus project at scrblnrd3.github.io/Javascript-CAS to shine if I can actually golf it Jun 17, 2014 at 14:33
• Should we assume there are no parens? Jun 17, 2014 at 14:59
• I've answered most of these questions in my edit. There is no scientific notation or product rule. Jun 17, 2014 at 17:08
# Perl - 121 122
(+2 for -p)
s/(?<![-\d.*^])-?[\d.]+(?![*^\d.])/0/g;s/(?<!\^)x(?!\^)/1/g;s/x\^(-?[\d.]+)/"$1*x^".($1-1)/ge;s/([\d.]+)\^x/ln($1)*$&/g
Test:
$perl -p diff.pl << EOF > -3 > 8.5 > x^0.5 > x^-7 > 0.5^x > 7^x > 3*x^5 > -0.1*0.3^x > -5*x^2+10-3^x > EOF 0 0 0.5*x^-0.5 -7*x^-8 ln(0.5)*0.5^x ln(7)*7^x 3*5*x^4 -0.1*ln(0.3)*0.3^x -5*2*x^1+0-ln(3)*3^x • Yet another reason for me to learn regex... Jun 17, 2014 at 15:27 • @KyleKanos Don't. Regex is bad, regex is terrific. Jun 17, 2014 at 15:35 • Meh, beat me to it. Not bad! (PS: regex is beautiful) Jun 17, 2014 at 15:36 • I have no idea what's going on here. +1 – qwr Jun 18, 2014 at 6:02 • Explanation: Constant --> 0, x --> 1, x^n --> n*x^(n-1), a^x --> ln(a)*a^x Jun 18, 2014 at 9:56 ## Wolfram 136134 109[Thanks to Calle for his comment below] Limited support for product and chain rules. n=n_?NumberQ;d[v_Plus]:=d/@v;d[v_]:=v/.{x_^n:>x^(n-1)d[x]n,n^x_:>Log[n]d[x]n^x,x_*y__:>d[x]y+d[y]x,n:>0,x:>1} Example: d[3^(x^2)*(x^3+2*x)^2] >> 2*3^x^2*(2+3*x^2)*(2*x+x^3) + 2*3^x^2*x*(2*x+x^3)^2*Log[3] Note that this does not use any "built-in functions to deal with equations or calculate derivatives": only pattern-matching is involved*. [*Well... technically the interpreter also parses and builds a sort of AST from the input too] Ungolfed: d[expr_Plus] := d /@ expr; d[expr_] := expr /. { Power[x_, n_?NumberQ] :> n Power[x, n - 1] d[x], Power[n_?NumberQ, x_] :> Log[n] Power[n, x] d[x], Times[x_, y__] :> d[x] y + d[y] x, n_?NumberQ :> 0, x :> 1 } • This is another version. You don't have to write Power, Times etc. IDK how much that will improve your golfed version though, but you have at least one Times in there so you can def. save some characters. Also note that in your ungolfed version it says d[expr_]:= v/.... – user11030 Jun 19, 2014 at 11:07 • @Calle "IDK how much that will improve your golfed version" -- 25 bytes! Cheers! Jun 19, 2014 at 11:22 ## Haskell 38 Chars The function d takes a function and returns a function. It is inputted in the form of a power series, and is outputted the same way (which is a type of whatever.) d=zipWith(*)[1..].tail For example, if we input x->x^2, we get x->2*x. λ <Prelude>: d [0,0,1] [0,2] And for the exponential function. λ <Prelude>: take 10 exp --exp redefined above to be in power series notation [1.0,1.0,0.5,0.16666666666666666,4.1666666666666664e-2,8.333333333333333e-3,1.388888888888889e-3,1.984126984126984e-4,2.48015873015873e-5,2.7557319223985893e-6] λ <Prelude>: let d=zipWith(*)[1..].tail in take 10$ d exp
[1.0,1.0,0.5,0.16666666666666666,4.1666666666666664e-2,8.333333333333333e-3,1.388888888888889e-3,1.984126984126984e-4,2.48015873015873e-5,2.7557319223985893e-6]
• But the OP doesn't know any maths! Can we expect him to express his exponential input as a power series? Jun 17, 2014 at 15:52
• Well he obviously knows notation. He just doesn't know how to do the derivative operation. Jun 17, 2014 at 15:56
• Can this handle 2^x? Jun 17, 2014 at 16:09
• What witchcraft is this? Jun 18, 2014 at 6:51
• I don't see where it "takes an expression (like 4*x^3-2) as input", as required by the OP.
– Gabe
Jun 18, 2014 at 12:53
## Prolog 176
d(N,0):-number(N).
d(x,1).
d(-L,-E):-d(L,E).
d(L+R,E+F):-d(L,E),d(R,F).
d(L-R,E-F):-d(L,E),d(R,F).
d(L*R,E*R+L*F):-d(L,E),d(R,F).
d(L^R,E*R*L^(R-1)+ln(L)*F*L^R):-d(L,E),d(R,F).
Supported operators: binary +, binary -, binary *, binary ^, unary -. Note that unary + is not supported.
Sample run:
49 ?- d(-3,O).
O = 0.
50 ?- d(8.5,O).
O = 0.
51 ?- d(x^0.5,O).
O = 1*0.5*x^ (0.5-1)+ln(x)*0*x^0.5.
52 ?- d(x^-7,O).
ERROR: Syntax error: Operator expected
ERROR: d(x
ERROR: ** here **
ERROR: ^-7,O) .
52 ?- d(x^ -7,O).
O = 1* -7*x^ (-7-1)+ln(x)*0*x^ -7.
53 ?- d(x,O).
O = 1.
54 ?- d(0.5^x,O).
O = 0*x*0.5^ (x-1)+ln(0.5)*1*0.5^x.
55 ?- d(7^x,O).
O = 0*x*7^ (x-1)+ln(7)*1*7^x.
56 ?- d(3*x^5,O).
O = 0*x^5+3* (1*5*x^ (5-1)+ln(x)*0*x^5).
57 ?- d(-0.1*0.3^x,O).
O = 0*0.3^x+ -0.1* (0*x*0.3^ (x-1)+ln(0.3)*1*0.3^x).
58 ?- d(-5*x^2+10-3^x,O).
O = 0*x^2+ -5* (1*2*x^ (2-1)+ln(x)*0*x^2)+0- (0*x*3^ (x-1)+ln(3)*1*3^x).
Prolog is confused when it runs into ^- sequence. A space must be inserted between ^ and - for it to parse the expression correctly.
Hope your teacher doesn't mind the mess of equation.
Crazy time:
59 ?- d(x^x,O).
O = 1*x*x^ (x-1)+ln(x)*1*x^x.
60 ?- d((x^2-x+1)*4^ -x,O).
O = (1*2*x^ (2-1)+ln(x)*0*x^2-1+0)*4^ -x+ (x^2-x+1)* (0* -x*4^ (-x-1)+ln(4)* - 1*4^ -x).
# C, 260 248 bytes
−12 bytes thanks to ceilingcat
Hey, I think I know your teacher! Isn't it that one who has the supernatural ability to detect students executing library pattern-matching functions in their head?
So, using sscanf is out of question... But don't worry:
#define P,s--||printf(
q=94,s,c,t;main(a){char*p,i[999],*e=p=i;for(gets(i);q=c=*p++,t=q^94|c^45?c%26-16?c/46:c%16/3:1,s=(a="30PCqspP#!C@ #cS #!cpp#q"[s*5+t])/16-3,p[-1]*=~a&1,!t?0 P"*0")P"/x")P"/x*%s",e)P"*ln(%s)",e),s=0:0,e=a&2?p:e,printf(&c););}
Running examples (input on stdin; output goes to stdout):
4*x^3-2
4*x^3/x*3-2*0
This format is much better than just 12*x^2, because this way your teacher can be sure that you calculated the answer yourself and didn't cheat by copying it from someone else!
x+2^x
x/x+2^x*ln(2)
The output has a slight domain problem at x=0, but it's correct almost everywhere!
For reference, here is an ungolfed, readable (by mere mortals) version. It uses a state machine with 5 states and 5 categories of input characters.
void deriv(char* input)
{
char* p = input; // current position
char* exp = p; // base or exponent
char q = '^'; // previous character
// State machine has 5 states; here are examples of input:
// state 0: 123
// state 1: 123*
// state 2: 123*x
// state 3: 123*x^456
// state 4: 123^x
int state = 0;
// Control bits for state machine:
// bit 0: special action: stop recording base or exponent
// bit 1: special action: start recording base or exponent
// bits 4-7: if first column, specify how to calculate the derivative:
// 3 - multiply the constant term by 0
// 4 - divide x by x
// 5 - divide x^n by x and multiply by n
// 6 - multiply n^x by ln(n)
// bits 4-7: if not first column, specify the next state
// (plus 3, to make the character printable)
const char* control =
"\x33\x30\x50\x43\x71"
"\x73\x70\x50\x23\x21"
"\x43\x40\x20\x23\x63"
"\x53\x60\x20\x23\x21"
"\x63\x70\x70\x23\x71";
for (;;) {
int c = *p++;
// Convert a char to a category:
// category 0: // - +
// category 3: // *
// category 2: // x
// category 4: // ^
// category 1: // numbers: 0...9 and decimal point
int category;
int action;
if (q == '^' && c == '-')
category = 1; // unary minus is a part of a number
else
category = c%26==16?c%16/3:c/46; // just does it
// Load new state and action to do
action = control[state * 5 + category];
if (action & 1)
p[-1] = 0;
state = (action >> 4) - 3;
if (category == 0)
{
if (state == 0)
printf("*0");
if (state == 1)
printf("/x");
if (state == 2)
printf("/x*%s", exp);
if (state == 3)
printf("*ln(%s)", exp);
state = 0;
}
if (action & 2)
exp = p;
if (c == 0 || c == '\n') // either of these can mark end of input
break;
putchar(c);
q = c;
}
}
P.S. Watch out for that gets function: it has a security vulnerability that can let your teacher execute a rootkit in your mind by providing too long input...
• @ceilingcat Fail on -x^-1 what does *0-x^/x*-1*0 mean?
– l4m2
Jun 5, 2020 at 7:12
• Oh you assume input 0-x^-1 but 0*0-x^/x*-1*0 still has a ^/
– l4m2
Jun 5, 2020 at 7:13
# Lua 296268 263
function d(a)l=""i=a:find"x" if i then if a:sub(i-1,i-1)=="^"then l="log("..a:sub(1,i-2)..")*"..a elseif a:sub(i+1,i+1)=="^"then l=a:sub(i+2).."*"..a:sub(1,i)p=a:sub(i+2)-1 if p~=1 then l= l..a:sub(i+1,i+1)..p end else l=a:sub(1,i-2)end else l="0"end return l end
Not very golfed and cannot currently handle multiple terms (you can just run it a few times, right?), but it can handle n^x, x^n and n as input.
Ungolfed...
function d(a)
l=""
i=a:find"x"
if i then
if a:sub(i-1,i-1)=="^" then
l="log("..a:sub(1,i-2)..")*"..a
elseif a:sub(i+1,i+1)=="^" then
l=a:sub(i+2).."*"..a:sub(1,i)
p=a:sub(i+2)-1 -- this actually does math here
if p~=1 then
l= l..a:sub(i+1,i+1)..p
end
else
l=a:sub(1,i-2)
end
else
l="0"
end
return l
end
• str.func(str,...) == str:func(...), that's why strings got the metatable after all... Jun 17, 2014 at 14:36
• @mniip: Still learning Lua. Thanks for the tip. Jun 17, 2014 at 14:38
• Since the OP is only looking for code "he can calculate in his head", I wouldn't bother with defining a function and declaring l local. Just expect the input to be stored in a and say the output will be stored in l. Jun 17, 2014 at 14:44
• You can omit parentheses in a:find("x"), also note that 1then only works in Lua 5.2 Jun 17, 2014 at 14:47
• @mniip: Whoa, that's pretty cool that () is optional. The 1then was just fixed as I don't have 5.2 (not doing any CPU updates until after dissertation is done b/c I don't want to mess anything up). Jun 17, 2014 at 14:49
## ECMAScript 6, 127 bytes
Here is my regex attempt (using a single regex and some logic in the replacement callback):
i.replace(/(^|[*+-])(\d+|(?:([\d.]+)\^)?(x)(?:\^(-?[\d.]+))?)(?![.*^])/g,(m,s,a,b,x,e)=>s+(b?'ln'+b+'*'+a:e?e--+'*x^'+e:x?1:0))
This expects the input string to be stored in i and simply returns the result. Try it out in an ECMAScript 6 compliant console (like Firefox's).
## sed, 110
Taking very literally "They don't need to be simplified at all or formatted exactly as shown above, as long as it's clear what the answer is saying":
s/.*/__&_/;s/x\^(-?[0-9.]+)/\1*x^(\1-1)/g;s/([0-9.]+)\^/ln\1*\1^/g;s/([^(][-+_])[0-9.]+([-+_])/\10\2/g;s/_//g
The byte count includes 1 for the r flag.
# Add underscores before and after the string, to help with solo-constant recognition
s/.*/__&_/
# Power rule: replace x^c with c*x^(c-1) where c is a number
s/x\^(-?[0-9.]+)/\1*x^(\1-1)/g
# Exponentials: replace c^ with lnc*c^ where c is a number
# (This assumes that there will be an x after the ^)
s/([0-9.]+)\^/ln\1*\1^/g
# Constants: replace ?c? with ?0? where c is a number and ? is +, -, or _
# Except if it's prededed by a parenthesis then don't, because this matches c*x^(c-1)!
s/([^(][-+_])[0-9.]+([-+_])/\10\2/g
# Get rid of the underscores
s/_//g
Sample run:
$cat derivatives.txt -3 8.5 x^0.5 x^-7 0.5^x 7^x 3*x^5 -0.1*0.3^x -5*x^2+10-3^x$ sed -re 's/.*/__&_/;s/x\^(-?[0-9.]+)/\1*x^(\1-1)/g;s/([0-9.]+)\^/ln\1*\1^/g;s/([^(][-+_])[0-9.]+([-+_])/\10\2/g;s/_//g' derivatives.txt
-0
0
0.5*x^(0.5-1)
-7*x^(-7-1)
ln0.5*0.5^x
ln7*7^x
3*5*x^(5-1)
-0.1*ln0.3*0.3^x
-5*2*x^(2-1)+0-ln3*3^x
I bet this could be golfed further; it's my first try at sed. Fun!
## Ruby, 152
...or 150 if you don't need to print... or 147 if you also are ok with an array that you need to join yourself.
run with ruby -nal
p gsub(/(?<!\^)([-+])/,'#\1').split(?#).map{|s|s[/x\^/]?$+$'+"x^(#{$'}-1)":s[/-?(.*)\^(.*)x/]?s+"*ln(#{$1}*#{$2[0]?$2:1})":s[/\*?x/]?($[0]?$:1):p}*''
ungolfed:
p gsub(/(?<!\^)([-+])/,'#\1').split(?#). # insert a # between each additive piece, and then split.
map{ |s|
if s[/x\^/] # if it's c*x^a
$ +$' + "x^(#{$'}-1)" # return c*ax^(a-1) elsif s[/-?(.*)\^(.*)x/] # if it's c*b^(a*x) ln =$1 + ?* + ($2[0] ?$2 : 1) # return c*b^(a*x)*ln(b*a)
s+"*ln(#{ln})"
elsif s[/\*?x/] # if it's c*x
($[0] ?$ : 1) # return c
else # else (constant)
nil # return nil
end
}*''
My main problem with this one is the number of characters proper splitting takes. The only other way I could think of was split(/(?<!\^)([-+])/) which gives + and - as their own results. Any hints for a better solution?
Also, is there a shorter way to return s if it's not empty, but otherwise return y? I've used s[0]?y:s? In JS I'd just do s||y, but "" is truthy in Ruby.
• Would a lookahead assertion help, like so: split(/(?<!\^)(?=[-+])/)`? Jun 19, 2014 at 0:43 | 5,367 | 14,450 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-18 | latest | en | 0.947774 |
https://gowanusballroom.com/what-is-purpose-of-mixing-of-signals-in-a-heterodyne-receiver/ | 1,675,457,716,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500074.73/warc/CC-MAIN-20230203185547-20230203215547-00008.warc.gz | 308,997,255 | 12,859 | ## What is purpose of mixing of signals in a heterodyne receiver?
To heterodyne means to mix to frequencies together so as to produce a beat frequency, namely the difference between the two. Amplitude modulation is a heterodyne process: the information signal is mixed with the carrier to produce the side-bands.
## What is purpose of mixing of signals in a heterodyne receiver?
To heterodyne means to mix to frequencies together so as to produce a beat frequency, namely the difference between the two. Amplitude modulation is a heterodyne process: the information signal is mixed with the carrier to produce the side-bands.
How do you calculate the image frequency of a superheterodyne receiver?
The image frequency is calculated as: fsi = fs + 2 I.F. Image frequency is given by fsi = fs + 2 I.F.
### Why if is 455 kHz?
455 was chosen because it was below the 540 end of the AM broadcast band. The 5 on the end was because mixing products of stations spaced 10 apart would not get into the IF and the local oscillator radiation would fall between stations.
How do you signal a heterodyne?
They do this by mixing the television signal frequency, fCH with a local oscillator at a much higher frequency fLO, creating a heterodyne at the sum fCH + fLO, which is added to the cable.
## Why is frequency needed?
Conversion to an intermediate frequency is useful for several reasons. When several stages of filters are used, they can all be set to a fixed frequency, which makes them easier to build and to tune. Lower frequency transistors generally have higher gains so fewer stages are required.
Armstrong called this new receiver (which used heterodyning to translate signals to a fixed, lower intermediate frequency for reception) the “superheterodyne” receiver, as shown in the block diagram in Figure 2. FIGURE 2. Architecture of Armstrong’s superheterodyne receiver.
### What is the architecture of Armstrong’s superheterodyne receiver?
Architecture of Armstrong’s superheterodyne receiver. Designing an AM superhet receiver for the commercial broadcast band is a good way to better understand the operation of Armstrong’s superheterodyne receiver. The AM broadcast band contains 117 10 kHz-wide channels spaced between 530–1,700 kHz.
What is superheterodyne design?
Almost a century after its introduction — except for sophisticated approaches such as software radio that involve advanced digital signal processing techniques — Armstrong’s “superheterodyne” or “superhet” design reigns supreme in communications electronics.
## How do I read a schematic diagram for a 6×2 receiver?
A schematic diagram like the one for the 6×2 receiver can seem very intimidating at first. The trick is to break down the circuit into its individual parts, and work on each part separately. Below you can either click on a link in the list or click on an area of the schematic diagram to take you to a page that discusses that individual circuit. | 633 | 2,951 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2023-06 | latest | en | 0.958136 |
https://physics.stackexchange.com/questions/30332/heat-of-vaporization-of-water-dependence-on-relative-humidity | 1,721,747,822,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518058.23/warc/CC-MAIN-20240723133408-20240723163408-00280.warc.gz | 379,594,670 | 39,954 | # Heat of vaporization of water - dependence on relative humidity?
Does the heat of vaporization of water depend strongly on the relative humidity of the gas into which it evaporates?
Some context: If we want to calculate the dew point of water, we find the temperature at which the partial pressure of the water lies on the liquid/vapor boundary of the water phase diagram. This is why water can evaporate from our bodies even though we do not heat it to anywhere near its boiling point.
The heat of vaporization should be pressure dependent (in addition to temperature dependent). Yet, when specifying the heat of vaporization, most references only specify the total ambient temperature, usually 1 atmosphere. Why is the total pressure used in this case instead of the partial pressure? And if the partial pressure is what matters after all, then wouldn't relative humidity be important when calculating heats of vaporization? Of course, relative humidity governs the rate and the total amount of evaporation, which is why we can't cool ourselves by sweating in humid weather, but that's not what my question is about.
Heat of vaporization is related to enthalpy change, while dew point is related to free energy change, i.e. enthalpy plus entropy. That's why they are very different concerning relative humidity.
The enthalpy of a gas is more-or-less independent of pressure or partial pressure, because gas molecules don't really interact with each other. At insanely-high pressures there would be some effect on enthalpy of course, but the effect at everyday pressures is very low. Pressure mainly affects a gas via entropy not enthalpy.
The enthalpy of a liquid is somewhat dependent on total pressure: A high pressure will push the molecules closer together and therefore change their interaction energies. But obviously the enthalpy of the liquid doesn't depend on what the gas partial pressures are, it can only depend on the liquid's own total internal pressure.
So the answer is: Heat of vaporization, being related to enthalpy not entropy, has essentially no dependence on relative humidity. (given a constant total air pressure)
-- UPDATE --
Oops, whenever I wrote "enthalpy" I should have said "enthalpy per molecule" or "enthalpy per mole" ["molar enthalpy"]. You can check for yourself that the enthalpy per molecule of an ideal gas is independent of pressure or partial pressure. For a real-world gas, it's approximately independent. The "per mole" quantities are what matter for dew point etc.
• +1 The partial pressure of water in the air mostly affects the entropy of the water molecules, and makes little if any difference in the energy needed to move a water molecule from the liquid to the air. Commented Jun 19, 2012 at 7:24
• Steve, thanks for the answer. I think I will accept it after another day grace period. I'm confused about your statement that "The enthalpy of a gas is more-or-less independent of pressure or partial pressure, because gas molecules don't really interact with each other." I thought enthalpy of a gas depended strongly on pressure - by definition $H = U + PV$, and I thought $U \approx PV$. Commented Jun 19, 2012 at 20:08
• Sorry, you're right. I put in an "update" section. Enthalpy of a gas PER MOLECULE is more-or-less independent of pressure or partial pressure. Commented Jun 20, 2012 at 1:10 | 724 | 3,358 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-30 | latest | en | 0.917854 |
https://virtualpsychcentre.com/definition-of-congruent-angle/ | 1,718,475,304,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861605.77/warc/CC-MAIN-20240615155712-20240615185712-00591.warc.gz | 555,618,762 | 11,637 | ## What is meant by congruent angles?
Congruent angles are two or more angles that are identical to each other. Thus, the measure of these angles is equal to each other. The type of angles does not make any difference in the congruence of angles, which means they can be acute, obtuse, exterior, or interior angles.
## What is the definition of congruent angles proof?
If two angles are supplements of the same angle (or congruent angles), then the two angles are congruent. Congruent Complements Theorem: If two angles are complements of the same angle (or congruent angles), then the two angles are congruent.
## Which is the best definition of congruence?
Definition of congruence
1 : the quality or state of agreeing, coinciding, or being congruent … the happy congruence of nature and reason …— Gertrude Himmelfarb. 2 : a statement that two numbers or geometric figures are congruent.
## What is the congruent symbol?
≅
Congruence is denoted by the symbol “≅”. From the above example, we can write ABC ≅ PQR. They have the same area and the same perimeter.
## Which types of angles are congruent?
Congruent angles have the same angle measure. For example, a regular pentagon has five sides and five angles, and each angle is 108 degrees. Regardless of the size or scale of a regular polygon, the angles will always be congruent.
## Does congruent mean equal?
Two angles are congruent if and only if they have equal measures. Two segments are congruent if and only if they have equal measures.
## What is a congruent shape?
Congruent shapes are shapes that are exactly the same. The corresponding sides are the same and the corresponding angles are the same. To do this we need to check all the angles and all the sides of the shapes. If two shapes are congruent they will fit exactly on top of one another. E.g.
## What lines are congruent?
When two line segments exactly measure the same, they are known as congruent lines. For example, two line segments XY and AB have a length of 5 inches and are hence known as congruent lines. When two angles exactly measure the same, they are known as congruent angles.
## How do you prove that two angles are congruent in a triangle?
The simplest way to prove that triangles are congruent is to prove that all three sides of the triangle are congruent. When all the sides of two triangles are congruent, the angles of those triangles must also be congruent. This method is called side-side-side, or SSS for short.
## What is SAS ASA and SSS congruence postulates?
The first two postulates, Side-Angle-Side (SAS) and the Side-Side-Side (SSS), focus predominately on the side aspects, whereas the next lesson discusses two additional postulates which focus more on the angles. Those are the Angle-Side-Angle (ASA) and Angle-Angle-Side (AAS) postulates.
## What is the meaning of SAS postulate?
Side-Angle-Side Postulate
If two sides and the included angle in one triangle are congruent to two sides and the included angle in another triangle, then the two triangles are congruent.
## What is the difference between SSS and SAS?
If all three pairs of corresponding sides are congruent, the triangles are congruent. This congruence shortcut is known as side-side-side (SSS). Another shortcut is side-angle-side (SAS), where two pairs of sides and the angle between them are known to be congruent.
## Why SSA is not congruent?
The SSA congruence rule is not possible since the sides could be located in two different parts of the triangles and not corresponding sides of two triangles. The size and shape would be different for both triangles and for triangles to be congruent, the triangles need to be of the same length, size, and shape.
## What is the SSA theorem?
The acronym SSA (side-side-angle) refers to the criterion of congruence of two triangles: if two sides and an angle not include between them are respectively equal to two sides and an angle of the other then the two triangles are equal.
## What does SSA mean in geometry?
side side angle postulate
SSA stands for side side angle postulate. In this postulate of congruence, we say that if two sides and an angle not included between them are respectively equal to two sides and an angle of the other triangle then the two triangles are equal.
## What means SAS in math?
first such theorem is the side-angle-side (SAS) theorem: If two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle, the triangles are congruent. | 1,024 | 4,531 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-26 | latest | en | 0.920772 |
https://www.programming-helper.com/snippet/0wL9xyzBvfAUtauwWDLa | 1,680,381,012,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950247.65/warc/CC-MAIN-20230401191131-20230401221131-00631.warc.gz | 1,041,086,880 | 6,097 | Generation
generate functionThu, 02 Feb 2023
# Write a program that implements the algorithm which determines the monthly wages of a salesperson. Every salesperson has a base salary. The salesperson also receives a bonus at the end of each month, based on the following criteria: If the salesperson has been with the store for five years or less, the bonus is \$10 for each year that he or she has worked there. If the salesperson has been with the store for more than five years, the bonus is \$20 for each year that he or she has worked there. The salesperson can earn an additional bonus as follows: If the total sales made by the salesperson for the month are at least \$5,000 but less than \$10,000, he or she receives a 3% commission on the sale. If the total sales made by the salesperson for the month are at least \$10,000, he or she receives a 6% commission on the sale. To calculate a salesperson’s monthly paycheck, you need to know the base salary, the number of years that the salesperson has been with the company, and the total sales made by the salesperson for that month. Suppose baseSalary denotes the base salary,noOfServiceYears denotes the number of years that the salesperson has been with the store,bonus denotes the bonus, totalSales denotes the total sales made by the salesperson for the month, and additionalBonus denotes the additional bonus.
baseSalary = 2000 noOfServiceYears = 3 bonus = noOfServiceYears * 10 totalSales = 8000 if totalSales >= 5000 && totalSales < 10000 { additionalBonus = totalSales * 0.03 } else if totalSales >= 10000 { additionalBonus = totalSales * 0.06 } | 378 | 1,613 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2023-14 | latest | en | 0.937869 |
https://math.stackexchange.com/questions/762933/why-does-euclids-algorithm-taken-one-step-past-the-gcd-seem-to-yield-the-lcm | 1,580,310,878,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251799918.97/warc/CC-MAIN-20200129133601-20200129163601-00088.warc.gz | 544,572,231 | 32,845 | # Why does Euclid's algorithm taken one step past the GCD seem to yield the LCM?
In class we are going over Euclid's Algorithm. For example, we learned that for integers $m$, $n$:$$\gcd(m,n) = sm + tn$$ Where $s$ and $t$ are integers that can be plugged in to satisfy the equation.
We also learned to step through the recursive algorithm like so: $$\gcd(m, n) = \gcd(m, n \bmod m)$$ to get pairs of coefficients $s$ and $t$ until we reached a remainder of zero ($n \bmod m = 0$), meaning the $s$ and $t$ at that point in the recursion are the ones that yield:$$sm + tn = \gcd(m,n)$$
What we observed but couldn't explain (professor included) was that if you take the algorithm one step further, getting $s'$ and $t'$ such that $s'm + t'n = 0$, it seems to be always true that $s'm = \operatorname{lcm}(m,n)$.
To clarify, I understand that the algorithm can't divide by zero, but if you continue the pattern:
$$s'' = s - s'*q$$ $$t'' = t - t'*q$$ you get an $s''$ and $t''$ such that: $$s''*m + t''*n = 0$$ Then taking this $s''$, multiplying it by the original $m$, you see that this is the $lcm(m, n)$. This seems to be true consistently, but I haven't been able to find a counterproof, contradiction, counterexample, etc, so I was wondering if anyone knew how to disprove it, or if there was a proof to verify this was always the case.
• What do you mean by "taking the algorithm one step further"? The algorithm can't be taken one step further, without dividing by zero. – Gerry Myerson Apr 21 '14 at 13:04
• This is evidently not true: $1000\cdot2-500\cdot 4=0$, but $2000$ is of course not the lowest common multiple of $2$ and $4$. It's trivial to see that $s'm$ is a common multiple of $m$ and $n$, though. – egreg Apr 21 '14 at 13:42
• @egreg yes, it is trivial to see, but not trivial to prove. I am trying to determine what the proof is, to see if it is true in all cases. – Steverino Apr 21 '14 at 13:48
• If $s'm=-t'n$, then obviously $s'm$ is a common multiple of $m$ and $n$. But quite certainly not the least common multiple, unless you choose the minimal positive $s'$ (which is $n/\gcd(m,n)$), assuming $m\ne0$ and $n\ne0$. – egreg Apr 21 '14 at 13:52
• @GerryMyerson I understand that the algorithm can't divide by zero, but if you continue the pattern s'' = s - s'q, t'' = t - t'*q, you get an s and t such that sm + t*n = zero. Then taking this s, multiplying it by m, you see that this is the LCM of m and n. This seems to be true consistently, I haven't been able to find a counterproof, contradiction, counterexample, etc, so I was wondering if anyone knew how to disprove it, or if there was a proof to verify this was always the case. – Steverino Apr 21 '14 at 13:53
Let $a,b$ be two positive integers. Euclid's algorithm produces unique sequences $(r_n),(s_n),(t_n)$ such that :
• $r_0=a$, $r_1=b$ and $r_{n+1} = r_{n-1} - q_n r_n$ (with usual conditions on $(r_{n+1},q_n)$).
• $s_0=1$, $s_1=0$, $t_0=1$, $t_1=1$ and $s_{n+1} = s_{n-1} - q_n s_n$, $t_{n+1} = t_{n-1} - q_n t_n$. This implies $r_n = s_n a_n + t_n b_n$.
Let $r_N$ be the last remainder. Then $r_N = s_N a + t_N b$ is Bezout's relation for $\gcd(a,b)$, and $0 = r_{N+1} = s_{N+1} a + t_{N+1} b$.
You are asking why $|r_{N+1} a| = |t_{N+1} b|$ is equal to ${\rm lcm}(a,b)$. Note that the sequence $w_n := s_{n} r_{n+1} - s_{n+1} r_{n}$ is constant and so $w_N=w_0$. But $$w_N = s_{N+1} r_N = s_{N+1}.\gcd(a,b)$$ and $$w_0 = b.$$ Combining with the fact that $\gcd(a,b).{\rm lcm}(a,b)=ab$, this implies $s_{N+1} a = {\rm lcm}(a,b)$.
• You definitely understood and clarified my question well! But I could not understand the steps taken to come to the answer. Where you wrote |rN+1a| did you mean |sN+1a|? Also, why is (s(n)r(n+1))−s(n+1)r(n)) constant? – Steverino Apr 21 '14 at 14:49
• Your are right there is a typo. About $w_n$, just compute $w_{n+1}$ with the induction formula (and in fact it is $(-1)^n w_n$ which is constant). – user10676 Apr 21 '14 at 17:21
• I suspect your answer is right since you understood what I was asking, but I don't understand the last two lines, in other words I don't understand why $$w_0 = b.$$ or why $$s_{N+1} a = {\rm lcm}(a,b)$$ is implied. – Steverino Apr 21 '14 at 19:05
• $w_n$ is defined to be $s_nr_{n+1}-s_{n+1}r_n$, so $w_0=s_0r_1-s_1r_0$. But $r_0=a$, $r_1=b$, $s_0=1$, $s_1=0$, so $w_0=b$. – Gerry Myerson Apr 22 '14 at 1:23 | 1,495 | 4,374 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2020-05 | latest | en | 0.933907 |
https://socratic.org/questions/how-do-you-evaluate-12-frac-1-3-5-frac-5-6 | 1,675,479,248,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500080.82/warc/CC-MAIN-20230204012622-20230204042622-00100.warc.gz | 532,277,043 | 6,207 | # How do you evaluate 12 \frac { 1} { 3} - 5\frac { 5} { 6}?
Mar 9, 2018
The answer is $\frac{13}{2}$
or $6 \frac{1}{2}$
#### Explanation:
First lets convert the mixed fractions into improper fractions (or whatever you want to call them) getting
$\frac{37}{3}$ - $- \frac{35}{6}$.
To get them both to have the same denominator we can multiply the first fraction by $\frac{2}{2}$ to get:
$\frac{74}{6}$ -$\frac{35}{6}$, giving us
$\frac{39}{6}$, or $\frac{13}{2}$
Mar 25, 2018
$6 \frac{1}{2}$
#### Explanation:
$12 \frac{1}{3} - 5 \frac{5}{6}$
$= 7 \text{ "+1/3-5/6" } \leftarrow$subtract the whole numbers
$= 7 \frac{2 - 5}{6} \text{ } \leftarrow$ find a common denominator
$= 6 \frac{6 + 2 - 5}{6} \text{ } \leftarrow$ convert a whole number: $1 = \frac{6}{6}$
$= 6 \frac{1}{2} \text{ } \leftarrow$ simplify the numerator | 322 | 839 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 16, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2023-06 | latest | en | 0.497879 |
https://www.coursehero.com/file/5918519/40-Taxes-on-salvage-163200-So-the-aftertax-salvage-value/ | 1,488,157,597,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501172156.69/warc/CC-MAIN-20170219104612-00331-ip-10-171-10-108.ec2.internal.warc.gz | 796,538,718 | 23,584 | Corporate_Finance_9th_edition_Solutions_Manual_FINAL0
40 taxes on salvage 163200 so the aftertax salvage
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Unformatted text preview: We will find the aftertax salvage value of the equipment first, which will be: Market value Taxes Total \$1,000,000 –340,000 \$660,000 184 Remember, to calculate the taxes on the equipment salvage value, we take the book value minus the market value, times the tax rate. Using the same method as the headache only pill, the cash flows each year for the headache and arthritis pill will be: Year 1 \$31,500,000 16,500,000 6,666,667 \$8,333,333 2,833,333 \$5,500,000 \$12,166,667 Year 2 \$31,500,000 16,500,000 6,349,206 \$8,650,794 2,941,270 \$5,709,524 \$12,058,730 Year 3 \$31,500,000 16,500,000 6,046,863 \$8,953,137 3,044,067 \$5,909,070 \$11,955,933 Sales Production costs Depreciation EBT Tax Net income OCF So, the NPV of the headache and arthritis pill is: NPV = –\$21,000,000 + \$12,166,667 / 1.13 + \$12,058,730 / 1.132 + (\$11,955,933 + 660,000) / 1.133 NPV = \$7,954,190.93 The company should manufacture the headache and arthritis remedy since the project has a higher NPV. 38. This is an in-depth capital budgeting problem. Since the project requires an initial investment in inventory as a percentage of sales, we will calculate the sales figures for each year first. The incremental sales will include the sales of the new table, but we also need to include the lost sales of the existing model. This is an erosion cost of the new table. The lost sales of the existing table are constant for every year, but the sales of the new table change every year. So, the total incremental sales figure for the five years of the project will be: Year 1 \$10,080,000 –1,125,000 \$8,955,000 Year 2 \$10,920,000 –1,125,000 \$9,795,000 Year 3 \$14,000,000 –1,125,000 \$12,875,000 Year 4 \$13,160,000 –1,125,000 \$12,035,000 Year 5 \$11,760,000 –1,125,000 \$10,635,000 New Lost sales Total Now we will calculate the initial cash outlay that will occur today. The company has the necessary production capacity to manufacture the new table without adding equipment today. So, the equipment will not be purchased today, but rather in two years. The reason is that the existing capacity is not being used. If the existing capacity were being used, the new equipment would be required, so it would be a cash flow today. The old equipment would have an opportunity cost if it could be sold. As there is no discussion that the existing equipment could be sold, we must assume it cannot be sold. The only initial cash flow is the cost of the inventory. The company will have to spend money for inventory in the new table, but will be able to reduce inventory of the existing table. So, the initial cash flow today is: New table Old table Total –\$1,008,000 112,500 –\$895,500 185 In year 2, the company will have a cash outflow to pay for the cost of the new equipment. Since the equipment will be purchased in two years rather than now, the equipment will have a higher salvage value. The book value of the equipment in five years will be the initial cost, minus the accumulated depreciation, or: Book value = \$16,000,000 – 2,288,000 – 3,920,000 – 2,800,000 Book value = \$6,992,000 The taxes on the salvage value will be: Taxes on salvage = (\$6,992,000 – 7,400,000)(.40) Taxes on salvage = –\$163,200 So, the aftertax salvage value of the equipment in five years will be: Sell equipment Taxes Salvage value \$7,400,000 –163,200 \$7,236,800 Next, we need to calculate the variable costs each year. The variable costs of the lost sales are included as a variable cost savings, so the variable costs will be: Year 1 \$4,536,000 –450,000 \$4,086,000 Year 2 \$4,914,000 –450,000 \$4,464,000 Year 3 \$6,300,000 –450,000 \$5,850,000 Year 4 \$5,922,000 –450,000 \$5,472,000 Year 5 \$5,292,000 –450,000 \$4,842,000 New Lost sales Variable costs Now we can prepare the rest of the pro forma income statements for each year. The project will have no incremental depreciation for the first two years as the equipment is not purchased for two years. Adding back depreciation to net income to calculate the operating cash flow, we get: Year 1 \$8,955,000 4,086,000 1,900,000 0 \$2,969,000 1,187,600 \$1,781,400 0 \$1,781,400 Year 2 \$9,795,000 4,464,000 1,900,000 0 \$3,431,000 1,372,400 \$2,058,600 0 \$2,058,600 Year 3 \$12,875,000 5,850,000 1,900,000 2,288,000 \$2,837,000 1,134,800 \$1,702,200 2,288,000 \$3,990,200 Year 4 \$12,035,000 5,472,000 1,900,000 3,920,000 \$743,000 297,200 \$445,800 3,920,000 \$4,365,800 Year 5 \$10,635,000 4,842,000 1,900,000 2,800,000 \$1,093,000 437,200 \$655,800 2,800,000 \$3,455,800 Sales VC Fixed costs Dep. EBT Tax NI +Dep. OCF 186 Next, we need to account for the changes in inventory each year. The inventory is a percentage of sales. The way we will calculate the change in inventory is the beginning of period inventory minus the end of period inventory. The sign of this calculation will tell us whether the inventory change is a cash inflow, or a cash outflow. The inventory each year, and the inventory change, will be: Year 1 \$1,008,000 1,092,000 –\$84,000 Year 2 \$1,092,000 1,400,000 –\$308,000 Year 3 \$1,400,000 1,316,000 \$84,000 Year 4 \$1,316,000 1,176,000 \$140,000 Year 5 \$1...
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This note was uploaded on 07/10/2010 for the course FIN 6301 taught by Professor Eshmalwi during the Spring '10 term at University of Texas-Tyler.
Ask a homework question - tutors are online | 1,676 | 5,603 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2017-09 | longest | en | 0.821039 |
http://www.carrawaydavie.com/how-many-fluid-ounces-are-in-1-gallon/ | 1,566,652,669,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027321140.82/warc/CC-MAIN-20190824130424-20190824152424-00406.warc.gz | 225,946,803 | 8,893 | ## What is a gallon how many fluid ounces are in 1 gallon?
Gallon is a royal as well as United States normal measurement system volume device. how many fluid ounces are in 1 gallon? There is one kind of gallon in the imperial system and also 2 kinds (fluid and completely dry) in the US customary dimension system. 1 US liquid gallon is specified as 231 cubic inches, 1 US dry gallon is 268.8 cubic inches and 1 royal gallon is 277.4 cubic inches. The abbreviation is gall. how many fluid ounces are in 1 gallon?
Gallons Conversion:
1 Gallon = 4 Quarts
1 Gallon = 8 Pints
1 Gallon = 16 Cups
1 Gallon = 256 Tablespoons
1 Gallon = 768 Teaspoons
how many fluid ounces are in 1 gallon
## What is a liquid ounce (fl oz)how many fluid ounces are in 1 gallon?
A liquid ounce is an imperial as well as US customary measurement system volume unit. 1 US fluid ounce equals to 29.5735 mL and 1 royal (UK) fluid ounce equals to 28.4131 mL. The acronym is fl oz. So, how many fluid ounces are in 1 gallon?
There are 8 fluid ounces in a United States mug and also 10 imperial liquid ounces in an imperial cup.
1 United States Fluid Gallon = 128 US Fluid Ounces
1 US Dry Gallon = 148.946 United States Fluid Ounces
1 Imperial Gallon (UK) = 160 Imperial Fluid Ounces (UK).
how many fluid ounces are in 1 gallon
## Liquid Ounce how many fluid ounces are in 1 gallon
Liquid Ounce is utilized for volume, Ounce for mass, and also they are various.
As an example, 1 liquid ounce of honey has a mass of about 1,5 ounces!
But also for water, 1 liquid ounce has a mass of about 1 ounce.
how many fluid ounces are in 1 gallon
If you imply an ounce of fluid says fluid ounce (fl oz).
In Summary:.
1 gallon = 4 quarts = 8 pints = 16 mugs = 128 fluid ounces.
how many fluid ounces are in 1 gallon | 484 | 1,776 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-35 | latest | en | 0.913987 |
https://schoollearningcommons.info/question/if-24-5-of-lenght-is-2m-45cm-what-is-the-lenght-24856481-84/ | 1,632,231,666,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057225.38/warc/CC-MAIN-20210921131252-20210921161252-00270.warc.gz | 536,576,898 | 13,686 | ## if 24.5% of lenght is 2m 45cm.what is the lenght
Question
if 24.5% of lenght is 2m 45cm.what is
the lenght
in progress 0
6 days 2021-09-15T10:06:02+00:00 2 Answers 0 views 0
lalalalallalalalalalalalaalalalalaalallalalaalalalalla
10 m
Step-by-step explanation:
Let the length be x cm then.
We know that 2m 45cm = 245cm
According to the question;
24.5% of x = 245
Hence, x = 1000cm = 1000/100 m = 10 m | 167 | 415 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2021-39 | latest | en | 0.768009 |
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