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https://www.savemyexams.co.uk/notes/a-level-chemistry-cie/5-physical-chemistry/5-3-principles-of-electrochemistry/5-3-4-measuring-the-standard-electrode-potential/ | 1,618,827,699,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038879305.68/warc/CC-MAIN-20210419080654-20210419110654-00431.warc.gz | 1,065,275,410 | 149,516 | # 5.3.4 Measuring the Standard Electrode Potential
### Measuring the Standard Electrode Potential
• There are three different types of half-cells that can be connected to a standard hydrogen electrode
• A metal / metal ion half-cell
• A non-metal / non-metal ion half-cell
• An ion / ion half-cell (the ions are in different oxidation states)
#### Metal/metal ion half-cell
Example of a metal / metal ion half-cell connected to a standard hydrogen electrode
• An example of a metal/metal ion half-cell is the Ag+/ Ag half-cell
• Ag is the metal
• Ag+ is the metal ion
• This half-cell is connected to a standard hydrogen electrode and the two half-equations are:
Ag+ (aq) + e⇌ Ag (s) E= + 0.80 V
2H+ (aq) + 2e⇌ H2 (g) E= 0.00 V
• Since the Ag+/ Ag half-cell has a more positive Evalue, this is the positive pole and the H+/H2 half-cell is the negative pole
• The standard cell potential (Ecell) is Ecell = (+ 0.80) – (0.00) = + 0.80 V
• The Ag+ ions are more likely to get reduced than the H+ ions as it has a greater Evalue
• Reduction occurs at the positive pole
• Oxidation occurs at the negative pole
#### Non-metal/non-metal ion half-cell
• In a non-metal/non-metal ion half-cell platinum wire or foil is used as an electrode to make electrical contact with the solution
• Like graphite, platinum is inert and does not take part in the reaction
• The redox equilibrium is established on the platinum surface
• An example of a non-metal/non-metal ion is the Br2/Br half-cell
• Br is the non-metal
• Br is the non-metal ion
• The half-cell is connected to a standard hydrogen electrode and the two half-equations are:
Br2 (l) + 2e⇌ 2Br (aq) E = +1.09 V
2H+ (aq) + 2e⇌ H2 (g) E = 0.00 V
• The Br2/Br half-cell is the positive pole and the H+/H2 is the negative pole
• The Ecellis: Ecell = (+ 1.09) – (0.00) = + 1.09 V
• The Br2 molecules are more likely to get reduced than H+ as they have a greater Evalue
Example of a non-metal / non-metal ion half-cell connected to a standard hydrogen electrode
#### Ion/Ion half-cell
• A platinum electrode is again used to form a half-cell of ions that are in different oxidation states
• An example of such a half-cell is the MnO4/Mn2+ half-cell
• MnO4 is an ion containing Mn with oxidation state +7
• The Mn2+ ion contains Mn with oxidation state +2
• This half-cell is connected to a standard hydrogen electrode and the two half-equations are:
MnO4 (aq) + 8H+ (aq) + 5e⇌ Mn2+ (aq) + 4H2O (l) E = +1.52 V
2H+ (aq) + 2e⇌ H2 (g) E= 0.00 V
• The H+ ions are also present in the half-cell as they are required to convert MnO4into Mn2+ ions
• The MnO4/Mn2+ half-cell is the positive pole and the H+/H2 is the negative pole
• The Ecell is Ecell = (+ 1.09) – (0.00) = + 1.09 V
Ions in solution half cell
### Standard Cell Potential: Direction of Electron Flow & Feasibility
#### Direction of electron flow
• The direction of electron flow can be determined by comparing the Evalues of two half-cells in an electrochemical cell
2Cl2 (g) + 2e⇌ 2Cl (aq) E = +1.36 V
Cu2+ (aq) + 2e⇌ Cu (s) E = +0.34 V
• The Cl2 more readily accept electrons from the Cu2+/Cu half-cell
• This is the positive pole
• Cl2 gets more readily reduced
• The Cu2+ more readily loses electrons to the Cl2/Cl half-cell
• This is the negative pole
• Cu2+ gets more readily oxidised
• The electrons flow from the Cu2+/Cu half-cell to the Cl2/Cl half-cell
• The flow of electrons is from the negative pole to the positive pole
The electrons flow through the wires from the negative pole to the positive pole
#### Feasibility
• The Evalues of a species indicate how easily they can get oxidised or reduced
• The more positive the value, the easier it is to reduce the species on the left of the half-equation
• The reaction will tend to proceed in the forward direction
• The less positive the value, the easier it is to oxidise the species on the right of the half-equation
• The reaction will tend to proceed in the backward direction
• A reaction is feasible (likely to occur) when the Ecell is positive
• For example, two half-cells in the following electrochemical cell are:
Cl2 (g) + 2e⇌ 2Cl (aq) E = +1.36 V
Cu2+ (aq) + 2e⇌ Cu (s) E = +0.34 V
• Cl2 molecules are reduced as they have a more positive E value
• The chemical reaction that occurs in this half cell is:
Cl2 (g) + 2e→ 2Cl (aq)
• Cu2+ ions are oxidised as they have a less positive E value
• The chemical reaction that occurs in this half cell is:
Cu (s) → Cu2+ (aq) + 2e
• The overall equation of the electrochemical cell is (after cancelling out the electrons):
Cu (s) + Cl2 (g) → 2Cl(aq) + Cu2+ (aq)
OR
Cu (s) + Cl2 (g) → CuCl2 (s)
• The forward reaction is feasible (spontaneous) as it has a positive E value of +1.02 V ((+1.36) – (+0.34))
• The backward reaction is not feasible (not spontaneous) as it has a negative Evalue of -1.02 ((+0.34) – (+1.36))
A reaction is feasible when the standard cell potential E is positive
#### Exam Tip
Remember that the electrons only move through the wires in the external circuit and not through the electrolyte solution.
### Redox Equations
• The redox equations of an electrochemical cell can be constructed using the relevant half-equations of the two half-cells
#### Constructing redox equations
• Step 1: Determine in which half-cell the oxidation and in which half-cell the reduction reaction takes place
Cl2 (g) + 2e⇌ 2Cl (aq) E = +1.36 V
Zn2+ (aq) + 2e⇌ Zn (s) E = -0.76 V
• Reduction occurs in the Cl2/Cl half-cell as it has the more positive Evalue
• Oxidation occurs in the Zn+/Zn half-cell as it has the least positive Evalue
• Step 2: Write down the half equations for each half-cell
• Half-equation of the Cl2/Cl half-cell
Cl2 (g) + 2e → 2Cl (aq)
• Half-equation of the Zn+/Zn half-cell
Zn (s) → Zn2+ (aq) + 2e
• Step 3: Balance the number of electrons in both half-equations
• The number of electrons is already balanced in both half-equations as they both contain two electrons
• Step 4 – Add up the two half-equations
Cl2 (g) + 2e → 2Cl (aq)
Zn (s) → Zn2+ (aq) + 2e
______________________________________ +
Cl2 (g) + Zn (s) + 2e → 2Cl (aq) + Zn2+ (aq) + 2e
• Step 5 – Cancel out the electrons (and H+ ions and H2O molecules if any present) to find the overall redox reaction
Cl2 (g) + Zn (s) → 2Cl (aq) + Zn2+ (aq)
OR
Cl2 (g) + Zn (s) → ZnCl2 (s)
#### Worked example: Constructing redox reactions
• Step 1:
Determine in which cell oxidation and in which cell reduction takes place
Reduction occurs in the Ag+/Ag half-cell as it has the most positive Evalue
Oxidation occurs in the MnO4/ Mn2+ half-cell as it has the least positive E value
• Step 2:
Write down the half-equations for each cell
The half-equation for the Ag+/Ag half-cell is:
Ag+ (aq) + e→ Ag (s)
The half-equation for the MnO4/ Mn2+ half-cell is:
Mn2+ (aq) + 4H2O (l) → MnO4 (aq) + 8H+ (aq) + 5e
• Step 3:
Balance the number of electrons in both half-equations
Multiply the half-equation for the Ag+/Ag half-cell by 5 so that both half-equations contain 5 electrons
This gives: 5Ag+ (aq) + 5e→ 5Ag (s)
• Step 4:
5Ag+ (aq) + 5e– + Mn2+ (aq) + 4H2O (l) → 5Ag (s) + MnO4 (aq) + 8H+ (aq) + 5e
• Step 5:
Cancel out the electrons to find the overall redox equation
5Ag+ (aq) + 5e + Mn2+ (aq) + 4H2O (l) → 5Ag (s) + MnO4 (aq) + 8H+ (aq) + 5e
The fully balanced redox equation is:
5Ag+ (aq) + Mn2+ (aq) + 4H2O (l) → 5Ag (s) + MnO4 (aq) + 8H+ (aq)
### Author: Francesca
Fran has taught A level Chemistry in the UK for over 10 years. As head of science, she used her passion for education to drive improvement for staff and students, supporting them to achieve their full potential. Fran has also co-written science textbooks and worked as an examiner for UK exam boards.
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Try a Free Sample of our revision notes as a printable PDF. | 2,430 | 7,911 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-17 | latest | en | 0.826139 |
https://www.freecodecamp.org/forum/t/freecodecamp-challenge-guide-diff-two-arrays/16008/60 | 1,571,487,937,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986693979.65/warc/CC-MAIN-20191019114429-20191019141929-00207.warc.gz | 914,028,223 | 7,412 | # freeCodeCamp Challenge Guide: Diff Two Arrays
freeCodeCamp Challenge Guide: Diff Two Arrays
0
Hi @sgenio
This is my simple solution
function diffArray(a, b){
c = a.concat(b)
d = [];
var diffarr = c.filter(function(c1){
if (a.indexOf(c1) === -1 || b.indexOf(c1) === -1){
d.push(c1);
}
});
return d;
}
diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5]);
Hi Friends
This is my solution
function diffArray(a, b){
c = a.concat(b)
d = [];
var diffarr = c.filter(function(c1){
if (a.indexOf(c1) === -1 || b.indexOf(c1) === -1){
d.push(c1)
}
})
return d
}
diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5]);
## .filter(el => !arr2.includes(el))
what does ! before arr mean?
Here’s My solution using indexOf() . Seems simple to understand. Well, hint helped me:sweat_smile:[spoiler]This text will be blurred
function diffArray(arr1, arr2) {
var newArr = [];
// Same, same; but different.
var x = arr1.concat(arr2);
for(var i = 0 ; i<x.length; i++){
if(arr1.indexOf(x[i]) === -1 || arr2.indexOf(x[i]) === -1 )
newArr.push(x[i]);
}
return newArr;
}
diffArray([1, "calf", 3, "piglet"], [1, "calf", 3, 4]);
[/spoiler]
Here’s My solution using indexOf() . Seems simple to understand. Well, hint helped me [spoiler]
function diffArray(arr1, arr2) {
var newArr = [];
// Same, same; but different.
var x = arr1.concat(arr2);
for(var i = 0 ; i<x.length; i++){
if(arr1.indexOf(x[i]) === -1 || arr2.indexOf(x[i]) === -1 )
newArr.push(x[i]);
}
return newArr;
}
diffArray([1, "calf", 3, "piglet"], [1, "calf", 3, 4]);
[/spoiler]
function diffArray(arr1, arr2) {
return [...arr1, ...arr2].filter(item => !arr1.includes(item) || !arr2.includes(item));
}
diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5]);
Had to read the hints to figure this out. Solution ended up being more or less Intermediate solution but not as optimized
function diffArray(arr1, arr2) {
var arr = [];
var newArr=[];
// Same, same; but different.
arr = arr1.concat(arr2);
newArr = arr.filter(function(x){
if((arr1.indexOf(x)===-1) || (arr2.indexOf(x)===-1)){
return x;
}
});
return newArr;
}
diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5]);
I was originally writing something basically like this,
1. merge and sort it out
2. Delete duplicates
3. Push to new array if conditions are met
but I couldn’t get the IF statement line to work quite right, I had most of it down ended up scraping it once i read the hints
e.g.
[1,1,2,2,3,3,4,5,5]
If the value [4] has no adjacent [4] next to it it must be unique
function diffArray(arr1, arr2) {
var newArr = [];
function resultDiffArray(a1,a2){
for(var i=0;i<a1.length;i++){
if(a2.indexOf(a1[i])==-1){
newArr.push(a1[i]);
}
}
}
resultDiffArray(arr1,arr2);
resultDiffArray(arr2,arr1);
return newArr;
}
So I’m really new to this, and to be honest I wasn’t expecting this code to work but it did…I’m not really sure if I fully understand why it worked.
function diffArray(arr1, arr2) {
var newArr = [];
var otherArr= [];
newArr = arr1.concat(arr2);
for (var i = 0; i < newArr.length; i++) {
otherArr = newArr.filter(function(i){
return arr1.indexOf(i) < 0 || arr2.indexOf(i) < 0;
});
}
return otherArr;
}
diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5]);
Simplified, with help from rmdawson71. This works!
function diffArray(arr1, arr2) {
return arr1.concat(arr2).filter(item=>!arr1.includes(item) || !arr2.includes(item)
);
}
my code
function diffArray(arr1, arr2) {
var newArr,temp,temp1;
// Same, same; but different.
temp=arr1.filter(function(el)
{
return arr2.indexOf(el) == -1;
});
temp1=arr2.filter(function(el)
{
return arr1.indexOf(el) == -1;
});
newArr=temp.concat(temp1);
return newArr;
}
diffArray([1, 2, 12, 3, 5], [1, 2, 3, 4, 5]);
Great works my friend!
1 Like
Since my understanding of filter function is really poor I came up with this:
function diffArray(arr1, arr2) {
var newArr = [];
var result = [];
// Same, same; but different.
newArr = arr1.concat(arr2);
for (var i=0; i<newArr.length; i++) {
if (arr1.includes(newArr[i]) && !arr2.includes(newArr[i]) || !arr1.includes(newArr[i]) && arr2.includes(newArr[i])) {
result.push(newArr[i]);
console.log(result);
}
}
return result;
}
diffArray(['a', 'b', 'c', 'e'], ['a', 'b', 'c', 'd', 'e']);
my solution is a bit different I guess, using concat and splice
function diffArray(arr1, arr2) {
var newArr = [];
newArr = arr1.concat(arr2).sort(); //combine the two arrays in one and sort them
for(i=0;i<newArr.length;i++){ //loop through the full new array
if(newArr[i] == newArr[i+1]){ //if two consecutive elements are similar, delete them both
newArr.splice(i,2);
i = i-2;//after deleting for example element 0 and 1 while the loop is at count 0, next when it goes to element 1 it will be the old element 3 because 2 elements are gone, so if we remove two we count two steps back
}
}
return newArr;
}
diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5]);
1 Like
function diffArray(arr1, arr2) {
// Same, same; but different.
// Setup temp array to hold arr1 since filter() will change contents of arr1
var tempArr1 = arr1;
return arr1.filter(item => arr2.indexOf(item) === -1).concat(arr2.filter(item => tempArr1.indexOf(item) === -1));
}
diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5]);
This was my solution. Thoughts?
1 Like
Hello,
Could anyone explain me the solution below? I am not sure if I get what exactly those 3 dots do. I think they are “spread syntax” but still don’t understand how it works.
function diffArray(arr1, arr2) {
return [
...diff(arr1, arr2),
...diff(arr2, arr1)
]
function diff(a, b) {
return a.filter(item => b.indexOf(item) === -1);
}
}
diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5]);
If you still have questions, create a new topic with your specific questions related to what you do not understand about using the spread syntax.
A post was split to a new topic: I have a problem with my code for Diff Two Arrays | 1,816 | 5,808 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2019-43 | latest | en | 0.439992 |
https://www.sarthaks.com/226771/a-simple-pendulum?show=226784 | 1,558,456,976,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256494.24/warc/CC-MAIN-20190521162634-20190521184634-00023.warc.gz | 927,512,073 | 11,329 | # A simple pendulum
+1 vote
195 views
A pendulum of length 1m is released from theta =60 degree . The rate of change of speed of the bob at theta = 30 degree is
We know that Rate change of speed, dv/dt= tangential acceleration =tangential force/mass
When θ=30, tangential force is mgsinθ
Therefore F=mgsin30
ma=mgsin30
a=mgsin30/m
=10sin30
=10×1/2
=5m/s2
So, The rate of change of speed of the bob at theta = 30 degree is 5m/s2
+1 vote
+1 vote | 167 | 461 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-22 | latest | en | 0.751697 |
https://physics.stackexchange.com/questions/237151/extending-the-reach-of-a-crocodile-clip-for-kelvin-water-dropper-experiment | 1,719,285,412,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865545.19/warc/CC-MAIN-20240625005529-20240625035529-00573.warc.gz | 407,656,823 | 40,334 | # Extending the reach of a crocodile clip for Kelvin Water Dropper experiment
We're doing a physics project that involves a version of the Kelvin Water dropper experiment. Everything is working great, and we have the following system:
In the image: two plastic containers (one in the bottom and one in the top). The bottom one is filled with water and has a water pump that pumps water to the top (using tubes for circulation). The water drip from two holes in the top box, go through two cans and into two (half) bottles. One side will become charged (say positively) and when we connect the cans and the bottles in an X-shape as seen in the image, the other side becomes negatively charged and thus we have a sort of "battery" (potential difference).
We would like it to work the same, only without connecting the cans and bottles in an X-shape immediately using crocodile clips (which requires fiddling with the cables inside the system). Instead we would like to connect it from "outside", meaning somehow extend the reach of the cables.
We've tried to connect two cables (a series of 2) on each side instead of one but that didn't work (also with thick cables that have less resistance), and we're stuck. We need to somehow figure how to make the same connection every time without putting hands inside the system, and would love to hear some ideas.
Thanks and advance! Also, we're sorry if this question seems too broad or not detailed enough - please ask if something is not clear.
• It would probably help to link to a simple description of the experiment (the version that you are doing if possible) because while I know I've heard of this bit I can't bring the details to mind. Second, the photograph is a useful reference, but having a labeled schematic as well is generally helpful for people trying to understand the photo (and to establish a uniformity of jargon in the answers). Commented Feb 14, 2016 at 21:08
• Unfortunately we can link only to this simple description en.wikipedia.org/wiki/Kelvin_water_dropper (only instead of spark gap we use it to create a potential difference elsewhere). We're actually in the process of making a labeled schematic but that might take a while. Thanks for the quick response! Commented Feb 14, 2016 at 21:14 | 490 | 2,268 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-26 | latest | en | 0.958452 |
http://wikien4.appspot.com/wiki/Affine_cipher | 1,555,951,370,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578558125.45/warc/CC-MAIN-20190422155337-20190422181337-00137.warc.gz | 186,610,807 | 17,640 | # Affine cipher
The affine cipher is a type of monoawphabetic substitution cipher, wherein each wetter in an awphabet is mapped to its numeric eqwivawent, encrypted using a simpwe madematicaw function, and converted back to a wetter. The formuwa used means dat each wetter encrypts to one oder wetter, and back again, meaning de cipher is essentiawwy a standard substitution cipher wif a ruwe governing which wetter goes to which. As such, it has de weaknesses of aww substitution ciphers. Each wetter is enciphered wif de function (ax + b) mod 26, where b is de magnitude of de shift.
## Description
In de affine cipher de wetters of an awphabet of size m are first mapped to de integers in de range 0 … m − 1. It den uses moduwar aridmetic to transform de integer dat each pwaintext wetter corresponds to into anoder integer dat correspond to a ciphertext wetter. The encryption function for a singwe wetter is
${\dispwaystywe {\mbox{E}}(x)=(ax+b){\bmod {m}},}$
where moduwus m is de size of de awphabet and a and b are de keys of de cipher. The vawue a must be chosen such dat a and m are coprime. The decryption function is
${\dispwaystywe {\mbox{D}}(x)=a^{-1}(x-b){\bmod {m}},}$
where a−1 is de moduwar muwtipwicative inverse of a moduwo m. I.e., it satisfies de eqwation
${\dispwaystywe 1=aa^{-1}{\bmod {m}}.}$
The muwtipwicative inverse of a onwy exists if a and m are coprime. Hence widout de restriction on a, decryption might not be possibwe. It can be shown as fowwows dat decryption function is de inverse of de encryption function,
${\dispwaystywe {\begin{awigned}{\mbox{D}}({\mbox{E}}(x))&=a^{-1}({\mbox{E}}(x)-b){\bmod {m}}\\&=a^{-1}(((ax+b){\bmod {m}})-b){\bmod {m}}\\&=a^{-1}(ax+b-b){\bmod {m}}\\&=a^{-1}ax{\bmod {m}}\\&=x{\bmod {m}}.\end{awigned}}}$
## Weaknesses
Since de affine cipher is stiww a monoawphabetic substitution cipher, it inherits de weaknesses of dat cwass of ciphers. The Caesar cipher is an Affine cipher wif a = 1 since de encrypting function simpwy reduces to a winear shift.
Considering de specific case of encrypting messages in Engwish (i.e. m = 26), dere are a totaw of 286 non-triviaw affine ciphers, not counting de 26 triviaw Caesar ciphers. This number comes from de fact dere are 12 numbers dat are coprime wif 26 dat are wess dan 26 (dese are de possibwe vawues of a). Each vawue of a can have 26 different addition shifts (de b vawue); derefore, dere are 12 × 26 or 312 possibwe keys. This wack of variety renders de system as highwy insecure when considered in wight of Kerckhoffs' Principwe.
The cipher's primary weakness comes from de fact dat if de cryptanawyst can discover (by means of freqwency anawysis, brute force, guessing or oderwise) de pwaintext of two ciphertext characters den de key can be obtained by sowving a simuwtaneous eqwation. Since we know a and m are rewativewy prime dis can be used to rapidwy discard many "fawse" keys in an automated system.
The same type of transformation used in affine ciphers is used in winear congruentiaw generators, a type of pseudorandom number generator. This generator is not a cryptographicawwy secure pseudorandom number generator for de same reason dat de affine cipher is not secure.
## Exampwes
In dese two exampwes, one encrypting and one decrypting, de awphabet is going to be de wetters A drough Z, and wiww have de corresponding vawues found in de fowwowing tabwe.
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
### Encrypting
In dis encrypting exampwe,[1] de pwaintext to be encrypted is "AFFINE CIPHER" using de tabwe mentioned above for de numeric vawues of each wetter, taking a to be 5, b to be 8, and m to be 26 since dere are 26 characters in de awphabet being used. Onwy de vawue of a has a restriction since it has to be coprime wif 26. The possibwe vawues dat a couwd be are 1, 3, 5, 7, 9, 11, 15, 17, 19, 21, 23, and 25. The vawue for b can be arbitrary as wong as a does not eqwaw 1 since dis is de shift of de cipher. Thus, de encryption function for dis exampwe wiww be y = E(x) = (5x + 8) mod 26. The first step in encrypting de message is to write de numeric vawues of each wetter.
pwaintext x A F F I N E C I P H E R 0 5 5 8 13 4 2 8 15 7 4 17
Now, take each vawue of x, and sowve de first part of de eqwation, (5x + 8). After finding de vawue of (5x + 8) for each character, take de remainder when dividing de resuwt of (5x + 8) by 26. The fowwowing tabwe shows de first four steps of de encrypting process.
pwaintext x (5x + 8) (5x + 8) mod 26 A F F I N E C I P H E R 0 5 5 8 13 4 2 8 15 7 4 17 8 33 33 48 73 28 18 48 83 43 28 93 8 7 7 22 21 2 18 22 5 17 2 15
The finaw step in encrypting de message is to wook up each numeric vawue in de tabwe for de corresponding wetters. In dis exampwe, de encrypted text wouwd be IHHWVCSWFRCP. The tabwe bewow shows de compweted tabwe for encrypting a message in de Affine cipher.
pwaintext x (5x + 8) (5x + 8) mod 26 ciphertext A F F I N E C I P H E R 0 5 5 8 13 4 2 8 15 7 4 17 8 33 33 48 73 28 18 48 83 43 28 93 8 7 7 22 21 2 18 22 5 17 2 15 I H H W V C S W F R C P
### Decrypting
In dis decryption exampwe, de ciphertext dat wiww be decrypted is de ciphertext from de encryption exampwe. The corresponding decryption function is D(y) = 21(y − 8) mod 26, where a−1 is cawcuwated to be 21, and b is 8. To begin, write de numeric eqwivawents to each wetter in de ciphertext, as shown in de tabwe bewow.
ciphertext y I H H W V C S W F R C P 8 7 7 22 21 2 18 22 5 17 2 15
Now, de next step is to compute 21(y − 8), and den take de remainder when dat resuwt is divided by 26. The fowwowing tabwe shows de resuwts of bof computations.
ciphertext y 21(y − 8) 21(y − 8) mod 26 I H H W V C S W F R C P 8 7 7 22 21 2 18 22 5 17 2 15 0 −21 −21 294 273 −126 210 294 −63 189 −126 147 0 5 5 8 13 4 2 8 15 7 4 17
The finaw step in decrypting de ciphertext is to use de tabwe to convert numeric vawues back into wetters. The pwaintext in dis decryption is AFFINECIPHER. Bewow is de tabwe wif de finaw step compweted.
ciphertext y 21(y − 8) 21(y − 8) mod 26 pwaintext I H H W V C S W F R C P 8 7 7 22 21 2 18 22 5 17 2 15 0 −21 −21 294 273 −126 210 294 −63 189 −126 147 0 5 5 8 13 4 2 8 15 7 4 17 A F F I N E C I P H E R
### Entire awphabet encoded
To make encrypting and decrypting qwicker, de entire awphabet can be encrypted to create a one-to-one map between de wetters of de cweartext and de ciphertext. In dis exampwe, de one-to-one map wouwd be de fowwowing:
wetter in de cweartext A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
number in de cweartext 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
(5x + 8) mod 26 8 13 18 23 2 7 12 17 22 1 6 11 16 21 0 5 10 15 20 25 4 9 14 19 24 3
ciphertext wetter I N S X C H M R W B G L Q V A F K P U Z E J O T Y D
### Programming exampwes
Using de Pydon programming wanguage, de fowwowing code can be used to create an encrypted awphabet using de Roman wetters A drough Z.
#Prints a transposition table for an affine cipher.
#a must be coprime to m=26.
def affine(a, b):
for i in range(26):
print(chr(i+65) + ": " + chr(((a*i+b)%26)+65))
#An example call
affine(5, 8)
Or in Java:
public void Affine(int a, int b){
for (int num = 0; num < 26; num++)
System.out.println(((char)('A'+num)) + ":" + ((char)('A'+(a*num + b)% 26)) );
}
Affine(5,8)
Or in Pascaw:
Procedure Affine(a,b : Integer);
begin
for num := 0 to 25 do
WriteLn(Chr(num+65) , ': ' , Chr(((a*num + b) mod 26) + 65);
end;
begin
Affine(5,8)
end.
In PHP:
function affineCipher($a,$b) {
for($i = 0;$i < 26; $i++) { echo chr($i + 65) . ' ' . chr(65 + ($a *$i + \$b) % 26) . '<br>';
}
}
affineCipher(5, 8);
## References
1. ^ Kozdron, Michaew. "Affine Ciphers" (PDF). Retrieved 22 Apriw 2014. | 2,792 | 7,803 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 4, "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} | 2.9375 | 3 | CC-MAIN-2019-18 | latest | en | 0.655216 |
https://www.tutorialspoint.com/minimum-characters-required-to-be-removed-to-sort-binary-string-in-ascending-order | 1,695,399,668,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506420.84/warc/CC-MAIN-20230922134342-20230922164342-00179.warc.gz | 1,171,485,874 | 18,349 | # Minimum characters required to be removed to sort binary string in ascending order
In computer science, string manipulation is an essential topic that involves operations such as concatenation, substring, reversing, and more. One common problem related to string manipulation is to remove all 0s from a binary string. In this article, we will discuss an algorithm to solve this problem using a minimum number of non-adjacent pair flips.
## Problem Statement
Given a binary string, we have to remove all 0s from the string using the minimum number of non-adjacent pair flips. A flip is defined as selecting two adjacent characters and swapping them. In other words, we need to find the minimum number of flips required to make all 0s in the string to the end of the string.
## Approach
We can solve this problem using a greedy approach. We can start from the left of the string and keep track of the last index where we have flipped a 0 to the end. For each 0 we encounter, we swap it with the last flipped 0 to move it to the end of the string. If we encounter a 1, we simply move to the next index.
Let's see the algorithm in detail −
• Initialize two variables, "lastFlipped" and "flipCount" to -1 and 0, respectively.
• Traverse the binary string from left to right.
• If the current character is '0', swap it with the character at the index "lastFlipped + 1" and increment the "lastFlipped" variable.
• Increment the "flipCount" variable for every swap operation.
• Once the traversal is completed, all the 0s will be at the end of the string, and the "flipCount" will contain the minimum number of flips required to remove all 0s.
### Example
Here's the C++ code to implement the above algorithm −
#include <iostream>
#include <string>
using namespace std;
int lastFlipped = -1;
int flipCount = 0;
for (int i = 0; i < s.length(); i++) {
if (s[i] == '0') {
swap(s[i], s[lastFlipped + 1]);
lastFlipped++;
flipCount++;
}
}
return flipCount;
}
int main() {
string s = "100101000";
cout << "Binary String: " << s << endl;
return 0;
}
### Output
Binary String: 100101000
## Explanation of Test Case
Let's take a binary string "100101000" as an example. We have to remove all 0s from this string using the minimum number of non-adjacent pair flips.
• Initially, "lastFlipped" and "flipCount" are set to -1 and 0, respectively.
• We start traversing the string from left to right.
• At index 1, we encounter a '0'. We swap it with the character at index "lastFlipped + 1" (which is index 0) and increment "lastFlipped" to 0. The string becomes "010101000". The "flipCount" is incremented to 1.
• At index 4, we encounter another '0'. We swap it with the character at index "lastFlipped + 1" (which is index 1) and increment "lastFlipped" to 1. The string becomes "011010000". The "flipCount" is incremented to 2.
• At index 5, we encounter a '1'. We simply move to the next index
## Conclusion
In this article, we have discussed an algorithm to remove all 0s from a binary string using the minimum number of non-adjacent pair flips. The approach used in this algorithm is greedy, which makes it efficient and easy to implement. We have also provided a C++ code to implement the algorithm along with an example test case.
This problem can also be solved using dynamic programming, but the greedy approach provides a simpler and faster solution. The time complexity of this algorithm is O(n), where n is the length of the binary string.
In conclusion, the minimum non-adjacent pair flips algorithm is a useful tool in string manipulation and can be applied in various contexts.
Updated on: 18-May-2023
71 Views | 890 | 3,643 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2023-40 | latest | en | 0.839754 |
https://rdrr.io/cran/joineR/man/simjoint.html | 1,669,938,708,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710870.69/warc/CC-MAIN-20221201221914-20221202011914-00687.warc.gz | 540,472,253 | 12,166 | # simjoint: Simulate data from a joint model In joineR: Joint Modelling of Repeated Measurements and Time-to-Event Data
## Description
This function simulates longitudinal and time-to-event data from a joint model.
## Usage
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 simjoint( n = 500, model = c("intslope", "int", "quad"), sepassoc = FALSE, ntms = 5, b1 = c(1, 1, 1, 1), b2 = c(1, 1), gamma = c(1, 0.1), sigu, vare = 0.01, theta0 = -3, theta1 = 1, censoring = TRUE, censlam = exp(-3), truncation = FALSE, trunctime = max(ntms), gridstep = 0.01 )
## Arguments
n the number of subjects to simulate data for. model a character string specifying the type of latent association. This defaults to the intercept and slope version as seen in Wulfsohn and Tsiatis (1997). For association via the random intercept only, choose model = "int", whereas for a quadratic association, use model = "quad". Computing times are commensurate with the type of association structure chosen. sepassoc logical value: if TRUE then the joint model is fitted with separate association, see Details. ntms the maximum number of (discrete) time points to simulate repeated longitudinal measurements at. b1 a vector specifying the coefficients of the fixed effects in the longitudinal sub-model. The order in each row is intercept, a continuous covariate, covariate, a binary covariate, the measurement time. b2 a vector of length = 2 specifying the coefficients for the time-to-event baseline covariates, in the order of a continuous covariate and a binary covariate. gamma a vector of specifying the latent association parameter(s) for the longitudinal outcome. It must be of length 1 if sepassoc = FALSE. sigu a positive-definite matrix specifying the variance-covariance matrix. If model = "int", the matrix has dimension dim = c(1, 1); if model = "intslope", the matrix has dimension dim = c(2, 2); else if model = "quad", the matrix has dimension dim = c(3, 3). If D = NULL (default), an identity matrix is assumed. vare a numeric value specifying the residual standard error. theta0, theta1 parameters controlling the failure rate. See Details. censoring logical: if TRUE, includes an independent censoring time. censlam a scale (>0) parameter for an exponential distribution used to simulate random censoring times for when censoring = TRUE. truncation logical: if TRUE, adds a truncation time for a maximum event time in the case of model = "int" or model = "intslope". trunctime a truncation time for use when truncation = TRUE. For model = "quad", trunctime is required, and defaults to max(ntms) if not specified. gridstep the step-size for the grid used to simulate event times when model = "quad". Default is gridstep = 0.01. See Details.
## Details
The function simjoint simulates data from a joint model, similar to that performed in Henderson et al. (2000). It works by first simulating longitudinal data for all possible follow-up times using random draws for the multivariate Gaussian random effects and residual error terms. Data can be simulated assuming either random-intercepts only (model = "int") in each of the longitudinal sub-models; random-intercepts and random-slopes (model = "intslope"); or quadratic random effects structures (model = "quad"). The failure times are simulated from proportional hazards time-to-event models, using the following methodologies:
model = "int"
The baseline hazard function is specified to be an exponential distribution with
λ_0(t) = \exp{θ_0}.
Simulation is conditional on known time-independent effects, and the methodology of Bender et al. (2005) is used to simulate the failure time.
model = "intslope"
The baseline hazard function is specified to be a Gompertz distribution with
λ_0(t) = \exp{θ_0 + θ_1 t}.
In the usual representation of the Gompertz distribution, θ_1 is the shape parameter, and the scale parameter is equivalent to \exp(θ_0). Simulation is conditional on on a predictable (linear) time-varying process, and the methodology of Austin (2012) is used to simulate the failure time.
model="quad"
The baseline hazard function is specified as per model="intslope". The integration technique used for the above two cases is complex in quadratic (and higher order) models, therefore we use a different approach. We note that hazard function can be written as
\lim_{dt \rightarrow 0} λ(t) dt = \lim_{dt \rightarrow 0} P[t ≤ T ≤ t + dt | T ≥ t].
In the simulation routine the parameter gridstep acts as dt in that we choose a nominally small value, which multiplies the hazard and this scaled hazard is equivalent to the probability of having an event in the interval (t, t + dt), or equivalently (t, t + gridstep). A vector of possible times is set up for each individual, ranging from 0 to trunctime in increments of dt (or gridstep). The failure probability at each time is compared to an independent U(0, 1) draw, and if the probability does not exceed the random draw then the survival time is set as trunctime, otherwise it is the generated time from the vector of candidate times. The minimum of these candidate times (i.e. when we deem the event to have first happened) is taken as the survival time.
## Value
A list of 2 data.frames: one recording the requisite longitudinal outcomes data, and one recording the time-to-event data.
Pete Philipson
## References
Austin PC. Generating survival times to simulate Cox proportional hazards models with time-varying covariates. Stat Med. 2012; 31(29): 3946-3958.
Bender R, Augustin T, Blettner M. Generating survival times to simulate Cox proportional hazards models. Stat Med. 2005; 24: 1713-1723.
Henderson R, Diggle PJ, Dobson A. Joint modelling of longitudinal measurements and event time data. Biostatistics. 2000; 1(4): 465-480.
## Examples
1 simjoint(10, sepassoc = TRUE)
### Example output
Loading required package: survival
90% experienced event
$longitudinal id Y time intercept ctsxl binxl ltime 1 1 3.02286464 0 1 0.53853824 0 0 2 1 2.56739365 1 1 0.53853824 0 1 3 2 0.37887665 0 1 -0.87010820 0 0 4 2 0.62445689 1 1 -0.87010820 0 1 5 2 0.79691682 2 1 -0.87010820 0 2 6 2 0.83935060 3 1 -0.87010820 0 3 7 2 1.10916654 4 1 -0.87010820 0 4 8 3 1.02812010 0 1 0.53117302 0 0 9 3 0.73377986 1 1 0.53117302 0 1 10 4 1.82009714 0 1 0.05398401 0 0 11 4 4.29024221 1 1 0.05398401 0 1 12 4 6.92207504 2 1 0.05398401 0 2 13 5 0.77686279 0 1 0.14287524 0 0 14 5 1.96727196 1 1 0.14287524 0 1 15 5 3.28054386 2 1 0.14287524 0 2 16 5 4.65646695 3 1 0.14287524 0 3 17 6 -0.58549813 0 1 -0.65451963 0 0 18 6 -0.32350394 1 1 -0.65451963 0 1 19 6 -0.04780583 2 1 -0.65451963 0 2 20 6 0.42113495 3 1 -0.65451963 0 3 21 7 -1.38670192 0 1 0.17176097 1 0 22 7 1.32335838 1 1 0.17176097 1 1 23 7 3.97374685 2 1 0.17176097 1 2 24 7 6.59571886 3 1 0.17176097 1 3 25 8 -1.07622025 0 1 -1.09697154 0 0 26 8 0.81536925 1 1 -1.09697154 0 1 27 8 2.64830669 2 1 -1.09697154 0 2 28 8 4.71079180 3 1 -1.09697154 0 3 29 8 6.51770420 4 1 -1.09697154 0 4 30 9 2.15771113 0 1 -1.27944478 1 0 31 9 1.57054830 1 1 -1.27944478 1 1 32 10 0.81335295 0 1 -1.85527181 1 0 33 10 1.93939639 1 1 -1.85527181 1 1 34 10 2.80191817 2 1 -1.85527181 1 2$survival
id survtime cens ctsx binx
1 1 1.194932 1 0.53853824 0
2 2 4.047252 1 -0.87010820 0
3 3 1.933055 0 0.53117302 0
4 4 2.202683 1 0.05398401 0
5 5 3.293996 1 0.14287524 0
6 6 3.974561 1 -0.65451963 0
7 7 3.749802 1 0.17176097 1
8 8 5.586596 1 -1.09697154 0
9 9 1.834924 1 -1.27944478 1
10 10 2.742520 1 -1.85527181 1
joineR documentation built on June 1, 2021, 5:06 p.m. | 2,539 | 7,620 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-49 | longest | en | 0.740616 |
https://www.edureka.co/community/95968/can-someone-please-help-me-with-python | 1,726,770,961,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652055.62/warc/CC-MAIN-20240919162032-20240919192032-00346.warc.gz | 701,980,851 | 25,241 | 1. Ask the user to enter two numbers
1. Display the result of the addition, multiplication, division, subtraction, Modulus, Floor division, and Exponent for these two numbers
2. If the result of any mathematical calculation equal 2 print the multiplication time table for it by using loop statement.
Nov 29, 2020 in Python 566 views
## 1 answer to this question.
Hello @ Anee,
Code:
```num1 = int(input("Enter First Number: "))
num2 = int(input("Enter Second Number: "))
print("Enter which operation would you like to perform?")
ch = input("Enter any of these char for specific operation +,-,*,/: ")
result = 0
if ch == '+':
result = num1 + num2
if result == 2:
//run your for loop for result
elif ch == '-':
result = num1 - num2
if result == 2:
//run your for loop for result
elif ch == '*':
result = num1 * num2
if result == 2:
//run your for loop for result
elif ch == '/':
result = num1 / num2
if result == 2:
//run your for loop for result
else:
print("Input character is not recognized!")
print(num1, ch , num2, ":", result)```
answered Nov 30, 2020 by
• 82,880 points
## Can someone help me understand timedelta in Python?
Because timedelta is defined like: class datetime.timedelta([days,] [seconds,] ...READ MORE
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You can simply the built-in function in ...READ MORE | 565 | 2,222 | {"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} | 2.703125 | 3 | CC-MAIN-2024-38 | latest | en | 0.813896 |
https://www.jiskha.com/questions/1201755/if-y-varies-inversely-as-x-and-y-12-when-x-6-what-is-k-the-variation-constant-a | 1,642,492,453,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300805.79/warc/CC-MAIN-20220118062411-20220118092411-00029.warc.gz | 893,058,940 | 4,847 | # math
If y varies inversely as x, and y = 12 when x = 6, what is K, the variation constant?
A. 1⁄3 C. 72
B. 2 D. 144
1. 👍
2. 👎
3. 👁
1. well, xy=k, so
12*6 = k
1. 👍
2. 👎
2. y=k/x
y=12,x=6
12=k/6
k=12*6
k=72
1. 👍
2. 👎
3. y varies inversely as x. y=12 when x=7. Find y when x=6.
1. 👍
2. 👎
4. If all men had identical body types, their weight would vary directly as the cube of their height. The tallest person reached a record height of 8 feet 11 inches (107 inches) before his death at age 22. If a man who is 6 feet 10 inches tall (82 inches) with the same body type as the tallest person weighs 190 pounds, what was the tallest person's weight shortly before his death?
1. 👍
2. 👎
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4. ### Mathematics
14. m varies inversely to n and directly to p. If m = 8 when n = 2 and p = 10, find p when y = 20 and n = 6. = | 820 | 2,391 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2022-05 | latest | en | 0.8946 |
https://bioinformatics.stackexchange.com/questions/695/count-genomic-ranges | 1,638,035,908,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358208.31/warc/CC-MAIN-20211127163427-20211127193427-00338.warc.gz | 211,951,130 | 33,858 | # Count genomic ranges
I have a set of genomic ranges that are potentially overlapping. I want to count the amount of ranges at certain positions using R.
I'm Pretty sure there are good solutions, but I seem to be unable to find them.
Solutions like cut or findIntervals don't achieve what I want as they only count on one vector or accumulate by all values <= break.
Also countMatches {GenomicRanges} doesn't seem to cover it.
Probably one could use Bedtools, but I don't want to leave R.
I could only come up with a hilariously slow solution
# generate test data
testdata <- data.frame(chrom = rep(seq(1,10),10),
starts = abs(rnorm(100, mean = 1, sd = 1)) * 1000,
ends = abs(rnorm(100, mean = 2, sd = 1)) * 2000)
# make sure that all end coordinates are bigger than start
# this is a requirement of the original data
testdata <- testdata[testdata$ends - testdata$starts > 0,]
# count overlapping ranges on certain positions
count.data <- lapply(unique(testdata$chrom), function(chromosome){ tmp.inner <- lapply(seq(1,10000, by = 120), function(i){ sum(testdata$chrom == chromosome & testdata$starts <= i & testdata$ends >= i)
})
return(unlist(tmp.inner))
})
# generate a data.frame containing all data
df.count.data <- ldply(count.data, rbind)
# ideally the chromosome will be columns and not rows
t(df.count.data)
• So you only want overlaps where the start position is not contained within your interval range? What's the actual biological context to this? Jun 13 '17 at 13:58
• You could start by optimizing the code by using data.table (and probably using vapply instead of lapply). Also if one nucleotide overlaps between ranges then you count that too?
– llrs
Jun 13 '17 at 14:02
• @DevonRyan no what i want is that the respective counting position is with in the range, which is a genomic deletion (hence the biological context) Jun 13 '17 at 14:55
• @Llopis yes i want that counted as well. The list contains overlapping areas and i want to know how many of them are overlapping a certain position. Which is then used later for plotting. Jun 13 '17 at 14:56
• Hi sargas, thanks for your question and welcome to Bioinformatics Stack Exchange. If you have any additional story/context associated with your question (such as a desire to count genomic deletions), it can be helpful to include that in your question. This makes it easier to understand the question, and helps answerers to solve the problem you have rather than the question you're asking (see the XY problem for more information).
– gringer
Jun 13 '17 at 19:27
GenomicRanges::countOverlaps seems to be what you’re after:
position_range = GRanges(position\$chrom, IRanges(position, position, width = 1)) | 663 | 2,687 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-49 | latest | en | 0.901779 |
http://clay6.com/qa/27065/solve-log-bigg-large-frac-frac-bigg-2x-5y | 1,481,422,822,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543782.28/warc/CC-MAIN-20161202170903-00250-ip-10-31-129-80.ec2.internal.warc.gz | 54,503,865 | 27,143 | Browse Questions
# Solve : $\log \bigg(\large\frac{1}{x}.\frac{dy}{dx}\bigg)$$=2x+5y (a)\;e^{5y}=e^{2x}-x+c \\ (b)\;e^{-5y}=1+2x+x \\ (c)\;e^{-5y}=\frac{5}{2}(e^{2x}-x-2^{2x})+c\\ (d)\;e^{-5y}=5e^{2x}(1-ex)+c Can you answer this question? ## 1 Answer 0 votes \large\frac{1}{x}.\frac{dy}{dx}$$=e^{2x}.e^{5y}$
$\large\frac{dy}{e^{5y}}$$=x.e^{2x}dx \large\frac{-e^{-5y}}{5}=\frac{x.e^{2x}}{2}-\frac{2^{2x}}{2} e^{-5y}=\large\frac{5}{2}$$(e^{2x}-x.e^{2x})$
Hence c is the correct answer. | 253 | 484 | {"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} | 4.03125 | 4 | CC-MAIN-2016-50 | longest | en | 0.138901 |
http://www.pinoybix.org/2016/06/fnd-the-third-proportional-of-the-two-integers.html | 1,508,469,818,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823630.63/warc/CC-MAIN-20171020025810-20171020045810-00326.warc.gz | 526,166,506 | 91,713 | # Solution: What is the third proportional of the square of 2 and the cube root of 8?
What is the third proportional of the square of 2 and the cube root of 8? How to find the third proportional of two numbers?
#### Problem Statement:
What is the third proportional of the square of 2 and the cube root of 8?
The third proportional of the square of 2 and the cube root of 8 is 1.
Solution:
### Latest Problem Solving in Clock, Variation, Progression and Miscelaneous Problems
More Questions in: Clock, Variation, Progression and Miscelaneous Problems
#### Search! Type it and Hit Enter
Accepting Contribution for Website Operation. Thanks! Option 1 : \$5 USD Option 2 : \$10 USD Option 3 : \$15 USD Option 4 : \$20 USD Option 5 : \$25 USD Option 6 : \$50 USD Option 7 : \$100 USD Option 8 : Other Amount | 204 | 813 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2017-43 | latest | en | 0.813821 |
https://www.experts-exchange.com/questions/27640534/Excel-Formula-to-Selectively-Sum-Data.html | 1,490,727,022,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189802.72/warc/CC-MAIN-20170322212949-00087-ip-10-233-31-227.ec2.internal.warc.gz | 902,620,117 | 26,656 | Solved
# Excel Formula to Selectively Sum Data
Posted on 2012-03-20
340 Views
I have a sheet that has text in column A, a amounts in column B and dates in column C. In row 1, columns O through Z I have the names of Months (January..December).
In the January column, row 2 (cell O2), I want to create a formula that will SUM all values in column B where MONTH(C#) = 1. Likewise in cell P2, I want to sum all rows in column B where MONTH(C#) = 2. And so on.
The result is 12 formulas, one for each month, that displays the total amount in colukmn B for that month (or 0 if nothing exists yet for the month).
0
Question by:dbbishop
• 2
• 2
LVL 50
Assisted Solution
barry houdini earned 250 total points
ID: 37743422
Try this formula in O2 copied across
=SUMPRODUCT((TEXT(\$C\$2:\$C\$100,"mmmm")=O\$1)+0,\$B\$2:\$B\$100)
assumes data in rows 2 to 100, adjust as required
regards, barry
0
LVL 7
Accepted Solution
leptonka earned 250 total points
ID: 37743530
You can use this formula in O2 and copy across columns:
=SUMPRODUCT(--(MONTH(\$C\$2:\$C\$6)=COLUMN()-COLUMN(\$O1)+1),\$B\$2:\$B\$6)
(rows from 2 to 6 - you can change)
(nb: "mmmm" in TEXT formula is language-dependent, it will not work in some non-english environment.)
Cheers,
Kris
0
LVL 15
Author Closing Comment
ID: 37743741
Both, in my case work and produce the same results. I've accepted Kris' as the best solution as he does take language into consideration, even though I am not considering running this code in a computer with Chinese set as the language any time in the near future, it may help others. I've split the points evenly. Thanks to both of you.
0
LVL 7
Expert Comment
ID: 37743834
Thank you!
The problem is not just Chinese but for example my Hungarian, where "hhhh" stands for "mmmm"
Cheers,
Kris (she :-)
0
LVL 50
Expert Comment
ID: 37744211
It's a fair point :)
You could possibly use COLUMNS function to avoid having 2 COLUMN functions, i.e.
=SUMPRODUCT(--(MONTH(\$C\$2:\$C\$6)=COLUMNS(\$O2:O2)),\$B\$2:\$B\$6)
regards, barry
0
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Excel styles will make formatting consistent and let you apply and change formatting faster. In this tutorial, you'll learn how to use Excel's built-in styles, how to modify styles, and how to create your own. You'll also learn how to use your custo… | 849 | 3,149 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2017-13 | longest | en | 0.879583 |
http://www.chegg.com/homework-help/differential-equations-and-linear-algebra-3rd-edition-chapter-8.4-solutions-9780130457943 | 1,475,236,251,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738662166.99/warc/CC-MAIN-20160924173742-00145-ip-10-143-35-109.ec2.internal.warc.gz | 399,381,278 | 16,808 | # Differential Equations and Linear Algebra (3rd Edition) View more editions Solutions for Chapter 8.4
• 3278 step-by-step solutions
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Chapter: Problem:
For Questions 1–4, decide if the given statement is true or false, and give a brief justification for your answer. If true, you can quote a relevant definition or theorem from the text. If false, provide an example, illustration, or brief explanation of why the statement is false.
For every function f with a continuous derivative on [0,), the Laplace transform of the derivative is given by L[f '] = sL[f] − f (0).
SAMPLE SOLUTION
Chapter: Problem:
• Step 1 of 1
Consider the following statement
“For every function with a continuous derivative on, the Laplace transform of the derivative is given by
Its need to determine whether the statement is true or false
Recall the theorem that “if is of exponential order on and that exists and is piecewise continuous on. Then exists and is given by
”therefore, for existence of, is must be of exponential order.
Hence, the given statement is false.
Corresponding Textbook
Differential Equations and Linear Algebra | 3rd Edition
9780130457943ISBN-13: 0130457949ISBN: Authors:
Alternate ISBN: 9780321996961 | 327 | 1,380 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2016-40 | latest | en | 0.838309 |
http://www.romannumerals.co/number-converter/4901-in-roman-numerals/ | 1,709,382,554,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475825.14/warc/CC-MAIN-20240302120344-20240302150344-00757.warc.gz | 63,259,109 | 14,355 | ## What is 4901 in Roman Numerals?
### A: IVCMI
4901 = IVCMI
Your question is, "What is 4901 in Roman Numerals?", and the answer is 'IVCMI'. Here we will explain how to convert, write and read the number 4901 in the correct Roman numeral figure format.
## How is 4901 converted to Roman numerals?
To convert 4901 to Roman Numerals the conversion involves you to split it up into place values (ones, tens, hundreds, thousands), like this:
Place ValueNumberRoman Numeral
Conversion4000 + 900 + 1IV + CM + I
Thousands4000IV
Hundreds900CM
Ones1I
## How to write 4901 in Roman numerals?
To write 4901 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:
IV+CM+I = IVCMI
## How do you read 4901 as Roman numerals
To correctly read the number 4901 as the Roman numeral IVCMI, 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 IVCMI in a text, unless it represents an ordinal value. In any other usage case it should be written in the normal format (arabic number) 4901.
## More from Roman Numerals.co
4902 in Roman numerals
Now you understand how to read and write 4901 in Roman Numerals, see how the number 4902 is written.
Convert Another Number | 366 | 1,419 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-10 | latest | en | 0.788089 |
http://mekesim.com/doc/Flotherm/MG541045/ | 1,679,785,055,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945376.29/warc/CC-MAIN-20230325222822-20230326012822-00617.warc.gz | 34,011,708 | 12,894 | # 重力方向
## SUMMARY
Where can I edit the value for acceleration due to gravity 'g' or turn it off if I wanted to?
I have my electronics box inclined at an angle (other than 90o) with vertical; how can I model that?
## DETAILS
It is not necessary to recreate geometry using inclined sloping blocks when performing thermal analysis on equipment inclined at an angle with gravity.
The default value of acceleration due to gravity set in the software is 9.81 m/s2 and is directed along negative Y direction. This can be reviewed under Model Setup>Gravity in the Project Manager. For certain applications, the user can edit this value (accelerating rocket) or turn it off completely (when in outer space).
The user can quickly change the gravity vector orientation to one of the normal directions X, Y,Z ( or -X, -Y or -Z) by selecting the appropriate option under the direction drop down.
In certain applications the electronic components or the entire chassis could be inclined at an arbitrary angle with vertical. A typical example is a telephone box placed on an inclined plank. The user can specify an angle (other than orthogonal directions) by selecting 'Angled' and specifying the directional cosines.
Gravity vector can be resolved into its components along the orthogonal directions and can be written in vector form as. $$g = |g|.(g_x i +g_y j +g_z k)$$ where $$g_x$$, $$g_y$$ & $$g_z$$ are normalized values of unit vector
If the gravity vector is inclined at angle of 45 deg (clockwise) with vertical the directional cosines are entered simply as below (user does not have to worry about normalizing the vector components). Gravity orientation is also available as an input variable in Command Center. The magnitude of gravity vector $$|g|$$ is still taken as default value of 9.81 m/s2 (unless user specified value is chosen).
Below is a simple example of an inclined heated plate modeled by rotating the gravity vector appropriately. | 420 | 1,952 | {"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": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-14 | latest | en | 0.906337 |
http://openstudy.com/updates/55f4f382e4b0cf7fa75ba893 | 1,495,481,517,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607046.17/warc/CC-MAIN-20170522190443-20170522210443-00611.warc.gz | 268,633,635 | 8,363 | • anonymous
You wonder how many cars there are in your area. You call up the Ford dealership and ask how many Fords they’ve sold. They report that they’ve sold a total of 15,000 cars. You then take a lawn chair and camp out on a busy road and count cars. You count 1200 cars and note that 700 of them are Fords. Estimate the number of cars in your area. Options are: 25,700 cars 32,000 cars 28,000 cars 22,400 cars
Mathematics
• Stacey Warren - Expert brainly.com
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Not the answer you are looking for? Search for more explanations. | 356 | 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} | 2.796875 | 3 | CC-MAIN-2017-22 | longest | en | 0.543218 |
https://developers.google.cn/earth-engine/reducers_grouping | 1,516,608,691,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891196.79/warc/CC-MAIN-20180122073932-20180122093932-00565.warc.gz | 678,741,918 | 11,817 | # Grouped Reductions and Zonal Statistics
You can get statistics in each zone of an `Image` or `FeatureCollection` by using `reducer.group()` to group the output of a reducer by the value of a specified input. For example, to compute the total population and number of housing units in each state, this example groups the output of a reduction of a counties `FeatureCollection` as follows:
```// Load a collection of US counties with census data properties.
var counties = ee.FeatureCollection('ft:1S4EB6319wWW2sWQDPhDvmSBIVrD3iEmCLYB7nMM');
// Compute sums of the specified properties, grouped by state name.
var sums = counties
.filter(ee.Filter.and(
ee.Filter.neq('Census 2000 Population', null),
ee.Filter.neq('Census 2000 Housing Units', null)))
.reduceColumns({
selectors: ['Census 2000 Population', 'Census 2000 Housing Units', 'StateName'],
reducer: ee.Reducer.sum().repeat(2).group({
groupField: 2,
groupName: 'state',
})
});
// Print the resultant Dictionary.
print(sums);
```
The `groupField` argument is the index of the input in the selectors array that contains the codes by which to group, the `groupName` argument specifies the name of the property to store the value of the grouping variable. Since the reducer is not automatically repeated for each input, the `repeat(2)` call is needed.
To group output of `image.reduceRegions()` you can specify a grouping band that defines groups by integer pixel values. This type of computation is sometimes called "zonal statistics" where the zones are specified as the grouping band and the statistic is determined by the reducer. In the following example, change in nightlights in the United States is grouped by land cover category:
```// Load a region representing the United States
var region = ee.Feature(
ee.FeatureCollection('ft:1tdSwUL7MVpOauSgRzqVTOwdfy17KDbw-1d9omPw')
.filter(ee.Filter.eq('Country', 'United States'))
.first());
// Load MODIS land cover categories in 2001.
var landcover = ee.Image('MODIS/051/MCD12Q1/2001_01_01')
// Select the IGBP classification band.
.select('Land_Cover_Type_1');
// Load nightlights image inputs.
var nl2001 = ee.Image('NOAA/DMSP-OLS/NIGHTTIME_LIGHTS/F152001')
.select('stable_lights');
var nl2012 = ee.Image('NOAA/DMSP-OLS/NIGHTTIME_LIGHTS/F182012')
.select('stable_lights');
// Compute the nightlights decadal difference, add land cover codes.
var nlDiff = nl2012.subtract(nl2001).addBands(landcover);
// Grouped a mean reducer: change of nightlights by land cover category.
var means = nlDiff.reduceRegion({
reducer: ee.Reducer.mean().group({
groupField: 1,
groupName: 'code',
}),
geometry: region.geometry(),
scale: 1000,
maxPixels: 1e8
});
// Print the resultant Dictionary.
print(means);
```
Note that in this example, the `groupField` is the index of the band containing the zones by which to group the output. The first band is index 0, the second is index 1, etc.
### 发送以下问题的反馈：
Google Earth Engine API | 751 | 2,933 | {"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} | 2.5625 | 3 | CC-MAIN-2018-05 | longest | en | 0.718096 |
https://socratic.org/questions/given-three-points-1-4-1-6-2-10-how-do-you-write-a-quadratic-function-in-standar | 1,725,929,637,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651164.37/warc/CC-MAIN-20240909233606-20240910023606-00329.warc.gz | 495,638,616 | 6,637 | # Given three points (-1,4) (1,6) (2,10) how do you write a quadratic function in standard form with the points?
May 2, 2017
Solve a system of three equations in three unknowns.
#### Explanation:
Let us assume that the axis of symmetry for the parabola is vertical.
The standard equation of a parabola with a vertical axis is $y = a {x}^{2} + b x + c$, where a, b, and c are real numbers -- with $a \ne 0$.
Using the coordinates of the above points, we know that...
$a {\left(- 1\right)}^{2} + b \left(- 1\right) + c = 4$
This simplifies to $a - b + c = 4$.
We also have:
$a {1}^{2} + b \left(1\right) + c = 6$
That is, $a + b + c = 6$.
And finally:
$a {\left(2\right)}^{2} + b \left(2\right) + c = 10$
$4 a + 2 b + c = 10$
The three equations give us the following system, which has a unique solution (a, b, c):
$a - b + c = 4$
$a + b + c = 6$
$4 a + 2 b + c = 10$.
While we may solve these using any method, it will be convenient to subtract equation (1) from equation (2).
$a + b + c = 6$
$a - b + c = 4$ (subtract)
Change the signs and add:
$a + b + c = 6$
$- a + b - c = - 4$ (add)
$2 b = 2$
$b = 1$.
Our system is now reduced to just two unknowns. Putting the known value of b into the three equations gives:
$a - 1 + c = 4$
$a + 1 + c = 6$
$4 a + 2 + c = 10$.
Simplify to:
$a + c = 5$
$4 a + c = 8$.
Subtract the first from the second to obtain the equation:
$3 a = 3$
$a = 1$
Substitute a = 1 into the equation $a + c = 5$ and obtain $c = 4$.
The equation of our parabola has a = 1, b = 1, and c = 4.
Therefore, it is
$y = {x}^{2} + x + 4$
Check, and you will find that all three points fit the equation. | 585 | 1,633 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 27, "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} | 4.8125 | 5 | CC-MAIN-2024-38 | latest | en | 0.868306 |
https://www.electronicspoint.com/forums/threads/subtraction-of-two-frequencys.2822/ | 1,606,957,434,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141717601.66/warc/CC-MAIN-20201203000447-20201203030447-00409.warc.gz | 456,905,853 | 12,584 | # subtraction of two frequencys
Discussion in 'Electronic Basics' started by Torbjörn Klemensson, Sep 24, 2003.
1. ### Torbjörn KlemenssonGuest
I have two frequencys f1= 30 MHz, f2=13 MHz that I want to get the
difference of (f=f1-f2). Is there a IC or a simple circuit that can
help me with the operation?
/Torbjorn
2. ### grahamkGuest
For a start have a look at
http://homepage.ntlworld.com/g.knott/elect113.htm
3. ### DboweyGuest
klemensson posted:
You will need, at least, an RF mixer into which you will send the two signals.
When the signals combine (multiply) in the mixer, the output will contain the
sum and difference of the two frequencies as well as other frequencies you may
need to filter. Diodes connected as a balanced mixer works well, but there are
other components that can be used.
You don't say what are the signal sources or what will be the recipient of the
difference signal so not much here can be said about what filtering you may
need. If the output is to a receiver, it might provide all that is needed.
In a google search you should probably look for balanced modulator as well as
balanced mixer. They are essentially the same. You will note that to make an
RF mixer work well, that one frequency (selected as the "carrier" frequency,
should be about ten times higher in level than the "modulating" frequency.
Don
5. ### John FieldsGuest
---
You need a double-balanced mixer and a filter. 30MHz goes in the RF (or
LO) port, 13MHz goes in the LO (or RF) port, and you'll get f1+f2 and
f1-f2 out of the IF port. If you want 17MHz you can get it by putting a
17MHz. bandpass (or lowpass) filter on the IF port which will let the
17MHz through and block the 43MHz.
Check http://www.minicircuits.com/dg03-92.pdf for some likely
candidates. | 475 | 1,777 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-50 | longest | en | 0.939841 |
https://www.coursehero.com/file/84140/Exam1-2005Spr-Solutions/ | 1,490,502,745,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189092.35/warc/CC-MAIN-20170322212949-00557-ip-10-233-31-227.ec2.internal.warc.gz | 880,806,872 | 57,569 | Exam1_2005Spr_Solutions
# Exam1_2005Spr_Solutions - Physics 8.02 F= Exam One...
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Physics 8.02 Exam One Solutions Spring 2005 F = 1 4 πε o q 1 q 2 r 2 23 ˆ 44 oo qq rr == Er G r G ˆ points source q observer r r r= from to G 2 0 1 ˆ 4 V dq r = G inside o closed surface Q κ ε ⋅= ∫∫ EdA G G w outside to inside from points A d G = = b a a b b to a from moving V V V s d E G G 0 = path closed s d E G G point charge 4 o q V r = many point charges 1 4 N i i oi q V = = G K = pairs all j i o j i q q U r r G G 4 2 1 2 ov UE ⎡⎤ = ⎢⎥ ⎣⎦ ∫∫∫ x o l d V E x = - V x , E y = - V y , E z = - V z r V E r =− for spherical symmetry Q C V = U = 1 2 C V 2 = Q 2 2C 1 parallel CC 2 C = + 12 series C = + Circumferences, Areas, Volumes: 1) The area of a circle of radius r is π r 2 Its circumference is 2 π r 2) The surface area of a sphere of radius r is 4 r 2 Its volume is 3 4 3 r 3) The area of the sides of a cylinder of radius r and height h is 2 r h Its volume is r 2 h Definition of trig functions sin is opposite/hypotenuse; cos is adjacent/hypotenuse; tangent is opposite over adjacent; Properties of 30, 45, and 60 degrees ( π /6, π /4, and π /3 radians) : sin( π /6) = cos( π /3) = 1/2, sin( π /3) = cos( π /6) = 2 / 3; sin( π /4) = cos( π /4) = 2 / 1 ; Integrals that may be useful a b dr b a = ) / ln( 1 a b dr r b a = = b a dr r b a 1 1 1 2
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MIT PHYSICS DEPARTMENT ` page 2 Problem 1: Five Short Questions. Circle your choice for the correct answer. Question A (5 points out of 25 points): The field line to the right is a field line of the vector field a) ˆˆ ( , ) sin( ) x yx =+ Fi G j CORRECT b) (, ) c o s () x =− G j c) s i n x G j d) c o s x G j 2 egative e) None of these I1 I2 O1 O2 Question B (5 points out of 25 points): Two hollow conducting shells are concentric. They carry no net charge. We place a negative point charge at the center of the shells. Let V be the potential at O1 and V be the potential at O2. Then 1 O 2 O a) and the surface charge at O2 is n 1 OO VV > b) ace charge at O2 is negative 12 < and the surf CORRECT c) e surface charge at O2 is negative = and th d) and the surface charge at O2 is positive > e) and the surface charge at O2 is positive < f) and the surface charge at O2 is positive =
MIT PHYSICS DEPARTMENT ` page 6 Question C (5 points out of 25 points): We have two electric dipoles. Each dipole consists of two equal and opposite point charges at the ends of an insulating rod of length d . The dipoles sit along the x- axis a distance r apart, oriented as shown below. The dipole ON THE LEFT : 1. will feel a force upwards and a torque trying to make it rotate clockwise 2. will feel a force upwards and a torque trying to make it rotate counterclockwise CORRECT 3. will feel a force upwards and no torque 4. will feel a force downwards and a torque trying to make it rotate clockwise.
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## This homework help was uploaded on 04/07/2008 for the course 8 8.02 taught by Professor Hudson during the Fall '07 term at MIT.
### Page1 / 11
Exam1_2005Spr_Solutions - Physics 8.02 F= Exam One...
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Ask a homework question - tutors are online | 1,122 | 3,520 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-13 | longest | en | 0.800853 |
http://tex.stackexchange.com/questions/28936/extra-space-between-number-and-variable-in-math-mode/104306 | 1,469,822,548,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257831771.10/warc/CC-MAIN-20160723071031-00089-ip-10-185-27-174.ec2.internal.warc.gz | 254,626,676 | 19,607 | # Extra space between number and variable in math mode
I find that if I typeset $3 q$, I find the 3 and the q a bit too close to each other. I know that I can manually add a thin space by typing: $3 \, q$.
1. However, I was wondering if there was a way to automate this, i.e. that every time a number precedes a variable, a thin space is added.
In other words I would like a thin space to be automatically added in $5 q$, but not in $5 7$.
2. And is it possible to have added in $p q$ too? Do you recommend such practice between two variables that are multiplied? (I don't want to use \cdot as it clutters the display.)
-
I recommend neither. Look at good quality math books. – egreg Sep 19 '11 at 19:12
@Caramdir I have opened a question on the same topic, with a different approach. It has received a satisfactory answer. You may want to take a look at it. – Pascal Romon Jun 24 '15 at 12:00
TeX has a finely balanced system of setting spaces between various types of math "atoms". For example, consider the following simple formula:
$y = b + cx$
Observe that the distance between y and = (and also between = and b) is slightly larger than that between b and +, which again exceeds that between c and x. This is done on purpose, of course, and the choices involved have proven their desirability over decades.
If you were to systematically increase the spacing between any two "atoms" that are multiplied together (such as c and x in the example above), you should also be willing to increase the spacing between all other types of "atoms" in order to preserve the overall balance. To claim that this would be a rather tricky enterprise would be a rather strong understatement.
In short, it's best not only to get used to TeX's way of typesetting mathematics but also to appreciate it for the high standard it sets. TeX's method is the standard against which all other systems for typesetting mathematics are judged and against which they, regrettably, almost invariably fail.
-
Ok thank you for answer! If I agree that it would be technically difficult to do so, I do not on the hand accept that one should NOT do it just on the grounds that this has been the way for thirty years. I have been using TeX for seven years and I do appreciate its general beauty. But nothing in this world is sacred. I do find the c and the x too close to each other to my tastes... but hey, if it's not changeable, it's a moot point. – Peutch Sep 20 '11 at 9:42
You may want to look into the jamtimes package (jam is short for Journal d'Analyse Mathematique), which uses a specially expanded form of Times New Roman for its fonts. – Mico Sep 20 '11 at 14:57
This is exactly the answer that I would not be looking for. 1. Very often (and the OP confirms this for his case) I am trying to intentionally deviate from some established style, for good reasons or sometimes for personal reasons. What I don't want is somebody else telling that my preferences shouldn't be implemented. You in fact don't know the OP's context. 2. You're not actually giving any arguments for the default style. The argument by tradition is unlikely to be convincing. 3. Praising LaTeX is best done by showing it can do anything. – Lover of Structure Mar 26 '13 at 3:17
... "The default does X, and this mathord/'atom'/whatever algorithm is why it's done this way, and it's well-respected within the community and established (and I like the default), so you likely won't go wrong with it. Changing X to Y is easy/possible/hard/impossible for reason Z (something relating to math character classes), but someone might attempt doing V." I find your wording, such as "best to get used to [it]" and "all other systems [...] almost invariably fail", too strong to read for a curious non-enthusiast (whether his thoughts about spacing are justified or not). ... – Lover of Structure Mar 27 '13 at 1:15
... Also, perhaps the OP is typesetting for a particular educational context or in another locale's tradition where the requested spacing is quite sensible. – Lover of Structure Mar 27 '13 at 1:16
Just type \ and press space bar between the numbers you want an extra space. for example: $(2,3,4)$ and I need extra space between numbers, then I would type: $(2,\ 3,\ 4)$ or if I need more extra space, then type: $(2,\ \ 3,\ \ 4)$. Hope this helps.
-
Welcome to TeX.sx! A tip: You can use backticks ` to mark your inline code as I did in my edit. – texenthusiast Mar 26 '13 at 2:33
Please just have a look to @Mico's answer. It is better to use the LaTeX-build-in-typesetting of mathematics than to try to do it by your own ... – Kurt Mar 26 '13 at 3:00 | 1,140 | 4,622 | {"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} | 3.015625 | 3 | CC-MAIN-2016-30 | latest | en | 0.971682 |
https://www.cravencountryjamboree.com/helpful-tips/how-do-you-calculate-iv-fluids-for-adults/ | 1,726,075,257,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651390.33/warc/CC-MAIN-20240911152031-20240911182031-00439.warc.gz | 689,131,262 | 14,418 | # How do you calculate IV fluids for adults?
## How do you calculate IV fluids for adults?
Maintenance Fluid Rate is calculated based on weight.
1. 4 mL / kg / hour for the first 10kg of body mass.
2. 2 mL / kg / hour for the second 10kg of body mass (11kg – 20kg)
3. 1 mL / kg / hour for any kilogram of body mass above 20kg (> 20kg)
How do you calculate pediatric fluid replacement?
1. For infants 3.5 to 10 kg the daily fluid requirement is 100 mL/kg.
2. For children 11-20 kg the daily fluid requirement is 1000 mL + 50 mL/kg for every kg over 10.
3. For children >20 kg the daily fluid requirement is 1500 mL + 20 mL/kg for every kg over 20, up to a maximum of 2400 mL daily.
### How do you fix fluid deficit?
The best treatment for fluid volume deficit is to address the electrolyte imbalance and water loss by using an oral rehydration solution. In severe cases, you may need nursing intervention and critical care support, including intravenous fluid therapy or a blood transfusion.
What is the 421 rule for IV fluids?
The 421 rule is used to calcuate the hourly infusion rate for maintenance fluids (generally just for pediatrics). 4 cc/hr for kg 1-10 .
#### What is the formula for maintenance fluid?
Formula Method. (100 ml for each of the first 10kg) + ( 50ml for each kg 11-20) + (20 ml for each additional kg) / 24hour. Example: Calculate the hourly maintenance fluid rate for a child who weighs 25kg. (100mL x 10kg) + (50mL x 10kg) + (20mL x 5kg) / 24hrs. (1000mL) + (500mL) + (100mL) = 1600mL / 24hrs = 66.7ml/hr.
How do you calculate maintenance IV fluids?
The rule for calculating maintenance fluids for a 24 hours period, is the 100/50/20 rule. IV administration of 100ml/kg for the first 10kg is given. Then 50ml/kg for the next 10kg and then 20ml/kg for every kg over 20. Once this has been figured, you divide the amount by 24 to get the hourly IV fluid flow rate.
## How do you calculate fluid requirement?
The other common way to calculate daily fluid needs is to base the fluid need on caloric intake. 1 milliliter of fluid for every calorie ingested. Converted to the household measurement of ounces, your body needs .034 ounces for every calorie that you ingest. | 595 | 2,204 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-38 | latest | en | 0.841103 |
https://bafybeiemxf5abjwjbikoz4mc3a3dla6ual3jsgpdr4cjr3oz3evfyavhwq.ipfs.dweb.link/wiki/Intersection_(Euclidean_geometry).html | 1,652,847,351,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662521041.0/warc/CC-MAIN-20220518021247-20220518051247-00397.warc.gz | 169,504,786 | 11,858 | # Intersection (Euclidean geometry)
In geometry, an intersection is a point, line, or curve common in two or more objects (such as lines, curves, planes, and surfaces). The most simple case in Euclidean geometry is the intersection points of two distinct lines, that is either one point or does not exist if lines are parallel.
intersection point of two lines
Determination of the intersection of flats is a simple task of linear algebra, namely a system of linear equations. In general the determination of an intersection leads to non-linear equations, which can be solved numerically, for example using a Newton iteration. Intersection problems between a line and a conic section (circle, ellipse, parabola, ...) or a quadric (sphere, cylinder, hyperboloid, ...) lead to quadratic equations that can be easily solved. Intersections between quadrics lead to quartic equations that can be solved algebraically.
## On a plane
Further information: plane (geometry) and two-dimensional space
### Two lines
For the determination of the intersection point of two non-parallel lines
one gets from Cramer's rule for the coordinates of the intersection point
(In case of the lines are parallel.)
If the lines are given by two points each, see next section.
### Two line segments
intersection of two line segments
For two non-parallel line segments and there is no need for an intersection point (see picture), because the intersection point of the corresponding lines need not to be contained in the line segments. In order to check the situation one uses parametric representations of the lines:
The line segments intersect only in a common point of the corresponding lines if the corresponding parameters fulfill the condition . The parametrs are the solution of the linear system
It can be solved using Cramer's rule (see above). If the condition is fulfilled one inserts or into the corresponding parametric representation and gets the intersection point .
Example: For the line segments and one gets the linear system
and . That means: the lines intersect at point .
Remark: Considering lines (not segments!) determined by pairs of points, each, condition can be skipped and the method yield the intersection point of the lines (see above).
line–circle intersection
### A line and a circle
For the intersection of
one solves the line equation for x or y and substitutes it into the equation of the circle and gets for the solution (using the formula of a quadratic equation) with
if If this condition holds with strict inequality, there are two intersection points; in this case the line is called a secant line of the circle, and the line segment connecting the intersection points is called a chord of the circle.
If holds, there exists only one intersection point and the line is tangent to the circle. If the weak inequality does not hold, the line does not intersect the circle.
If the circle's midpoint is not the origin, see.[1] The intersection of a line and a parabola or hyperbola may be treated analogously.
### Two circles
circle–circle intersection
circle–ellipse intersection
The determination of the intersection points of two circles
can be reduced to the previous case of intersecting a line and a circle. By subtraction of the two given equations one gets the line equation:
The intersection of two disks (the interiors of the two circles) forms a shape called a lens.
### Two conic sections
The problem of intersection of an ellipse/hyperbola/parabola with another conic section leads to a system of quadratic equations, which can be solved in special cases easily by elimination of one coordinate. Special properties of conic sections may be used to obtain a solution. In general the intersection points can be determined by solving the equation by a Newton iteration. If a) both conics are given implicitly (by an equation) a 2-dimensional Newton iteration b) one implicitly and the other parametrically given a 1-dimensional Newton iteration is necessary. See next section.
### Two curves
A transversal intersection of two curves
touching intersection (left), touching (right)
Two curves in , which are continuously differentiable (i.e. there is no sharp bend), have an intersection point, if they have a point of the plane in common and have at this point
a: different tangent lines (transversal intersection), or
b: the tangent line in common and they are crossing each other (touching intersection, s. picture).
If both the curves have a point S and the tangent line there in common but do not cross each other, they are just touching at point S.
Because touching intersection appears rarely and is difficult to deal with, the following considerations omit this case. In any case below all necessary differential conditions are presupposed. The determination of intersection points always lead to 1 or 2 non-linear equations which can be solved by a Newton iteration. A list of the appearing cases follows:
intersection of a parametric curve and an implicit curve
intersection of two implicit curves
• If both curves are explicitly given: , equalizing yields the equation
• If both curves are parametrically given:
Equalizing yields two equations for two variables:
• If one curve is parametrically and the other implicitly given:
This is beside the explicit case the simplest case. One has to insert the parametric representation of into the equation of curve and one gets the equation:
• If both curves are implicitly given:
Here, an intersection point is a solution of the system
Any Newton iteration needs convenient starting values, which can be derived by a visualization of both the curves. A parametrically or explicitly given curve can easily be visualized, because to any parameter t or x respectively it is easy to calculate the corresponding point. For implicitly given curves this task is not as easy. In this case one has to determine a curve point with help of starting values and an iteration. See .[2]
Examples:
1: and circle (s. picture).
The Newton iteration for function
has to be done. As startvalues one can choose −1 and 1.5.
The intersection points are: (−1.1073, −1.3578), (1.6011, 4.1046)
2:
(s. picture).
The Newton iteration
has to be performed, where is the solution of the linear system
at point . As starting values one can choose(−0.5, 1) and (1, −0.5).
The linear system can be solved by Cramer's rule.
The intersection points are (−0.3686, 0.9953) and (0.9953, −0.3686).
### Two polygons
intersection of two polygons: window test
If one wants to determine the intersection points of two polygons, one can check the intersection of any pair of line segments of the polygons (see above). For polygons with a lot of segments this method is rather time consuming. In praxis one accelerates the intersection algorithm by using window tests. In this case one divides the polygons into small sub-polygons and determines the smallest window (rectangle with sides parallel to the coordinate axes) for any sub-polygon. Before starting the time consuming determination of the intersection point of two line segments any pair of windows is tested for common points. See.[3]
## In space (three dimensions)
Further information: three-dimensional space
In 3-dimensional space there are intersection points (common points) between curves and surfaces. In the following sections we consider transversal intersection only.
### A line and a plane
Line–plane intersection
The intersection of a line and a plane in general position in three dimensions is a point.
Commonly a line in space is represented parametrically and a plane by an equation . Inserting the parameter representation into the equation yields the linear equation
for parameter of the intersection point .
If the linear equation has no solution, the line either lies on the plane or is parallel to it.
### Three planes
If a line is defined by two intersecting planes and should be intersected by a third plane , the common intersection point of the three planes has to be evaluated.
Three planes with linear independent normal vectors have the intersection point
For the proof one should establish using the rules of a scalar triple product. If the scalar triple product equals to 0, then planes either do not have the triple intersection or it is a line (or a plane, if all three planes are the same).
### A curve and a surface
intersection of curve with surface
Analogously to the plane case the following cases lead to non-linear systems, which can be solved using a 1- or 3-dimensional Newton iteration.[4]
• parametric curve and
parametric surface
• parametric curve and
implicit surface
Example:
parametric curve und
implicit surface (s. picture).
The intersection points are: (−0.8587, 0.7374, −0.6332), (0.8587, 0.7374, 0.6332).
A line–sphere intersection is a simple special case.
Like the case of a line and a plane, the intersection of a curve and a surface in general position consists of discrete points, but a curve may be partly or totally contained in a surface.
### Two surfaces
Main article: Intersection curve
Two transversally intersecting surfaces give an intersection curve. The most simple case the intersection line of two non-parallel planes. | 1,936 | 9,240 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2022-21 | latest | en | 0.890124 |
https://admin.clutchprep.com/chemistry/practice-problems/66128/a-1-00-l-flask-is-filled-with-1-00-g-of-argon-at-25-176-c-a-sample-of-ethane-vap | 1,601,149,012,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400244353.70/warc/CC-MAIN-20200926165308-20200926195308-00770.warc.gz | 229,668,757 | 27,870 | # Problem: A 1.00 L flask is filled with 1.00 g of argon at 25°C. A sample of ethane vapor is added to the same flask until the total pressure is 1.250atm .Part AWhat is the partial pressure of argon, PAr, in the flask? Express your answer to three significant figures and include the appropriate units.PAr = _________________PART BWhat is the partial pressure of ethane, Pethane, in the flask? Express your answer to three significant figures and include the appropriate UNITS.Pethane =_________________
🤓 Based on our data, we think this question is relevant for Professor Zimmer-De Iuliis and Zhang's class at TORONTO.
###### Problem Details
A 1.00 L flask is filled with 1.00 g of argon at 25°C. A sample of ethane vapor is added to the same flask until the total pressure is 1.250atm .
Part A
What is the partial pressure of argon, PAr, in the flask? Express your answer to three significant figures and include the appropriate units.
PAr = _________________
PART B
What is the partial pressure of ethane, Pethane, in the flask? Express your answer to three significant figures and include the appropriate UNITS.
Pethane =_________________
What scientific concept do you need to know in order to solve this problem?
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What professor is this problem relevant for?
Based on our data, we think this problem is relevant for Professor Zimmer-De Iuliis and Zhang's class at TORONTO. | 473 | 2,014 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.921246 |
https://ask.sagemath.org/question/57577/thue-mahler-equation/?sort=oldest | 1,632,614,160,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057787.63/warc/CC-MAIN-20210925232725-20210926022725-00238.warc.gz | 163,273,190 | 13,654 | # Thue-Mahler equation
Let $F \in \mathbb{Z}[X,Y]$ be an homogenous polynomial. An equation of the form $$F(X,Y) = d p_1^{a_1}\cdots p_n^{a_n},$$ with $p_1, \cdots, p_n$ are prime numbers and $d \in \mathbb{Z}$ is called a Thue-Mahler equation. Is there an explicit function in sage that can solve this kind of equations even if is a simpler case such as F(x,y) = p^a ?. I have been looking for this with no result.
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1
It seems not. There is Magma code in the master's thesis Implementation of a Thue-Mahler equation solver by Kyle Hambrook.
( 2021-06-15 21:29:41 +0200 )edit
Sort by ยป oldest newest most voted
One can rely on the Thue equation solver present in PARI. Here is a sample code for solving the equation $f(x,y)=n$:
def solvethue(f,n):
assert f.is_homogeneous()
t = gp.thueinit( f.subs({f.variables()[1]:1}) )
return gp.thue(t,n)
R.<x,y> = PolynomialRing(QQ)
print( solvethue(x^3+y^3,1) )
more
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https://byjus.com/question-answer/for-what-values-of-c-76c-5-leaves-remainder-2-c-2c-3c-7c-4/ | 1,720,912,607,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514517.94/warc/CC-MAIN-20240713212202-20240714002202-00055.warc.gz | 137,407,882 | 25,422 | 1
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Question
# For what values of C, 76C ÷ 5 leaves remainder 2?
A
C = 2
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B
C = 3
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C
C = 7
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D
C = 4
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Solution
## The correct options are A C = 2 C C = 7Divisibility of 5 states that a number is divisible by 5 if it has 0 or 5 in its ones place. 76C is divisible by 5 if C = 0 or 5 Since, it leaves a remainder 2, C = 0 + 2 or 5 + 2. ⇒ C = 2 or 7 ∴ For C = 2 and 7, 76C ÷ 5 will leave remainder 2.
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Join BYJU'S Learning Program | 317 | 908 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-30 | latest | en | 0.856102 |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-7-section-7-1-systems-of-linear-equations-in-two-variables-exercise-set-page-818/16 | 1,618,814,171,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038878326.67/warc/CC-MAIN-20210419045820-20210419075820-00210.warc.gz | 893,230,528 | 12,699 | ## Precalculus (6th Edition) Blitzer
$(-1,-2)$
Here, we have $2x=3y+4 \implies x=\dfrac{3y+4}{2}$ Now, $2(\dfrac{3y+4}{2})+6y=5y-4$ This gives: $4y=-8$ or, $y=-2$ Thus, $y=\dfrac{3(-2)+4}{2}=-1$ Solution is: $(x,y)=(-1,-2)$ | 118 | 224 | {"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} | 4.15625 | 4 | CC-MAIN-2021-17 | latest | en | 0.424956 |
https://www.assignmenthelper.my/questions/bafb4124-investment-analysia-assignment-uiu-malaysia/ | 1,717,016,434,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059408.76/warc/CC-MAIN-20240529200239-20240529230239-00899.warc.gz | 555,205,505 | 128,629 | # BAFB4124: Investment Analysis Assignment, UIU, Malaysia Small Investors Berhad, a large foreign mutual fund, is considering investing in stocks on Bursa Malaysia. One of the portfolio
University UNTAR International University (UIU) Subject BAFB4124: Investment Analysis
Question 1
Small Investors Berhad, a large foreign mutual fund, is considering investing in stocks on Bursa Malaysia. One of the portfolio managers, Mr. Ramy, has identified two stocks, Sore and Durby, which have great potential and are considered blue-chip stocks. As a new investor in the local market, Mr. Ramy approached you to advise him on which one of the two stocks should be selected. He provides you with the following information for analysis.
• Calculate the expected return and standard deviation for both stocks. Advise Mr. Ramy on the investment he should choose.
• Calculate the covariance between the two stocks.
• Calculate the expected return and standard deviation of a portfolio consisting of 60% Sore and 40% Durby.
Question 2
Amanda Tan cannot decide whether to invest in Stock Hewlett or Stock Packard or in a portfolio that is a combination of both stocks. She has approached OSK Securities and the firm has provided her with the following information:
Using these stocks, she has identified two investment portfolio alternatives :
a) Calculate the portfolio return and standard deviation for each alternative.
b) Based on the findings above, which of the two alternatives would she choose? Explain your answer.
Question 3
John currently has a portfolio of shares giving a return of 20% with a risk of 10%. He is considering a new investment which gives a return of 20% with a risk of 12%. The coefficient of correlation of the new investment with his existing portfolio is +0.2. The new investment will comprise 40% of his enlarged portfolio. Should he invest in the new investment?
Question 4
Redesign Corp is considering a new strategy that would increase its expected return from 12% to 13.9%, but would also increase its beta from 1.2 to 1.8. If the risk-free rate is 5% and the return on the market is expected to be 10%, should Redesign change its strategy?
Question 5
Qin Bhd has a systematic risk of 6%. The market is giving a return of 12% with a risk of 4%. The risk-free rate is 5%. What will be the required return from Qin Bhd?
Question 6
Tuck Bhd is giving a return of 20%. The stock exchange as a whole is giving a return of 25% with a risk of 8%, and the return on government securities is 8%. What is the β of Tuck Bhd, and what is the systematic risk of Tuck Bhd?
## Get Help By Expert
Seeking assistance with your BAFB4124 Investment Analysis assignment? Look no further than Assignment Helper MY, the leading Malaysian assignment helpers. Our team of experienced writers specializes in investment analysis and can provide you with top-quality assignment solutions. Whether it's analyzing financial statements, evaluating investment opportunities, or conducting risk assessments, our experts will deliver well-researched and comprehensive assignments tailored to your requirements. Besides, this our experts also provide Programming Assignment Help.
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Hire a Writer Now | 919 | 4,250 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-22 | latest | en | 0.951673 |
https://www.hackmath.net/en/example/1462 | 1,519,253,399,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813818.15/warc/CC-MAIN-20180221222354-20180222002354-00698.warc.gz | 890,446,410 | 6,681 | # Arithmetic average
The arithmetic mean of the five numbers is exactly 8. The sum of these four numbers is 30. What is the fifth number?
Result
x = 10
#### Solution:
(30+x)/5 = 8
x = 10
x = 10
Calculated by our simple equation calculator.
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
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#### To solve this example are needed these knowledge from mathematics:
Do you have a linear equation or system of equations and looking for its solution? Or do you have quadratic equation? Looking for a statistical calculator? Looking for help with calculating arithmetic mean?
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Kim needs 3/4 cup of flour to make 12 cookies. How much flour would she need to make 60 cookies? | 909 | 3,518 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2018-09 | longest | en | 0.959484 |
https://www.slideserve.com/avak/twins | 1,511,215,427,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806225.78/warc/CC-MAIN-20171120203833-20171120223833-00210.warc.gz | 873,453,087 | 16,084 | 1 / 51
# Twins - PowerPoint PPT Presentation
Twins. Peter and Paul are twin brothers. One of them (we don’t know which) always lies. The other one always tells the truth. I ask one of them: “Is Paul the one that lies?” “Yes,” he answers. Did I speak to Peter or Paul?. Twins. I spoke to Peter. Average Speed.
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Peter and Paul are twin brothers. One of them (we don’t know which) always lies. The other one always tells the truth. I ask one of them:
“Is Paul the one that lies?”
“Yes,” he answers.
Did I speak to Peter or Paul?
I spoke to Peter
On the way to his grandmother’s house, John travels at an average of 30 mph. On the way back, he travels at an average of 40 mph. What was his average speed for the entire trip?
Suppose the distance is 120 miles.
That means 4hrs there, 3hrs back.
That’s 240 miles / 7 hrs = 34.3 mph
Review of Last Couple of Weeks
Linear or Exponential?
Roach population grows 5% every 2 days.
Every year, 15 more students attend Matt Forte High School.
There has been a \$100 increase in wage every 6 months.
Review of Last Couple of Weeks
Linear or Exponential?
Review of Last Couple of Weeks
In how many years will a river expanding at 17.35% every year be three times its original size? Use Excel.
### Making and Interpreting Graphs
Week 3
• To better communicate information
• Newspapers
• Magazines
• Work meetings
• Pie chart
• XY Graph
• Bar Chart
• Pie Chart
• Limited applicability
• Can be used only when you have a quantitative variable associated with a list of categories where both the categories and the quantities each add up to a whole.
• The categories must also be disjoint (no overlap)
• Common error: to use a pie chart on a set of categories that do not make a whole and to use a pie chart when the categories overlap.
• Use percentage label when using a pie chart
• AKA line graph (not necessarily a line)
• used when you have "a lot" of data points
• when categories along the x-axis are numerical
• show trends in data clearly
• enable the viewer to make predictions about the results of data not yet recorded
• use whenever there is a quantitative variable associated with the a categorical variable
• For limited amount of data
• Succinct
• Easy to make comparisons within categories and across categories
• Disadvantage: sometimes presents far too much info.
• hard to make a single, clear point with them
• a. What is the purpose of making a graph from this data?
• b. What type of graph should you make? pie bar x-y scatter (line)
• c. Decide on a title and consider the the W's (who, what, where and when)
• d. Legend: yes no
• e. Descriptive x-axis label (if applicable)
• f. Descriptive y-axis label (if applicable)
• g. Scale (if applicable)
• h. Source
• People can use graphs to tell you skewed, misleading information.
• Go to the qrc homepage. Under excel files, find IL_Pop_By_Race.xls and open it.
• Add a column which contains the percentage of total population for each racial category.
• Make an effective graph.
• a. What is the purpose of making a graph from this data?
• b. What type of graph should you make? pie bar x-y scatter (line)
• c. Decide on a title and consider the the W's (who, what, where and when)
• d. Legend: yes no
• e. Descriptive x-axis label (if applicable)
• f. Descriptive y-axis label (if applicable)
• g. Scale (if applicable)
• h. Source
• Go to the qrc homepage. Under excel files, find DePaulMajors04.xls and open it.
• Make an effective graph.
• a. What is the purpose of making a graph from this data?
• b. What type of graph should you make? pie bar x-y scatter (line)
• c. Decide on a title and consider the the W's (who, what, where and when)
• d. Legend: yes no
• e. Descriptive x-axis label (if applicable)
• f. Descriptive y-axis label (if applicable)
• g. Scale (if applicable)
• h. Source
Change the scale of the y-axis.
• Go to the qrc homepage. Under excel files, under the folder Chicago, find ChicagoPopulation1830-2000.xls and open it.
• Make an effective graph.
• a. What is the purpose of making a graph from this data?
• b. What type of graph should you make? pie bar x-y scatter (line)
• c. Decide on a title and consider the the W's (who, what, where and when)
• d. Legend: yes no
• e. Descriptive x-axis label (if applicable)
• f. Descriptive y-axis label (if applicable)
• g. Scale (if applicable)
• h. Source
• What is wrong with this chart? | 1,355 | 5,375 | {"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} | 3.9375 | 4 | CC-MAIN-2017-47 | latest | en | 0.937014 |
https://blog.larrydavidson.com/2017/08/26/if-it-aint-broke-dont-fix-it-rethinking-quadrilaterals/ | 1,679,630,318,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945242.64/warc/CC-MAIN-20230324020038-20230324050038-00673.warc.gz | 160,971,769 | 25,403 | # “If it ain’t broke, don’t fix it?” (Rethinking quadrilaterals)
Why on earth would we spend two whole days rewriting our Honors Geometry quadrilaterals unit? Our textbook, after all, contains a perfectly serviceable sequence of four lessons on this topic:
These lessons are adequate. In the words of the standard phrase, they “ain’t broke.” So why fix them?
The answer is that “adequate” isn’t good enough. The lessons in the textbook have a very low cognitive demand. They just don’t require honors-level thinking. And where are the Big Ideas?
Actually, there are a couple of Big Ideas buried in these four lessons:
• The concept of a hierarchy is certainly present in the first of the four lessons. However, it’s merely taught to the students didactically, with no exploration required — it’s just a list of definitions, like “A rectangle is a parallelogram in which at least one angle is a right angle.”
• The related distinction bettwen necessary and sufficient is implied in the second and third lessons, though it’s never made explicit.
So that’s a start. But for the most part the chapter promotes memorization rather than thinking. It emphasizes tedious lists like this one, excerpted from the middle of a list of quadrilateral properties:
There’s no sense of being a mathematician here! How does an honors math student react to such a list of 28 items (in the full list)? Well, she can simply memorize the list, or she can give up and say it’s an unreasonable task, or she can learn a few of the more important ones and ignore the rest. The textbook claims that “with some effort you will soon learn them all.”
Bleh. We can do better.
In our workshop we replaced these four textbook-based lessons with five investigations (packaged together into a sequence of 18 problems, most of which contain multiple sub-problems):
1. This investigation is an exploration of definitions of various quadrilaterals. Definitions can partition a larger set (integers are positive, negative, or zero); they can be hierarchical (equilateral triangles are isosceles); they can be subjects of significant dispute (the definition of trapezoid, for instance). Also, definitions have consequences. They can be equivalent (specifying the same set of objects) or not; they can make it easier or harder to prove something.
2. The second investigation introduces the use of coordinates in this context and distinguishes necessary from sufficient conditions. In all cases it requires reasoning and defense of one’s reasoning, sometimes to the point of being a proof.
3. The third investigation pursues the necessary/sufficient distinction by presenting a list of 12 proposed properties without any didactic teaching: each one might be necessary for a particular type of quadrilateral, or it might be sufficient, or both necessary and sufficient, or neither. Exploring these possibilities starts to lead students to the idea of moving up or down the hierarchy in the next investigation.
4. And here in the fourth investigation we assemble a bunch of observations in order to construct that hierarchy ourselves. We also pursue coordinate reasoning some more.
5. Finally, the fifth investigation explores the creation of one quadrilateral from another. For instance, if you connect the midpoints of all four sides of a rectangle, what do you get? Can you prove it? What if you bisect the angles and connect the intersection points? What if you construct equilateral triangles from the sides (pointing outward) and then connect the outermost vertices of these triangles?
This newly rewritten unit still needs some tweaking, but our hope is that we have a much richer sequence that truly challenges honors students to do their best.
Categories: Math, Teaching & Learning, Weston | 773 | 3,767 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2023-14 | latest | en | 0.919517 |
https://en.wanweibaike.com/wiki-Orders_of_magnitude_(entropy) | 1,624,187,149,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487660269.75/warc/CC-MAIN-20210620084505-20210620114505-00106.warc.gz | 228,233,922 | 8,915 | # Orders of magnitude (entropy)
The following list shows different orders of magnitude of entropy.
Factor (J K−1) Value Item
10−24 9.5699×10−24 J K−1 Entropy equivalent of one bit of information, equal to k times ln(2)[1]
10−23 1.381×10−23 Planck entropy
1 5.74 J K−1 Standard entropy of 1 mole of graphite[2]
1033 ≈ 1035 J K−1 Entropy of the Sun (given as ≈ 1042 erg K−1 in Bekenstein (1973))[3]
1054 1.5 × 1054 J K−1 Entropy of a black hole of one solar mass (given as ≈ 1060 erg K−1 in Bekenstein (1973))[3]
1081 4.3 × 1081 J K−1 One estimate of the theoretical maximum entropy of the universe[4][5]
## References
1. ^ Jean-Bernard Brissaud (14 February 2005). "The Meaning of Entropy" (PDF). Entropy, 2005, 7[1], 68–96. Retrieved 2010-04-21. page 72 (page 5 of pdf)
2. ^ Chung Chieh. "Entropy: A Study Guide". Retrieved 2010-04-21.
3. ^ a b Jacob D. Bekenstein (1973). "Black Holes and Entropy" (PDF). Physical Review D. 7 (8): 2333–2346. Bibcode:1973PhRvD...7.2333B. doi:10.1103/PhysRevD.7.2333. Archived from the original (PDF) on 2010-05-23.
4. ^ Chas A. Egan; Charles H. Lineweaver (25 January 2010). "A Larger Estimate of the Entropy of the Universe". The Astrophysical Journal. 710 (2): 1825–1834. arXiv:0909.3983. Bibcode:2010ApJ...710.1825E. doi:10.1088/0004-637X/710/2/1825. 3.1 x 10^104k
5. ^ Calculated: 3.1e104 * k = 3.1e104 * 1.381e-23 J/K = 4.3e81 J/K | 541 | 1,373 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-25 | latest | en | 0.473907 |
https://www.farsleyspringbank.co.uk/summer-1-week-1-23421/ | 1,723,370,331,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640983659.65/warc/CC-MAIN-20240811075334-20240811105334-00503.warc.gz | 607,993,033 | 49,724 | # Summer 1 Week 1 23.4.21
Hello! Welcome back after the Easter break. We hope you all had a lovely time. And here we are in the Summer term already! Where has the time gone? We have been really lucky this week with some beautiful weather so let’s hope that continues.
This week the children have been working hard in their phonics groups. Thank you for your continued support in your child’s phonics journey, listening to them read and helping them to practise sounds and red words. For three mornings a week, we have a picture of the day which the children write about. We have been so impressed with the children’s independent writing and how confident they are becoming in using the sounds they have learnt to write new words and sentences.
In maths, we have started our topic ‘To 20 and beyond’. This week we have been learning how to build numbers beyond 10. The children have explored number patterns from 11-20, they have been matching numerals and pictorials and they have used a range of resources including numicon and ten frames to make up numbers. The children have identified that the 1 in a teen number represents 10. They have played a range of games including matching pairs and bingo to help consolidate this learning. At home, you can support this learning by practising counting to 20, recognising numbers on walks or at home and counting objects such as sweets or pebbles.
In understanding the world, we have had lots of fun learning about plants. We have learnt what a plant is and have learnt the names of the parts of a plant. We have identified plants in the school garden and the children took photos using ipads of what they found. We have been learning about what plants need to grow and we set up an experiment which we will be observing over the next few weeks. The children have been busy in the garden, preparing the soil and planting seeds. We hope to have lots of yummy produce!
Next week, we will continue with our maths topic ‘To 20 and beyond’, building numbers, recognising numerals and pictorials and exploring patterns in number.
In understanding the world, we will continue to learn about plants, identifying leaves and different plants and learning about the life cycle of a plant.
Our five favourite books this half-term are Lost and Found, Where the Wild things Are, On Sudden Hill, Owl Babies and The Runaway Pea. If you have any of these at home, you may want to share these with your child. These are books which we read and re-read throughout the half-term. This enables the children to get to know the text really well which encourages them to join in with the story language and read these books independently in the book corner.
You may like to enjoy the new Cbeebies series together which links perfectly to our plants topic: ‘Maddie, the plants and you’. Here is a link: https://www.bbc.co.uk/iplayer/episode/m000trpr/maddie-the-plants-and-you-series-1-1-the-flowers
Top | 633 | 2,931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.973792 |
http://wintechnology.net/pdf/computational-methods-in-ordinary-differential-equations-introductory | 1,611,320,983,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703529331.99/warc/CC-MAIN-20210122113332-20210122143332-00609.warc.gz | 106,692,428 | 8,695 | # Computational Methods in Ordinary Differential Equations by J. D. Lambert
By J. D. Lambert
Read Online or Download Computational Methods in Ordinary Differential Equations (Introductory mathematics for scientists & engineers) PDF
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The Asymptotic Solution of Linear Differential Systems: Applications of the Levinson Theorem
The trendy thought of linear differential platforms dates from the Levinson Theorem of 1948. it's only in additional contemporary years, despite the fact that, following the paintings of Harris and Lutz in 1974-7, that the importance and diversity of purposes of the theory became preferred. This ebook provides the 1st coherent account of the vast advancements of the final 15 years.
Extra info for Computational Methods in Ordinary Differential Equations (Introductory mathematics for scientists & engineers)
Sample text
Oo (~/n = 0 if and only if 1',1"; 1, ' it is clear that the method will not be convergent if any of the roots (, has modulus greater than one. Also, consider the case when e, is a reaCroot of pm with mUltiplicity m > 1. 3n(n - 1) + ... mn(n - 1)". (n - m + 2)](~. Since for q :;;" 1, lim hnq(: = x lim nq-Ie~ = 0 if and only if h ..... O nh=x n-oo 1',1 < 1, it is clear that the method will not be convergent if p(() has a multiple root of modulus greater than or equal to one. The argument extends to the case when the roots of p(() are complex, and motivates our next definition.
J h2A2 = [exp (nhA)] { exp (hA) - I - hA - ""2 - thA[exp(hA) - I - hAJ} = [exp(nhA)][(1 - 1hA) exp (hA) - (I Also Y•• I - Y. = 1hA(y•• I + yJ + 1hA)]. , / 30 Computational methods in ordinary differential equations + thA) exp (nhA), or (1 - 4hA)y = (1. • "+ 1 Y. + 1] = by the localizing assumption (l - thA) exp [In = [exp(nhA)][(! '. = + I)hA]- that + thA) exp (nhA) thA)exp(hA) - (1 + 1M)] (1 -2'[y{x,); h], verifying (24). 8 Consistency and zero-stability Definition The linear multistep method (2) is said to be consistent order p ;;.
1). l is not critical. l -:- 0 since this causes two coefficients, namely ct: 1 and 11:3, to vaDlsh. Another sunphfylOg choice is Jl = 1\. which causes P2 to-vanish; the resulting method turns out to be Quade's method, defined in exercise 5 (page 27). Exercise 11. Find the range of a. (Y,+2 - y,+,) - Y. +,) is zero-stable. Show that there exists a value of a. t if the method is to be zero~stable, its order cannot exceed 2. 10 Specification oflinear multistep methods In the days of desk computation, it was common practice to write the right-hand side of a linear multistep method in terms of a power series in a difference operator. | 770 | 2,651 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.809413 |
https://www.thejuliagroup.com/blog/livebinders-20-day-blogging-challenge-day-two/ | 1,723,254,024,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640782288.54/warc/CC-MAIN-20240809235615-20240810025615-00439.warc.gz | 793,486,412 | 16,975 | # Livebinders: 20-day blogging challenge, day two
Today I’m on day two of the 20-day blogging challenge, the brain child of Kelly Hines and a great way to find new, interesting bloggers. The second day prompt was to share an organizational tip from your classroom, one thing that works for you.
The latest tool I’ve been using is livebinders . Remember when you were in college having a binder full of notes, handouts from the professor, maybe even copies of tests to study for the final? Well, livebinders appears to be designed more for clipping websites and including media from the web but personally I am using it to create binders for teaching statistics. I’ve just started with one but I’m sure this will eventually split off into several binders.
I’m always writing notes to myself but I have them everywhere – I used Google notebook until they got rid of that, evernote, I’ve got notepads on my laptop, desktop, iPad, phone and even paper notebooks around the place. I even have a PadsX program The Invisible Developer wrote years ago just for me (yes, he loves me).
Still, I’m thinking livebinders is going to be really useful for me to organize all of these notes into one spot.
Why do I want to do that, you might ask?
Well, statistics is a big field, and I have taught a lot of it, from advanced multivariate statistics to psychometrics to biostatistics and a lot of special topics courses. It seems to me that we often assume students have a solid grasp of certain concepts, such as variance or standardization, when I’m sure many of them do not. As I read books and articles, I’m trying to note what these assumptions are. My next step is to have pages in the binders where students can get greater explanation of, say, what does a confidence interval really mean. Right now, I feel that universities are trying to cut costs by combining information into fewer and fewer courses. We say that students learned Analysis of Variance in a course, but did they really? The basic statistics I took in graduate school consisted of a descriptive statistics class (I tested out of that). It ended with a brief introduction to hypothesis testing and a discussion of t-tests, z-scores, t-tests and correlation. The inferential statistics course reviewed hypothesis testing, t-tests and correlation, then focused on regression and ANOVA. The multivariate statistics course covered techniques like cluster analysis, canonical correlation and discriminant function analysis. Psychometric statistics covered factor analysis and various types of reliability and validity. These four courses were the BASICS, what everyone in graduate school took. (People like me who specialized in applied statistics took a bunch more classes on top of that.) Oh, yes, and each class came with a three-hour computer lab AFTER the three-hour lecture, to teach you enough programming so you could do the analyses yourself. Now, many textbooks try to include all of this in one course, which is just a joke, and ends up with students concluding that they “are just not very good at math”.
I can’t change the curriculum, but what I at least can do is provide some type of resource where every time a student feels he or she needs to back up and understand some concept, there is an explanation of that something.
I plan to have this done by the time I teach Data Mining in August.
Suggestions for what to include are welcome.
## One Comment
1. I love, love, love Livebinders! I create them for standards, for lesson plans, for student resources, and more. They feed my need for visual order, and they are so easy to re-arrange, like I would do with a physical object. | 769 | 3,658 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-33 | latest | en | 0.968626 |
http://www.algebra.com/algebra/homework/word/travel/Travel_Word_Problems.faq.question.316048.html | 1,369,134,599,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699899882/warc/CC-MAIN-20130516102459-00011-ip-10-60-113-184.ec2.internal.warc.gz | 318,000,551 | 4,859 | # SOLUTION: A ball is thrown upward with an initial velocity of 32 ft/sec from a height of six feet. The function s(t)=-16t^2+32t+6 gives the height of the ball t seconds after release. Determ
Algebra -> Algebra -> Customizable Word Problem Solvers -> Travel -> SOLUTION: A ball is thrown upward with an initial velocity of 32 ft/sec from a height of six feet. The function s(t)=-16t^2+32t+6 gives the height of the ball t seconds after release. Determ Log On
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Word Problems: Travel and Distance Solvers Lessons Answers archive Quiz In Depth
Question 316048: A ball is thrown upward with an initial velocity of 32 ft/sec from a height of six feet. The function s(t)=-16t^2+32t+6 gives the height of the ball t seconds after release. Determine the time when the ball is at maximun height, and find that height.Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!s(t) is a downward opening parabola ("a" is negative) ; so the maximum is at the vertex , which is on the axis of symmetry t = -b / 2a = -32 / 2(-16) = 1 s(1) = -16(1^2) + 32(1) + 6 = 22 | 367 | 1,326 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2013-20 | latest | en | 0.837821 |
https://www.measuringknowhow.com/convert-100-fahrenheit-to-celsius-quick-guide/ | 1,713,812,510,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818337.62/warc/CC-MAIN-20240422175900-20240422205900-00074.warc.gz | 786,064,084 | 20,259 | # Convert 100 Fahrenheit to Celsius – Quick Guide
Did you know that 100 degrees Fahrenheit is equivalent to 37.8 degrees Celsius?
To convert 100 degrees Fahrenheit to Celsius, you can use the Celsius to Fahrenheit formula: (°F – 32) / 1.8 = °C. Plugging in 100 for °F gives us (100 – 32) / 1.8 = 37.8 °C. This means that 100 degrees Fahrenheit is equivalent to 37.8 degrees Celsius.
### Key Takeaways:
• To convert 100 degrees Fahrenheit to Celsius, use the formula (°F – 32) / 1.8 = °C.
• 100 degrees Fahrenheit is equivalent to 37.8 degrees Celsius.
• Understanding how to convert between Fahrenheit and Celsius is useful in various contexts.
## Celsius vs Fahrenheit: Key Differences
When it comes to temperature measurements, the Celsius and Fahrenheit scales are commonly used around the world. However, there are some significant differences between these two temperature scales.
The Celsius temperature scale (°C) is the most widely used scale and is adopted by all countries except for five. It is part of the International System of Units (SI) and is based on the freezing and boiling points of water. The freezing point of water on the Celsius scale is 0°C, while the boiling point is 100°C.
On the other hand, the Fahrenheit temperature scale (°F) is primarily used in a few countries, including the United States. It is part of the Imperial system and was developed by Daniel Gabriel Fahrenheit. Unlike the Celsius scale, the Fahrenheit scale has a wider range of temperatures and higher values for freezing and boiling points. The freezing point of water on the Fahrenheit scale is 32°F, while the boiling point is 212°F.
To illustrate the differences between the Celsius and Fahrenheit scales, here’s a table summarizing their freezing and boiling points:
Temperature ScaleFreezing PointBoiling Point
Celsius (°C)0°C100°C
Fahrenheit (°F)32°F212°F
“The Celsius temperature scale is widely used internationally, while the Fahrenheit scale is mostly used in the United States. Understanding the freezing and boiling points on each scale is important for accurate temperature measurement.”
In conclusion, the Celsius and Fahrenheit temperature scales differ in their usage, range of temperatures, and freezing and boiling points. Familiarizing yourself with these differences can help ensure accurate temperature conversions and enhance your understanding of temperature measurements.
## Celsius to Fahrenheit Formula
Converting temperatures from Celsius to Fahrenheit is a simple process that can be done using the Celsius to Fahrenheit formula. This formula provides a precise calculation to convert any temperature in Celsius to its Fahrenheit equivalent.
The Celsius to Fahrenheit formula is:
(°C * 1.8) + 32 = °F
To apply this formula, we multiply the temperature in Celsius by 1.8, then add 32 to the result. This will give us the temperature in Fahrenheit.
For example, let’s convert 18 °C to Fahrenheit using the formula:
(18 * 1.8) + 32 = 64.4 °F
Therefore, 18 °C is equivalent to 64.4 °F.
### Quick Tip: Celsius to Fahrenheit Conversion
If you’re looking for a quick way to estimate the Fahrenheit temperature when given a Celsius value, you can use the following trick:
1. Multiply the Celsius temperature by 2.
2. Add 30 to the result.
While this method may not provide an exact conversion, it can give you a close approximation for everyday use.
### Example:
Let’s use the quick trick to estimate the Fahrenheit temperature for 15 °C:
1. 15 * 2 = 30
2. 30 + 30 = 60 °F
So, 15 °C is roughly equivalent to 60 °F using the quick trick.
Now, let’s move on to the next section to learn about the formula and methods for converting temperatures from Fahrenheit to Celsius.
## Celsius to Fahrenheit Conversion Chart
Need to convert temperatures between Celsius and Fahrenheit? Look no further! Check out our handy conversion chart below for quick and easy temperature conversions:
Celsius (°C)Fahrenheit (°F)
032
1050
2068
3086
40104
50122
As you can see from the chart, if the temperature is 10 degrees Celsius, it will be 50 degrees Fahrenheit. On the other hand, if it’s 30 degrees Celsius, it will be 86 degrees Fahrenheit. This conversion chart provides a simple reference for converting temperatures between the Celsius and Fahrenheit scales.
## How to Convert Celsius to Fahrenheit: Quick Trick
When you need to convert Celsius to Fahrenheit quickly and don’t have a calculator on hand, there’s a simple trick that can help you get an estimated temperature in Fahrenheit. By following these steps, you can convert Celsius to Fahrenheit in your head:
1. Multiply the temperature in degrees Celsius by 2.
2. Add 30 to the result.
Let’s take an example of converting 15 °C to Fahrenheit using this quick trick:
Celsius (°C)Fahrenheit (°F) (Estimated)
15(15 * 2) + 30 = 60 °F
As seen in the table above, multiplying 15 by 2 gives us 30. Adding 30 to 30 results in an estimated temperature of 60 degrees Fahrenheit.
Keep in mind that this method provides a close estimation rather than an exact conversion. It can be a handy trick when you need a quick temperature estimate and don’t have access to a calculator. For more precise conversions, it’s recommended to use the Celsius to Fahrenheit formula discussed in earlier sections.
Inserting the image above for reference, we can visually demonstrate the conversion process:
### Convert Celsius to Fahrenheit Example:
We want to convert 15 °C to Fahrenheit using the quick trick:
1. Multiply 15 by 2: 15 * 2 = 30
2. Add 30 to the result: 30 + 30 = 60
Therefore, 15 °C is approximately equal to 60 °F using the quick trick.
## How to Convert Fahrenheit to Celsius: Quick Trick
If you need to convert Fahrenheit to Celsius quickly and mentally, there’s a simple trick you can use. Just follow these steps:
1. Start by subtracting 30 from the temperature in degrees Fahrenheit.
2. Next, divide the result by 2.
3. The final value is the estimated temperature in Celsius.
Let’s take an example. If the temperature is 84 °F:
Step 1: 84 – 30 = 54
Step 2: 54 ÷ 2 = 27
So, the estimated temperature in Celsius would be 27 °C. It’s important to note that this quick trick provides an estimation and may not yield an exact conversion. However, it can be handy when you need to convert temperatures on the go without relying on a calculator.
Converting temperatures mentally can save time and be a useful skill in various situations. By practicing and using this quick trick, you can easily convert Fahrenheit to Celsius without the need for external tools. Whether you’re traveling, cooking, or simply curious about temperature conversions, this technique will come in handy.
## Converting Fahrenheit to Celsius: Formula and Chart
When you need to convert temperatures from Fahrenheit to Celsius, you can use a simple formula that provides accurate results. The formula to convert Fahrenheit to Celsius is:
(°F – 32) / 1.8 = °C
This formula allows you to convert any temperature in Fahrenheit to Celsius. To convert a specific temperature, subtract 32 from the Fahrenheit temperature, then divide the result by 1.8. The resulting value is the temperature in Celsius.
For example, let’s convert 90 degrees Fahrenheit to Celsius:
(90°F – 32) / 1.8 = 32.2°C
So, 90 degrees Fahrenheit is equivalent to 32.2 degrees Celsius.
In addition to the conversion formula, you can also refer to a Fahrenheit to Celsius conversion chart for quick reference. This chart provides a list of common Fahrenheit temperatures and their corresponding values in Celsius, making it easier to convert temperatures without performing calculations.
Here is an example of a Fahrenheit to Celsius conversion chart:
FahrenheitCelsius
32°F0°C
50°F10°C
68°F20°C
86°F30°C
104°F40°C
Using the conversion chart, you can find the Celsius equivalent of any Fahrenheit temperature by locating the Fahrenheit value and reading the corresponding Celsius value.
Converting temperatures from Fahrenheit to Celsius becomes easier with the formula and the conversion chart. Whether you prefer performing calculations or referring to a chart, you can confidently convert temperatures between the Fahrenheit and Celsius scales.
## Understanding Fahrenheit and Celsius
The Fahrenheit and Celsius scales are two commonly used temperature scales in the world. They have their own distinct characteristics and are used in different parts of the world. Let’s take a closer look at these two scales and explore their differences.
### Fahrenheit Temperature Scale
The Fahrenheit scale, named after the German physicist Daniel Gabriel Fahrenheit, was developed in the early 18th century. It is primarily used in the United States, as well as a few other countries such as the Bahamas, Belize, and the Cayman Islands. The Fahrenheit scale divides the range between the freezing and boiling points of water into 180 equal divisions.
### Celsius Temperature Scale
The Celsius scale, originally called centigrade, is used worldwide and is the most commonly used temperature scale. It was developed by the Swedish astronomer Anders Celsius in the mid-18th century. The Celsius scale is based on the freezing and boiling points of water, dividing the range between these two points into 100 equal divisions.
### Differences Between Fahrenheit and Celsius
The Fahrenheit and Celsius scales differ in several ways:
1. Starting Point: The Fahrenheit scale has a starting point of 32 degrees, which is the freezing point of water. In contrast, the Celsius scale has a starting point of 0 degrees, which is also the freezing point of water.
2. Degree Size: The Fahrenheit scale has a smaller degree size compared to the Celsius scale. Each degree on the Fahrenheit scale is equivalent to 0.556 degrees on the Celsius scale.
3. Range of Temperatures: The Fahrenheit scale has a wider range of temperatures compared to Celsius. It spans from -459.67 degrees Fahrenheit (absolute zero) to 1,000 degrees Fahrenheit, while the Celsius scale spans from -273.15 degrees Celsius (absolute zero) to 100 degrees Celsius.
Here’s a visual representation of the differences between the Fahrenheit and Celsius scales:
FahrenheitCelsius
-459.67°F-273.15°C
32°F0°C
212°F100°C
1,000°F537.78°C
As you can see, the Fahrenheit and Celsius scales have different starting points, degrees, and temperature ranges. It’s important to keep these differences in mind when converting temperatures between the two scales or when comparing temperature readings.
## Historical Background of Fahrenheit and Celsius
The Fahrenheit temperature scale was invented by Daniel Gabriel Fahrenheit in 1714. He based the scale on the freezing and boiling points of water, as well as the average human body temperature. Fahrenheit, a German physicist, developed this scale to provide a more accurate and precise measurement of temperature. His goal was to create a scale that was more easily reproducible and less subjective than previous temperature scales.
Anders Celsius, a Swedish astronomer, introduced the Celsius temperature scale in 1742. Originally known as the centigrade scale, Celsius based his scale on the freezing and boiling points of water. The freezing point of water was set at 0 degrees Celsius, while the boiling point of water was set at 100 degrees Celsius.
Over time, the Celsius scale became more widely adopted, particularly with the rise of the metric system. The metric system, which uses Celsius as its primary temperature scale, gained global recognition and became the international standard for scientific and everyday temperature measurements. Today, Celsius is used by the vast majority of countries around the world, while Fahrenheit remains the official temperature scale in a few countries, including the United States.
### Comparison of Fahrenheit and Celsius Scales
FahrenheitCelsius
Based on the freezing and boiling points of water and the average human body temperature.Based on the freezing and boiling points of water.
32 degrees is the freezing point of water.0 degrees is the freezing point of water.
212 degrees is the boiling point of water.100 degrees is the boiling point of water.
Has a wider range of temperatures compared to Celsius.Has a more limited range of temperatures compared to Fahrenheit.
Used officially in a few countries, including the United States.Used officially and internationally by the majority of countries.
## Conclusion
Converting temperatures between Fahrenheit and Celsius is a common task that can be accomplished using formulas or a quick conversion trick. Understanding the differences between the two temperature scales and knowing how to convert between them can be helpful in various contexts, such as scientific research, cooking, and weather forecasting.
By following the provided formulas, you can easily convert temperatures from Fahrenheit to Celsius and vice versa. The Celsius to Fahrenheit formula is (°C * 1.8) + 32 = °F, while the Fahrenheit to Celsius formula is (°F – 32) / 1.8 = °C. Utilizing these formulas, you can accurately convert any temperature.
Additionally, if you prefer a more convenient approach, there are online Fahrenheit Celsius calculators available that can quickly convert temperatures for you. These handy tools eliminate the need for manual calculations, allowing for efficient and accurate temperature conversions.
Whether you’re a scientist analyzing data, a chef following a recipe, or simply checking the weather forecast, understanding how to convert temperatures between Fahrenheit and Celsius is essential. With the use of the conversion formulas and tools at your disposal, you can confidently navigate between the two temperature scales and make accurate temperature conversions.
## FAQ
### How do I convert 100 Fahrenheit into Celsius?
To convert 100 degrees Fahrenheit to Celsius, you can use the Celsius to Fahrenheit formula: (°F – 32) / 1.8 = °C. Plugging in 100 for °F gives us (100 – 32) / 1.8 = 37.8 °C. This means that 100 degrees Fahrenheit is equivalent to 37.8 degrees Celsius.
### What is the Celsius to Fahrenheit conversion formula?
The Celsius to Fahrenheit formula is (°C * 1.8) + 32 = °F. This formula allows you to convert any temperature in Celsius to Fahrenheit. For example, if we want to convert 18 °C to Fahrenheit: (18 * 1.8) + 32 = 64.4 °F.
### What is the Fahrenheit to Celsius conversion formula?
The Fahrenheit to Celsius formula is (°F – 32) / 1.8 = °C. This formula allows you to convert any temperature in Fahrenheit to Celsius.
### Is there a conversion chart for Celsius to Fahrenheit?
Yes, there is a Celsius to Fahrenheit conversion chart available that lists common Celsius temperatures and their equivalent values in Fahrenheit. This chart can be a useful reference for converting temperatures between the two scales.
### Can I convert Celsius to Fahrenheit without a calculator?
Yes, if you need to quickly convert Celsius to Fahrenheit without a calculator, you can use a simple trick. Multiply the temperature in degrees Celsius by 2, then add 30 to get an estimated temperature in degrees Fahrenheit.
### How can I convert Fahrenheit to Celsius in my head?
To convert Fahrenheit to Celsius in your head, subtract 30 from the temperature in degrees Fahrenheit, then divide by 2. This quick trick can help you estimate temperatures without the need for a calculator.
### How can I convert Fahrenheit to Celsius using a formula?
To convert Fahrenheit to Celsius, you can use the formula (°F – 32) / 1.8 = °C. This formula allows you to convert any temperature in Fahrenheit to Celsius. Additionally, there is a Fahrenheit to Celsius conversion chart available that lists common Fahrenheit temperatures and their corresponding values in Celsius.
### What are the differences between Fahrenheit and Celsius?
The Fahrenheit temperature scale is primarily used in the United States and a few other countries, while the Celsius scale is used worldwide. The two scales have different starting points and degrees, with Fahrenheit having a wider range of temperatures compared to Celsius.
### Who invented the Fahrenheit and Celsius temperature scales?
The Fahrenheit scale was invented by Daniel Gabriel Fahrenheit in 1714, while the Celsius scale was introduced by Anders Celsius in 1742. Fahrenheit based his scale on the freezing and boiling points of water and the average human body temperature, while Celsius based his scale solely on the freezing and boiling points of water.
### Why is it important to understand Fahrenheit and Celsius conversions?
Converting temperatures between Fahrenheit and Celsius is essential in various contexts, such as scientific research, cooking, and weather forecasting. Understanding how to convert between the two scales allows for better communication and ensures accurate temperature readings. | 3,492 | 16,852 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-18 | latest | en | 0.800045 |
https://www.kilomegabyte.com/19-gb-to-kibit | 1,726,061,190,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651387.63/warc/CC-MAIN-20240911120037-20240911150037-00700.warc.gz | 786,695,496 | 5,705 | #### Data Units Calculator
###### Gigabyte to Kibibit
Online data storage unit conversion calculator:
From:
To:
The smallest unit of measurement used for measuring data is a bit. A single bit can have a value of either zero(0) or one(1). It may contain a binary value (such as True/False or On/Off or 1/0) and nothing more. Therefore, a byte, or eight bits, is used as the fundamental unit of measurement for data storage. A byte can store 256 different values, which is sufficient to represent standard ASCII table, such as all numbers, letters and control symbols.
Since most files contain thousands of bytes, file sizes are often measured in kilobytes. Larger files, such as images, videos, and audio files, contain millions of bytes and therefore are measured in megabytes. Modern storage devices can store thousands of these files, which is why storage capacity is typically measured in gigabytes or even terabytes.
# 19 gb to kibit result:
19 (nineteen) gigabyte(s) is equal 148437500 (one hundred and forty-eight million four hundred and thirty-seven thousand five hundred) kibibit(s)
#### What is gigabyte?
The gigabyte is a multiple of the unit byte for digital information. The prefix giga means 10^9 in the International System of Units (SI). Therefore, one gigabyte is 1000000000 bytes. The unit symbol for the gigabyte is GB.
#### What is kibibit?
The kibibit is a multiple of the bit, a unit of digital information storage, using the standard binary prefix kibi, which has the symbol Ki, meaning 2^10. The unit symbol of the kibibit is Kibit. 1 kibibit = 2^10 bits = 1024 bits
#### How calculate gb. to kibit.?
1 Gigabyte is equal to 7812500 Kibibit (seven million eight hundred and twelve thousand five hundred kibit)
1 Kibibit is equal to 0.000000128 Gigabyte (zero point zero × 6 one hundred and twenty-eight gb)
1 Gigabyte is equal to 8000000000 bits (eight billion bits)
1 Kibibit is equal to 1024 bits (one thousand and twenty-four bits)
19 Gigabyte is equal to 152000000000 Bit (one hundred and fifty-two billion bit)
Gigabyte is greater than Kibibit
Multiplication factor is 0.000000128.
1 / 0.000000128 = 7812500.
19 / 0.000000128 = 148437500.
Maybe you mean Gibibyte?
19 Gigabyte is equal to 17.695128917694091796875 Gibibyte ( gib) convert to gib
### Powers of 2
gb kibit (Kibibit) Description
1 gb 7812500 kibit 1 gigabyte (one) is equal to 7812500 kibibit (seven million eight hundred and twelve thousand five hundred)
2 gb 15625000 kibit 2 gigabyte (two) is equal to 15625000 kibibit (fifteen million six hundred and twenty-five thousand)
4 gb 31250000 kibit 4 gigabyte (four) is equal to 31250000 kibibit (thirty-one million two hundred and fifty thousand)
8 gb 62500000 kibit 8 gigabyte (eight) is equal to 62500000 kibibit (sixty-two million five hundred thousand)
16 gb 125000000 kibit 16 gigabyte (sixteen) is equal to 125000000 kibibit (one hundred and twenty-five million)
32 gb 250000000 kibit 32 gigabyte (thirty-two) is equal to 250000000 kibibit (two hundred and fifty million)
64 gb 500000000 kibit 64 gigabyte (sixty-four) is equal to 500000000 kibibit (five hundred million)
128 gb 1000000000 kibit 128 gigabyte (one hundred and twenty-eight) is equal to 1000000000 kibibit (one billion)
256 gb 2000000000 kibit 256 gigabyte (two hundred and fifty-six) is equal to 2000000000 kibibit (two billion)
512 gb 4000000000 kibit 512 gigabyte (five hundred and twelve) is equal to 4000000000 kibibit (four billion)
1024 gb 8000000000 kibit 1024 gigabyte (one thousand and twenty-four) is equal to 8000000000 kibibit (eight billion)
2048 gb 16000000000 kibit 2048 gigabyte (two thousand and forty-eight) is equal to 16000000000 kibibit (sixteen billion)
4096 gb 32000000000 kibit 4096 gigabyte (four thousand and ninety-six) is equal to 32000000000 kibibit (thirty-two billion)
8192 gb 64000000000 kibit 8192 gigabyte (eight thousand one hundred and ninety-two) is equal to 64000000000 kibibit (sixty-four billion)
### Convert Gigabyte to other units
gb System Description
19 gb 152000000000 bit 19 gigabyte (nineteen) is equal to 152000000000 bit (one hundred and fifty-two billion)
19 gb 19000000000 b 19 gigabyte (nineteen) is equal to 19000000000 byte (nineteen billion)
19 gb 19000000 kb 19 gigabyte (nineteen) is equal to 19000000 kilobyte (nineteen million)
19 gb 19000 mb 19 gigabyte (nineteen) is equal to 19000 megabyte (nineteen thousand)
19 gb 0.019 tb 19 gigabyte (nineteen) is equal to 0.019 terabyte (zero point zero × 1 nineteen)
19 gb 18554687.5 kib 19 gigabyte (nineteen) is equal to 18554687.5 kibibyte (eighteen million five hundred and fifty-four thousand six hundred and eighty-seven point five)
19 gb 18119.81201171875 mib 19 gigabyte (nineteen) is equal to 18119.81201171875 mebibyte (eighteen thousand one hundred and nineteen point eighty-one billion two hundred and one million one hundred and seventy-one thousand eight hundred and seventy-five)
19 gb 17.695128917694091796875 gib 19 gigabyte (nineteen) is equal to 17.695128917694091796875 gibibyte
19 gb 0.0172803993336856365203857422 tib 19 gigabyte (nineteen) is equal to 0.0172803993336856365203857422 tebibyte
19 gb 152000000 kbit 19 gigabyte (nineteen) is equal to 152000000 kilobit (one hundred and fifty-two million)
19 gb 152000 mbit 19 gigabyte (nineteen) is equal to 152000 megabit (one hundred and fifty-two thousand)
19 gb 152 gbit 19 gigabyte (nineteen) is equal to 152 gigabit (one hundred and fifty-two)
19 gb 0.152 tbit 19 gigabyte (nineteen) is equal to 0.152 terabit (zero point one hundred and fifty-two)
19 gb 148437500 kibit 19 gigabyte (nineteen) is equal to 148437500 kibibit (one hundred and forty-eight million four hundred and thirty-seven thousand five hundred)
19 gb 144958.49609375 mibit 19 gigabyte (nineteen) is equal to 144958.49609375 mebibit (one hundred and forty-four thousand nine hundred and fifty-eight point forty-nine million six hundred and nine thousand three hundred and seventy-five)
19 gb 141.561031341552734375 gibit 19 gigabyte (nineteen) is equal to 141.561031341552734375 gibibit (one hundred and forty-one point five hundred and sixty-one quadrillion thirty-one trillion three hundred and forty-one billion five hundred and fifty-two million seven hundred and thirty-four thousand three hundred and seventy-five)
19 gb 0.1382431946694850921630859375 tibit 19 gigabyte (nineteen) is equal to 0.1382431946694850921630859375 tebibit | 1,880 | 6,414 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.865543 |
https://islam.stackexchange.com/questions/21628/is-allah-hidden-in-the-ayah-frequences-in-the-koran | 1,720,805,584,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514450.42/warc/CC-MAIN-20240712161324-20240712191324-00403.warc.gz | 259,515,061 | 37,233 | # Is "Allah" hidden in the ayah frequences in the Koran?
I found a hidden name of Allah while I was experimenting with the numerical part of the Koran's order. The results can been seen in the attached diagram:
What have I done? In short, nothing special. I just put each number value of each surah and ayah in a coordinate system and joined the outer points with a line. Now, with a little imagination you can see the name of Allah. But the more interesting aspect is that all points (all chapters) are located within the corpus of Allah's name (btw: the green points are muhkam verses, the red points mutashabih verses).
Is it already known that such a name exists? Is this a coincidence? Could that be the reason why the Koran is ordered by length of chapters? Or is it only an imagination?
• Could you show us the exact methodology you used to get the ayahs and suras? Commented Mar 12, 2017 at 8:59
• I am curius what made you draw the alif (A) . Because it is not following the rules given to join the outer points as the other points. Another question is how you determined the x and y offset. in x (sura number ) we see that 0 - 120 is way longer than Y (aya number). How did you determine that length? Also the diagram displays an extra L (lam) . What is that supposed to be? On the other hand I dont think the message of the Quran is to hide hidden messages. It's clear guidance not hidden guidance. Commented Mar 12, 2017 at 11:53
• Pretty good demonstration of the human mind's inclination to seek patterns, whether there are any to be found or not. Commented Mar 21, 2017 at 12:32
Maasha Allah. Hidden wonders are many in Quran. This shows the perfection in every way. As far as your question is concerned, It may or may not be the reason why length of chapters are arranged in order. Until Allah subhanu wa ta'la says, we cannot come to a conclusion. For He knows the best. Allah says in Quran
“This Qur’an is not such as can be produced by other than Allah, on the contrary it is a confirmation of (revelations) that were sent before it, and a full explanation of the book, wherein there is not doubt from the Lord of the worlds”. (10:37)
“... A book which We have sent down unto you with full of blessings so that they may meditate on its signs and that man of understanding may receive admonition”. (38:29).
The one you found could be one of the signs
“We have revealed for you (O men) a book in which there is a message for you, will you then not understand”. (21:10).
The message could be hidden like how you found out or straight forward.
All praise be to Allah.. Lord of the worlds. | 634 | 2,614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.948491 |
https://outgrow.co/blog/ecommerce-cro?amp | 1,638,515,788,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362605.52/warc/CC-MAIN-20211203060849-20211203090849-00049.warc.gz | 482,597,666 | 24,024 | General marketing
# The eCommerce Conversion Rate Optimization Guide [Tips+Examples]
eCommerce conversion rate optimization is the process of improving the customer experience on an eCommerce website to boost conversion.
But, what is a conversion?
When a visitor completes a desired action on your website, you can call it as a conversion. For example: purchase, email signup, video click, etc.
Now, conversions are broadly divided into two categories – micro conversion and macro conversion. Improving the micro-conversion ultimately results in better macro conversion.
In the eCommerce domain, the macro conversion is generally referred to as the sale of a product or service. And micro-conversions include:
1. A user adding a product to their cart
2. A user adding a product to a wishlist
4. Social media shares
5. Product video clicks
6. Comparing products
7. Other indicators that influence macro conversion
Must Read: 45 Conversion Rate Optimization Statistics Every Marketer Should Know
## How to Calculate Conversion Rate for eCommerce?
By definition, the conversion rate for an eCommerce company is the percentage of the website traffic that has taken the desired action on your website.
By formula: Number of Conversions divided by Website Traffic multiplied by 100.
For example:
Your eCommerce website has monthly traffic of 1000 and out of that 100 users bought your product. So, your macro conversion rate will be calculated as:
100 / 1000 x 100 = 10%
Similarly, for calculating micro-conversions, divide the total number of users who have taken the action you consider as a micro-conversion factor by total traffic and multiply it by 100.
Must Read: Conversion Rate Optimization Tools You Must Have in 2021
## What Is the Average eCommerce Conversion Rate?
The average conversion rate in the eCommerce industry ranges between 1-2%. This may vary depending on the type of product or service you are offering. However, a 2%+ conversion rate is a good benchmark for any eCommerce business.
Here’s an illustration that shows the average micro conversion rate of an eCommerce website. Notice the change in micro conversions as users move down the sales funnel.
Source: Smart Insights
According to the infographic, nearly 44% of users who visit an eCommerce portal move to the product page, and out of that nearly 15% add the product to their cart. And finally, only 3% end up buying the product.
Now, to improve the conversion rates you need to run a conversion rate optimization (CRO) campaign on your website. You can optimize all the pages of your eCommerce website that serve as a customer touchpoint. In other words, you have to optimize your website for each stage of the funnel.
But before you start optimizing, have a quick look at the various stages of your customer’s journey down the conversion funnel. Once you optimize for each stage, you will start generating more conversions.
## Ecommerce Conversion Funnel
The eCommerce conversion funnel illustrates the various stages of a customer’s journey based on their awareness, interest, and perception of your brand or product. Typically, the eCommerce conversion funnel has four stages:
### • Awareness
This is the first stage of the conversion funnel where the consumer becomes familiar with your brand or product. They are actively searching for a solution to their problem on the internet and might stumble upon your website. So, at this point, greet them with free educational content like blogs, webinars, podcasts, ebooks, courses, etc., and seize this opportunity to generate leads.
Lead generation is a key micro-conversion metric. Hence, keep track of the micro-conversion rate at this point and optimize for better lead generation.
### • Interest
Now that your customers are aware of your brand or product, it’s time to keep them hooked.
Remember, so far your customers were only searching for a solution to their problem. But they have not yet developed an interest in your brand. Therefore, you must keep them engaged with more educational and entertaining content so that they develop an interest in your brand. Try offering them some free interactive tools, checklists, demo videos, etc. to catch their attention.
### • Desire
At this stage, you should try to convert their interest in your brand into desire. Highlight the benefits of using your product. Show your audience how you are different from your competitors and how your product can add value. ROI calculators and quizzes can easily do the job here!
### • Action
This is the final stage of the funnel and by now your users have almost made up their mind to make the purchase. Typically, at this stage, your customers will add products to the cart, fill in the required information, select the payment method and make the final purchase. Your job is to make this experience as smooth and easy as possible.
Examine your cart pages and find out possible causes of distractions and obstacles that your customers might face at this stage. Try to address these issues and keep optimizing unless you reach your conversion goal.
## How to Use Interactive Content at Every Stage of the Funnel to Increase Your Conversion Rate?
Interactive content is attractive, engaging, and highly customizable. And these make interactive content the perfect weapon to increase conversion rate in the eCommerce domain. You can customize your interactive content for each stage of the conversion funnel.
### I. Awareness stage
In the awareness stage, customers look for educational content and as we mentioned, it’s a great opportunity to generate leads. So, here you can create interactive quizzes that will help your customers figure out their pain points. Create separate landing pages targeted to capture users in the awareness stage and embed your quizzes strategically.
Suppose you run an online book store and your customers are looking for English grammar books. So you can greet these customers with an interactive assessment like an “English Grammar Quiz”. And before showing them the results, you can ask for their contact details. This will not only help you generate leads but also build awareness about your brand.
### II. Interest stage
In the interest stage, retarget the users who have completed the quizzes at the previous stage. Tools like Outgrow not only help you create interactive content but also let you retarget the audience using cookies.
Related Post
### III. Desire stage
For the desire stage, you have to tell your audience how your product is different and how it adds value. So, create content that clearly shows how your users can benefit from using your product instead of the alternatives.
For example, this interactive calculator shows how much a person can make by leisurely driving an Uber. It’s perfect for attracting drivers in the ‘desire’ stage looking to join an online cab service.
### IV. Final stage
In the final stage of the conversion funnel, customers often face the problem of choice and are unable to take the final decision of which product to buy. Again, you can use interactive content to address this issue.
If you are selling sunglasses online, then you can embed a product recommendation quiz like “Which sunglasses should you buy” on your cart page. The tool will recommend the perfect pair of sunglasses based on the user’s preference. Also, don’t forget to add links to the product purchase page offering the sunglass to make the navigation smoother.
## Tips for eCommerce Conversion Rate Optimization
Apart from using interactive content to boost conversions, you can also leverage these tips and tricks.
### 1. Use Urgency Messaging
Limited time offers and count-down timers can work miracles. When your customers see that time is running out, a sense of urgency will drive them to take a quick decision.
You can send notifications with exclusive limited-time discounts to customers with abandoned products in their cart. Most of your customers wouldn’t like to miss this opportunity and might come back to their cart page, avail the special discount within time, and proceed to check out!
Here’s a giveaway example of how you can use countdown timers to entice your customers into taking the action that you want them to take.
### 2. Offer and Highlight ‘Free Shipping’
Nowadays, customers expect brands to maintain a certain level of the standard set by big eCommerce portals like Amazon. ‘Free Shipping’ is an absolute must in today’s scenario. Even if you need to increase your product price to cover the shipping cost, you must.
Did you know that high shipping charges lead to cart abandonment?
### 3. Use Chatbots
Chatbots can make interactions with your customers fun and seamless. And the best part is that chatbots can help you generate qualified leads easily. Not only that, but it also captures visitor insights that you can later extract and analyze for future communications. So, if you haven’t incorporated this tip in your CRO strategy yet, why not try creating a chatbot without any coding?
Check out: 10 Best Website Chatbot Examples of 2021 [Real-Life Examples]
### 4. Display Social Proof and Brand Authenticity
Consumers are often wary about the authenticity of products on e-commerce websites. So it is crucial to display customer reviews, ratings, and brand endorsements. Amazon displays customer reviews and ‘Amazon Fulfilled’ alongside its products. This builds trust and confidence about the product among customers.
### 5. Payment Security Is a Must
If you are new to the eCommerce industry, then it’s obvious that your customers will think twice before making a payment on your site. So you must comfort them with the best-in-class payment security protocols.
### 6. Provide a 360° View of Products
It is a known fact that big and clear images increase conversion rates. Most of the eCommerce brands follow this technique.
So how can you stand out?
Well, go a step further and provide a 360° view of your products.
### 7. Let Users Add Products to “WishList” Without Registration
Customers often hesitate to register and can bounce off without adding products to their wishlist. To avoid such a scenario, allow your users a registration-free wishlist. It will increase user engagement, return visits, and conversions.
### 8. Optimize for Mobile
Did you know that the global eCommerce revenue from mobile is expected to reach \$3.56 trillion by 2021? Want a fair share of the booming mobile shopping industry? We bet you do! Even Google is constantly updating to empower mobile-friendly websites. So, optimize your eCommerce portal for all types of mobile devices and cut through the competition.
Videos help people imagine how they will feel when they have the product in their hands. So shoot videos with someone using the product and add them to your product page.
Especially for lifestyle and fashion accessories, real-life videos are a must.
### 10. Promote Social Share Post-purchase
We often ignore the ‘Thank you’ pages while talking about conversion rate optimization. But thank you pages are crucial to get some extra shares! A customer who has just bought an exclusive product might want to flaunt it on their social handles. So why lose the opportunity? Introduce social share buttons on your thank you page and get some extra shares!
## Conclusion
Well, we tried to cover all the tips and tricks for eCommerce conversion rate optimization. But, it’s a never-ending discussion and it varies from business to business. So, we recommend that you read our guide on conversion rate optimization and figure out what fits best for your brand.
Also, don’t try to implement these tips randomly on your website. There’s a process that marketers should follow for their eCommerce conversion rate optimization campaigns. Here’s a blog where we have discussed the conversion rate optimization process in detail. Give it a quick read!
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Soumodip Roy
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2 weeks ago | 2,598 | 13,032 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-49 | latest | en | 0.88344 |
https://undergroundmathematics.org/trigonometry-triangles-to-functions/r8136/suggestion | 1,537,806,116,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160568.87/warc/CC-MAIN-20180924145620-20180924170020-00189.warc.gz | 630,797,302 | 5,184 | Review question
How many solutions does this equation involving $\sin^8x$ have? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R8136
Suggestion
In the range $0 \leq x < 2\pi$ the equation $(3 + \cos x)^2 = 4 - 2\sin^8x$ has
1. $0$ solutions,
2. $1$ solution,
3. $2$ solutions,
4. $3$ solutions.
What is the range of values that the left hand side can take? | 131 | 453 | {"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": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-39 | latest | en | 0.873082 |
http://physicsgoeasy.blogspot.com/2010_07_26_archive.html | 1,490,618,128,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189472.3/warc/CC-MAIN-20170322212949-00604-ip-10-233-31-227.ec2.internal.warc.gz | 270,856,193 | 22,185 | ## Pages
### IITJEE/AIEEE/PMT tips and tricke in electrostatics
1. The force with which two charges interact is not changed by the presence of the other charges.
2. Net force on any charge F=F1+F2+F3+F4+.....
3. Electric field lines extend away from the positive charge and towards thge negative charges.
4. Electric field produces the force so if a charge q is placed in the electric field E the force experience by the charge is F=qE
5. Principle of superposition also applies to electric field
so E=E1+E2+E3+E4+......
6. E is the electric field present due to all charges in the system not just the charge inside in the Gauss law.
7. Flux crossing a closed surface does not depend on the shapes and size of gaussian surface.
8. ∫E.dl over closed path is zero.
9. Electric Potential is scalar quantity.
10. Potential at point due to system of charges will be obtained by the summation of potential of each charge at that point V=V1+V2+V3+V4+..........
For more such tips and tricks click the link given below | 250 | 1,023 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-13 | longest | en | 0.893662 |
https://www.storyboardthat.com/storyboards/eeb02795/biologia | 1,603,195,494,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107872686.18/warc/CC-MAIN-20201020105000-20201020135000-00202.warc.gz | 918,249,184 | 64,193 | Biología
Updated: 10/14/2020
This storyboard was created with StoryboardThat.com
#### Storyboard Description
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#### Storyboard Text
• Pero con ella puedes entender algunos deportesVen y te enseño
• Aveces pienso que esas clases de los Movimientos no sirven para nada ¿No te pasa?
• Por ejemplo mira ese chico, la pelota hace un movimiento parabólico.
• O esta chica, falló su tiro y su pelota realizó un movimiento semiparabólico
• Mira las ruedas de la bicicleta de esa chica, hacen un movimiento circular uniforme
• Y por último mira el deportista recorriendo el circuito para no cansarse realiza un movimiento rectilíneo uniforme
• Denada amigo! Siempre estoy para ayudarte.
• Ahora si pude ver varios ejemplos de movimientos en distintos deportes, Muchas Gracias Pana! | 312 | 1,596 | {"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 | 3 | CC-MAIN-2020-45 | latest | en | 0.933385 |
https://thefloatingschoolid.org/and-pdf/283-inverse-laplace-transform-problems-and-solutions-pdf-704-266.php | 1,653,201,920,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545090.44/warc/CC-MAIN-20220522063657-20220522093657-00284.warc.gz | 653,466,904 | 10,122 | # Inverse Laplace Transform Problems And Solutions Pdf
File Name: inverse laplace transform problems and solutions .zip
Size: 1035Kb
Published: 18.04.2021
Crystal wineglass or order differential equation application with solutions pdf preview is. Wish to differential application problems with solutions pdf navigation and a little integration on using variation of the solution that of mathematics along the eigenvalues for all the other.
## Differential Equation Application Problems With Solutions Pdf
Inverse laplace transform program, program for forming table for inverse laplace transform, program for calculating numerical solution of inverse laplace transform, and inverse laplace transform device. USA1 ja. EPA1 ja. JPA ja. CAA1 ja.
## 8.2: The Inverse Laplace Transform
The Laplace transform is an integral transform perhaps second only to the Fourier transform in its utility in solving physical problems. The Laplace transform is particularly useful in solving linear ordinary differential equations such as those arising in the analysis of electronic circuits. The unilateral Laplace transform not to be confused with the Lie derivative , also commonly denoted is defined by. The unilateral Laplace transform is almost always what is meant by "the" Laplace transform, although a bilateral Laplace transform is sometimes also defined as. Oppenheim et al.
## Laplace transform Solved Problems 1 - Semnan University
What we would like to do now is go the other way. As you will see this can be a more complicated and lengthy process than taking transforms. So, we take the inverse transform of the individual transforms, put any constants back in and then add or subtract the results back up.
### Laplace Transforms: Theory, Problems, and Solutions
Solved Problems ON. Laplace transform. Samir Al-Amer November Many mathematical Problems are Solved using transformations. The idea is to transform the problem into another problem that is easier to solve. Once a solution is obtained, the inverse transform is used to obtain the solution to the original problem.
The transform has many applications in science and engineering because it is a tool for solving differential equations. In particular, it transforms differential equations into algebraic equations and convolution into multiplication. The Laplace transform is named after mathematician and astronomer Pierre-Simon Laplace , who used a similar transform in his work on probability theory. Laplace's use of generating functions was similar to what is now known as the z-transform , and he gave little attention to the continuous variable case which was discussed by Niels Henrik Abel. The current widespread use of the transform mainly in engineering came about during and soon after World War II, [10] replacing the earlier Heaviside operational calculus.
The next theorem enables us to find inverse transforms of linear combinations of transforms in the table. We omit the proof. In such cases you should refer to the table of Laplace transforms in Section 8. Using the Laplace transform to solve differential equations often requires finding the inverse transform of a rational function. The next two examples illustrate this. The next theorem states this method formally. For a proof and an extension of this theorem, see Exercise 8.
#### Some Properties of the Inverse Laplace Transform
We first saw these properties in the Table of Laplace Transforms. Our question involves the product of an exponential expression and a function of s , so we need to use Property 4 , which says:. For the sketch, recall that we can transform an expression involving 2 trigonometric terms. We complete the square on the denominator first:. We observe that the Laplace inverse of this function will be periodic , with period T.
ФЭГ и экологи так и не смогли установить, какая из двух версий соответствует истине, и средства массовой информации в конце концов устали от всей этой истории и перешли к другим темам. Каждый шаг Стратмора был рассчитан самым тщательным образом. Строя свои планы, Стратмор целиком полагался на собственный компьютер. Как и многие другие сотрудники АНБ, он использовал разработанную агентством программу Мозговой штурм - безопасный способ разыгрывать сценарий типа Что, если?. на защищенном от проникновения компьютере. Мозговой штурм был своего рода разведывательным экспериментом, который его создатели называли Симулятором причин и следствий.
Очень. Двухцветный застыл на месте и зашелся в истерическом хохоте. - Ты хочешь сказать, что это уродливое дерьмовое колечко принадлежит. Глаза Беккера расширились. - Ты его. Двухцветный равнодушно кивнул.
- Я хотел сказать… - Чертовщина. - Если бы вы согласились мне помочь. Это так важно. - Извините, - холодно ответила женщина. - Все совсем не так, как вы подумали.
- Это плохо. Это очень и очень плохо. - Спокойствие, - потребовал Фонтейн. - На какие же параметры нацелен этот червь.
Она в конце концов перестала протестовать, но это продолжало ее беспокоить. Я зарабатываю гораздо больше, чем в состоянии потратить, - думала она, - поэтому будет вполне естественным, если я буду платить. Но если не считать его изрядно устаревших представлений о рыцарстве, Дэвид, по мнению Сьюзан, вполне соответствовал образцу идеального мужчины. Внимательный и заботливый, умный, с прекрасным чувством юмора и, самое главное, искренне интересующийся тем, что она делает. Чем бы они ни занимались - посещали Смитсоновский институт, совершали велосипедную прогулку или готовили спагетти у нее на кухне, - Дэвид всегда вникал во все детали.
- Я думаю, я поняла, что вам от меня. - Она наклонилась и принялась рыться в сумке. Беккер был на седьмом небе.
Но, посмотрев на распростертую на простынях громадную тушу, почувствовала облегчение. То, что она увидела пониже его живота, оказалось совсем крошечным. Немец схватил ее и нетерпеливо стянул с нее рубашку. Его толстые пальцы принялись методично, сантиметр за сантиметром, ощупывать ее тело.
Она ждет уже целый час. Очевидно, Анонимная рассылка Америки не слишком торопится пересылать почту Северной Дакоты. Сьюзан тяжело вздохнула. Несмотря на все попытки забыть утренний разговор с Дэвидом, он никак не выходил у нее из головы.
### Related Posts
1 Response
1. Millie S.
The final stage in that solution procedure involves calulating inverse Laplace transforms. In this section we look at the problem of finding inverse Laplace. | 1,569 | 6,428 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2022-21 | latest | en | 0.888772 |
https://scoop.eduncle.com/please-explain-ex-dm-kt-where-m-is-the-by-dt-dv-dm-dt-mg-motion-of-the-rain-drop-is-m-dm-dt-t0-with-zero | 1,618,637,953,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038101485.44/warc/CC-MAIN-20210417041730-20210417071730-00306.warc.gz | 621,987,777 | 14,065 | IIT JAM Follow
January 11, 2021 10:15 pm 30 pts
please explain Ex. dm kt, where m is the by dt dv dm dt = mg motion of the rain drop is m dm dt t0. with zero initial velocity and initial mass m, (m, = 2 gm, k = 12gm/ If the drop starts s and g= 1000 cm/s*), the velocity v of the drop after one second is IT-JAM 2011 (A) 250 cm/s (B) 500 cm/s (C) 750 cm/s (D) 1000 cm/ls
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option b is correct see the attachment | 161 | 448 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-17 | latest | en | 0.81473 |
https://books.ifpenergiesnouvelles.fr/ebooks/thermodynamics/03/exo_03_01.html | 1,550,414,837,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247481994.42/warc/CC-MAIN-20190217132048-20190217154048-00348.warc.gz | 493,529,862 | 4,506 | # Example 3-1: Vapour pressure predictions of methane using different formulas
Compare the results of the various correlations with the DIPPR correlation.
In the following tables, the characteristic properties necessary for the evaluation of each expression is given. Remember that the normal boiling point is measured at 101.325 kPa. The various parameters of the equations are:
Table 1: Properties of methane used as parameters in vapour pressure equations
Parameters Tc Pc ω Tb T3 P3
Value 190.564 4599.0 0.0115478 111.66 90.694 11.696
Table 2: Parameters for various vapour pressure equations for Methane
Parameter A B C D E
Antoine-dimensionless, equation (3.12b) 5.3157 -5.3008 -0.0063
Antoine-API, equation (3.12) 8.6775 -911.234 -6.34
DIPPR, equation (3.13) 38.664 -1314.7 -3.3373 3.0155E-05 2
Wagner, equation (3.14) -6.00435 1.1885 -0.834082 -1.22833
(Temperature in K - Pressure in Pa)
We recall some equations of chapter 3:
Antoine's equation:
(3.12)
Dimensionless Antoine's equation:
(3.12b)
DIPPR equation:
(3.13)
Wagner's equation:
(3.14)
with
## Analysis:
• Pressure and temperature are the basic properties of the system.
• Methane is a well-known light hydrocarbon.
• Phases present are vapour and liquid all along the saturation curve.
## Solution:
### See complete results in file (xls):
Some help on nomenclature and tips to use this file can be found here.
Tables for each equation are constructed and compared with the DIPPR set of calculated values. Note that the DIPPR equation does not predict exactly the critical point, the normal bubble point and the triple point; there are relative deviations around 0.2 to 0.6 %. Additionally, the acentric factor (‘omega point’ shown in figure 1) predicted by this equation is not exactly the same as that given in the database.
Relative differences are plotted in figure 1 over the full temperature range, from triple point to critical point for six different equations:
• the 3 points equation (3.11) that uses critical point, normal boiling point and triple point;
• the 2 points equation (3.8) that uses the critical point and the triple point. Note that the Wilson equation is identical but uses the critical point and the ‘omega point’. It is not shown in figure 1.
• The corresponding states equations of Ambrose (equation 3.16) and Lee-Kesler (equation (3.17) are also shown: they provide almost identical results above the ‘omega point’, but the more complex Ambrose equation extrapolates better to low reduced temperatures.
• The Antoine equation (3.12) is used either with a regression over the entire temperature domain, or using the parameters that are recommended by API Riazi (2005), but no validity range is associated with it It is clear that the difference becomes very high beyond 150 K, clearly indicating that this equation must be used only over the range 90-140 K.
• The Wagner equation (3.14) is not shown in the plot. It behaves very well if fitted on the entire temperature range.
An additional curve (not shown in the book) can be constructed based on a basic formulation (Antoine's equation in this application but any other expression can be used). In this case, the parameter C is a parameter to be optimized while A and B have to satisfy the equations on 2 selected points (for example triple and critical points). Thus, the following equations have to be solved:
We find that:
Both parameters can be automatically determined once the value of C is fixed. Then an error minimization procedure can be executed on parameter C to obtain the optimum value while satisfying the restriction criteria.
## References
R. L. Rowley, W. V. Wilding, J. L. Oscarson, Y. Yang, N. A. Zundel, T. E. Daubert, R. P. Danner, DIPPR® Data Compilation of Pure Compound Properties, Design Institute for Physical Properties, AIChE, New York, NY (2003).
Riazi, M. R. Characterization and Properties of Petroleum Fluids; American Society for Testing and Materials: Philadelphia, 2005. | 984 | 3,983 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-09 | latest | en | 0.831428 |
http://mathgardenblog.blogspot.com/2013/10/basic-compass-straightedge-construction.html | 1,556,001,775,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578593360.66/warc/CC-MAIN-20190423054942-20190423080942-00327.warc.gz | 112,736,784 | 16,781 | ## Pages
### Compass-and-straightedge construction
Today we will start a series of posts on compass-and-straightedge construction. Believe it or not, there are a few construction problems that sound simple but it had required more than two thousand years to settle! The most famous ones are the problem of regular polygon construction and the problem of angle trisection. These problems were known in ancient times but it was not until the late 18th-19th centuries that mathematicians could finally solve them using the most modern tools of mathematics.
In this first post, we will learn about some basic compass-and-straightedge constructions. These constructions are assumed as obvious and can be employed in more complex construction problems.
Basic constructions
Construction of the perpendicular bisector of a line segment
Given a line segment $AB$. To construct the perpendicular bisector of $AB$, we follow the steps below:
• Construct two circles of equal radius with centres at $A$ and $B$ so that they intersect at two points.
• The line connecting the two intersection points is the perpendicular bisector of $AB$.
Construction of the midpoint of a line segment
Given a line segment $AB$. To construct the midpoint of the line segment $AB$, we follow the steps below:
• Construct the perpendicular bisector of $AB$.
• The perpendicular bisector meets $AB$ at the midpoint $M$ of $AB$.
Through a point, construct a line perpendicular to a given line
Given a line $\ell$ and a point $A$. To construct a line passing though $A$ and perpendicular to $\ell$, we follow the steps below:
• Construct a circle with centre $A$ so that it intersects with the line $\ell$ at two points $B$ and $C$.
• Construct the perpendicular bisector of $BC$, this is the line passing through $A$ and perpendicular to $\ell$.
Through a point, construct a line parallel to a given line
Given a line $\ell$ and a point $A$. To construct a line passing though $A$ and parallel to $\ell$, we follow the steps below:
• Construct the line $t$ passing through $A$ and perpendicular to $\ell$.
• Construct the line $u$ passing through $A$ and perpendicular to $t$, the line $u$ is the line that passes through $A$ and is parallel to $\ell$.
Construction of the angle bisector
Given an angle $\angle xOy$, to construct the angle bisector we follow the steps below:
• Construct a circle with centre $O$ that intersects $Ox$ and $Oy$ at $A$ and $B$, respectively.
• Construct the perpendicular bisector of $AB$, this is the required angle bisector of $\angle xOy$.
Construction of an equal angle to a given angle
Given an angle $\angle xOy$ and a ray $A \ell$, to construct an angle equal to $\angle xOy$ that has $A \ell$ as one side, we follow the steps below:
• Take a point $B$ on the ray $A \ell$.
• Construct the circle with centre $O$ and radius equal to $AB$, this circle meets $Ox$ and $Oy$ at $D$ and $C$, respectively.
• Construct the circle with centre $A$ and radius $AB$, and construct the circle with centre $B$ and radius $CD$, these two circles intersect at $E$ and $F$.
• The two angles $\angle EA \ell$ and $\angle FA \ell$ are both equal to $\angle xOy$.
Through a point, construct a tangent line to a circle
Given a circle with centre $O$ and a point $A$ lying outside of the circle, to construct a line passing though $A$ and tangent to the circle $(O)$, we follow the steps below:
• Construct the midpoint $B$ of $OA$.
• Construct the circle with centre $B$ and radius equal to $AB$, this circle intersects with the circle $(O)$ at two points $C$ and $D$.
• The two lines $AC$ and $AD$ are the two required tangent lines to the circle $(O)$.
Some construction problems
Example 1. Given a line segment $AB$. Use compass and straightedge to divide this line segment into five equal parts.
Solution:
• Through $A$ construct a ray, and on this ray, use compass to subsequently construct the points $C_1$, $C_2$, $C_3$, $C_4$, $C_5$ so that $AC_1 = C_1C_2=C_2C_3=C_3C_4=C_4C_5$.
• Connect $BC_5$.
• Construct the four lines passing through $C_1$, $C_2$, $C_3$, $C_4$, respectively, and parallel to $BC_5$ which meet $AB$ at $D_1$, $D_2$, $D_3$, $D_4$.
• We have $AD_1 = D_1D_2=D_2D_3=D_3D_4=D_4B$.
Example 2. Given an angle $xOy$ and a point $M$. Use compass and straightedge to construct a point $A$ on $Ox$ and a point $B$ on $Oy$ so that $M$ is the midpoint of $AB$.
Solution:
• Draw the line $OM$ and use compass to construct the point $N$ on $OM$ such that $OM=MN$.
• Construct a line passing through $N$ and parallel to $Oy$ which meets $Ox$ at $A$.
• Construct a line passing through $N$ and parallel to $Ox$ which meets $Oy$ at $B$.
• $OANB$ is a parallelogram so the midpoint $M$ of the diagonal $ON$ is also the midpoint of the diagonal $AB$.
Example 3. Given a triangle $ABC$. Use compass and straightedge to construct a square $PQRS$ so that the vertex $Q$ lies on the side $AB$ of the triangle, the vertex $R$ lies on the side $AC$, and the two vertices $P$, $S$ lie on the side $BC$.
Solution:
• Take a point $U$ on $AB$.
• Construct the line $UV$ perpendicular to $BC$.
• Use compass to construct the point $F$ on the ray $VC$ so that $VF=VU$.
• Construct the square $UVFE$.
• Construct the intersection point $R$ of the two lines $BE$ and $AC$.
• Construct the line $RS$ perpendicular to $BC$.
• Use compass to construct the point $P$ on the ray $SB$ so that $SP=SR$.
• Construct the square $PQRS$.
Let us stop here for now. We will continue this topic of compass-and-straightedge construction in the next post. Hope to see you again there.
Homework.
1. Give compass-and-straightedge constructions of an equilateral triangle, a square, a regular hexagon (6 sides), and a regular octagon (8 sides).
2. Given two circles, use compass and straightedge to construct all common tangents to these two circles.
3. Given two line segments with lengths $a$ and $b$, respectively, use compass and straightedge give a construction of a line segment of length $\sqrt{ab}$.
4. Prove that $$\cos{\frac{\pi}{5}} = \frac{1 + \sqrt{5}}{4}$$
and use this formula to derive a compass-and-straightedge construction of a regular pentagon.
5. Given a quadrilateral $ABCD$ and four points $M_1$, $M_2$, $M_3$, $M_4$, in this order, on $AB$. Using compass and straightedge, show how to construct four points $N_1$, $N_2$, $N_3$, $N_4$ on $CD$ so that the four line segments $M_1 N_1$, $M_2 N_2$, $M_3 N_3$ and $M_4 N_4$ divide the quadrilateral into 5 small pieces of equal area.
6. Given two points $A$ and $B$, only use compass (straightedge is not allowed), show how to construct four points $D_1$, $D_2$, $D_3$, $D_4$ on the line segment $AB$ so that they divide the line segment into five equal parts. | 1,882 | 6,727 | {"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} | 4.90625 | 5 | CC-MAIN-2019-18 | latest | en | 0.88368 |
http://www.catonmat.net/blog/derivation-of-ycombinator | 1,547,943,436,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583688396.58/warc/CC-MAIN-20190120001524-20190120023524-00437.warc.gz | 272,023,432 | 21,496 | Computer Science February 19, 2010
# Deriving the Y-Combinator
In this post I'll derive the Y-combinator and explain all the steps taken. The key to understanding the Y-combinator is knowing precisely what it does. I am sure many people never understand Y-combinator because of this very reason, or because the introduction is too lengthy to comprehend. Therefore I'll give the shortest possible explanation of Y-combinator before I derive it so that you know what the end result should be.
The Y-combinator allows an anonymous function to call itself, that is, it allows anonymous function recursion. Since anonymous functions don't have names, they can't refer to themselves easily. The Y-combinator is a clever construct helps them to refer to themselves. That's it. That's all the Y-combinator does. Remember that.
I'll use the Scheme programming language to derive the Y-combinator. My derivation is based on the one in "The Little Schemer" book. I took the examples from the book, filled in the missing steps and explained the steps in more details.
Here it is.
Suppose we want to write the `length` function, which, given a list, returns the number of elements in it. It's really easy if we can give the function a name:
```(define length
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (length (cdr list)))))))
```
But now suppose that we can't give names - we can only use anonymous functions:
```(lambda (list)
(cond
((null? list) 0)
(else
(add1 (??? (cdr list)))))))
```
Suddenly there is no way this anonymous function can refer to itself. What do we put in `???`? We can't refer to the name `length` anymore because it's an anonymous function, which doesn't have any name. One thing we can try is to put the function itself in the place of `???`:
```(lambda (list)
(cond
((null? list) 0)
(else
(add1 (
(lambda (list) ; the
(cond ;
((null? list) 0) ; function
(else ;
(add1 (??? (cdr list)))))) ; itself
(cdr list))))))
```
This is not much better, we are still left with `???`. But there is a bright side to it, this function can determine lengths of lists with one element and no elements. Let's try inserting the same anonymous function in place of `???` again:
```(lambda (list)
(cond
((null? list) 0)
(else
(add1 (
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (??? (cdr list))))))
(cdr list))))))
(cdr list))))))
```
Now this function can determine lengths of lists with 0, 1 and 2 elements. If we continue this way, we can construct the `length` function for lists with a certain number of elements. But this is not what we want, we the real `length` function that works for lists with any number of elements.
As we all know, repeating code is not a good thing. Let's try to factor out the repetitions and rewrite the original anonymous function slightly. Instead of leaving `???` in the code, let's pass it to an anonymous function via an argument called `length`.
```((lambda (length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (length (cdr list)))))))
???)
```
Notice that this code invokes the first `lambda (length)` function and passes it `???` as the `length` argument. This function in turn returns the original anonymous function:
```(lambda (list)
(cond
((null? list) 0)
(else
(add1 (??? (cdr list))))))
```
Now let's try constructing the previous two length functions that work for lists with a maximum of one and two elements. First the function for lists with a maximum of one element:
```((lambda (f)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (f (cdr list)))))))
((lambda (g)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (g (cdr list)))))))
???))
```
Here `???` first gets passed to `lambda (g)`, which returns the original anonymous function, which then gets passed to `lambda (f)`, which in turn returns the function that works for lists with one element:
```(lambda (list)
(cond
((null? list) 0)
(else
(add1 (
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (??? (cdr list))))))
(cdr list))))))
```
Similarly, the following code returns a function that works for lists with a maximum of two elements:
```((lambda (f)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (f (cdr list)))))))
((lambda (g)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (g (cdr list)))))))
((lambda (h)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (h (cdr list)))))))
???)))
```
Since the argument names `f`, `g`,`h` in `(lambda (f) ...)`, `(lambda (g) ...)`, `(lambda (h) ...)` are independent, we can rename all of them to `length`, to make it look more similar to the `length` function:
```((lambda (length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (length (cdr list)))))))
((lambda (length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (length (cdr list)))))))
((lambda (length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (length (cdr list)))))))
???)))
```
We are still repeating code. The obvious thing we can do about it is create an anonymous function called `mk-length` (make length) that creates this code for us:
```((lambda (mk-length)
(mk-length ???))
(lambda (length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (length (cdr list))))))))
```
This is pretty tricky. Observe how `(lambda (mk-length) ...)` gets the `(lambda (length) ...)` function passed as the `mk-length` argument, which in turn accepts `???` as an argument and returns our original anonymous function:
```(lambda (list)
(cond
((null? list) 0)
(else
(add1 (??? (cdr list))))))
```
Now let's try constructing length functions for lists of length one and two. For the list of length one, we just need to apply `mk-length` on itself:
```((lambda (mk-length)
(mk-length
(mk-length ???)))
(lambda (length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (length (cdr list))))))))
```
Let's go through this code. First `(lambda (length) ...)` gets passed to the `lambda (mk-length)` function as the `mk-length` argument. Then it applies the result of `(mk-length ???)` (which is the original anonymous function) to the `mk-length`. This produces our well known function that works on lists with one or none elements:
```(lambda (list)
(cond
((null? list) 0)
(else
(add1 (
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (??? (cdr list))))))
(cdr list))))))
```
Similarly, by adding another call to `mk-length`, we get the function that works for lists with two or less elements:
```((lambda (mk-length)
(mk-length
(mk-length
(mk-length ???)))
(lambda (length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (length (cdr list))))))))
```
Now, since we don't know what `???` is, let's just call it `mk-length` and see what happens:
```((lambda (mk-length)
(mk-length mk-length))
(lambda (length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (length (cdr list))))))))
```
Nothing bad actually happens, we still get the same original anonymous function, except in place of `???` we have the function:
```(lambda (length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (length (cdr list)))))))
```
Notice also that argument names `mk-length` and `length` in `lambda (mk-length)` and `lambda (length)` are independent. Therefore we can rename `length` to `mk-length` to remind that the first argument to `mk-length` is `mk-length`:
```((lambda (mk-length)
(mk-length mk-length))
(lambda (mk-length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (mk-length (cdr list))))))))
```
And now comes the key trick. We replace `mk-length` with a self-application `(mk-length mk-length)`:
```((lambda (mk-length)
(mk-length mk-length))
(lambda (mk-length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 ((mk-length mk-length) (cdr list))))))))
```
If you try this code out, you'll see that it returns the length of a list for any list. Here is an example of applying it to a list of 10 elements:
```(((lambda (mk-length)
(mk-length mk-length))
(lambda (mk-length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 ((mk-length mk-length) (cdr list))))))))
'(a b c d e f g h i j))
```
The function works because it keeps adding recursive uses by passing `mk-length` to itself, just as it is about to expire. This is not yet the Y-combinator, but we have successfully managed to recursively call an anonymous function. Now we need to massage the code a bit to separate the anonymous function from the self-applicative code.
The first step is to move the self-applicative code out as much as possible:
```((lambda (mk-length)
(mk-length mk-length))
(lambda (mk-length)
((lambda (length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (length (cdr list)))))))
(lambda (x)
((mk-length mk-length) x)))))
```
Now the code in bold looks very much like the `length` function we need! Let's move it outside as well:
```((lambda (le)
((lambda (mk-length)
(mk-length mk-length))
(lambda (mk-length)
(le (lambda (x)
((mk-length mk-length) x))))))
(lambda (length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (length (cdr list))))))))
```
There we have it! The anonymous `length` function has been separated from the self-applicative call. The code in bold is the `length` function, and the code not-in-bold is the Y-combinator!
Now let's rename `mk-length` to `f` and join the lines to make it more compact:
```((lambda (le)
((lambda (f) (f f))
(lambda (f)
(le (lambda (x) ((f f) x))))))
(lambda (length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (length (cdr list))))))))
```
And we are done. We can now write down the definition of Y-combinator:
```(define Y
(lambda (le)
((lambda (f) (f f))
(lambda (f)
(le (lambda (x) ((f f) x)))))))
```
And we can apply it to our anonymous `length` function:
```((Y (lambda (length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (length (cdr list))))))))
'(a b c d e f g h i j))
; ==> 10
```
This is one of many derivations of the applicative-order Y-combinator. See the publication "The Why of Y" for another derivation.
### Comments
Great post!
I think this approach to deriving the Y combinator really appeals to the hackers. I'll share with my students.
I teach a compilers class, and I wrote up an alternate derivation for my students that starts with fixed points; you might be interested:
s/But this is now what we want, we a real length function that works for lists with any number of elements./But this is not what we want, we want a real length function that works for lists with any number of elements./
most programming languages that support anonymous functions can also name the anonymous function(might be an extra line of code but that's fine).
When is a situation where we "cannot give names" to an anonymous function?
tommy > there is a quirk in the Erlang syntax, wich makes that you can't have a local declaration of a recursive function (inside an expression) : actually, there is no syntax for named function declaration inside expressions, so you have to use anonymous functions. Hence, you can't declare a local recursive function without such a fixpoint operator. This is actually a bad idea, as you end up reimplementing a part of the runtime system in a probably slow and certainly cryptic way, but that was just for the example.
Peter > this combinator is just one among the different fixpoint combinators available. There is for example the Turing Fixpoint combinator, wich is also derivable. Do you know where in your derivation you made a decision that lead you to this combinator instead of another ? I have a gut feeling that the derivation is not "canonical" and that one could probably use a slightly different succession of software-engineering-and-common-sense steps that would naturally lead to another combinator.
Also, I'm not sure I agree with your presentation of the Y-combinator. You describe it as something about recursion and anonymous functions. I believe it is much more general than that : fixpoint combinators allow you to control recursion. It is all about parametrizing the function you call over at your recursion sites, instead of hardcoding recursion using the language syntax (or lack of). From a software engineering point of view, you could say that it decouples two aspects of your function : the implementation (with domain-specific knowledge and all) and the "looping" / "tying the knot" process, wich is deferred to the fixpoint combinator.
You can do quite interesting (but somewhat hacky) things once you've decoupled those two separate concerns. Matt mentioned memoization (automagic memoization of inner recursive calls, wich have been exposed by the decoupling process), there are quite a few folk examples, see for example the following article That About Wraps it Up.
It's a funny hack, for example, to write in your favorite language a function that take any derecursified/decoupled function and, assuming the original function was tail-recursive, produce a tail-recursive implementation of it, using an explicit call stack, wich doesn't depend on the implementation of tail recursion in the compiler/runtime.
most programming languages that support anonymous functions can also name the anonymous function(might be an extra line of code but that’s fine).
When is a situation where we “cannot give names” to an anonymous function?
When we need to explain how the Y combinator works.
tommy i don't think you understand the point of this.
you need to realize that there are many applications of the Y combinator, many times the language you THINK "names" anon functions doesn't actually name them.
my only critique on this post is that you don't talk about omega, which to me is the easiest way to understand why Y works
This is nice! Just like the http://www.ece.uc.edu/~franco/C511/html/Scheme/ycomb.html explanation from some time in 1999. Good to see a refresh!!
Matt, thanks for noticing the typo. Gonna fix it.
Tommy, I can only think of lambda calculus that, if I understand correctly, uses Y-combinator to recurse.
Bluestorm, I am also familiar with Y! (Y-bang) combinator (from The Seasoned Schemer book). But I am not exactly sure how it's different from Y-combinator. I am still thinking about it. The derivation is similar to Y-combinator's but it's a bit different. The book says that it works the same as Y-combinator but then gives an example which Y-combinator processes but Y! combinator goes into an infinite loop. And thanks for the in-depth comment!
I wrote the Y! out when I was reading that chapter. I posted the code here: (define Y-bang ...) and example where Y-bang fails, but Y works.
Bruce, nice explanation! Thanks for sharing!
s/And now comes they they key trick/And now comes the key trick/
Paulo, fixed. :)
You didn't close a span:
(which is the original anonymous function) to the <code><span style="color:green">mk-length</code></code>.
Also, this: "Now the code in bold looks very much like the length function we need!" refers to nonexisting boldness.
Aside from that, an interesting read.
Roman, closed now, thanks for noticing! Also put the code in blog there. :)
The problem with the Y-Combinator is, that it cannot be typed. We have Y = \s.(\x.s(x x))(\x.s(x x)), such that Ys reduces to (\x.s(x x))(\x.s (x x)) which reduces to s((\x.s(x x))(\x.s(x x))) = s Ys, thus, having Ys reducing to all sssss...sYs, which means, Ys is not strongly normalizing - and since all typed lambda terms are strongly normalizing, it cannot be typed (actually, in the same way, no fixpoint combinator can be typed).
Anyway, even strongly typed programming languages like Haskell have recursion, even though mostly through calling the function by its name. In the theory of typed lambda terms, you therefore mostly axiomatically define recursion operators, like R:(a->a)->(a->bool)->a, which applies the first argument to a until (a->bool) gets true (there are a lot of other possibilities, though). Anyway, these cannot be expressed as terms, they must be defined axiomatically.
dasuxullebt, thanks for the insightful comment.
Wow, that's a really great post! After reading The Little Schemer I had problems with understanding Y-combinator, but now everything is clear :)
very good explanation, thanks. There is a recent talk worth watching about this esoteric exercises: http://www.infoq.com/presentations/Y-Combinator
I'm among readers of 'The Little Schemer' too :-)
I just completed the functional programming course in Scala offered by Martin Odersky on Coursera and find this concepts really delicious
Good read! You see all this clearly, and I dare ask for more :-)
One intriguing thing about chapter 9 is to discover that the first way that comes to head for simplifying the value function (by factoring out (make-length make-length)) results in a function which enters an infinite loop BEFORE even starting the actual computation.
Have you got an explanation for this (apart from just saying "hey, it happens!" as the book does)? Is this something general about factoring out f(f)? or does it happen just here?
I trust it is a general problem with f(f), but I've thought long about it without coming to a conclusion. Your view will be greatly appreciated.
Thank you and have a good day.
Peter - 5 years on and your article is still really useful. I had coped with The Little Schemer up until the Y combinator - you got me through it.
For entertainment value only you might want to see my attempt in Javascript at http://jsbin.com/loromo/edit.
Thanks!
There's also an account of deriving the Y combinatory in something approximating the untyped lambda calculus in "An Introduction to Functional Programming Through Lambda Calculus" (Addison-Wesley, 1989) in Chapter 4.
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## Kyle discusses special values in primitive types: NaN, Infinity, null, undefined (void), and negative and positive zero (+0, -0).
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Transcript from the "Special Values" Lesson
[00:00:00]
>> Kyle Simpson: Now there's some values within the primitive types that we should pay attention to. Some values that we wanna look at. Specifically we're gonna look at NaN, which is transliterated as not a number. Although we'll find out in just a moment. That's a really terrible name for that.
[00:00:15] Both infinity and -infinity, not all languages have that but JavaScript definitely has a representation of both infinity and -infinity. Some of your intuition about those might not be the same like I know we have some math folks in the crowd. It may anger you to think how JavaScript interprets infinity and things like that, but like for example, you can technically add up numbers, and when you overflow the 54 bits, it just goes to infinity.
[00:00:40] That's probably super annoying to math folks. [COUGH] But these are special values that we should pay attention to. Null is a value, as well as a type. It's the only value in the null type. Undefined is the only value in the undefined type. And I mentioned here, the void operator.
[00:01:01] That's an important operator, but it doesn't get any attention. That's how you can take any value and produce the undefined value from it. So it takes whatever you put in. Say void true, void function, whatever. You're gonna get back undefined. Now there's not a lot of uses for that, but there's a couple places that I've used that in the past.
[00:01:18] And then again, something that's gonna annoy the math folks in the audience. There's both a +0 and -0. Now you may think, well, this is crazy. Right off the bat, see, we have proof that JavaScript was poorly defined. Why on Earth would they come up with this kind of craziness?
[00:01:33] None of this is JavaScript invented. Everything that I just talked about, the + and -0, and the positive and negative infinity, that is true of all IEEE 754 compliant languages. Which include Ruby, Python, and about a list of 12 or 15,000 other languages that are out there. Most languages actually these days are created to be IEEE 754 compliant.
[00:01:59] And JavaScript is just one of them. So whenever we complain about 0.1 plus 0.2 is not equal to 0.3 or whatever, has nothing to do with JavaScript. That's on IEEE. They chose the floating point representation that can't correctly represent numbers in memory that are floating point numbers. That's on IEEE, and all JavaScript did was say we're gonna be compliant to IEEE.
[00:02:22] The exact same thing is true. Some languages try to hide it a little bit more than others. But this is just all basic computing science, discrete mathematics stuff, okay? So -0, even though that sounds strange and it's counterintuitive from a mathematic perspective, is something that's required by IEEE.
[00:02:40] And it turns out there's at least one or two small used cases where it's helpful, so we're gonna look at that in just a moment. NaN. Supposedly, NaN stands for not a number. Let me tell you where NaN actually comes from. Any time you try to take a value and convert it to a number, and that conversion fails, the resultant value is to tell you this was an invalid number.
[00:03:08] That is the only place that NaN comes from. So if we look at this line 1, we see the divide operator. The divide operator specifically designed to do mathematic division, which means it needs both of its operands to be numbers. If you give it something that's not a number, we kick in this thing that we're gonna talk about a lot more.
[00:03:28] This implicit coercion thing, it's gonna try to convert it to a number. And when you try to convert "a" to a number, because we're not telling it what base, it's gonna assume base ten. And "a" is not a valid character in base ten, so it's going to fail to convert it.
[00:03:45] Which means it's going to produce this special value. We have a term in computing science we call the sentinel value. A value that represents some special characteristic. We need to give it a special label, if you will, so that you know there's something special about it. In this case, the specialness is, it's invalid.
[00:04:08] So what are some other options that we could have used? If we were to try to go back and backseat drive on design and language, what could he have done here? If you try to convert something to a number and it didn't properly convert to a number, what could we have done?
[00:04:22] You could have resulted in false. But that would be super weird, that you tried to do a numeric operation and you ended up with a Boolean value. That doesn't make any sense. Null or undefined are similarly weird results from that type of operation because they are, in themselves, distinct different primitive types.
[00:04:43] Really, the only resulting type that a number operation should produce is a number. Would we all agree with that? I hope so. If you're gonna do a numeric operation, and you need a result, by well better be a number. Otherwise, what are we doing here? Why do we even programming?
[00:05:01] Okay? Was it just like programming by guessing? No, that would be terrible design. So what's the only other option if it needs to be a number but it failed to be a number? What's the only other option besides something like the special NaN? Well we could have thrown an exception.
[00:05:16] As a matter of fact, many languages do. If you try to convert "a" to an integer in c, you get an exception. If you try to do it in Java, you get an exception. Why didn't JavaScript choose to make an exception here? Well, you may not know about the history of JavaScript, but JavaScript didn't get the try catch and exception handling until four years later.
[00:05:40] First four years of JavaScript did not have a try catch or any notion of handling exceptions. So this would have been a fatal, crash your program, kind of an error, unrecoverable. That's kind of unfriendly design. So JavaScript chose as a design aesthetic, early on, to try to be as forgiving as possible.
[00:05:57] We're gonna see this design aesthetic over and over and over again. That there was an intentional decision at the beginning of the language to say, let's try not to just frustrate the crap out of people where there's just error after error after error and there's no hope of getting anything done.
[00:06:12] JavaScript in a sense was playing kind of a marketing game. They were trying to convince people in real programming languages. Hey, come try out this toy language called JavaScript. One of the best ways I can think of to attract people is to not frustrate them by they're being a nuance like you need a semicolon here or you need a comma there or whatever.
[00:06:31] Try to be as forgiving as we possibly can while still being sane. And implicit coercion is a nod to that, as a matter of fact. It's a nod to saying instead of throwing an exception and just slapping somebody down and saying sorry, try again. We're saying let's figure out what's the best that we can do.
[00:06:46] Is there any kind of guess that we could make? You did something dumb, but could we try to recover it? Instead of garbage in, garbage out, it's like garbage in, recycled material out, right? Let's try to figure out the best that we can do here. So, we needed a value to represent that this number conversion failed.
[00:07:03] And NaN is that value. NaN is the value that says, number conversion failed. Okay? So it hasn't even happened that the divide has occurred on line 1. It's just the case that we tried to convert and that failed. So now we have a NaN divided by 2 and NaN with anything else is gonna produce NaN.
[00:07:22] NaN times NaN, NaN divided by, whatever you do, if NaN is part of it, NaN is gonna come out. It's just like the invalidating [COUGH] of all values. So of course a has the value NaN on line 3. And if we ask for the type of a, we get "number" because that's the only sensible result is that we did a numeric operation, we got a number back.
[00:07:44] But now we find the conflict with calling NaN not a number, because we say the type of not a number is number, and that makes no sense. The only proper response here, is to stop calling it not a number. That's not what it should be referred to. If I could go back and change it, I'd call it the invalid number, cuz that's what it is.
[00:08:05] I call it IN instead of NaN. I'd call it the invalid number, okay? That's what it really is. Now, NaN, again because of IEEE, this is all required by IEEE. NaN, [COUGH] there's actually a bunch of NaN values and the way you get at them differently, like the bit patterns form can be different depending on your scenario, whatever.
[00:08:28] So there's a requirement in IEEE that NaN is never equal to itself. Which sounds bizarre because it's literally the only value that I'm aware of in all of computing science, that does not have the identity property. It is not equal to itself. We're gonna actually make use of that identity property in just a moment, but NaN is never equal to itself.
[00:08:49] So if you wanna test for NaN, you're gonna need a special privileged utility. And they gave us isNaN. It's built into the global space. We get isNaN or window.isNaN or global.isNaN. Supposedly, it just tests to see whether or not the thing you pass in is the NaN value.
[00:09:04] Unfortunately, there's a fatal flaw in that original utility. A bug that we can't fix, cuz we already just discussed, as soon as we make mistakes. Let's write them in stone. We can't fix this bug. Because if I pass in isNaN and I pass in "foo". Now let me ask you a question, is "foo" a number?
[00:09:25] No. Is it not a number? You see the difference? See the problem with using not a number as our description? It is not the special, sentinel, invalid number type, and yet we're gonna get true back from isNaN because you know what isNaN does? If you pass something that's not already a number to isNaN, guess what it does?
[00:09:47] It tries to convert it to a number first. And then check to see if it resulted in the NaN. Here is a place where that conversion was a pretty terrible idea. We should never have tried to convert it. You should have just taken it and checked that and be done with it, right?
[00:10:03] But unfortunately they automatically convert first before checking so by the time it checks it, we've already converted "foo" to the NaN value and it says sure, that's a NaN. Fatal flaw. So this is a bug in isNaN. Now, thankfully, we finally decided to fix it by making a new isNaN in a different space, Number.isNaN.
[00:10:25] Number.isNaN does exactly what we want, which is it checks only the value, doesn't allow that coercion. This is what we call the polyfill pattern. You notice I've got an if statement that checks to make sure it's not already defined. And that code would only be true in older environments, so we're basically patching an older environment to have something that's been added in newer environments, like for example, ES6.
[00:10:46] And you notice what is NaN does here, it just says typeof num == "number". So at first, checks to make sure it's already of that number type and then it uses the isNaN. So really the true NaN is only gonna pass if both of those are the case.
[00:11:05] So that's an easy way of polyfilling. The new built-in Number.isNaN. If you're using regular isNaN in your checks, switch to using Number.isNaN so you don't have those bugs, okay? Cuz if you're not already accounting for this problem, the fact is you have a bug in your current JavaScript programs, even if you've never been bitten by it.
[00:11:24] Just a bug that's been built into JavaScript. Turns out there's an even easier, and perhaps even cleaner way of testing for NaN. NaN is the only value that's never equal to itself. So if we test the value to be equal to itself, NaN is the only one where that will fail.
[00:11:41] All right, so that is NaN and that's how we deal with the NaN. Again, we check to see if these values, not equal to itself is the only one without that identity property. | 2,976 | 12,503 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-21 | longest | en | 0.932725 |
https://math.answers.com/questions/What_is_the_greatest_common_factor_of_45_and_59 | 1,713,323,577,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817128.7/warc/CC-MAIN-20240417013540-20240417043540-00749.warc.gz | 357,637,597 | 47,725 | 0
# What is the greatest common factor of 45 and 59?
Updated: 8/30/2023
Wiki User
8y ago
The greatest common factor of 14 and 59 is 1.
Wiki User
13y ago
Wiki User
7y ago
The greatest common factor (GCF) refers to a factor that is COMMON to two or more numbers. You have only one number in the question! The greatest factor of any number is itself.
Wiki User
8y ago
The GCF is 1.
Wiki User
9y ago
GCF(45, 59) = 1.
Wiki User
6y ago
GCF(14, 55) = 1.
Wiki User
7y ago
It is: 1
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Q: What is the greatest common factor of 45 and 59?
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1.
The GCF is 1.
The GCF is 1.
### What is the greatest common factor of 42 59 and 74?
It is 1 because 59 is a prime number
### What is the greatest common factor of 58 and 59?
1 and that's it 177
Related questions
### What is the greatest common factor of 59 and 48?
The greatest common factor of 59 , 48 = 1
### What is the greatest common factor of 100 and 59?
1 is the greatest common factor.
The GCF is 59.
### What is the greatest common prime numbers between 45 and 60?
The prime numbers between 45 and 60 are 47, 53, and 59. The common factors of 45 and 60 are 1, 3, 5, and 15. The greatest common factor of 45 and 60 is 15.
### What is the greatest common factor of 59 and 75?
The GCF is 1... since 59 is a prime number.
1
It is: 1
1.
The GCF is 1.
The GCF is 1.
The GCF is 1.
The GCF is 1. | 463 | 1,457 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-18 | latest | en | 0.925922 |
https://www.physicsforums.com/threads/practice-latex.532700/ | 1,547,977,646,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583705091.62/warc/CC-MAIN-20190120082608-20190120104608-00206.warc.gz | 884,004,823 | 13,435 | # Practice latex
1. Sep 22, 2011
### X89codered89X
$$y_2 = y_1 * \int^t ( 1/y_1^2(s) * exp(- integral[.. to (s)]( p_1(z)dz)))$$
is there a page on this site on how to make this crap work?
Last edited by a moderator: Sep 22, 2011
2. Sep 22, 2011
### micromass
$$y_2 = y_1 * \int^t ( 1/y_1^2(s) * exp(- \int[.. to (s)]( p_1(z)dz)))$$
Click "quote" to see what I did.
Be sure not to use the wrong slash. You need / in the last tex-bracket, and no slash in the first.
3. Sep 22, 2011
### estro
I suggest looking at this thread:
https://www.physicsforums.com/showthread.php?t=386951 [Broken]
Last edited by a moderator: May 5, 2017
4. Sep 22, 2011
### Fredrik
Staff Emeritus
That post is in need of some updates. I wrote some comments about that in this post. The best way to learn is however to do what micromass suggested: Just hit the quote button next to a post with latex, to see how they did it.
5. Sep 22, 2011
### estro
Also if you're running Mac OS X, there is a cool piece of software that does exactly this, its called "LatexIt" [it is also very useful when composing texts with mathematical expressions].
http://dl.dropbox.com/u/27412797/screenshot%20copy.png [Broken]
Last edited by a moderator: May 5, 2017
6. Sep 23, 2011
### Hootenanny
Staff Emeritus
You can practice and test code here: https://www.physicsforums.com/mathjax/test/preview.html [Broken]
Last edited by a moderator: May 5, 2017
7. Sep 23, 2011
### Staff: Mentor
Also, right clicking on the LaTeX image and selecting Show Source can help.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook | 501 | 1,618 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2019-04 | latest | en | 0.882195 |
https://www.alphabetworksheetsfree.com/trace-numbers-1-10-worksheet/ | 1,701,645,958,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100518.73/warc/CC-MAIN-20231203225036-20231204015036-00507.warc.gz | 728,307,220 | 12,978 | # Trace Numbers 1-10 Worksheet
Trace Numbers 1-10 Worksheet – There is a lot of evidence to suggest that number worksheets can help children learn math. This article will focus on what is important about number worksheets for children. We will look at the advantages and various types of number worksheets.
We’ll also take a look at two case studies that show how number worksheets assisted children in improving their math skills in the shortest period.
## Purpose of Using a Numbers Worksheet and How It Helps Educators
A numbers worksheet is used to assist students with the math basics they learned in class. Students can use it as a way to practice individual practice or group activities. Students are also able to use it to evaluate their understanding of the topic.
A number worksheet can be used by educators to provide a quick and easy way to evaluate students’ understanding of specific math abilities. Teachers can also make use of these worksheets in order to ensure that students are following along to their learning objectives and make adjustments as necessary.
## 5 Effective Ways You Can Use a Numbers Worksheet to Teach Children Math
A worksheet with numbers is a sheet of paper that has rows and columns used in teaching maths to children. These worksheets are generally used in elementary school. This article will present five practical ways to use math worksheets to teach kids math.
The first option is inviting the child to copy numbers from the top row into the corresponding column. Another method is coloring each number that matches the color of its corresponding column to the left side. The third method is making a count loudly as they fill in each row either on their own or with the assistance of an adult. The fourth method is by using a number line and filling in each number which is in the same position on this line, starting with zero, continuing until they get to nine.
## Final Thoughts on the Numbers Worksheet
We hope that this blog has helped you understand the worksheet for numbers and how to use it to improve your business.
Trace Numbers 1 10 Worksheet Uploaded by admin on Friday, June 3rd, 2022. We have 3 great pictures of Trace Numbers 1 10 Worksheet. Find AlphabetWorksheetsFree.com on category Numbers.
Here we have 3 great printables about Trace Numbers 1 10 Worksheet. We hope you enjoyed it and if you want to download the pictures in high quality, simply just click the image and you will be redirected to the download page of Trace Numbers 1 10 Worksheet. | 498 | 2,527 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2023-50 | longest | en | 0.947803 |
https://number.academy/4010486 | 1,716,850,807,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059045.34/warc/CC-MAIN-20240527205559-20240527235559-00043.warc.gz | 360,987,209 | 11,244 | # Number 4010486 facts
The even number 4,010,486 is spelled 🔊, and written in words: four million, ten thousand, four hundred and eighty-six, approximately 4.0 million. The ordinal number 4010486th is said 🔊 and written as: four million, ten thousand, four hundred and eighty-sixth. The meaning of the number 4010486 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 4010486. What is 4010486 in computer science, numerology, codes and images, writing and naming in other languages
## What is 4,010,486 in other units
The decimal (Arabic) number 4010486 converted to a Roman number is (M)(M)(M)(M)(X)CDLXXXVI. Roman and decimal number conversions.
#### Time conversion
(hours, minutes, seconds, days, weeks)
4010486 seconds equals to 1 month, 2 weeks, 4 days, 10 hours, 1 minute, 26 seconds
4010486 minutes equals to 8 years, 3 months, 1 week, 6 days, 1 hour, 26 minutes
### Codes and images of the number 4010486
Number 4010486 morse code: ....- ----- .---- ----- ....- ---.. -....
Sign language for number 4010486:
Number 4010486 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
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#### Is Prime?
The number 4010486 is not a prime number.
#### Factorization and factors (dividers)
The prime factors of 4010486 are 2 * 67 * 173 * 173
The factors of 4010486 are 1, 2, 67, 134, 173, 346, 11591, 23182, 29929, 59858, 2005243, 4010486.
Total factors 12.
Sum of factors 6141012 (2130526).
#### Powers
The second power of 40104862 is 16.083.997.956.196.
The third power of 40104863 is 64.504.648.627.352.674.304.
#### Roots
The square root √4010486 is 2002,619784.
The cube root of 34010486 is 158,878697.
#### Logarithms
The natural logarithm of No. ln 4010486 = loge 4010486 = 15,204423.
The logarithm to base 10 of No. log10 4010486 = 6,603197.
The Napierian logarithm of No. log1/e 4010486 = -15,204423.
### Trigonometric functions
The cosine of 4010486 is -0,475681.
The sine of 4010486 is -0,879618.
The tangent of 4010486 is 1,849176.
## Number 4010486 in Computer Science
Code typeCode value
4010486 Number of bytes3.8MB
Unix timeUnix time 4010486 is equal to Monday Feb. 16, 1970, 10:01:26 a.m. GMT
IPv4, IPv6Number 4010486 internet address in dotted format v4 0.61.49.246, v6 ::3d:31f6
4010486 Decimal = 1111010011000111110110 Binary
4010486 Decimal = 21112202100112 Ternary
4010486 Decimal = 17230766 Octal
4010486 Decimal = 3D31F6 Hexadecimal (0x3d31f6 hex)
4010486 BASE64NDAxMDQ4Ng==
4010486 MD58056d653536a46146893de2f78abd6e9
4010486 SHA1d9011e4b87c079418cd6c72fd34d77a6dee98938
4010486 SHA2245c0ed8839b012b313850b82a2c8579acb36b00272dd77a1023406e3d
4010486 SHA256e42c302631c32b6df8b30339ff2ff35e0c28f5a1bedfae44e7816471bcd92204
4010486 SHA38497e56750855fb700f865ee6d7ed67efceffb42f193a1c801194d6f355a93c94fe18644ddb09aafca39e769e6f52e855a
More SHA codes related to the number 4010486 ...
If you know something interesting about the 4010486 number that you did not find on this page, do not hesitate to write us here.
## Numerology 4010486
### Character frequency in the number 4010486
Character (importance) frequency for numerology.
Character: Frequency: 4 2 0 2 1 1 8 1 6 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 4010486, the numbers 4+0+1+0+4+8+6 = 2+3 = 5 are added and the meaning of the number 5 is sought.
## № 4,010,486 in other languages
How to say or write the number four million, ten thousand, four hundred and eighty-six in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 4.010.486) cuatro millones diez mil cuatrocientos ochenta y seis German: 🔊 (Nummer 4.010.486) vier Millionen zehntausendvierhundertsechsundachtzig French: 🔊 (nombre 4 010 486) quatre millions dix mille quatre cent quatre-vingt-six Portuguese: 🔊 (número 4 010 486) quatro milhões e dez mil, quatrocentos e oitenta e seis Hindi: 🔊 (संख्या 4 010 486) चालीस लाख, दस हज़ार, चार सौ, छियासी Chinese: 🔊 (数 4 010 486) 四百零一万零四百八十六 Arabian: 🔊 (عدد 4,010,486) أربعة ملايين و عشرة آلاف و أربعمائة و ستة و ثمانون Czech: 🔊 (číslo 4 010 486) čtyři miliony deset tisíc čtyřista osmdesát šest Korean: 🔊 (번호 4,010,486) 사백일만 사백팔십육 Danish: 🔊 (nummer 4 010 486) fire millioner titusinde og firehundrede og seksogfirs Hebrew: (מספר 4,010,486) ארבעה מיליון ועשרת אלפים ארבע מאות שמונים ושש Dutch: 🔊 (nummer 4 010 486) vier miljoen tienduizendvierhonderdzesentachtig Japanese: 🔊 (数 4,010,486) 四百一万四百八十六 Indonesian: 🔊 (jumlah 4.010.486) empat juta sepuluh ribu empat ratus delapan puluh enam Italian: 🔊 (numero 4 010 486) quattro milioni e diecimilaquattrocentottantasei Norwegian: 🔊 (nummer 4 010 486) fire million ti tusen fire hundre og åttiseks Polish: 🔊 (liczba 4 010 486) cztery miliony dziesięć tysięcy czterysta osiemdziesiąt sześć Russian: 🔊 (номер 4 010 486) четыре миллиона десять тысяч четыреста восемьдесят шесть Turkish: 🔊 (numara 4,010,486) dörtmilyononbindörtyüzseksenaltı Thai: 🔊 (จำนวน 4 010 486) สี่ล้านหนึ่งหมื่นสี่ร้อยแปดสิบหก Ukrainian: 🔊 (номер 4 010 486) чотири мільйони десять тисяч чотириста вісімдесят шість Vietnamese: 🔊 (con số 4.010.486) bốn triệu mười nghìn bốn trăm tám mươi sáu Other languages ...
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The content of the comments is the opinion of the users and not of number.academy. It is not allowed to pour comments contrary to the laws, insulting, illegal or harmful to third parties. Number.academy reserves the right to remove or not publish any inappropriate comment. It also reserves the right to publish a comment on another topic. Privacy Policy. | 2,059 | 5,991 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-22 | latest | en | 0.78156 |
http://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-6th-edition/chapter-6-section-6-5-solving-equations-containing-rational-expressions-vocabulary-readiness-video-check-page-377/9 | 1,480,773,333,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540932.10/warc/CC-MAIN-20161202170900-00066-ip-10-31-129-80.ec2.internal.warc.gz | 501,970,855 | 40,554 | ## Intermediate Algebra (6th Edition)
Published by Pearson
# Chapter 6 - Section 6.5 - Solving Equations Containing Rational Expressions - Vocabulary, Readiness & Video Check: 9
#### Answer
$a. (x+4)(x-4)$
#### Work Step by Step
The lowest common denominator for this problem can be found by comparing the three denominators. 1. $(x-4)$ 2. $(x+4)(x-4)$ [which is $(x^{2}-16)$ factored] 3. $(x+4)$ If we multiply the first denominator by $(x+4)$ and third denominators by $(x-4)$ we will arrive at a LCD of $(x+4)(x-4)$.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 196 | 686 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2016-50 | longest | en | 0.810702 |
https://openenergymonitor.org/forum-archive/node/12365.html | 1,631,986,554,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056548.77/warc/CC-MAIN-20210918154248-20210918184248-00571.warc.gz | 495,559,881 | 5,858 | ### Measure DC voltage and current at PV panel..
Hello..i mechanical student..i need to get the information about the pv panel to complete my assignmet..from the picture that i attach, it is a true way to get the dc voltage and current??can anybody help me and correct my way if it wrong...thanks..
1 picture = The value that i measure is to get direct dc voltage or just voc??if voc, how to get dc voltage...
2 picture = i need to measure value of current using acs 712 so that i get the sensor value to show in matlab through arduino..it is a right way??how to get measure value of current??i want to compare value between sensor and measuremet value..
### Re: Measure DC voltage and current at PV panel..
First, is your P.V. panel connected to anything that is not shown on your diagram? I ask because here, P.V. panels are almost always connected to an inverter. I shall assume that as nothing is shown, there is no other connection. If there is another connection, then you must consider what happens when you connect your Arduino, which will probably be earthed via the computer you are using to program it. Get it wrong and you could destroy both Arduino and computer.
Your output voltage is 21.8 V. What is the maximum voltage that the ADC input on your Arduino can accept? Is that a problem, and if so, what are you going to do about it?
Secondly, in your second diagram (IMG_20160319_153523.jpg), what voltage are you actually measuring? I am sure it is not the voltage that you want.
There is always a problem when trying to measure both voltage and current. Either the current you measure includes the current that flows in whatever you measure the voltage with, or the voltage you measure includes the voltage dropped across whatever you measure the current with. You can either make sure that the error is too small to matter, or you can calculate the error and subtract it to give the true value.
### Re: Measure DC voltage and current at PV panel..
thanks..
i attach here a switch box of standalone system of my pv panel..
firstly, i want to measure vdc..right now, i take the terminal + and - from the pv(at incoming part below) and measure it..i get about 18 v dpend on sun..from this terminal, i connected to voltage divider so that it will drop to 5v and that value the arduino can read it..so far, the matlab software can read this value..i still confuse this way i get value vdc or voc??..i will take the sensor value and measure value to make analysis..
secondly, how i can get value of idc??it is simple just i take that terminal + and - and connected to current sensor acs 712(30amp) and to the arduino??
### Re: Measure DC voltage and current at PV panel..
That photograph makes me think that you DO have an inverter connected to the system. If an inverter is connected, it is not a standalone system and therefore my warning about creating a path for a fault stands.
What do you think you mean by "vdc" and "voc"? In normal use, Vo.c. means the open-circuit voltage, and that will be the voltage from your P.V. array with no load. It is sometimes called the "no-load voltage". It has nothing to do with whether the voltage is alternating or direct. (note: we use "V", not "v" for voltage as the unit comes from the name of Alessandro Volta - like Ampere and Kelvin, but don't confuse Kelvin with kilo.) When we need to be clear that the voltage is alternating, we write Va.c. and when it is direct, we write Vd.c. You can have and you can measure the open-circuit voltage on both a.c. and d.c. systems.
You should look at the data sheet for the ACS712 to see how to use this device. Application 4 on page 12 is almost exactly what you need, replace the diode D1 with a direct connection and omit resistor R1 and capacitor C1. If your reading is not steady, then see the notes on page 13.
### Re: Measure DC voltage and current at PV panel..
Thanks..
From you explanation, my understanding is i can divided into 2 part..for d.c system, i can get value for Vo.c,Is.c and Pmax..same like d.c system, in a.c system i can get value for Vo.c,Is.c and Pmax right??
When i do the measurement of Is.c for d.c system, there is a spark when i connected + and - multimeter to the pv terminal..i already setting multimeter to current. I already follow the step from the youtube but still has spark..What the right way to measure Is.c in d.c systems from the pv terminal??
### Re: Measure DC voltage and current at PV panel..
I think I understand now what you are doing. Are you measuring voltage and current under varying load conditions in order to determine the maximum power available from the array?
When you are connecting a circuit and the prospective current is anything more than a few mA, you will get a spark. That is exactly what happens inside a switch, relay, contactor, circuit breaker or indeed any mechanical switching device. If you do not want a spark, cover the panels to prevent the light from falling on them.
You can measure the short-circuit current of your P.V. panels because a photo-voltaic cell is, roughly speaking, a variable current source - it will deliver a current proportional to the light falling upon it.
You must never try to measure the short-circuit current of the inverter's a.c. output. Hopefully, it will sense that there is no connection to the electricity supply and refuse to work. If it is one that is designed to work independently of the mains supply - 'off-grid', then there is a strong possibility that you will blow fuses because it will behave as a voltage source and there is nothing to limit the current that might flow. | 1,318 | 5,619 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2021-39 | latest | en | 0.931365 |
https://www.aqua-calc.com/one-to-one/density/long-ton-per-liter/pound-per-cubic-inch/82 | 1,603,554,180,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107883636.39/warc/CC-MAIN-20201024135444-20201024165444-00211.warc.gz | 624,428,128 | 8,986 | # 82 long tons per liter [long tn/l] in pounds per cubic inch
## long tons/liter to pound/inch³ unit converter of density
82 long tons per liter [long tn/l] = 3 009.98 pounds per cubic inch [lb/in³]
### long tons per liter to pounds per cubic inch density conversion cards
• 82
through
106
long tons per liter
• 82 long tn/l to lb/in³ = 3 009.98 lb/in³
• 83 long tn/l to lb/in³ = 3 046.68 lb/in³
• 84 long tn/l to lb/in³ = 3 083.39 lb/in³
• 85 long tn/l to lb/in³ = 3 120.1 lb/in³
• 86 long tn/l to lb/in³ = 3 156.8 lb/in³
• 87 long tn/l to lb/in³ = 3 193.51 lb/in³
• 88 long tn/l to lb/in³ = 3 230.22 lb/in³
• 89 long tn/l to lb/in³ = 3 266.93 lb/in³
• 90 long tn/l to lb/in³ = 3 303.63 lb/in³
• 91 long tn/l to lb/in³ = 3 340.34 lb/in³
• 92 long tn/l to lb/in³ = 3 377.05 lb/in³
• 93 long tn/l to lb/in³ = 3 413.75 lb/in³
• 94 long tn/l to lb/in³ = 3 450.46 lb/in³
• 95 long tn/l to lb/in³ = 3 487.17 lb/in³
• 96 long tn/l to lb/in³ = 3 523.87 lb/in³
• 97 long tn/l to lb/in³ = 3 560.58 lb/in³
• 98 long tn/l to lb/in³ = 3 597.29 lb/in³
• 99 long tn/l to lb/in³ = 3 634 lb/in³
• 100 long tn/l to lb/in³ = 3 670.7 lb/in³
• 101 long tn/l to lb/in³ = 3 707.41 lb/in³
• 102 long tn/l to lb/in³ = 3 744.12 lb/in³
• 103 long tn/l to lb/in³ = 3 780.82 lb/in³
• 104 long tn/l to lb/in³ = 3 817.53 lb/in³
• 105 long tn/l to lb/in³ = 3 854.24 lb/in³
• 106 long tn/l to lb/in³ = 3 890.94 lb/in³
• 107
through
131
long tons per liter
• 107 long tn/l to lb/in³ = 3 927.65 lb/in³
• 108 long tn/l to lb/in³ = 3 964.36 lb/in³
• 109 long tn/l to lb/in³ = 4 001.07 lb/in³
• 110 long tn/l to lb/in³ = 4 037.77 lb/in³
• 111 long tn/l to lb/in³ = 4 074.48 lb/in³
• 112 long tn/l to lb/in³ = 4 111.19 lb/in³
• 113 long tn/l to lb/in³ = 4 147.89 lb/in³
• 114 long tn/l to lb/in³ = 4 184.6 lb/in³
• 115 long tn/l to lb/in³ = 4 221.31 lb/in³
• 116 long tn/l to lb/in³ = 4 258.01 lb/in³
• 117 long tn/l to lb/in³ = 4 294.72 lb/in³
• 118 long tn/l to lb/in³ = 4 331.43 lb/in³
• 119 long tn/l to lb/in³ = 4 368.14 lb/in³
• 120 long tn/l to lb/in³ = 4 404.84 lb/in³
• 121 long tn/l to lb/in³ = 4 441.55 lb/in³
• 122 long tn/l to lb/in³ = 4 478.26 lb/in³
• 123 long tn/l to lb/in³ = 4 514.96 lb/in³
• 124 long tn/l to lb/in³ = 4 551.67 lb/in³
• 125 long tn/l to lb/in³ = 4 588.38 lb/in³
• 126 long tn/l to lb/in³ = 4 625.08 lb/in³
• 127 long tn/l to lb/in³ = 4 661.79 lb/in³
• 128 long tn/l to lb/in³ = 4 698.5 lb/in³
• 129 long tn/l to lb/in³ = 4 735.21 lb/in³
• 130 long tn/l to lb/in³ = 4 771.91 lb/in³
• 131 long tn/l to lb/in³ = 4 808.62 lb/in³
• 132
through
156
long tons per liter
• 132 long tn/l to lb/in³ = 4 845.33 lb/in³
• 133 long tn/l to lb/in³ = 4 882.03 lb/in³
• 134 long tn/l to lb/in³ = 4 918.74 lb/in³
• 135 long tn/l to lb/in³ = 4 955.45 lb/in³
• 136 long tn/l to lb/in³ = 4 992.16 lb/in³
• 137 long tn/l to lb/in³ = 5 028.86 lb/in³
• 138 long tn/l to lb/in³ = 5 065.57 lb/in³
• 139 long tn/l to lb/in³ = 5 102.28 lb/in³
• 140 long tn/l to lb/in³ = 5 138.98 lb/in³
• 141 long tn/l to lb/in³ = 5 175.69 lb/in³
• 142 long tn/l to lb/in³ = 5 212.4 lb/in³
• 143 long tn/l to lb/in³ = 5 249.1 lb/in³
• 144 long tn/l to lb/in³ = 5 285.81 lb/in³
• 145 long tn/l to lb/in³ = 5 322.52 lb/in³
• 146 long tn/l to lb/in³ = 5 359.23 lb/in³
• 147 long tn/l to lb/in³ = 5 395.93 lb/in³
• 148 long tn/l to lb/in³ = 5 432.64 lb/in³
• 149 long tn/l to lb/in³ = 5 469.35 lb/in³
• 150 long tn/l to lb/in³ = 5 506.05 lb/in³
• 151 long tn/l to lb/in³ = 5 542.76 lb/in³
• 152 long tn/l to lb/in³ = 5 579.47 lb/in³
• 153 long tn/l to lb/in³ = 5 616.17 lb/in³
• 154 long tn/l to lb/in³ = 5 652.88 lb/in³
• 155 long tn/l to lb/in³ = 5 689.59 lb/in³
• 156 long tn/l to lb/in³ = 5 726.3 lb/in³
• 157
through
181
long tons per liter
• 157 long tn/l to lb/in³ = 5 763 lb/in³
• 158 long tn/l to lb/in³ = 5 799.71 lb/in³
• 159 long tn/l to lb/in³ = 5 836.42 lb/in³
• 160 long tn/l to lb/in³ = 5 873.12 lb/in³
• 161 long tn/l to lb/in³ = 5 909.83 lb/in³
• 162 long tn/l to lb/in³ = 5 946.54 lb/in³
• 163 long tn/l to lb/in³ = 5 983.24 lb/in³
• 164 long tn/l to lb/in³ = 6 019.95 lb/in³
• 165 long tn/l to lb/in³ = 6 056.66 lb/in³
• 166 long tn/l to lb/in³ = 6 093.37 lb/in³
• 167 long tn/l to lb/in³ = 6 130.07 lb/in³
• 168 long tn/l to lb/in³ = 6 166.78 lb/in³
• 169 long tn/l to lb/in³ = 6 203.49 lb/in³
• 170 long tn/l to lb/in³ = 6 240.19 lb/in³
• 171 long tn/l to lb/in³ = 6 276.9 lb/in³
• 172 long tn/l to lb/in³ = 6 313.61 lb/in³
• 173 long tn/l to lb/in³ = 6 350.32 lb/in³
• 174 long tn/l to lb/in³ = 6 387.02 lb/in³
• 175 long tn/l to lb/in³ = 6 423.73 lb/in³
• 176 long tn/l to lb/in³ = 6 460.44 lb/in³
• 177 long tn/l to lb/in³ = 6 497.14 lb/in³
• 178 long tn/l to lb/in³ = 6 533.85 lb/in³
• 179 long tn/l to lb/in³ = 6 570.56 lb/in³
• 180 long tn/l to lb/in³ = 6 607.26 lb/in³
• 181 long tn/l to lb/in³ = 6 643.97 lb/in³
#### Foods, Nutrients and Calories
ORIGINAL HERKIMER, CHUTTER, HEAT and POUR CHEESE SPREAD, CHEDDAR, UPC: 041711022707 contain(s) 321 calories per 100 grams or ≈3.527 ounces [ price ]
#### Gravels, Substances and Oils
CaribSea, Freshwater, Eco-Complete Cichlid, Sand weighs 1 473.7 kg/m³ (92.00009 lb/ft³) with specific gravity of 1.4737 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Difluorodibromomethane [CBr2F2] weighs 2 306.3 kg/m³ (143.97761 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-407A, liquid (R407A) with temperature in the range of -51.12°C (-60.016°F) to 60°C (140°F)
#### Weights and Measurements
The grain per square angstrom surface density measurement unit is used to measure area in square angstroms in order to estimate weight or mass in grains
Torque can be defined as a turning or twisting action of the force F upon an object.
oz/yd³ to long tn/yd³ conversion table, oz/yd³ to long tn/yd³ unit converter or convert between all units of density measurement.
#### Calculators
Calculate volume of a right circular cone and its surface area | 2,591 | 6,173 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.372056 |
https://math.stackexchange.com/questions/527811/prove-that-if-a-is-a-set-with-m-elements-b-is-a-set-with-n-elements-an | 1,713,200,277,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817002.2/warc/CC-MAIN-20240415142720-20240415172720-00810.warc.gz | 349,307,572 | 36,024 | # Prove that if $A$ is a set with $m$ elements, $B$ is a set with $n$ elements, and $A \cap B = \emptyset$, then $A \cup B$ has $m+n$ elements.
I'm working through a real analysis textbook, so I don't want the full answer. I'm only looking for a hint on this problem.
The book starts the proof like this: let $f$ be a bijection of $\mathbb{N}_m$ onto $A$, and let $g$ be a bijection of $\mathbb{N}_n$ onto $B$. Define $h$ on $\mathbb{N}_{m+n}$ as $h(i):=f(i)$ for $i=1, ..., m$, and $h(i):=g(i)$ for $i=m+1,...,m+n$. Show that $h$ is a bijection from $\mathbb{N}_{m+n}$ onto $A \cup B$.
Ok, so with that, I started out proving that $h$ is an injection into $A \cup B$. The usual method for this is to prove that if $h(i)=h(j)$, where $i,j=1,...,m+n$, then $i=j$. I think I proved this by looking at four cases.
1. $1 \leq i,j \leq m$. In this case, $h(i)=f(i)$ and $h(j)=f(j)$, so because $f$ is a bijection, we know that $f(i)=f(j)$ implies that $i=j$.
2. $1 \leq i \leq m$ and $m+1 \leq j \leq m+n$. We know that $i \neq j$ (obviously), so we want to prove that in this case, $h(i) \neq h(j)$. By definition, $h(i)=f(i)$ and $h(j)=g(j-m)$. Because $f$ is a bijection into $A$ and $g$ is a bijection into $B$, and $A \cap B = \emptyset$, we know that $f(i) \neq g(j-m)$, so $h(i) \neq h(j)$.
3. This case is identical to case 2, except that $m+1 \leq i \leq m+n$ and $1 \leq j \leq m$, so the logic is reversed.
4. The final case, where $m+1 \leq i,j \leq m+n$, is similar to case 1, because since $h(i)=g(i)$ and $h(j)=g(j)$, we can use the fact that $g$ is a bijection to prove that $g(i)=g(j)$ implies that $i=j$.
This is where I got stuck. I know I need to prove that $h$ is surjective, but am I on the right track for proving that it's injective? I think proving that $h$ is a surjection shouldn't be too hard, but I don't know if I'm approaching this part of the proof correctly.
• I think you should have $h(i) = g(i-m)$. Oct 16, 2013 at 0:05
• $a \unites b$ will definitely give you a set of $|a|+|b|$ elements since no elements are common. (Their intersection is null). Isn't this an explanation? I'm only a high schooler; please correct me if I'm wrong. Mar 19, 2020 at 16:04
Your function $h$ isn’t well-defined: the domain of $g$ is $N_n$, not $\{m+1,\ldots,m+n\}$. You could, however, let $h(i)=g(i-m)$ for $i\in\{m+1,\ldots,m+n\}$. This is the definition that you actually use in Case 2, but in Case 4 you revert to the incorrect definition. Once you repair that error, your proof that $h$ is injective will be fine. (And surjectivity of $h$ won’t be much of a problem, since $f$ and $g$ are surjective.) | 920 | 2,629 | {"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} | 4.21875 | 4 | CC-MAIN-2024-18 | latest | en | 0.895128 |
https://proxieslive.com/what-is-the-signifigance-of-the-dot-product-in-world-to-local-transformations/ | 1,603,551,246,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107883636.39/warc/CC-MAIN-20201024135444-20201024165444-00014.warc.gz | 477,533,172 | 7,011 | What is the signifigance of the Dot Product in World to Local Transformations?
Looking for help understanding why this World to Local Space function works.
I’m working my way through Buckland’s Programming Game AI By Example The following function is used in the book to convert a point from world space to an agent’s local coordinate space (its i hat and j hat are AgentHeading and AgentSide respectively)
`` inline Vector2D PointToLocalSpace(const Vector2D &point, Vector2D &AgentHeading, Vector2D &AgentSide, Vector2D &AgentPosition) { //make a copy of the point Vector2D TransPoint = point; //create a transformation matrix C2DMatrix matTransform; double Tx = -AgentPosition.Dot(AgentHeading); double Ty = -AgentPosition.Dot(AgentSide); //create the transformation matrix matTransform._11(AgentHeading.x); matTransform._12(AgentSide.x); matTransform._21(AgentHeading.y); matTransform._22(AgentSide.y); matTransform._31(Tx); matTransform._32(Ty); //now transform the vertices matTransform.TransformVector2Ds(TransPoint); return TransPoint; } ``
I’ve been learning about vectors and matrices for the last three weeks and still I’m having a lot of trouble understanding why this works in terms of vectors and matrices. What is the reason for using a a dot product in the translation coordinates? Why is it the dot product of the agent’s position and its heading? What does this dot product correspond to geometrically?
Why not just translate by the Agent’s position?
Are there any simpler ways of representing this algorithm? | 357 | 1,669 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-45 | latest | en | 0.704947 |
http://www.edupil.com/question/find-difference-present-age/ | 1,518,895,764,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891807660.32/warc/CC-MAIN-20180217185905-20180217205905-00368.warc.gz | 430,649,768 | 15,958 | # Find The Difference Of Their Present Age
Two years ago, Rahu was three times as old as his son and two years hence, after two years, twice his age will be equal to five times that of his son. Difference of their present age
1. 24 years
2. 22 years
3. 38 years
4. 28 years
Anurag Mishra Professor Asked on 20th July 2015 in
Explanation:-
Let be the age of the Rahu = X years
And the age of his son = Y years
According to the question,
In first condition-
X – 2 = 3 (Y – 2)
X – 2 = 3Y – 6
X – 3 Y = – 6 + 2
X – 3Y = – 4 ………………………. (1)
In second condition-
2 (X + 2) = 5 (Y + 2)
2X + 4 = 5Y + 10
2X – 5Y = 10 – 4
2 X – 5 Y = 6 …………………………… (2)
Solve equation (1) & (2)
Y = 14
Put the value of Y in equation (1)
X – 3 x 14 + – 4
X – 42 = – 4
X = – 4 + 42
X = 38
Thus, the present age of Rahu = 38 years & the present age of his son = 14 years
The difference their age = 38 – 14 = 24 years
Hence, the answer is (1) 24 years.
Anurag Mishra Professor Answered on 20th July 2015. | 389 | 1,021 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2018-09 | latest | en | 0.948021 |
https://brainmass.com/economics/investments/51031 | 1,398,240,163,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1398223201753.19/warc/CC-MAIN-20140423032001-00488-ip-10-147-4-33.ec2.internal.warc.gz | 1,031,794,596 | 7,831 | # GDP/Equilibrium and MPC
GDP/Equilibrium and MPC questions:
---
C = 250 Billion +. 80GDP
Please use equation above for following questions:
1. If the planned investment is \$200 billion, the equilibrium level of GDP is:
2. If the equilibrium is \$2000 billion, autonomous investment is:
--------------------------------------------------------------------------------------
C = \$100 + .80 (GDP); I = \$20
3. Private sector equilibrium occurs at GDP of:
4. The equation for private sector equilibrium can be expressed as:
5. What is the value of the marginal propensity to consume?
6. What is the value of the multiplier
7. If planned investment increases by \$10, by how much will equilibrium GDP increase?
8. The new equilibrium GDP after the \$10 increase in investment is:
9. Assume government decided to spend \$30. Equilibrium GDP at investment of \$30 and government spending of \$30 is:
---
#### Solution Summary
What is the value of the marginal propensity to consume? | 211 | 993 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2014-15 | latest | en | 0.826776 |
http://physics.stackexchange.com/questions/60357/beta-decay/60382 | 1,469,414,081,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824201.56/warc/CC-MAIN-20160723071024-00196-ip-10-185-27-174.ec2.internal.warc.gz | 198,828,951 | 20,422 | # $\beta^+$ decay
We've been discussing radioactive decay at school, and I grasped everything except for $\beta +$ decay. When I googled radioactive decay, I immediately found out they dumbed down radioactive decay for us, which is probably why they didn't care to explain what they did, they just showed some calculations. We have never discussed neutrino' s and antineutrino's, they leave that out of the equation, which is no problem since they have negligible mass and no charge.
So we've been taught that a proton and electron form a neutron, which I have also discovered is not true (I'm discovering a lot of new things :P). I learned that this is caused by the spontaneous (?) change from up to down quarks and vice versa.
However, at school I must keep to the 'rules' and by those rules I don't really understand $\beta +$ decay. I see $\beta -$ decay as follows:
An electron flees a neutron and leaves a proton. That's why you get an atom with a higher atomic number.
However, how would this work with beta + decay? Can it even be dumbed down to this kind of high school thinking?
-
As a kid, I believed that a neutron "was" composed of a proton, electron, and a (negligible) antineutrino as well. But that's really wrong. The decay genuinely transforms the identity of a neutron to that of proton. The neutron and proton are equally elementary and equally composite, in fact. Microscopically, a down-quark gets changed to an up-quark (in proton) and electron and antineutrino. Again, the down-quark isn't "composed" of the three products: it's equally elementary as the up-quark.
The $\beta^+$ decay is a bit of a problem for the wrong "composite neutron" picture. But the only "coherent" picture is that you must simultaneously imagine a proton to be a composite of a neutron, positron, and a neutrino. This picture sort of contradicts the original $\beta^-$ composite picture for the neutron (unless you are ready to tolerate an infinite hierarchy of compositeness) but because there is really a sort of symmetry between the proton and the neutron, and between the $\beta^+$ and $\beta^-$ decay, the composite proton (containing a neutron and a positron) is the only legitimate picture you may add for the $\beta^-$ decay.
-
The problem for these composite models is that due to the Heisenberg uncertainty principle, the electron would have a kinetic energy greater than the rest mass of the neutron. – Ben Crowell Apr 7 '13 at 19:44
This is an excellent question and here is a school level fudgy answer.
Observe that $\beta^{+}$ decay never occurs for free particles, but for matter inside a nucleus. In this way, the lighter proton can effectively steal any energy that it might need to decay to the heavier set of neutron/lepton/neutrino. Protons are very, very stable particles. Their direct decay has never been observed. So $\beta^+$ and $\beta^-$ decay are essentially different.
-
It works with the same ideas, however there is the condition that since the proton is a lighter particle it needs more energy to be able to emit a positron(this might be in kinetic energy, or more likely, in some nucleus it may have lots of excess energy (its excited)). You should remember that everything wants to lose energy to be in a more stable state, so a ball at the top of the hill has an excess of gravitational potential energy and loses it when it rolls down. Similarly a neutron has a tiny more more energy than it would like and so would like to lose this by beta minus decay when it decays to the very stable proton. For the proton, if for some reason it is given lots of energy, it will have more energy than is contained in a neutron, so it wants to lose energy to be in a more stable state (in case of beta plus a neutron).
-
It is not true that nuclei that decay by $\beta^+$ re necessarily excited. It is enough that they can get to a lower state by doing so. – dmckee Apr 8 '13 at 3:27
But when the proton does decay, it does have an excess of energy? – Dmist Apr 8 '13 at 8:40
You can't really say that the proton does or does not have an excess of energy at all. The parent nucleus has more energy (mass) that the daughter nucleus plus the positron plus the neutrino (and including their kinetic energy). – dmckee Apr 8 '13 at 12:23
Cheers that makes sense – Dmist Apr 8 '13 at 13:52
$\beta^+$-decay can be explained as a result of nuclear protons collision:
$p^+ p^+\to p^+ p^+ W^- W^+ \to (p^+ W^-)\: p^+ W^+\to n\: p^+ e^+ \nu_e$
Compare to $\beta^-$-decay:
$n \to p^+ W^- \to p^+ e^- \bar{\nu}_e$
- | 1,103 | 4,545 | {"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} | 2.9375 | 3 | CC-MAIN-2016-30 | latest | en | 0.962521 |
https://www.mint.com/education/lesson2/ | 1,398,384,010,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1398223207046.13/warc/CC-MAIN-20140423032007-00223-ip-10-147-4-33.ec2.internal.warc.gz | 1,013,253,771 | 6,308 | # Tools to learn
Our program teaches middle school students the value of
worksheets, games and handouts below and your students
will be on their way to financial literacy.
education@mint.com
## Lesson 2: Under the Bed or in the Bank?
How banks function is at the core of Lesson 2, with students using straightforward computation to learn about savings, simple interest, and more.
### Objective
Students will learn about savings, simple interest, compound interest, and the function of banks.
### Materials
Student Worksheet 2, pen or pencil, calculator
### Directions
1. Distribute a copy of Student Worksheet 2 to each student.
2. Tell students that after they purchase the mountain bike, they have money left over from their weekly paycheck.
3. Have students solve question 1 and share answers with the class. Ask the class: "What would be a better way to save money than putting it in a box under your bed?" (Answers: put it in a bank; invest; buy real estate.)
4. Ask students why it's a good idea for them to put money in a bank. Explain to students that putting money in the bank means that they can earn interest. Interest is an amount of money the bank will pay you for letting it hold your money. The amount of interest you earn depends on the bank and the amount of money you put in the bank.
5. Complete question 2 as a class to help students understand how to compute interest. Depending on your students' experience you may need to review decimals and how they relate to percentages.
6. Once students are comfortable calculating interest, direct them to question 3. Explain that the factors to consider when saving money in a bank are: how much money you put in, how long you leave it there, and the interest rate. Have students determine how much interest they would earn from each bank.
7. Tell students there is something called compound interest, where a bank pays interest on the money you started with and on the interest you've earned. So if Bank D pays 1% interest every month, you will have \$1,020.10 after two months. Next month, the bank will pay interest on \$1,020.10. This means that the amount earned from interest will go up every month!
8. Have students review question 4 and complete the problem. Encourage students to work in pairs if they need help solving the problem. Once they have finished, ask students to complete Part II of the worksheet. When they are done, have students to continue their financial adventures through an online game!
##### Answer key for Student Worksheet 2: Under the Bed or in the Bank?
1. \$244.12 - \$125.79 = \$118.33 saved per week x
4 weeks = \$473.32 total saved.
2. \$1,000 x .01 = \$10 interest + \$1,000 = \$1,010.
3. Bank A: \$1,000 x .005 = \$5;
Bank B: \$1,000 x .02 = \$20;
Bank C: \$1,000 x .015 = \$15.
4. At the end of one year, students would have \$1,126.83 in their account.
5. Review: Interest is an amount of money that a bank will pay you for letting it hold your money. Banks pay interest to you because they use your money to make their own investments. It's like they're "renting" your money. | 719 | 3,089 | {"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} | 4.0625 | 4 | CC-MAIN-2014-15 | latest | en | 0.947463 |
https://www.math-only-math.com/to-measure-the-length-of-a-line-segment.html | 1,720,884,258,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514510.7/warc/CC-MAIN-20240713150314-20240713180314-00096.warc.gz | 707,392,715 | 13,026 | # To Measure the Length of a Line-segment
Here we will learn to measure the length of a line-segment.
Let there be a line-segment AB.
We have to measure its length.
The scale is placed along the line-segment putting its zero (0) mark at A. We see the end B is at the 3 cm mark of the scale. So the length of the line-segment AB = 3 cm.
To draw a line-segment of a given length say;
(i) 6 cm,
(ii) 6 cm 5 mm,
(iii) 8 cm 6 mm
(i) 20 cm scale is placed flat on the paper.
A point A is marked in front of the zero mark of the scale. Now in front of the 6 cm mark, another point B is marked. A line-segment AB is drawn from A to B.
Now the scale is removed.
AB is the required line-segment.
(ii) Similarly the. line-segments of length 6 cm 5 mm and 8 cm 6 mm are drawn. (Try to draw to draw the line)
Related Concepts
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## Recent Articles
1. ### Types of Fractions |Proper Fraction |Improper Fraction |Mixed Fraction
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4. ### Conversion of Improper Fractions into Mixed Fractions |Solved Examples
Jul 12, 24 12:59 AM
To convert an improper fraction into a mixed number, divide the numerator of the given improper fraction by its denominator. The quotient will represent the whole number and the remainder so obtained… | 544 | 2,192 | {"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} | 4.4375 | 4 | CC-MAIN-2024-30 | latest | en | 0.897765 |
https://www.lexic.us/definition-of/Newton%27s_second_law | 1,542,514,051,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743963.32/warc/CC-MAIN-20181118031826-20181118053826-00519.warc.gz | 912,239,473 | 9,019 | ### Definition of Newton's second law
1. Noun. The rate of change of momentum is proportional to the imposed force and goes in the direction of the force.
### Definition of Newton's second law
1. Noun. (physics) Newton's observation that the rate of change of the momentum of a body is directly proportional to, and in the same direction as, the net force acting on it. ¹
¹ Source: wiktionary.com
### Newton's Second Law Pictures
Click the following link to bring up a new window with an automated collection of images related to the term: Newton's Second Law Images
### Lexicographical Neighbors of Newton's Second Law
NewportNewport NewsNewquayNewspeakNewtonNewton's cradleNewton's diskNewton's first lawNewton's first law of motionNewton's flaming laser sword Newton's interference coloursNewton's lawNewton's law of gravitationNewton's law of motionNewton's ringsNewton's second lawNewton's second law of motionNewton's theory of gravitationNewton's third lawNewton's third law of motion Newton hearingNewtonianNewtonian constant of gravitationNewtonian fluidNewtonian mechanicsNewtonian reflectorNewtonian telescopeNewtonmasNewtonmasesNewtown Wonder
### Literary usage of Newton's second law
Below you will find example usage of this term as found in modern and/or classical literature:
1. Introductory Course of Natural Philosophy for the Use of High Schools and by Adolphe Ganot, William Guy Peck, Levi Sumner Bubank, James Irvin Hanson (1881)
"Give Newton-s Second Law. What three elements determine a force '. Define each. How represented ? 39. Define simple and compound motion. ..."
2. ELements of Physics by George Arthur Hoadley (1908)
"Newton-s Second Law means that any force acting upon a body produces its own effect, ... From Newton-s Second Law it is seen that the measurement of a force ..."
3. Mechanics, Molecular Physics and Heat: A Twelve Weeks' College Course by Robert Andrews Millikan (1903)
"The equation/= ma is then merely the statement of Newton,s Second Law. Note, however, that it holds in this form only ..."
4. The Bibliographer's Manual of English Literature: Containing an Account of by William Thomas Lowndes, Henry George Bohn (1865)
"... Newton s second Law of Motion, by the Rev. William Lud- ktm. Lond. 1780, 8vo. An Illustration of Sir Isaac Newton • Method of Reasoning, ..."
5. A Treatise on the Screw Propeller: With Various Suggestions of Improvement by John Bourne (1852)
"In Newton-s " Second Law of Motion" it is maintained that " the change or alteration of motion produced in a body by the action of any external force, ..."
6. Modern Practice in Mining by Richard Augustine Studdert Redmayne (1911)
"And, as by Newton-s second law of motion, change of motion is proportional to the acting force, and takes place in the direction of the straight line in ..." | 634 | 2,827 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2018-47 | longest | en | 0.862673 |
http://www.blogdesignsbydani.com/topic/world-time-gmt | 1,582,602,709,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146004.9/warc/CC-MAIN-20200225014941-20200225044941-00043.warc.gz | 167,622,281 | 7,678 | # World Time gmt
Universal time gmt
The Coordinated Universal Time (UTC) is used as the official world reference for time. I' m astrology: Calculation of the graph - Greenwich Mean Time GMT
Also, it may be the case that if an astrologer does not spend much time creating maps by hands, he will never fully grasp it. Information you need is the date, time and place of delivery. Place of nativity is given in degrees of longitudes and latitudes. The Universal Time, or Greenwich Mean Time, is used to determine the position of the planet.
Universal time? World Time is the time in Greenwich Mean Time for the specified civilian time, taking summer time into consideration. Mary, for example, was conceived on May 7, 1982 at 11:17 a.m. AM, New Berlin, New York, New York, USA. Which means it's the meridian is 75 deg, 20 min due West of Greenwich.
north of the equator. Notice: From Greenwich, the time differential is one extrahour for each 15 longitudes. The Eastern Standard Time (EST) for New York is 75 degree longitudinally long. If it' s 7:00 in New York, it' s 12:00 in Greenwich because New York is just east of Greenwich.
In the case of westerly longitude, sum the time differential with the locale time. And for the easterly longitude, deduct the time differential. We' ll have to find the time in Greenwich (GMT) to compute the graph. Sunrise in the East, the time in the east is sooner than in Greenwich, and the time in the west is later than in Greenwich.
When you look at the above graph, you will see that the solar radiation is in the east. Greenwich has not yet achieved it, so time in the East is sooner than time in Greenwich. Greenwich will be hit by the rays of the light before it arrives at the places to the west, so the places to the west are a little later than Greenwich.
So, if you are living east of Greenwich, the dawn is early and you get your lunch before Greenwich. So if you are in the West, you will have your lunch later than Greenwich. Therefore, the Greenwich Mean Time (GMT) is later for those in the West and sooner for those in the East.
The GMT of delivery is also referred to as universal time. If you were thus birthed at 7 o'clock in the morning at latitude 75 degrees W, it would be midday in Greenwich (7 o'clock +5 hours). Similarly, if you were birthed at 17:00 on latitude 75 East, it would be midday in Greenwich (17:00 - 17 hours).
In our example Date of Birth, we need to put in five additional lessons to get the GMT: So the time of delivery, so far, is: It is not GMT yet, as the watches may have been converted, so we need to verify DST before we can get GMT time.
There is no star changing when we switch our watches, so any daylight saving time (or other changes) must be taken into account. Sometimes in some coutries the time is modified to use the lighting, and this is referred to as summer time. If this is in effect, you will often have to deduct an extra 1h ( but sometimes 1/2h or 2h ).
Daylight saving watches are brought forward one extra hour, so we have to take this off if there is daylight saving time. For our example, daylight saving time existed, i.e. the GMT time of birth: These are the times in Greenwich when Mary was firstborn. Whether the daylight saving time was in use, we can find out by means of a specific map or this URL:
I need you to know the time zones for childbirth. UT is obtained by addition of the time zones if the childbirth took place west of Greenwich, or by subtraction of the time zones if the childbirth took place east of Greenwich. You' ll also have to deduct summer time. Latitude is 178E01 and time is 12.
Summer time was not used. Childbirth was in the East, so we're subtracting the time area. This means the world time was 2 a.m. (time on the Greenwich meridian). Childbirth took place in the east, so that the time zones are deducted from the time of childbirth (civil time):
Put in 24hrs so we can deduct an hour: This means that on 21 May 1924 at 23:15 - the world time here is the date before the date of childbirth. Childbirth took place on Tuesday, June 26, 1990 at 20:03 in New York, NY, United States of America. Timezone is 5 and summer time was running.
Given the west birthing, the time zones are deducted. Considering the UT rescues the baby in summer: Since the time is greater than 24, we deduct 24 and sum one to the date. That' the UT's time of birth: This means that the UT in Greenwich is 3 min after 12:00 on June 27, 1990, the following one!
It'?s the exact time of day. For every two hours, the Earth's rays move 15º. That'?s how it goes one grade in four-minute time. Therefore it travels 4_? minutes of time per 20 seconds of sheet. Your average mean LOCT is therefore: Converting to GMT, we are adding the addition for the West meridian and subtracting it for the East meridian.
Conversely, when switching to LTC, we deduct the birthrate to the west of the time zones and sum it for birthrates east of the time zones. Mary's GMT birthday is therefore the above time with the time zones added, or: This means: 15 hrs 15 min and 40 sec (7 May 1982); This is the Greenwich period in which Maria was conceived.
May 7, 1982 at 11:17 AM, New Berlin, New York, New York, USA. It'?s 75° long, 20 min due west of Greenwich. north of the Equator. It'?s 75 and 20 min. longitude: Now, for each grade the Earth is traveling, the time changes by 4 min.
Thus, the distance between Marias and Greenwich is 75 x 4 min. That' 300 mins or 5 hrs. And the other "bit" is the 20 minute bow, or a third of a grade. That'?s a third of four mins. So, the time differential is:
In order to obtain GMT, we sum this number to their natal time and take into consideration daylight saving time (summer time, daylight saving time, daylight saving time). That makes 3:18:20, that's the time in Greenwich when she was birthed. It is not yet ready because we have to take into consideration the fact that Mary was not exactly birthed on the time axis (75 degrees), but on a latitude of 75 degree 20 min.
Now she was birthed with 10 h rs 7 min (taking into account summer time), but she was birthed with 20 min arcs westward of the time zones (at 75 degrees). At the time she was conceived, time on the 75th meridian was the time she indicated. That was the time at 75 deg, but 20 min bow westwards of this area, it was only: 10 h 15 min and 40 sec.
Every single hours the temperature of the moon rises 15°C. It'?ll leave in a minute, 60/15 or 4 mins. In order to drive 20 min. of sheets you need 4 x 20/60 min. or 1 min. 20 seconds. That' her native time of nativity. In order to get the time in Greenwich, we have to insert the time zones, which in this case is 5hrs.
That' how her GMT was born: locally: Fifteen hour, fifteen minute, forty second. Average time locally is similar to UT except that average time locally is calculated from the precise degree of longevity of the child's childbirth and not from the time area. Mean delivery time is the adjusted universal time of delivery.
UT is used to determine the planet's location and UT is used to determine the house's increment and hump. | 1,762 | 7,119 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | longest | en | 0.925065 |
https://docs.scipy.org/doc/scipy-0.8.x/reference/tutorial/interpolate.html | 1,679,479,862,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943809.22/warc/CC-MAIN-20230322082826-20230322112826-00542.warc.gz | 258,897,035 | 10,353 | # Interpolation (scipy.interpolate)¶
There are two general interpolation facilities available in SciPy. The first facility is an interpolation class which performs linear 1-dimensional interpolation. The second facility is based on the FORTRAN library FITPACK and provides functions for 1- and 2-dimensional (smoothed) cubic-spline interpolation. There are both procedural and object-oriented interfaces for the FITPACK library.
## Linear 1-d interpolation (interp1d)¶
The interp1d class in scipy.interpolate is a convenient method to create a function based on fixed data points which can be evaluated anywhere within the domain defined by the given data using linear interpolation. An instance of this class is created by passing the 1-d vectors comprising the data. The instance of this class defines a __call__ method and can therefore by treated like a function which interpolates between known data values to obtain unknown values (it also has a docstring for help). Behavior at the boundary can be specified at instantiation time. The following example demonstrates it’s use.
>>> import numpy as np
>>> from scipy import interpolate
>>> x = np.arange(0,10)
>>> y = np.exp(-x/3.0)
>>> f = interpolate.interp1d(x, y)
>>> xnew = np.arange(0,9,0.1)
>>> import matplotlib.pyplot as plt
>>> plt.plot(x,y,'o',xnew,f(xnew),'-')
## Spline interpolation in 1-d: Procedural (interpolate.splXXX)¶
Spline interpolation requires two essential steps: (1) a spline representation of the curve is computed, and (2) the spline is evaluated at the desired points. In order to find the spline representation, there are two different ways to represent a curve and obtain (smoothing) spline coefficients: directly and parametrically. The direct method finds the spline representation of a curve in a two- dimensional plane using the function splrep. The first two arguments are the only ones required, and these provide the and components of the curve. The normal output is a 3-tuple, , containing the knot-points, , the coefficients and the order of the spline. The default spline order is cubic, but this can be changed with the input keyword, k.
For curves in -dimensional space the function splprep allows defining the curve parametrically. For this function only 1 input argument is required. This input is a list of -arrays representing the curve in -dimensional space. The length of each array is the number of curve points, and each array provides one component of the -dimensional data point. The parameter variable is given with the keword argument, u, which defaults to an equally-spaced monotonic sequence between and . The default output consists of two objects: a 3-tuple, , containing the spline representation and the parameter variable
The keyword argument, s , is used to specify the amount of smoothing to perform during the spline fit. The default value of is where is the number of data-points being fit. Therefore, if no smoothing is desired a value of should be passed to the routines.
Once the spline representation of the data has been determined, functions are available for evaluating the spline (splev) and its derivatives (splev, spalde) at any point and the integral of the spline between any two points ( splint). In addition, for cubic splines ( ) with 8 or more knots, the roots of the spline can be estimated ( sproot). These functions are demonstrated in the example that follows.
>>> import numpy as np
>>> import matplotlib.pyplot as plt
>>> from scipy import interpolate
Cubic-spline
>>> x = np.arange(0,2*np.pi+np.pi/4,2*np.pi/8)
>>> y = np.sin(x)
>>> tck = interpolate.splrep(x,y,s=0)
>>> xnew = np.arange(0,2*np.pi,np.pi/50)
>>> ynew = interpolate.splev(xnew,tck,der=0)
>>> plt.figure()
>>> plt.plot(x,y,'x',xnew,ynew,xnew,np.sin(xnew),x,y,'b')
>>> plt.legend(['Linear','Cubic Spline', 'True'])
>>> plt.axis([-0.05,6.33,-1.05,1.05])
>>> plt.title('Cubic-spline interpolation')
>>> plt.show()
Derivative of spline
>>> yder = interpolate.splev(xnew,tck,der=1)
>>> plt.figure()
>>> plt.plot(xnew,yder,xnew,np.cos(xnew),'--')
>>> plt.legend(['Cubic Spline', 'True'])
>>> plt.axis([-0.05,6.33,-1.05,1.05])
>>> plt.title('Derivative estimation from spline')
>>> plt.show()
(png, pdf)
Integral of spline
>>> def integ(x,tck,constant=-1):
>>> x = np.atleast_1d(x)
>>> out = np.zeros(x.shape, dtype=x.dtype)
>>> for n in xrange(len(out)):
>>> out[n] = interpolate.splint(0,x[n],tck)
>>> out += constant
>>> return out
>>>
>>> yint = integ(xnew,tck)
>>> plt.figure()
>>> plt.plot(xnew,yint,xnew,-np.cos(xnew),'--')
>>> plt.legend(['Cubic Spline', 'True'])
>>> plt.axis([-0.05,6.33,-1.05,1.05])
>>> plt.title('Integral estimation from spline')
>>> plt.show()
(png, pdf)
Roots of spline
>>> print interpolate.sproot(tck)
[ 0. 3.1416]
Parametric spline
>>> t = np.arange(0,1.1,.1)
>>> x = np.sin(2*np.pi*t)
>>> y = np.cos(2*np.pi*t)
>>> tck,u = interpolate.splprep([x,y],s=0)
>>> unew = np.arange(0,1.01,0.01)
>>> out = interpolate.splev(unew,tck)
>>> plt.figure()
>>> plt.plot(x,y,'x',out[0],out[1],np.sin(2*np.pi*unew),np.cos(2*np.pi*unew),x,y,'b')
>>> plt.legend(['Linear','Cubic Spline', 'True'])
>>> plt.axis([-1.05,1.05,-1.05,1.05])
>>> plt.title('Spline of parametrically-defined curve')
>>> plt.show()
(png, pdf)
## Spline interpolation in 1-d: Object-oriented (UnivariateSpline)¶
The spline-fitting capabilities described above are also available via an objected-oriented interface. The one dimensional splines are objects of the UnivariateSpline class, and are created with the and components of the curve provided as arguments to the constructor. The class defines __call__, allowing the object to be called with the x-axis values at which the spline should be evaluated, returning the interpolated y-values. This is shown in the example below for the subclass InterpolatedUnivariateSpline. The methods integral, derivatives, and roots methods are also available on UnivariateSpline objects, allowing definite integrals, derivatives, and roots to be computed for the spline.
The UnivariateSpline class can also be used to smooth data by providing a non-zero value of the smoothing parameter s, with the same meaning as the s keyword of the splrep function described above. This results in a spline that has fewer knots than the number of data points, and hence is no longer strictly an interpolating spline, but rather a smoothing spline. If this is not desired, the InterpolatedUnivariateSpline class is available. It is a subclass of UnivariateSpline that always passes through all points (equivalent to forcing the smoothing parameter to 0). This class is demonstrated in the example below.
The LSQUnivarateSpline is the other subclass of UnivarateSpline. It allows the user to specify the number and location of internal knots as explicitly with the parameter t. This allows creation of customized splines with non-linear spacing, to interpolate in some domains and smooth in others, or change the character of the spline.
>>> import numpy as np
>>> import matplotlib.pyplot as plt
>>> from scipy import interpolate
InterpolatedUnivariateSpline
>>> x = np.arange(0,2*np.pi+np.pi/4,2*np.pi/8)
>>> y = np.sin(x)
>>> s = interpolate.InterpolatedUnivariateSpline(x,y)
>>> xnew = np.arange(0,2*np.pi,np.pi/50)
>>> ynew = s(xnew)
>>> plt.figure()
>>> plt.plot(x,y,'x',xnew,ynew,xnew,np.sin(xnew),x,y,'b')
>>> plt.legend(['Linear','InterpolatedUnivariateSpline', 'True'])
>>> plt.axis([-0.05,6.33,-1.05,1.05])
>>> plt.title('InterpolatedUnivariateSpline')
>>> plt.show()
LSQUnivarateSpline with non-uniform knots
>>> t = [np.pi/2-.1,np.pi/2-.1,3*np.pi/2-.1,3*np.pi/2+.1]
>>> s = interpolate.LSQUnivariateSpline(x,y,t)
>>> ynew = s(xnew)
>>> plt.figure()
>>> plt.plot(x,y,'x',xnew,ynew,xnew,np.sin(xnew),x,y,'b')
>>> plt.legend(['Linear','LSQUnivariateSpline', 'True'])
>>> plt.axis([-0.05,6.33,-1.05,1.05])
>>> plt.title('Spline with Specified Interior Knots')
>>> plt.show()
## Two-dimensional spline representation: Procedural (bisplrep)¶
For (smooth) spline-fitting to a two dimensional surface, the function bisplrep is available. This function takes as required inputs the 1-D arrays x, y, and z which represent points on the surface The default output is a list whose entries represent respectively, the components of the knot positions, the coefficients of the spline, and the order of the spline in each coordinate. It is convenient to hold this list in a single object, tck, so that it can be passed easily to the function bisplev. The keyword, s , can be used to change the amount of smoothing performed on the data while determining the appropriate spline. The default value is where is the number of data points in the x, y, and z vectors. As a result, if no smoothing is desired, then should be passed to bisplrep .
To evaluate the two-dimensional spline and it’s partial derivatives (up to the order of the spline), the function bisplev is required. This function takes as the first two arguments two 1-D arrays whose cross-product specifies the domain over which to evaluate the spline. The third argument is the tck list returned from bisplrep. If desired, the fourth and fifth arguments provide the orders of the partial derivative in the and direction respectively.
It is important to note that two dimensional interpolation should not be used to find the spline representation of images. The algorithm used is not amenable to large numbers of input points. The signal processing toolbox contains more appropriate algorithms for finding the spline representation of an image. The two dimensional interpolation commands are intended for use when interpolating a two dimensional function as shown in the example that follows. This example uses the numpy.mgrid command in SciPy which is useful for defining a “mesh-grid “in many dimensions. (See also the numpy.ogrid command if the full-mesh is not needed). The number of output arguments and the number of dimensions of each argument is determined by the number of indexing objects passed in numpy.mgrid.
>>> import numpy as np
>>> from scipy import interpolate
>>> import matplotlib.pyplot as plt
Define function over sparse 20x20 grid
>>> x,y = np.mgrid[-1:1:20j,-1:1:20j]
>>> z = (x+y)*np.exp(-6.0*(x*x+y*y))
>>> plt.figure()
>>> plt.pcolor(x,y,z)
>>> plt.colorbar()
>>> plt.title("Sparsely sampled function.")
>>> plt.show()
Interpolate function over new 70x70 grid
>>> xnew,ynew = np.mgrid[-1:1:70j,-1:1:70j]
>>> tck = interpolate.bisplrep(x,y,z,s=0)
>>> znew = interpolate.bisplev(xnew[:,0],ynew[0,:],tck)
>>> plt.figure()
>>> plt.pcolor(xnew,ynew,znew)
>>> plt.colorbar()
>>> plt.title("Interpolated function.")
>>> plt.show()
(png, pdf)
## Two-dimensional spline representation: Object-oriented (BivariateSpline)¶
The BivariateSpline class is the 2-dimensional analog of the UnivariateSpline class. It and its subclasses implement the FITPACK functions described above in an object oriented fashion, allowing objects to be instantiated that can be called to compute the spline value by passing in the two coordinates as the two arguments.
## Using radial basis functions for smoothing/interpolation¶
Radial basis functions can be used for smoothing/interpolating scattered data in n-dimensions, but should be used with caution for extrapolation outside of the observed data range.
### 1-d Example¶
This example compares the usage of the Rbf and UnivariateSpline classes from the scipy.interpolate module.
>>> import numpy as np
>>> from scipy.interpolate import Rbf, InterpolatedUnivariateSpline
>>> import matplotlib.pyplot as plt
>>> # setup data
>>> x = np.linspace(0, 10, 9)
>>> y = np.sin(x)
>>> xi = np.linspace(0, 10, 101)
>>> # use fitpack2 method
>>> ius = InterpolatedUnivariateSpline(x, y)
>>> yi = ius(xi)
>>> plt.subplot(2, 1, 1)
>>> plt.plot(x, y, 'bo')
>>> plt.plot(xi, yi, 'g')
>>> plt.plot(xi, np.sin(xi), 'r')
>>> plt.title('Interpolation using univariate spline')
>>> # use RBF method
>>> rbf = Rbf(x, y)
>>> fi = rbf(xi)
>>> plt.subplot(2, 1, 2)
>>> plt.plot(x, y, 'bo')
>>> plt.plot(xi, fi, 'g')
>>> plt.plot(xi, np.sin(xi), 'r')
>>> plt.title('Interpolation using RBF - multiquadrics')
>>> plt.show()
### 2-d Example¶
This example shows how to interpolate scattered 2d data.
>>> import numpy as np
>>> from scipy.interpolate import Rbf
>>> import matplotlib.pyplot as plt
>>> from matplotlib import cm
>>> # 2-d tests - setup scattered data
>>> x = np.random.rand(100)*4.0-2.0
>>> y = np.random.rand(100)*4.0-2.0
>>> z = x*np.exp(-x**2-y**2)
>>> ti = np.linspace(-2.0, 2.0, 100)
>>> XI, YI = np.meshgrid(ti, ti)
>>> # use RBF
>>> rbf = Rbf(x, y, z, epsilon=2)
>>> ZI = rbf(XI, YI)
>>> # plot the result
>>> n = plt.normalize(-2., 2.)
>>> plt.subplot(1, 1, 1)
>>> plt.pcolor(XI, YI, ZI, cmap=cm.jet)
>>> plt.scatter(x, y, 100, z, cmap=cm.jet)
>>> plt.title('RBF interpolation - multiquadrics')
>>> plt.xlim(-2, 2)
>>> plt.ylim(-2, 2)
>>> plt.colorbar() | 3,348 | 12,982 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2023-14 | latest | en | 0.865247 |
https://currencyconverts.com/cny/bgn/2185 | 1,686,409,606,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657720.82/warc/CC-MAIN-20230610131939-20230610161939-00030.warc.gz | 228,438,120 | 11,229 | # 2185 CNY to BGN
2,185Chinese Yuan (CN¥)
=
556.15Bulgarian Lev (лв)
## 2,185 Chinese Yuan (CNY) to Bulgarian Lev (BGN)
You have converted 2185 Chinese Yuan (CNY) to Bulgarian Lev (BGN). Exchange rate for the conversion is 556.14805 for 10 June 2023, Saturday. How much is 2185 Chinese Yuan to Bulgarian Lev? 556.15 Bulgarian Lev's.
## How much Two Thousand One Hundred Eighty-five (2185) Chinese Yuan in Bulgarian Lev?
Today, 2,185 (two thousand one hundred eighty-five) Chinese Yuan are worth 556.15 Bulgarian Lev. That's because the current exchange rate, to BGN, is 0.255. So, to make Chinese Yuan to Bulgarian Lev conversion, you just need to multiply the amount in CN¥ by 0.255.
• Name Chinese Yuan
• Money2185
• Country China
• Symbol CN¥
• ISO 4217 CNY
• Name Bulgarian Lev
• Money556.14805
• Country Bulgaria
• Symbol BGN
• ISO 4217 BGN
This page provides the exchange rate of two thousand one hundred eighty-five Chinese Yuan to Bulgarian Lev, sale and conversion rate. We added the list of the most popular conversions for visualization and the history table with exchange rate diagram for 2185 Chinese Yuan to Bulgarian Lev from Saturday, 10/06/2023 till Tuesday, 30/05/2023.
DateChinese Yuan (CN¥)Bulgarian Lev (BGN)
10 June 2023, Saturday2,185 CNY556.148 BGN
09 June 2023, Friday2,185 CNY557.535525 BGN
08 June 2023, Thursday2,185 CNY557.557375 BGN
07 June 2023, Wednesday2,185 CNY560.057015 BGN
06 June 2023, Tuesday2,185 CNY561.245655 BGN
05 June 2023, Monday2,185 CNY561.55374 BGN
04 June 2023, Sunday2,185 CNY562.543545 BGN
03 June 2023, Saturday2,185 CNY563.08761 BGN
02 June 2023, Friday2,185 CNY563.57705 BGN
01 June 2023, Thursday2,185 CNY560.043905 BGN
31 May 2023, Wednesday2,185 CNY562.4627 BGN
30 May 2023, Tuesday2,185 CNY562.587245 BGN
This graph show how much is 2185 Chinese Yuan in Bulgarian Lev = 556.14805 BGN, according to actual pair rate equal 1 CNY = 0.25453 BGN. Yesterday this currency exchange rate plummeted on -0.4632 and was BGN 557.5355 Bulgarian Levs for CN¥ 1. On this graph you can see trend of change 2,185 CNY to BGN. And average currency exchange rate for the last week was BGN 0.2565 BGN for CN¥1 CNY.
``<a href="https://currencyconverts.com/cny/bgn/2185" title="2,185 Chinese Yuan(CNY) to Bulgarian Lev(BGN) Currency Rates Today">2,185 Chinese Yuan(CNY) to Bulgarian Lev(BGN)</a>`` | 746 | 2,345 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-23 | latest | en | 0.857536 |
forex-free-training.com | 1,397,769,897,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609532128.44/warc/CC-MAIN-20140416005212-00436-ip-10-147-4-33.ec2.internal.warc.gz | 86,203,738 | 5,336 | How to calculate a PIP value in forex online trading?
Before one will learn how to calculate a pip value in forex online trading, a forex trader must understand what pip is. Surprisingly, we encounter “experienced” forex traders that after a year or more of online trading still make confusion on what PIP is and how to calculate its value, and this is the basic tool to manage your risks correctly during forex online trading.
In forex online trading, the PIP is always the forth digit after the decimal point in the currency exchange rate, for example in the price quote 1.30010 of the EURUSD pair, the pip is the forth digit after the decimal point. In the forex online trading market, there are brokers which provide their customers quotes with only four digits after the point, reasoning that kind of choice by convenience and comfort to a trader, the real reason behind this choice of the forex broker is that by not showing you the last digit, which is a decimal of a PIP, the broker can cut a bit of the trader’s profit and increase a bit his loss.
So think twice before choosing a four digit quotes broker!
So how do we calculate a PIP value in forex online trading?
There are few methods; our preferred one is by a simple formula, which will help you avoiding the confusions!
Just take the trade size, divide it by 10,000 and you will get the PIP value in base currency
Example 1: A trade of 1 Lot (100,000\$) of EUR/USD = 100,000/ 10,000 = 10\$
Example 2: A trade of 0.1 lot (10,000\$ of EUR/USD = 10,000/10,000 = 1\$
As we said before, the PIP is the forth digit after the point, so let’s assume we opened a Long position on EUR/USD of 1 lot, on 1.30000, and closed this position on 1.30050, we earned 5 PIPs, which are 50\$.
Remember, do not open any trades in forex online trading without understanding how much money you will pay / earn on every pip the market moves in your direction or against you.
By understanding this you will make first step towards the right risk management in your forex online trading. | 484 | 2,027 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2014-15 | latest | en | 0.924457 |
https://www.jiskha.com/questions/1008922/which-deal-gives-you-more-pizza-two-six-inch-pizzas-or-one-ten-inch-pizza | 1,575,949,525,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540525781.64/warc/CC-MAIN-20191210013645-20191210041645-00452.warc.gz | 757,074,037 | 5,487 | # algebra 1
which deal gives you more pizza: two six inch pizzas or one ten inch pizza
1. 👍 0
2. 👎 0
3. 👁 181
1. A = pi * r^2
A = 3.14 * 9
A = 28.26 sq. inches in one 6-inch pizza
A = 3.14 * 25
A = 78.5 sq. inches in one 10-inch pizza
1. 👍 0
2. 👎 0
2. THANKS
1. 👍 0
2. 👎 0
3. You're welcome
1. 👍 0
2. 👎 0
4. witch one is it
1. 👍 0
2. 👎 0
posted by mon
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### XII Physics : Sample Practical Activities Part 1
For Aga Khan Board
Contains:
Topic 11: Electrostatics
1. 11.7.3 Determine the relation between current and capacitance when different capacitors are used in AC circuit using different series and parallel combinations of capacitors.AC milliammeter, AC voltmeter,capacitors of different, capacitances,step-down transformer, sand paper,connecting wires.
Topic 12: Current Electricity
3. 12.2.5 Determine resistance of voltmeter by drawing graph between (R) and (I/V).Voltmeter, resistance box, two keys,sand paper, connecting wires, graphpaper.
4. 12.3.5 Investigate the relationship between current passing through a tungsten filament lamp and the potential applied across it. 36W, 12 Volt car bulb, bulb holder, 12 volt battery, high resistance rheostat,voltmeter, ammeter, key, sand paper,connecting wires.
Topic 17: Electronics
10. 17.4.2 Verify truth table for logic gates. DC power supply, OR, AND, NOR, NAND, NOT gates, LED indicator module, two key plugs, connecting wires.
11. 17.4.3 Make burglar alarm using NAND gate. Two NAND gates, two resistance of
100k , electronic bell, connecting wires, power supply 5V DC, key plugs.
Topic 18: Dawn of Modern Physics
13. 18.3.5 Study of the variation of electric current with intensity of light using a photocell. Photocell, galvanometer, battery, rheostat, key, electric bulb with case, connecting wires.
### COMMON COLLECTOR CONFIGURATION OF A TRANSISTOR
COMMON COLLECTOR CONNECTION
In this configuration the input is applied between the base and the collector and the output is taken from the collector and the emitter. Here the collector is common to both the input and the output circuits as shown in Fig.
Common Collector Transistor Circuit
In common collector configuration the input current is the base current IB and the output current is the emitter current IE. The ratio of change in emitter current to the change in the base current is called current amplification factor.
It is represented by
COMMON COLLECTOR CIRCUIT
A test circuit for determining the static characteristic of an NPN transistor is shown in Fig. In this circuit the collector is common to both the input and the output circuits. To measure the base and the emitter currents, milli ammeters are connected in series with the base and the emitter circuits. Voltmeters are connected across the input an…
### Solution Manual : Mathematical methods for physicists 5th edition Arfken and Weber
Download
DJ VU Reader
Book Description Now in its 7th edition, Mathematical Methods for Physicists continues to provide all the mathematical methods that aspiring scientists and engineers are likely to encounter as students and beginning researchers. This bestselling text provides mathematical relations and their proofs essential to the study of physics and related fields. While retaining the key features of the 6th edition, the new edition provides a more careful balance of explanation, theory, and examples. Taking a problem-solving-skills approach to incorporating theorems with applications, the book's improved focus will help students succeed throughout their academic careers and well into their professions. Some notable enhancements include more refined and focused content in important topics, improved organization, updated notations, extensive explanations and intuitive exercise sets, a wider range of problem solutions, improvement in the placement, and a wider ra… | 816 | 3,563 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | latest | en | 0.567104 |
https://discuss.codechef.com/questions/122221/codeforces-500b-help-needed | 1,553,103,681,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202450.64/warc/CC-MAIN-20190320170159-20190320192159-00103.warc.gz | 467,759,638 | 11,833 | ×
# WRONG
solution
I used priority queue to keep shortest at top of the graph and print them according to group number, as group number will be the smallest element in the individual graph, and initially each pq contains one element i.e. itself, and group number as itself, if a link with some other vertex is found with smaller group number, its all elements are transferred to that pq, and in case of duplicate , all of them are removed/ as it failed at test case 8 and 22nd th element is different , and other aren't visible to me and even if were, making a matrices on copy and solving by pen would be tough , can you please help me figure out whats wrong. Thanks in advance :) CODEFORCES 500B LINK
875
accept rate: 0%
# RIGHT
Though no one helped me, finally solved it myself, though shalln't answer my own question (maybe), still if it helps someone while googling, i'll be glad.
>>1) Create graphs of all connected components (priority queue, with group number of that individual graph as the minimum edge in that particular graph)
>>2) as using priority queue i.e. pq, all graphs will be sorted.
>>3) read each edge in permutation and print the minimum in that individual graph (via group number), though all may be connected to single, i.e. in 2nd case.
>>{help} as if edges have same group numbers (already lie in same graph), will not consider push, this will avoid duplicate and infinite loop i.e. putting and extracting from same graph. and poping out also help avoid duplicate.
# Thanks to me!
875
accept rate: 0%
1
why dont u participate in codechef contests ?
(31 Jan '18, 01:42)
actually am still scared of them, actually i started with codechef itself, the statements were so tricky, i was scared on asked someone on linkedin for help, he guided me to use codeforces to start, besides test cases are visible (highly helpful for a beginner least). And codechef has a very good friendly community, so i always ask questions here, i tried participatin in contests but couldn't solve any, once i get average with coding i will surely on all websites like topcoder, codechef etc, though codechef is best, can you please guide me as am alone, what contests to participate as beginner.
(31 Jan '18, 03:10)
1
Obviously the long contests in codechef.
My case is totally opposite,i am afraid of codeforces(as i felt them too hard),but probably would start doing on it nowonwards.
During long if u dont get something try to google as much as u can.After it ends see the editorials(i prefer unofficial editorials as they are simple).After that solve problems on spoj for the remaining month (it will help u learn fast (really fast) ).Then maybe u can practice dp where i found codeforces the best.
(31 Jan '18, 13:29)
1
@ashigahlawat try Long challenges .. codeforces has very good questions but for a beginner they are quite hard to solve in 2 hours ..
(31 Jan '18, 13:50)
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last updated: 31 Jan '18, 13:50 | 841 | 3,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-13 | latest | en | 0.948783 |
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## Sun position in "texcoord space"
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### #1belfegor Members
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Posted 18 November 2012 - 07:41 AM
Let me explain my problem:
I have a sun, and let say it is at position 100, 50, 20
And then i use this code to obtain sun position in "texcoord space"
viewProj = cam.view * cam.proj;
...
D3DXVECTOR2 sunPosCS;
D3DXVECTOR4 clipSpaceLightPos;
D3DXVec4Transform(&clipSpaceLightPos, &D3DXVECTOR4(sunPos.x, sunPos.y, sunPos.z, 1.f), &viewProj);
D3DXVECTOR4 screenSpaceLightPos = clipSpaceLightPos / clipSpaceLightPos.w;
sunPosCS = D3DXVECTOR2(screenSpaceLightPos.x, screenSpaceLightPos.y);
sunPosCS.x = ((sunPosCS.x + 1.0f) / 2.0f);
sunPosCS.y = ((-sunPosCS.y + 1.0f) / 2.0f);
...
effect->SetValue(hSunPosCS, sunPosCS, sizeof(D3DXVECTOR2));
...
Then i render fullscreen quad and use this pixel shader to be able to see where the sun is:
OUT.Color = distance(IN.TexCoord0, LightPosCS) < 0.05f ? float4(1.0f, 0.0f, 0.0f, 1.0f) : float4(0.0f, 0.0f, 0.0f, 1.0f);
When i face my camera to look at the sun, and then just move left or right, the red circle (representing the sun) is also moving but it should not, because sun in real life is quite far away from the viewer and its movement on the screen should not be noticeable. If you understand what i mean? Any suggestions?
Thank you for your time.
Edited by belfegor, 18 November 2012 - 07:41 AM.
### #2Waterlimon Members
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Posted 18 November 2012 - 07:54 AM
Do something like:
1.Define the sun as a direction instead of a position
2.Transform the point into camera space
3.If Z is negative, sun is not visible. Then throw the Z away.
4.now you have x,y, multiply that by some FoV and stuff and check if the resulting point is on the screen, then render sun there if it is.
o3o
### #3belfegor Members
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Posted 18 November 2012 - 08:00 AM
I just tried something:
sunPos = sunDir * 100000.0f;// move it far away
and it is barely seen to move.
@Waterlimon
1 What you mean by "define as direction"? Just take sun direction?
2. Transform sun direction by view matrix?
D3DXVec4Transform(&out, &D3DXVECTOR4(sunDir.x, sunDir.y, sunDir.z, 0.0f, &view);
### #4Waterlimon Members
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Posted 18 November 2012 - 10:40 AM
By sun direction, since the sun is practically infinitely far away, i mean take the direction from the origin of the world to where on the sky you want the sun to be.
Not sure about the matrix. I mean take the world-space sun direction, and somehow transform it into camera-space (like the camera was in the origin, facing somewhere, and the direction was a point 1 unit away from the origin, in the direction of the sun, then move it into camera-space)
1.Take direction of the sun from the origin of the world (0,1,0) would be directly above
2.Move that direction to the object space of the camera. So if camera faced down, the direction from cameras perspective would be 0,-1,0 if in world space it was 0,1,0
I dont exactly remember how this matrix stuff works, but it MIGHT go like:
a)Take camera rotation (throw position away)
b)Take inverse of that
c)Multiply the world space sun direction with that (not sure about multiplication order)
3.Now make sure Z of the resulting vector is positive/negative to make sure the sun is not behind the camera
4.Throw Z away, leaving you with X and Y
5.Multiply or divide X and Y with some values related to the FoV, and you have the position of the sun on the screen. (make sure its on the screen, might be outside) Also you might need to adjust it to get correct scale and origin...
o3o
### #5EsorU Members
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Posted 19 November 2012 - 03:10 PM
Maybe this >Math for computing relative sun direction< could help.
Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic. | 1,181 | 4,142 | {"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-2016-50 | latest | en | 0.726738 |
https://www.skinscanapp.com/essay-writing-blog/how-do-you-score-a-whist-contract/ | 1,686,252,598,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655092.36/warc/CC-MAIN-20230608172023-20230608202023-00575.warc.gz | 1,082,205,286 | 17,762 | # How do you score a whist contract?
How do you score a whist contract?
## How do you score a whist contract?
At the end of each round, one point is scored for each trick taken, and ten points are scored if you took the number of tricks stated at the beginning. So someone who stated they would take 3 and wins 4 would get 4 points, whereas someone who stated they wouldn’t win any and proceeds to do so would get 10 points.
## How do you keep score in bid whist?
If the bidding team makes their bid, they score one point for every book they took after the initial six. If they fail to take enough books for their bid, they score nothing for their books, and instead their bid is subtracted from their score. A team may win more than they bid, but may lose only what was bid.
How do you score in solo whist?
When a bid target is achieved, the bidding player or partnership receives the points once for each player opposing them. If, for instance, you exceed a solo by two tricks, you score 14 points for each opposing player, 42 in total.
### What does Whisted mean?
intransitive verb. dialectal British. : to be silent : hush —often used interjectionally to enjoin silence. whist.
### Why is it called a Boston in Bid Whist?
As a result, many Bid Whist terms come from trains and cross-country travel. The phrase “running a Boston” is thought to come from the all night card games played on the longest routes. If you were the big winner, you could brag, “I won all the way from New Orleans to Boston!”
What does no trump mean in Bid Whist?
Variations. Some play that, in a no trump bid, the jokers must be placed out of play in the kitty. If you win the bid in no trumps, then any jokers you find in the kitty must be left there, and you must discard any jokers that were in your original hand.
#### Can you play whist with 3 players?
Three-Handed Whist, also known as Widow Whist, is a variant of the trick-taking game Whist. “Widow” whist is named because of an extra hand that is dealt just to the left of the dealer. This extra hand is called the “widow” and players may have a chance to use the widow instead of their own hand.
#### Can you go alone in 500?
Five-handed 500 The player who wins the bidding may then choose to “go it alone” (go on his/her own – without a partner for this hand) or gets to choose a card (the joker cannot be chosen) to select a partner.
How many cards do you get in Bid Whist?
To set up for Bid Whist the deck, including the two jokers will be shuffled. The twelve cards will be dealt to each player by the dealer. The remaining cards make up the kitty and will be the first trick won by the winner of the bid.
## What is a Quist?
noun plural quists or quist. West Midland and Southwest English dialect a wood pigeon. | 663 | 2,776 | {"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} | 2.59375 | 3 | CC-MAIN-2023-23 | latest | en | 0.973735 |
http://medical-dictionary.thefreedictionary.com/Beer-Lambert-Bouguer+law | 1,547,770,636,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659417.14/warc/CC-MAIN-20190117224929-20190118010929-00577.warc.gz | 148,731,941 | 11,179 | # Beer-Lambert law
(redirected from Beer-Lambert-Bouguer law)
Also found in: Encyclopedia.
## Beer-Lam·bert law
(bēr lam'bert),
the absorbance of light is directly proportional to the thickness of the media through which the light is being transmitted multiplied by the concentration of absorbing chromophore; that is, A = εbc where A is the absorbance, ε is the molar extinction coefficient, b is the thickness of the solution, and c is the concentration.
[August Beer, Johann Heinrich Lambert]
## Beer’s law
A law stating that the concentration of an analyte is directly proportional to the amount of light absorbed, or inversely proportional to the logarithm of the transmitted light.
Beer’s law
A = abc = log(100/%T) 2 - log %T
where:
A = absorbance
a = absorptivity
b = light path of the solution in cm
c = concentration of the substance of interest
%T = per cent transmittance—the ratio of transmitted light to incident light
## Beer,
August, German physicist, 1825-1863.
Beer-Lambert law - the absorbance of light is directly proportional to the thickness of the ligand through which the light is being transmitted multiplied by the concentration of absorbing chromophore.
Beer law - the intensity of a color or of a light ray is inversely proportional to the depth of liquid through which it is transmitted.
## Lambert,
Johann Heinrich, German mathematician and physicist, 1728-1777.
Beer-Lambert law - see under Beer, August
Lambert cosine law - mathematical measure of the intensity of radiation.
References in periodicals archive ?
The experimental dependences of transmittance on the relative density of fog for LEDs of different colour were found to be exponential and corresponded to the Beer-Lambert-Bouguer law.
However, for the practical exercise presented in this article, it is not necessary to calculate the immersion factor for the recommended UV meter, as only relative measurements will need to be obtained and analysed in order to investigate the Beer-Lambert-Bouguer Law.
The following practical exercise aims to provide the Year 11 and Year 12 Physics students with a step by step appreciation into how the Beer-Lambert-Bouguer Law can be applied underwater and how handheld solar UV measurement instrumentation can be employed to provide rapid estimations for both [K.
Site: Follow: Share:
Open / Close | 521 | 2,341 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-04 | latest | en | 0.912754 |
https://www.convertunits.com/from/kilometers/to/fall+%5BScotland%5D | 1,670,519,859,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711344.13/warc/CC-MAIN-20221208150643-20221208180643-00102.warc.gz | 742,261,410 | 12,984 | ## ››Convert kilometre to fall [Scotland]
kilometers fall [Scotland]
Did you mean to convert kilometers to fall [English] fall [Scotland]
How many kilometers in 1 fall [Scotland]? The answer is 0.00567.
We assume you are converting between kilometre and fall [Scotland].
You can view more details on each measurement unit:
kilometers or fall [Scotland]
The SI base unit for length is the metre.
1 metre is equal to 0.001 kilometers, or 0.17636684303351 fall [Scotland].
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between kilometres and falls.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of kilometers to fall [Scotland]
1 kilometers to fall [Scotland] = 176.36684 fall [Scotland]
2 kilometers to fall [Scotland] = 352.73369 fall [Scotland]
3 kilometers to fall [Scotland] = 529.10053 fall [Scotland]
4 kilometers to fall [Scotland] = 705.46737 fall [Scotland]
5 kilometers to fall [Scotland] = 881.83422 fall [Scotland]
6 kilometers to fall [Scotland] = 1058.20106 fall [Scotland]
7 kilometers to fall [Scotland] = 1234.5679 fall [Scotland]
8 kilometers to fall [Scotland] = 1410.93474 fall [Scotland]
9 kilometers to fall [Scotland] = 1587.30159 fall [Scotland]
10 kilometers to fall [Scotland] = 1763.66843 fall [Scotland]
## ››Want other units?
You can do the reverse unit conversion from fall [Scotland] to kilometers, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Kilometer
A kilometre (American spelling: kilometer, symbol: km) is a unit of length equal to 1000 metres (from the Greek words khilia = thousand and metro = count/measure). It is approximately equal to 0.621 miles, 1094 yards or 3281 feet.
## ››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! | 567 | 2,262 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.818043 |
https://snippets.cacher.io/snippet/7fbb727c92435674890d | 1,702,333,831,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679518883.99/warc/CC-MAIN-20231211210408-20231212000408-00458.warc.gz | 587,656,102 | 4,269 | MariaSzubski
7/31/2016 - 3:27 PM
## Compare two equations to see if they are ever equal. #congruence #Hackerrank
Compare two equations to see if they are ever equal. #congruence #Hackerrank
``````/*
Solution for HackerRank > Algorithms > Implementation > Kangaroo
https://www.hackerrank.com/challenges/kangaroo
In this example, 0 <= x1 <= x2
*/
function kangaroos() {
var x1 = 43;
var v1 = 5;
var x2 = 49;
var v2 = 3;
var done = false;
// Test for simple solutions
switch (true){
case (x1 == x2 && v1 == v2) :
done = true;
console.log("YES");
break;
case ((x1 + v1) < v2) || ((x2 + v2) < v1) :
done = true;
console.log("NO");
break;
case ((x1 < x2 && v1 <= v2) || (x2 < x1 && v2 <= v1)) :
done = true;
console.log("NO");
break;
}
// If all switch cases are false, test for congruence
if (!done){
var sm, lg;
if (Math.abs(x1 - x2) <= Math.abs(v2 -v1)) {
lg = v2 - v1;
sm = x1 - x2;
} else {
lg = x1 - x2;
sm = v2 - v1;
}
// Is is possible for roo1 to be congruent to roo2?
(lg % sm == 0) ? console.log("YES") : console.log("NO");
}
}`````` | 378 | 1,048 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.442948 |
https://gre.magoosh.com/forum/832-of-the-20-lightbulbs-in-a-box | 1,590,641,764,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347396495.25/warc/CC-MAIN-20200528030851-20200528060851-00583.warc.gz | 374,968,233 | 16,059 | Source: Official Guide Revised GRE 1st Ed. Part 8; Section 6; #24
3
# Of the 20 lightbulbs in a box
Of the 20 lightbulbs in a box, 2 are defective. An inspector will select 2 lightbulbs simultaneously and at random from the box. What is the probability that neither of the lightbulbs selected will be defective? Give your answer as a fraction.
### 4 Explanations
1
Drasti Chaudhari
Why did you multiply and not add? Also, why is it not 1/18 and then 1/17? Isn't he picking just 1 bulb each time
Jun 13, 2018 • Comment
Sam Kinsman, Magoosh Tutor
Hi Drasti,
In probability, when we have two events that both need to happen, we have to multiply the probabilities of each event together. In other words, we need BOTH the first lightbulb to not be defective, AND the second lightbulb to not be defective. That means we have to multiply the two events together. You can read more about this rule here: https://magoosh.com/gmat/2012/gmat-math-probability-rules/
When the person picks the first lightbulb, the probability that it will not be defective is (# of non-defective bulbs) / (total # of bulbs). So we have: 18/20.
When the person picks the second lightbulb, there are only 19 lightbulbs left in the box. And, assuming that the first lightbulb that was picked was not defective, there are 17 nondefective lightbulbs in the box. So (# of non-defective bulbs) / (total # of bulbs) = 17/19.
I hope this helps! :)
Best,
Sam
1
MIshal Patel
Apr 8, 2016 • Comment
Cydney Seigerman, Magoosh Tutor
Hi there :) This is the 1st Edition of the Official Guide, which can be purchased in bookstores or sites such as amazon.com:
http://www.amazon.com/Official-Guide-revised-General-Test/dp/0071700528/ref=sr_1_1?ie=UTF8&qid=1318979846&sr=8-1
This question also appears in 2nd Edition of the OG (page 344, question 24).
Hope this helps :)
3
mugen hassei
If I use the P(not a) rule(i.e probability of selecting a defective piece)-first try (2/20) probability that one gets a defective piece and 2nd try (1/19) is the probability that one gets a defective piece.Overall (2/20)(1/19)=1/190 is the probability that one gets a defective piece.Then probability of not selecting a defective piece is (1-1/190)=189/190.Where have I gone wrong?
Aug 7, 2015 • Comment
Cydney Seigerman, Magoosh Tutor
Hi Mugen :)
The solution you've proposed only considers the probability of choosing 2 defective bulbs. However, in order to use 1 - P(not A) = P(A), where A is choosing 2 working bulbs, we must consider all of the possible situations in which we would not choose 2 working bulbs, or in other words, situations in which we would choose at least 1 defective bulb. If we consider the selection in two stages, we can write out the possible ways of not selecting 2 working bulbs:
1. 1st: defective; 2nd: defective
2. 1st: defective; 2nd: working
3. 1st: working; 2nd: defective
As you can see, there are three different ways to not select two working bulbs. P(not A) is the sum of the probabilities that these sets of events occur. We can determine the probabilities of each set of events separately. I will use D for defective and W for working:
1. P(1.D 2.D) = 2/20*1/19 = 1/190
2. P(1.D 2.W) = 2/20*18/19 = 18/190
3. P(1.W 2.D) = 18/20*2/19 = 18/190
P(not A) = (1+18+18)/190 = 37/190.
Therefore, 1-P(not A) = 1 - 37/190 = 153/190.
This is the same answer that we would get if we only considered selecting 2 working bulbs. And as you can see, for this question, it is actually less complicated to find P(A) versus 1-P(not A). I hope this helps! :)
4
Chris Lele, Magoosh Tutor
Oct 11, 2012 • Comment | 1,036 | 3,608 | {"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} | 4.28125 | 4 | CC-MAIN-2020-24 | latest | en | 0.921981 |
https://www.physicsforums.com/threads/general-relativity-bending-spacetime-gravity-question.999998/ | 1,674,973,774,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499700.67/warc/CC-MAIN-20230129044527-20230129074527-00232.warc.gz | 940,113,025 | 19,451 | # General Relativity - bending spacetime - gravity - Question
• I
manolo-mm
Hi everybody
I saw quite a nice Youtube vid about general relativity and how gravity bends spacetime and therefor redirects angular momentum into the center of gravity. I thought the first time I begun to understand the concept but immediatly the questions poped up.
The video basically says that curved spacetime redirects angular momentum into the gravitational center. Ok there is no no difference in gravity on northpol to southpol. so all external angular momentum forces are redirected in full. And all objects in our solar system are moving in the same speed along with the sun through the universe. And there is no gravitational differences on other bodys in our solar system as far as we know. So they all redirect the angular momentum of traveling through the universe to their center of gravity. To why do we have different amounts of gravity on the different bodys in our solar system? The only answer i can imagine is that its not angular momentum that is redirekted through curved space time but another force. But that's just a guess. Anybody has a solid answer for me?
## Answers and Replies
Mentor
how gravity bends spacetime
No, stress-energy bends spacetime; "gravity" is just another name for the effects of bending spacetime.
and therefor redirects angular momentum into the center of gravity
This sounds like nonsense to me. I have not watched the video but I am skeptical that it is a reliable source. [Edit: It looks like the problem is with the OP's misunderstanding something, not the video itself--see follow-up post below.] YouTube videos in general are not a good place to be looking if you want to learn actual physics. You should be looking at textbooks and peer-reviewed papers.
vanhees71
Mentor
The video basically says that curved spacetime redirects angular momentum into the gravitational center.
And now having watched the video, I don't see where it says this anywhere. What it does say actually looks OK to me, though of course it only barely scratches the surface of GR. But I don't see it saying anything like this. Where are you getting this from?
vanhees71 and Motore
manolo-mm
Thats how I understood and interpreted this vid.
weirdoguy and Motore
Homework Helper
Gold Member
2022 Award
Thats how I understood and interpreted this vid.
Which of the seven levels mentioned angular momentum?
Gold Member
2022 Award
YouTube videos in general are not a good place to be looking if you want to learn actual physics. You should be looking at textbooks and peer-reviewed papers.
I think, what are pretty reliable videos is if they come from lectures at well-known universities, often also appearing at youtube (e.g., the online lectures from MIT). It's of course easier to produce reliable videos on the natural sciences when addressed to students of the subject rather than to the public. To "explain science as simple as possible but not simpler" is among the most difficult tasks for a scientist!
Mentor
I think, what are pretty reliable videos is if they come from lectures at well-known universities, often also appearing at youtube (e.g., the online lectures from MIT).
Yes, these are something of a special case, since as course materials they have to meet certain standards that a random YouTube video does not.
vanhees71 | 700 | 3,357 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-06 | latest | en | 0.934579 |
https://webeduclick.com/tree-terminology/ | 1,716,095,964,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057684.4/warc/CC-MAIN-20240519035827-20240519065827-00409.warc.gz | 547,792,455 | 55,478 | # Tree Terminology in Data Structure
A tree is a finite set of vertices connected by edges such that-
i. There is one specially designated vertex called root.
ii. The remaining vertices are partitioned into a collection of sub-trees T1, T2, T3,…, Tn each of which is a tree.
## Basic Tree Terminology:
Root: A node without any parent is called a root node.
Node: Each data item in a tree is called a node.
Edge: The line drawn from one node to another node is called an edge.
Internal Node: A node with at least one child is called an internal node.
External Node: A node without any children is called an external node or leaf node.
Degree of a node: The number of sub-trees of a node is called its degree.
Degree of a tree: It is the maximum degree of nodes in a tree.
Siblings: The nodes who share the same parent are called siblings.
Ancestor and Descendant: A node n is an ancester of node m and m is descendant of n if n is either the father of m.
Level: Number of ancestors of nodes is called its level. Each node in a tree is assigned a level number as follows:
1. The root of the tree is at level 0.
2. Level of other node=level of its parent +1.
Depth: The depth of a tree T can be defined as follows:
depth of T = maximum level number of T+1
Path and Path length: It is a sequence of consecutive edges from the source node to the destination node. The number of edges in a path is called path length.
Branch: Path ending in a leaf node is called as Branch.
Internal Path length: Internal path length of a tree can be defined as the sum of the levels of all the internal nodes in the tree.
External Path length: External path length of a tree can be defined as the sum of the levels of all the external nodes in the tree.
Forest: A set of disjoint trees are called a forest. | 421 | 1,802 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.955405 |
http://de.metamath.org/mpegif/refsum2cnlem1.html | 1,618,461,677,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038083007.51/warc/CC-MAIN-20210415035637-20210415065637-00052.warc.gz | 23,909,926 | 22,198 | Mathbox for Glauco Siliprandi < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > refsum2cnlem1 Structured version Visualization version Unicode version
Theorem refsum2cnlem1 37421
Description: This is the core Lemma for refsum2cn 37422: the sum of two continuous real functions (from a common topological space) is continuous. (Contributed by Glauco Siliprandi, 20-Apr-2017.)
Hypotheses
Ref Expression
refsum2cnlem1.1
refsum2cnlem1.2
refsum2cnlem1.3
refsum2cnlem1.4
refsum2cnlem1.5
refsum2cnlem1.6
refsum2cnlem1.7 TopOn
refsum2cnlem1.8
refsum2cnlem1.9
Assertion
Ref Expression
refsum2cnlem1
Distinct variable groups: ,, , , ,, ,, ,
Allowed substitution hints: () (,) () ()
Proof of Theorem refsum2cnlem1
StepHypRef Expression
1 refsum2cnlem1.4 . . 3
2 refsum2cnlem1.5 . . . . . . . . 9
3 nfmpt1 4485 . . . . . . . . 9
42, 3nfcxfr 2610 . . . . . . . 8
5 nfcv 2612 . . . . . . . 8
64, 5nffv 5886 . . . . . . 7
7 nfcv 2612 . . . . . . 7
86, 7nffv 5886 . . . . . 6
98a1i 11 . . . . 5
10 nfcv 2612 . . . . . . . 8
114, 10nffv 5886 . . . . . . 7
1211, 7nffv 5886 . . . . . 6
1312a1i 11 . . . . 5
14 1cnd 9677 . . . . 5
15 2cnd 10704 . . . . 5
16 1ex 9656 . . . . . . . . . . 11
1716prid1 4071 . . . . . . . . . 10
18 refsum2cnlem1.8 . . . . . . . . . . 11
19 refsum2cnlem1.9 . . . . . . . . . . 11
2018, 19ifcld 3915 . . . . . . . . . 10
21 eqeq1 2475 . . . . . . . . . . . 12
2221ifbid 3894 . . . . . . . . . . 11
2322, 2fvmptg 5961 . . . . . . . . . 10
2417, 20, 23sylancr 676 . . . . . . . . 9
25 eqid 2471 . . . . . . . . . 10
2625iftruei 3879 . . . . . . . . 9
2724, 26syl6eq 2521 . . . . . . . 8
2827adantr 472 . . . . . . 7
2928fveq1d 5881 . . . . . 6
30 eqid 2471 . . . . . . . . . . 11
31 eqid 2471 . . . . . . . . . . 11
3230, 31cnf 20339 . . . . . . . . . 10
3318, 32syl 17 . . . . . . . . 9
34 refsum2cnlem1.7 . . . . . . . . . . . 12 TopOn
35 toponuni 20019 . . . . . . . . . . . 12 TopOn
3634, 35syl 17 . . . . . . . . . . 11
3736eqcomd 2477 . . . . . . . . . 10
38 refsum2cnlem1.6 . . . . . . . . . . . . 13
3938unieqi 4199 . . . . . . . . . . . 12
40 uniretop 21861 . . . . . . . . . . . 12
4139, 40eqtr4i 2496 . . . . . . . . . . 11
4241a1i 11 . . . . . . . . . 10
4337, 42feq23d 5734 . . . . . . . . 9
4433, 43mpbid 215 . . . . . . . 8
4544anim1i 578 . . . . . . 7
46 ffvelrn 6035 . . . . . . 7
47 recn 9647 . . . . . . 7
4845, 46, 473syl 18 . . . . . 6
4929, 48eqeltrd 2549 . . . . 5
50 2ex 10703 . . . . . . . . . . 11
5150prid2 4072 . . . . . . . . . 10
5218, 19ifcld 3915 . . . . . . . . . 10
53 eqeq1 2475 . . . . . . . . . . . 12
5453ifbid 3894 . . . . . . . . . . 11
5554, 2fvmptg 5961 . . . . . . . . . 10
5651, 52, 55sylancr 676 . . . . . . . . 9
57 1ne2 10845 . . . . . . . . . . 11
5857nesymi 2700 . . . . . . . . . 10
5958iffalsei 3882 . . . . . . . . 9
6056, 59syl6eq 2521 . . . . . . . 8
6160adantr 472 . . . . . . 7
6261fveq1d 5881 . . . . . 6
6330, 31cnf 20339 . . . . . . . . . 10
6419, 63syl 17 . . . . . . . . 9
6537, 42feq23d 5734 . . . . . . . . 9
6664, 65mpbid 215 . . . . . . . 8
6766anim1i 578 . . . . . . 7
68 ffvelrn 6035 . . . . . . 7
69 recn 9647 . . . . . . 7
7067, 68, 693syl 18 . . . . . 6
7162, 70eqeltrd 2549 . . . . 5
7257a1i 11 . . . . 5
73 fveq2 5879 . . . . . . 7
7473fveq1d 5881 . . . . . 6
7574adantl 473 . . . . 5
76 fveq2 5879 . . . . . . 7
7776fveq1d 5881 . . . . . 6
7877adantl 473 . . . . 5
799, 13, 14, 15, 49, 71, 72, 75, 78sumpair 37419 . . . 4
8029, 62oveq12d 6326 . . . 4
8179, 80eqtrd 2505 . . 3
821, 81mpteq2da 4481 . 2
83 prfi 7864 . . . 4
8483a1i 11 . . 3
85 eqid 2471 . . . . . . . . . 10
8685ax-gen 1677 . . . . . . . . 9
87 refsum2cnlem1.1 . . . . . . . . . . . 12
88 nfcv 2612 . . . . . . . . . . . 12
8987, 88nffv 5886 . . . . . . . . . . 11
90 refsum2cnlem1.2 . . . . . . . . . . 11
9189, 90nfeq 2623 . . . . . . . . . 10
92 fveq1 5878 . . . . . . . . . . 11
9392a1d 25 . . . . . . . . . 10
9491, 93ralrimi 2800 . . . . . . . . 9
95 mpteq12f 4472 . . . . . . . . 9
9686, 94, 95sylancr 676 . . . . . . . 8
9796adantl 473 . . . . . . 7
98 retopon 21862 . . . . . . . . . . . . 13 TopOn
9938, 98eqeltri 2545 . . . . . . . . . . . 12 TopOn
10099a1i 11 . . . . . . . . . . 11 TopOn
101 cnf2 20342 . . . . . . . . . . 11 TopOn TopOn
10234, 100, 18, 101syl3anc 1292 . . . . . . . . . 10
103 ffn 5739 . . . . . . . . . 10
104102, 103syl 17 . . . . . . . . 9
10590dffn5f 5935 . . . . . . . . 9
106104, 105sylib 201 . . . . . . . 8
107106adantr 472 . . . . . . 7
10897, 107eqtr4d 2508 . . . . . 6
10918adantr 472 . . . . . 6
110108, 109eqeltrd 2549 . . . . 5
111110adantlr 729 . . . 4
112 refsum2cnlem1.3 . . . . . . . . . . 11
11389, 112nfeq 2623 . . . . . . . . . 10
114 fveq1 5878 . . . . . . . . . . 11
115114a1d 25 . . . . . . . . . 10
116113, 115ralrimi 2800 . . . . . . . . 9
117 mpteq12f 4472 . . . . . . . . 9
11886, 116, 117sylancr 676 . . . . . . . 8
119118adantl 473 . . . . . . 7
120 cnf2 20342 . . . . . . . . . . 11 TopOn TopOn
12134, 100, 19, 120syl3anc 1292 . . . . . . . . . 10
122 ffn 5739 . . . . . . . . . 10
123121, 122syl 17 . . . . . . . . 9
124112dffn5f 5935 . . . . . . . . 9
125123, 124sylib 201 . . . . . . . 8
126125adantr 472 . . . . . . 7
127119, 126eqtr4d 2508 . . . . . 6
12819adantr 472 . . . . . 6
129127, 128eqeltrd 2549 . . . . 5
130129adantlr 729 . . . 4
131 simpr 468 . . . . . . . 8
13218, 19ifcld 3915 . . . . . . . . 9
133132adantr 472 . . . . . . . 8
1342fvmpt2 5972 . . . . . . . 8
135131, 133, 134syl2anc 673 . . . . . . 7
136 iftrue 3878 . . . . . . 7
137135, 136sylan9eq 2525 . . . . . 6
138137orcd 399 . . . . 5
139135adantr 472 . . . . . . 7
140 neeq2 2706 . . . . . . . . . . . 12
14157, 140mpbiri 241 . . . . . . . . . . 11
142141necomd 2698 . . . . . . . . . 10
143142neneqd 2648 . . . . . . . . 9
144143adantl 473 . . . . . . . 8
145144iffalsed 3883 . . . . . . 7
146139, 145eqtrd 2505 . . . . . 6
147146olcd 400 . . . . 5
148 elpri 3976 . . . . . 6
149148adantl 473 . . . . 5
150138, 147, 149mpjaodan 803 . . . 4
151111, 130, 150mpjaodan 803 . . 3
1521, 38, 34, 84, 151refsumcn 37414 . 2
15382, 152eqeltrrd 2550 1
Colors of variables: wff setvar class Syntax hints: wn 3 wi 4 wo 375 wa 376 wal 1450 wceq 1452 wnf 1675 wcel 1904 wnfc 2599 wne 2641 wral 2756 cif 3872 cpr 3961 cuni 4190 cmpt 4454 crn 4840 wfn 5584 wf 5585 cfv 5589 (class class class)co 6308 cfn 7587 cc 9555 cr 9556 c1 9558 caddc 9560 c2 10681 cioo 11660 csu 13829 ctg 15414 TopOnctopon 19995 ccn 20317 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1677 ax-4 1690 ax-5 1766 ax-6 1813 ax-7 1859 ax-8 1906 ax-9 1913 ax-10 1932 ax-11 1937 ax-12 1950 ax-13 2104 ax-ext 2451 ax-rep 4508 ax-sep 4518 ax-nul 4527 ax-pow 4579 ax-pr 4639 ax-un 6602 ax-inf2 8164 ax-cnex 9613 ax-resscn 9614 ax-1cn 9615 ax-icn 9616 ax-addcl 9617 ax-addrcl 9618 ax-mulcl 9619 ax-mulrcl 9620 ax-mulcom 9621 ax-addass 9622 ax-mulass 9623 ax-distr 9624 ax-i2m1 9625 ax-1ne0 9626 ax-1rid 9627 ax-rnegex 9628 ax-rrecex 9629 ax-cnre 9630 ax-pre-lttri 9631 ax-pre-lttrn 9632 ax-pre-ltadd 9633 ax-pre-mulgt0 9634 ax-pre-sup 9635 ax-addf 9636 This theorem depends on definitions: df-bi 190 df-or 377 df-an 378 df-3or 1008 df-3an 1009 df-tru 1455 df-fal 1458 df-ex 1672 df-nf 1676 df-sb 1806 df-eu 2323 df-mo 2324 df-clab 2458 df-cleq 2464 df-clel 2467 df-nfc 2601 df-ne 2643 df-nel 2644 df-ral 2761 df-rex 2762 df-reu 2763 df-rmo 2764 df-rab 2765 df-v 3033 df-sbc 3256 df-csb 3350 df-dif 3393 df-un 3395 df-in 3397 df-ss 3404 df-pss 3406 df-nul 3723 df-if 3873 df-pw 3944 df-sn 3960 df-pr 3962 df-tp 3964 df-op 3966 df-uni 4191 df-int 4227 df-iun 4271 df-iin 4272 df-br 4396 df-opab 4455 df-mpt 4456 df-tr 4491 df-eprel 4750 df-id 4754 df-po 4760 df-so 4761 df-fr 4798 df-se 4799 df-we 4800 df-xp 4845 df-rel 4846 df-cnv 4847 df-co 4848 df-dm 4849 df-rn 4850 df-res 4851 df-ima 4852 df-pred 5387 df-ord 5433 df-on 5434 df-lim 5435 df-suc 5436 df-iota 5553 df-fun 5591 df-fn 5592 df-f 5593 df-f1 5594 df-fo 5595 df-f1o 5596 df-fv 5597 df-isom 5598 df-riota 6270 df-ov 6311 df-oprab 6312 df-mpt2 6313 df-of 6550 df-om 6712 df-1st 6812 df-2nd 6813 df-supp 6934 df-wrecs 7046 df-recs 7108 df-rdg 7146 df-1o 7200 df-2o 7201 df-oadd 7204 df-er 7381 df-map 7492 df-ixp 7541 df-en 7588 df-dom 7589 df-sdom 7590 df-fin 7591 df-fsupp 7902 df-fi 7943 df-sup 7974 df-inf 7975 df-oi 8043 df-card 8391 df-cda 8616 df-pnf 9695 df-mnf 9696 df-xr 9697 df-ltxr 9698 df-le 9699 df-sub 9882 df-neg 9883 df-div 10292 df-nn 10632 df-2 10690 df-3 10691 df-4 10692 df-5 10693 df-6 10694 df-7 10695 df-8 10696 df-9 10697 df-10 10698 df-n0 10894 df-z 10962 df-dec 11075 df-uz 11183 df-q 11288 df-rp 11326 df-xneg 11432 df-xadd 11433 df-xmul 11434 df-ioo 11664 df-icc 11667 df-fz 11811 df-fzo 11943 df-seq 12252 df-exp 12311 df-hash 12554 df-cj 13239 df-re 13240 df-im 13241 df-sqrt 13375 df-abs 13376 df-clim 13629 df-sum 13830 df-struct 15201 df-ndx 15202 df-slot 15203 df-base 15204 df-sets 15205 df-ress 15206 df-plusg 15281 df-mulr 15282 df-starv 15283 df-sca 15284 df-vsca 15285 df-ip 15286 df-tset 15287 df-ple 15288 df-ds 15290 df-unif 15291 df-hom 15292 df-cco 15293 df-rest 15399 df-topn 15400 df-0g 15418 df-gsum 15419 df-topgen 15420 df-pt 15421 df-prds 15424 df-xrs 15478 df-qtop 15484 df-imas 15485 df-xps 15488 df-mre 15570 df-mrc 15571 df-acs 15573 df-mgm 16566 df-sgrp 16605 df-mnd 16615 df-submnd 16661 df-mulg 16754 df-cntz 17049 df-cmn 17510 df-psmet 19039 df-xmet 19040 df-met 19041 df-bl 19042 df-mopn 19043 df-cnfld 19048 df-top 19998 df-bases 19999 df-topon 20000 df-topsp 20001 df-cn 20320 df-cnp 20321 df-tx 20654 df-hmeo 20847 df-xms 21413 df-ms 21414 df-tms 21415 This theorem is referenced by: refsum2cn 37422
Copyright terms: Public domain W3C validator | 5,487 | 9,996 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-17 | latest | en | 0.11968 |
https://www.instructables.com/id/FIRE-SENSOR/ | 1,591,054,701,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347419639.53/warc/CC-MAIN-20200601211310-20200602001310-00390.warc.gz | 760,058,258 | 23,188 | # FIRE SENSOR
198
1
## Introduction: FIRE SENSOR
Hello everyone!
Fire sensor is a sensor designed to detect and respond to the presence of a flame or fire. Here, it is a PIN diode based fire sensor that activates when it detects fire. Thermistor based fire alarms have a drawback; the alarm turns on only if the fire heats the thermistor in close vicinity.
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## Step 1: Hardware Required
• CA3140 OP-AMP - 1
• CD4060 COUNTER - 1
• BC547 NPN TRANSISTOR - 2
• BPW34 PIN photodiode
• LED 5 mm - 3
• PIEZO BUZZER-1
• 9V BATTERY-1
• 0.22uf ceramic disk capacitor-1
• 1M ohm resistor- 3
• 1k ohm resistor - 2
• 100-ohm resistor - 3
## Step 2: Circuit Diagram
Circuit diagram of the PIN diode based fire sensor is shown above in the image. It is built around 9V battery, PIN diode BPW34, op-amp CA3140(IC1), counter CD4060(IC2), transistors BC547, a piezo buzzer and a few other components.
In the circuit, PIN photodiode BPW34 is connected to the inverting and non-inverting inputs of op-amp IC1 in reverse-biased mode to feed photocurrent into the input of the op-amp. CA3140 is a 4.5MHz BiMOs op-amp with MOSFET inputs and bipolar output.
Gate-protected MOSFET (PMOS) transistors in the input circuit provide very high input impedance, typically around 1.5T ohms. The IC requires very low input current, as low as 10pA, to change output status to high or low.
In the circuit, IC1 is used as a transimpedance amplifier to act as a current-to-voltage converter. IC1 amplifies and converts the photocurrent generated in the PIN diode to the corresponding voltage in its output. The non-inverting input is connected to the ground and anode of the photodiode, while the inverting input gets photocurrent from the PIN diode.
## Step 3: Circuit Operation
Large-value feedback resistor R1 sets the gain of the transimpedance amplifier since it is in inverting configuration. Connection of non-inverting input to ground provides low impedance load for the photodiode, which keeps the photodiode voltage low.
The photodiode operates in the photovoltaic mode with no external bias. Feedback of the op-amp keeps the photodiode current equal to the feedback current through R1. So the input offset voltage due to the photodiode is very low in this self-biased photovoltaic mode. This permits a large gain without any large-output offset voltage. This configuration is selected to get large gain in low-light conditions.
Normally, in ambient light condition, photocurrent from the PIN diode is very low; it keeps the output of IC1 low. When the PIN diode detects visible light or IR from fire, its photocurrent increases and transimpedance amplifier IC1 converts this current to the corresponding output voltage. High output from IC1 activates transistor T1 and LED1 glows. This indicates that the circuit has detected fire. When T1 conducts, it takes reset pin 12 of IC2 to ground potential and CD4060 starts oscillating.
IC2 is a binary counter with ten outputs that turn high one by one when it oscillates due to C1 and R6. Oscillation of IC2 is indicated by the blinking of LED2. When output Q6 (pin 4) of IC2 turns high after 15 seconds, T2 conducts and activates piezo buzzer PZ1, and LED3 also glows. The alarm repeats again after 15 seconds if fire persists.
You can also turn on an AC alarm that produces a loud sound by replacing PZ1 with a relay circuitry (not shown here). The AC alarm is activated through contacts of the relay used for this purpose.
## Step 4: Schematic and Layout Design
A PCB for PIN-based fire sensor is designed using EAGLE. The schematic and board-layout are shown above in the image.
## Step 5: Sending Gerber Files to the Manufacturer
After exporting my GERBER files from EAGLE I'm uploading them on LIONCIRCUITS to get my board manufactured. I usually order my PCBs from them only. They provide low-cost prototyping only within 6 days.
## Step 6: Fabricated Boards
I have received my board from LIONCIRCUITS and I am sharing my Gerber files with you in case anyone needs the board to be manufactured.
## Step 7: Assembling and Testing
After assembling my board with components it looks like this.
Testing the circuit is simple. Normally, when there is no fire flame near the PIN diode, the piezo buzzer does not sound. When a fire flame is sensed by the PIN diode, the piezo buzzer sounds an alarm. Its detection range is around two meters.
## Recommendations
17 3.2K
161 12K
94 9.2K
Large Motors Class
14,642 Enrolled | 1,159 | 4,616 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-24 | latest | en | 0.829713 |
https://istopdeath.com/write-the-equation-in-standard-form-7x27y2-28x42y-350/ | 1,674,854,335,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764495012.84/warc/CC-MAIN-20230127195946-20230127225946-00852.warc.gz | 338,244,876 | 15,811 | # Write the Equation in Standard Form 7x^2+7y^2-28x+42y-35=0
Move to the right side of the equation because it does not contain a variable.
Divide both sides of the equation by .
Complete the square for .
Use the form , to find the values of , , and .
Consider the vertex form of a parabola.
Substitute the values of and into the formula .
Simplify the right side.
Cancel the common factor of and .
Factor out of .
Cancel the common factors.
Factor out of .
Cancel the common factor.
Rewrite the expression.
Divide by .
Multiply by .
Find the value of using the formula .
Simplify each term.
Cancel the common factor of and .
Rewrite as .
Apply the product rule to .
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Substitute for in the equation .
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Substitute the values of and into the formula .
Cancel the common factor of and .
Factor out of .
Cancel the common factors.
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Cancel the common factor.
Rewrite the expression.
Divide by .
Find the value of using the formula .
Simplify each term.
Raise to the power of .
Multiply by .
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Substitute the values of , , and into the vertex form .
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Move to the right side of the equation by adding to both sides.
Simplify . | 384 | 1,627 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.878335 |
http://mathhelpforum.com/statistics/16610-need-help-stats-print.html | 1,529,287,474,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267859923.59/warc/CC-MAIN-20180618012148-20180618032148-00323.warc.gz | 199,333,065 | 3,274 | # need help stats
• Jul 8th 2007, 01:51 AM
harry
need help stats
Confidence interval for a population proportion.
In a random sample of 400 students at a university, 332 stated that they were nonsmokers. Based on this sample, compute a 95% confidence interval for the proportion of all students at the university who are nonsmokers. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places
(a)What is the lower limit of the confidence interval?
(b) What is the upper limit of the confidence interval?
• Jul 8th 2007, 03:58 AM
rualin
Confidence interval for the population proportion: $\displaystyle \hat{P} - E < p < \hat{P} + E$, where $\displaystyle E = z_c \sqrt{\frac{\hat{p} \cdot \hat{q}}{n}}$.
Make a list of the given information for easier reference.
$\displaystyle n=400$
$\displaystyle x=332$
$\displaystyle c=95\%=0.95$
$\displaystyle z_c=1.96$ by looking at the standard normal distribution table
$\displaystyle \hat{p} = \frac{x}{n} = \frac{332}{400} = 0.83$
$\displaystyle \hat{q} = 1 - \hat{p} = 1 - 0.83 = 0.17$
Solving for E, we get:
$\displaystyle E = 1.96 \sqrt{\frac{0.83 \cdot 0.17}{400}} \approx 0.0368$
And substituting back into the confidence interval formula:
$\displaystyle 0.83 - 0.0368 < p < 0.83 + 0.0368$
$\displaystyle \boxed{0.793 < p < 0.867}$
• Jul 8th 2007, 05:38 AM
CaptainBlack
Quote:
Originally Posted by harry
Confidence interval for a population proportion.
In a random sample of 400 students at a university, 332 stated that they were nonsmokers. Based on this sample, compute a 95% confidence interval for the proportion of all students at the university who are nonsmokers. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places
(a)What is the lower limit of the confidence interval?
(b) What is the upper limit of the confidence interval?
Just to be arkward I will observe that you cannot get such a confidence
interval using this data. A significant number of respondents lied, giving
the answer that they thought would make them look good in the eyes
of the survey organisers.
All you can do is obtain a 95% confidence interval for the proportion of
students at the university who would have claimed to be non-smokers
on a survey worded and conducted like the one in question.
RonL | 651 | 2,411 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 4.15625 | 4 | CC-MAIN-2018-26 | latest | en | 0.815586 |
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9. Why aren’t these in stores yet? Or if they are where can I pick one up?
• Hi! We are currently struggling to keep up with the demand for our puzzles. Each one is made from start to finish at our studio in Somerville, MA. Once we catch up on orders we will start selling them to some of our stockists which are small game stores. For now, you can purchase them in our web shop. The link is the in the blog post above. thanks!
10. As an avid puzzler, I think this is the best invention since sliced bread. As I have no friends this would be the best thing to do whilst I’m sat at home in my dark dingy room, watching antiques roadshow. Why don’t people make useful things like this more often instead of making stupid things like designer clothes and what not. I may even quit may job and have considered entering infinity puzzle competitions along with my fellow avid puzzlers. | 1,356 | 6,328 | {"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} | 2.75 | 3 | CC-MAIN-2017-26 | longest | en | 0.936276 |
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1. This bundle contains 7 sets of Task cards to focus on reviewing major CCSS standards for Algebra: Algebra 1: Arithmetic of Polynomials & Rational Expressions Task Cards w QR Codes Algebra 1: Linear and Exponential Model Task Cards with QR Codes Algebra 1: Building Functions Task Cards with QR C
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This set of 24 task cards is made up of 6 sets of 4 cards. Each set of exercises has a different focus on the Interpreting Functions
Standards (A.IF)
Set 1. Understand the concept of a function and use function notation.
Set 2. Use function notation in terms of a context.
Set 3: Recognize that sequences are functions, sometimes defined recursively, whose domain is a subset of the integers.
Set 4: Interpret key features of graphs, equations and tables in terms of the
quantities used.
Set 5: Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. Estimate the rate of change from a graph.
Set 6: Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases.
After students have solved a question on their recording sheet, they can use a QR reader to scan the QR code. This will show the students a complete solution to the question. These solutions show students good techniques for solving the problem. They might be able to apply these techniques to solving similar questions.
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# Autarky and Economics Questions and Answers
Info: 4030 words (16 pages) Essay
Published: 25th Jul 2017
Question 1
(a). Write down the problem of an agent that maximizes ex-ante utility in autarky. Find the conditions that characterise the allocation in autarky. Explain how the allocation changes with β.
Autarky is a situation where no trading takes place between agents. Each agent needs to provide for his own needs in an autarky, ie he independently chooses the amount of I that he wants to invest in the long run technology. The issue of liquidity insurance arises here.
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Every agent wants to maximise his ex-ante utility but the problem is that at time t=0 he does not know about his type whether he wants to consume early at t=1 or late at t=2 resulting in asymmetric information. Hence, there is a risk that more than is optimal may be invested.
The conditions that characterise the allocation in autarky are bounded by the constraints of C1 and C2. If agent decides to consume early, he will get savings (1-I) and liquidated investment (ÉI).
C1 = 1 – I + ÉI = 1 – I (1-É)
If agent decides to consume late, he will obtain savings (1-I) and returns from investment (RI).
C2 = 1 – I + RI = 1 + I(R-1)
Agent will choose his consumer profile (C1, C2) that will maximise his ex-ante utility U based on the above constraints.
However, the allocation is not efficient in autarky as shown in the next part of the question.
Max U(C1,C2) = u(C1) + βu(C2) = [1 – I + ÉI]+ [1 – I + RI]= 2+ ÉI + RI
We set up the lagrangian method to explain the allocation changes in β where the constraint in the below equation is the maximum utility.
L = πu(C1) + (1-π)βu(C2) + λ[2+ ÉI + βRI]
= π + λÉI = 0
= (1- π)β + λRI = 0
= 2+ ÉI + βRI = 0
Complementary Slackness Condition: λ*[2+ ÉI + βRI] = 0
If values were given for the variables, we could even have solved and get the value of β. If a value close to zero is obtained for β, it means agent is impatient anda value close to one indicates that agent is patient.
This argument is further supported by the marginal rate of substitution concept where = R. If β=0, no returns obtained as the agent wants to consume immediately. If β=1, returns will result for the patient agent. Hence, it shows that the discount factor β will not change the basic results of the model.
(b) Write down the conditions that characterise the Pareto-optimal allocation. Show that autarky is not efficient. Explain how the allocation changes with β.
The conditions that characterise the allocation in autarky are bounded by the constraints of C1 and C2.
π1C1 = 1 – I => C1 =
(1-π)C2= RI => C2 =
The constraints can be combined in a single one.
π1C1 + (1-π)= 1
The key result is that allocation is inefficient in autarky as shown below:
Recall in autarky: C1 = 1 – I + ÉI = 1 – I (1-É)
C2 = 1 – I + RI = 1 + I(R-1)
If C1 < 1 (unless I = 0) and C2 < R (unless I = 1), then combining these two facts we obtain
π1C1 + π2 < 1 which states that efficiency is not reached. It is true as less money and fewer resources exist in an autarky than in Pareto optimal allocation as no trade occurs. Therefore, consumption level is lower in autarky.
Max U(C1,C2) = u(C1) + βu(C2) = + β
We set up the lagrangian method to explain the allocation changes in β where the constraint in the below equation is the maximum utility.
L = πu(C1) + (1-π)βu(C2) + λ[ + β ]
= π + λ = 0
= (1- π)β + λ β = 0
= + β = 0
Complementary Slackness Condition: λ*[ + β ] = 0
If values were given for the variables, we could even have solved and get the value of β. If a value close to zero is obtained for β, it means agent is impatient anda value close to one indicates that agent is patient.
The argument of marginal rate of substitution is also applicable here where = R. If β=0, no returns obtained as the agent wants to consume immediately. If β=1, returns will result for the patient agent. Hence, it shows that the discount factor β will not change the basic results of the model.
(c) Assume the agents are now infinitely risk-averse. That is U(c1,c2) = min{c1,c2}. What is the Pareto-optimal allocation?
Pareto optimal is an allocation of resources where it is impossible to distribute resources without making at least one consumer worst off. Pareto optimal is the best outcome that could result in an economy with trade taking place and thus there is higher consumption level. It is like a desired state where assets are increased for patient people and consumption is increased for impatient people.
The Pareto optimal allocation for risk neutral agents satisfies the following first order condition:
Uʹ(C1) / Uʹ(C2) = R
which means that agents would like to equate the marginal rate of substitution between consumption levels at t=1 and t=2 with the returns on the long run technology.
When U(c1,c2) = min[c1,c2], it shows agents’ attitude to risk aversion.
The pareto optimal allocation for the risk averse agent is u(C1) + πβu(C2G) + (1-π)βu(C2B) where the superscripts G and B denote good and bad state respectively.
L = u(C1) + πβu(C2G) + (1-π)βu(C2B) + λ[u(C1)]
The concaved utility function states that agents prefer to consume more to less and shows how consumption is smoothed out over time and across states in the future. The agent is risk averse in the sense that he does not want consumption in the bad state at t=2 to be too much different from consumption at t=1.
Question 2
(a) Write down the incentive constraint of the bank. How does collateral affect the repayment R the bank can promise?
Banks, regarded as information sharing coalitions, can easily overcome the problem of asymmetric information of investors. It is assumed that banks will use the signaling tool to invest in high quality projects which will benefit the investors. Banks are expected to behave in such a way that will maximise investors’ interest.
The firm chooses the good project if
pH(y-Ru-Rm) > pL(y-Ru-Rm) + b => Ru + Rm < y-
The bank must also be encouraged to monitor the project:
pHRm– C > pLRm => Rm >
The bank will borrow only least possible amount from banks as bank finance is more expensive than direct finance.
Im = Im (β) ≡ = where β denotes expected rate of return.
The bank will collect get the remaining finance Iu = from uninformed investors. Hence, the bank’s incentive constraint binds.
Using the incentive constraints we have: Ru < y- which states: Iu < [y – ] indicating that the project will only be financed if:
A + Iu + Im > 1 => A > (β,r) ≡ 1 – Im(β) – [y – ]
Other constraints would include a lack of monitoring from the bank giving rise to the probability of non-monitoring pL and the inability to dispose the collateral, ie if the collateral appreciates, the bank will not be able to sell it until loan to investors has been repaid.
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The collateral, usually in the form of assets, plays the role of a guarantee that banks give to investors as a security in case of failure of the project. Collateral is also seen as an alternative to monitoring as it saves efforts and reduces the risk of the bank. ϵ ∈ (0,1) can be interpreted as if K is close to one, bank will be able to refund the money to investors whereas if K is close to zero, bank will be unable to repay back the loan.
A better collateral equals better chance of getting money back as the bank will prefer to behave or else it will lose the collateral.
If the project is successful with expected probability p, the bank will gain returns X which will be used to refund the loan to investors and claim back the collateral. The higher the returns from the project, the bank will be able to distribute partly between the investors and keep partly as its own profits.
In case of failure of the project, the bank will obtain zero returns and is then unable to repay R to the investors. The latter will seize the collateral and will liquidate it to gain maximum money from it as refund of their investment in the unsuccessful project.
(b) Suppose investors have all the bargaining power. Write down their objective, find the optimal contract and their equilibrium profits.
If investors have all the bargaining power, they will be able to influence the project financing process significantly and dictate their terms. The objective of investors is to obtain maximum returns X from the project. They will want to have full details about the project to ensure that the bank is choosing a high quality project (θ) rather than making an adverse selection. Investors delegate the monitoring of the project to the bank as the latter has comparative advantage in monitoring activities hence monitoring costs will be reduced. Investors will use monitoring and auditing as tools to be free from asymmetric information and to improve efficiency. They will expect close monitoring and continuous feedback on the project from the bank.
The optimal contract for investors will be where lending will be most profitable and the below equation is taken from the Diamond Model (1984):
E[y] > 1 + r + C = E[y] > 1 + 1 + C = E[y] > 2+ C
where E[y] = Returns from investment
r = risk free rate, equal to 1 in the question
C = monitoring costs
The optimal contract is bounded by the break-even constraint of uninformed investors implying an upper bound on Iu:
pHRu > (1 + r) => Iu < < [y – ]
Equilibrium profits of the investors will be at a feasible break-even point, usually where demand equals to supply:
A + Iu + Im > 1 => A > (β,r) ≡ 1 – Im(β) – [y – ]
(c) For which value of K can the bank borrow and invest?
The value of the collateral must be either equal or slightly higher than the investment in project (I) and monitoring costs (C) to encourage investors to finance the project as a lower value of the collateral will not attract them.
K = I + C or K > I + C
Ideally if K > I + C, this will attract more investors to finance the project and in turn banks will be able to borrow from them and invest in the project.
Question 3
(a) If A ≥ A3, the firm issues high-quality public debt (public debt that has a high probability of being re-paid)
We will discuss circumstances when the entrepreneur can issue high quality public debt:
• Well-capitalised firms [A > ] can issue direct debt as they possess high capital.
• Low credit risk – High quality public debt refers that the entrepreneur is likely to meet payment obligations. This type of public debt is an attractive investment vehicle as it has a low risk of default.
• High dilution costs
• Good reputed firms can issue direct debt only if πs > where πs is the probability of repayment at t=2, conditionally on success at t=0 and given all firms are monitored at t=0.
• It is assumed that monitoring cost c is small such that in the credit market at equilibrium. The entrepreneur has incentive to issue high quality public debt at a rate of when as the latter equation means high probability of success. The economic interpretation is when project is successful, returns (R) are obtained. The entrepreneur cannot ask for more than R as the firm will also keep some profits for itself. Every party in the transaction is happy and is in equilibrium when a good project is undertaken.
(b) If A3 > A ≥ A2, the firm borrows from a monitor (and from uninformed investors)
We will analyse circumstances when the firm borrows from a monitor and uninformed investors:
• Firms with medium capital [(β,r) < A < ] borrow from banks.
• Firms borrow from banks when they suffer from high credit risk and high dilution costs because banks can provide efficient renegotiation in case of default and can limit dilution costs though there will be an intermediation cost involved.
• Uninformed investors are ready to invest Iu in exchange of return Ru upon successful project. Firms must be encouraged to choose good project pH(y – Ru) > pL (y- Ru) + B <==> Ru < y –
• When the firm falls short of capital to issue a direct debt, it can borrow Im from banks (with return Rm if project succeeds) and Iu from uninformed investors (with return Ru if project succeeds).
The firm chooses the good project if
pH(y-Ru-Rm) > pL(y-Ru-Rm) + b => Ru + Rm < y-
The bank must also be encouraged to monitor the project:
pHRm– C > pLRm => Rm >
The bank will borrow only least possible amount from banks as bank finance is more expensive than direct finance.
Im = Im (β) ≡ = where β denotes expected rate of return.
The bank will collect get the remaining finance Iu = from uninformed investors
Hence, the bank’s incentive constraint binds.
• Two conditions are necessary for bank lending to be in equilibrium in credit market:
(i) Monitoring cost must be less than the returns of the good project
pH G – 1 > c
(ii) Direct lending which is cheaper must be impossible.
pHRc < 1
Firm should borrow from a monitor (for example a bank) and from uninformed investors at intermediate probability of success when pH ] at a rate of R = .
(c) If A2 > A ≥ A1, the firm issues junk bonds (public debt that has a low probability of success)
We will discuss circumstances when the firm issues junk bonds:
• It is possible that firms with medium capital [(β,r) < A < ] issue junk bonds.
• High credit risk- Junk bonds refer to bonds with low credit quality and high default risk. They are attractive to risk seeker investors due to their high yielding returns.
• Low dilution costs as it limits exposure to bad firms but involves inefficient bankruptcy costs for good firms.
• The zero profit condition for investors is:
1 = pR + (1- p) A
This nominal return R is feasible (R < y) if py + (1- p) A > 1 and the expected profit of good firms is then:
πB = p (y- R)+ py
By substituting R, we will obtain: πB = 2py – 1 + (1- p) A
• When the monitoring element c is added, the monitor can reduce the entrepreneur’s private benefit of misbehaving from B to b.
pH > c >(pHpL) R−pH
If R > Rc, the firm will issue junk bonds with low probability of success. This states that the firm is indebted and have too much risk associated with it. The economic interpretation out of it is that the entrepreneur will ask for higher returns but the firm will not afford to provide it. This will lead the entrepreneur to choose the bad project and disequilibrium occurs. Hence, such a combination is not feasible because the maximum repayment is K.
(d) If A1 > A, the firm does not invest
We will analyse circumstances when the firm cannot invest:
• Firms with low capital [A < (β,r)] can neither invest nor borrow. Venture capitalists are the only solution for such firms.
• When monitoring costs are added, if pH < it means there is a small probability of success. The equilibrium consists of no trade taking place and the credit market collapses because good projects cannot be funded and bad projects have a negative net present value. Hence, the firm should not invest as there is no trade equilibrium.
References
Frexias X. and Rochet J-C., (2006) Microeconomics of Banking, 2nd Edition
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Question 1 of 1
ID: GMAT-DS-10
Section: Quantitative Reasoning - Data Sufficiency
Topics: Geometry (3D); Cubes and Cuboid
Difficulty level: Hard
(Practice Mode: Single selected Question » Back to Overview)
A rectangular carton has integer dimensions measuring $a$ centimeters, $b$ centimeters and $c$ centimeters. Is the maximum number of 64 cubic centimeter cubes that can fit in the carton 12?
1. $9 ≤ a < b < c ≤ 12$
2. Only one of the three dimensions of the carton is a prime number.
AStatement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient to answer the question asked.
BStatement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient to answer the question asked.
CBOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked. | 230 | 890 | {"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} | 2.921875 | 3 | CC-MAIN-2021-39 | longest | en | 0.86895 |
https://www.assignmentessays.com/regression-outcome-staistics-assignment-2015/ | 1,579,398,750,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594101.10/warc/CC-MAIN-20200119010920-20200119034920-00368.warc.gz | 761,047,363 | 12,793 | # Regression Outcome/ Staistics Assignment 2015
August 30, 2017
Question
Regression
As a consumer of research, you know that relationships are of critical importance. You must first know if a relationship exists between two variables before you can determine if one variable may account for another. In this week’s readings, you focused on correlations that are used to tell you if two variables are related to one another, but you also now know that you cannot infer causation from a significant correlation alone. That is, you might find that years of education and salary are related, but that does not tell you if more education causes your salary to increase. Correlations also do not allow you to predict a participant’s score on one variable, based on his or her score on another variable. One way to predict one score from another is by using regression. For example, if you wanted to know what salary you could expect in your field if you went back to school for another 2 years, regression could help you make that prediction.
In this Discussion you will apply regression to a research scenario of your choosing.
To prepare: Imagine a situation in which you would like to predict an outcome. Think about why you would choose to use regression rather than correlation. Why is prediction more important than simply describing a relationship?
Post by Day 3 a description of a scenario where you would like to predict an outcome based on a predictor variable. Describe how regression would help you make your prediction. Apply the following terms to your scenario (making sure to fully explain each concept in relation to your example): criterion, predictor, linear regression line, correlation (positive or negative), and proportion of variance accounted for (R2).
Respond by Day 6 to at least one of your peers in one of the following ways:
Challenge key points in the scenario in which they use regression.
Support key points in the scenario in which they use regression.
Resources:
Heiman, G. (2015). Behavioral sciences STAT 2 (2nd ed). Stamford, CT: Cengage.
Chapter 10, “Describing Relationships Using Correlation and Regression” (pp.162-181)
Chapter 13, “Chi Square and Nonparametric Procedures” (pp.218-229 only)
Chapter 10 Review Card (p. 10.4)
Chapter 13 Review Card (p. 13.4)
Media
Note: This video demonstrates calculating and interpreting chi-square analyses in SPSS.
Ludwig, T. E. (n.d.a). Correlation [Interactive media]. Retrieved June 11, 2013, from http://bcs.worthpublishers.com/WebPub/Psychology/psychsim5/PsychSim5%20Tutorials/Correlation/Correlation.htm
Note: This site offers additional information about correlations, including interactive media examples.
This video explains linear regression, including predictor and response variables.
This video shows how to calculate Pearson’s r in SPSS. In addition, a hypothesis test is conducted to determine if the Pearson’s r is significant.
Optional Resources
BBC (Producer). (2010). The joy of stats [Video series]. Retrieved from http://www.bbc.co.uk/programmes/p00cgkfk
“Hans Rosling’s 200 Countries, 200 Years, 4 Minutes”
University of South Carolina. (n.d.a). Regression applet. Retrieved June 11, 2013, from http://www.stat.sc.edu/~west/javahtml/Regression.html
University of South Carolina. (n.d.b). Understanding correlation. Retrieved June 11, 2013, from http://www.stat.tamu.edu/~west/applets/rplot.html
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View Full Version : Asteroid 2012 DA14
lrh_geh
2012-Apr-02, 06:05 PM
What will be the velocity (relative to Earth) when asteroid 2012 DA14 passes near Earth in Feb. 2013? Does DA14 orbit the sun in the same direction as Earth, or in the opposite direction? Could its relative velocity be less than Earth's escape velocity, causing it to veer into collision?
lrh_geh
antoniseb
2012-Apr-02, 06:24 PM
It looks like it will be in the outbound part of its orbit around the Sun, and will pass by the Earth at about 50,000 miles/hr. It is impossible for its relative velocity to be slow enough to cause capture or collision for this, or any asteroid. It orbits in about the same direction as the Earth.
Rhaedas
2012-Apr-02, 06:32 PM
It's a shame its delta-v is so high. A slower moving NEO would have more potential for some type of mission, even if it's just a thrown together one.
lrh_geh
2012-Apr-02, 09:31 PM
50,000 mile/hr / 3600 secs/hr = 13.89 miles/sec. Why is its relative velocity so high? Its period is only one day longer, so I presumed that its orbital velocity would be very nearly the same as Earth's. Its orbit is more eccentric, and not exactly in the same plane as Earth's, but still ....
lrh_geh
It looks like it will be in the outbound part of its orbit around the Sun, and will pass by the Earth at about 50,000 miles/hr. It is impossible for its relative velocity to be slow enough to cause capture or collision for this, or any asteroid. It orbits in about the same direction as the Earth.
Chew
2012-Apr-03, 06:43 AM
50,000 mile/hr / 3600 secs/hr = 13.89 miles/sec. Why is its relative velocity so high?
It's not that high. It's relative velocity at closest approach will be 7.17 km/s (16,040 mph). See NEO Earth Close-Approaches table. (http://neo.jpl.nasa.gov/cgi-bin/neo_ca?type=NEO&hmax=all&sort=date&sdir=ASC&tlim=far_future&dmax=5LD&max_rows=20&action=Display+Table&show=1)
Most of the relative velocity is from being out of plane with the Earth's orbit. It is true DA14's orbit is very similar to Earth's when viewed from above. You are thinking in two dimensions, padawan. Go to the JPL Small-Body Database Browser (http://ssd.jpl.nasa.gov/sbdb.cgi?sstr=2012%20DA14;orb=1;cov=0;log=0;cad=1# orb)
and rotate the view until you are looking at the orbits edge on and you will understand better.
lrh_geh
2012-Apr-03, 12:35 PM
It's not that high. Its relative velocity at closest approach will be 7.17 km/s (16,040 mph).
OK. Thanks. So, its delta-V is roughly twice the velocity required to maintain a geosynchronous circular orbit, but less than the ground-to-orbit escape velocity. The theoretical impact velocity figures I have seen evidently include both the orbital delta-V and the added velocity for a fall from infinity.
Rhaedas
2012-Apr-03, 02:23 PM
A rendezvous mission would be difficult, but what if we put a probe right where the asteroid will pass. Certainly there's a lot we could learn from the impact and its debris field, and how much it changes the orbit.
NEOWatcher
2012-Apr-03, 02:27 PM
A rendezvous mission would be difficult, but what if we put a probe right where the asteroid will pass. Certainly there's a lot we could learn from the impact and its debris field, and how much it changes the orbit.
Perhaps you are at a loss for words, but I think you mean to intercept rather than rendezvous.
antoniseb
2012-Apr-03, 02:32 PM
A rendezvous mission would be difficult, but what if we put a probe right where the asteroid will pass. Certainly there's a lot we could learn from the impact and its debris field, and how much it changes the orbit.
I imagine that getting such a mission together in such a short time would be expensive if possible at all, but more broadly, this kind of opportunity comes fairly often, so building and designing a mission to a yet undetermined close-flyby asteroid seems workable. I don't know if it would save any money... but it would make the mission time-line very short.
Rhaedas
2012-Apr-03, 02:49 PM
Perhaps you are at a loss for words, but I think you mean to intercept rather than rendezvous.
I guess rendezvous is only used for ships in particular (Rama was a ship, after all). Intercept would be correct, but I was thinking a more open ended type of mission that matching velocities would allow, such as an orbit, landing, etc. Encounter maybe?
Chew
2012-Apr-03, 05:30 PM
OK. Thanks. So, its delta-V
I understand your use of "delta-V" here to mean "difference in velocity" but please be aware delta-V has a different meaning in orbital mechanics and astrodynamics. (http://en.wikipedia.org/wiki/Delta-V) When people who know orbital mechanics speak of delta-v it is always in the context of propulsion and changing orbits. A better term for what you want to describe is "relative velocity".
is roughly twice the velocity required to maintain a geosynchronous circular orbit, but less than the ground-to-orbit escape velocity.
Yes, but it's the escape velocity at the distance of its closest approach that matters. The escape velocity at 2014 DA14's nominal approach distance of 57,787 km is only 3.71 km/s.
The theoretical impact velocity figures I have seen evidently include both the orbital delta-V and the added velocity for a fall from infinity.
The infinite velocity is the velocity the asteroid would have at its closest approach to Earth if the Earth had no mass, i.e. the Earth were not pulling it in faster. The infinite velocity is used to calculate the impact velocity if it were in fact to hit us. It is analogous to computing the length of the hypotenuse of a right triangle, with the Earth's escape velocity (11.14 km/s) as one leg, the infinite velocity as the other leg, and the impact velocity as the hypotenuse. Vimpact˛ = Vescape velocity˛ + Vinfinite˛
The relative velocity is the velocity the asteroid will have at closest approach and includes the gravitational pull of the Earth. You will notice the relative velocity is slightly higher than the infinite velocity and the closer the approach to the Earth the bigger the difference in velocities. The NEO close approach table shows DA14 will make the closest approach of all the asteroids listed and it also has the largest increase in speed.
NEOWatcher
2012-Apr-03, 08:13 PM
I guess rendezvous is only used for ships in particular (Rama was a ship, after all). Intercept would be correct, but I was thinking a more open ended type of mission that matching velocities would allow, such as an orbit, landing, etc. Encounter maybe?
Actually; I didn't word my response too well either.
I was thinking "intercept" instead of "put a probe right there".
When I said rendezvous I was kind of rewording what you said rather than explaining my substitution.
I think of rendezvous as being able to stay with an object instead of just whiz by or crash into which I apply to the more broader "intercept". | 1,679 | 6,875 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-35 | latest | en | 0.92255 |
https://encyclopedia2.thefreedictionary.com/free+fall | 1,591,502,422,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348523476.97/warc/CC-MAIN-20200607013327-20200607043327-00349.warc.gz | 326,064,155 | 14,370 | free fall
Also found in: Dictionary, Thesaurus, Acronyms, Idioms, Wikipedia.
free fall,
in physics, the state of a body moving solely under the influence of gravitational forces (see gravitationgravitation,
the attractive force existing between any two particles of matter. The Law of Universal Gravitation
Since the gravitational force is experienced by all matter in the universe, from the largest galaxies down to the smallest particles, it is often called
). A body falling freely toward the surface of the earth undergoes an accelerationacceleration,
change in the velocity of a body with respect to time. Since velocity is a vector quantity, involving both magnitude and direction, acceleration is also a vector. In order to produce an acceleration, a force must be applied to the body.
due to gravity of 32 ft/sec2 (9.8 m/sec2), which is symbolized by g.
Free fall
The accelerated motion toward the center of the Earth of a body acted on by the Earth's gravitational attraction and by no other force. If a body falls freely from rest near the surface of the Earth, it gains a velocity of approximately 9.8 m/s every second. Thus, the acceleration of gravity g equals 9.8 m/s2 or 32.16 ft/s2. This acceleration is independent of the mass or nature of the falling body. For short distances of free fall, the value of g may be considered constant. After t seconds the velocity vt of a body failing from rest near the Earth is given by Eq. (1).
(1)
If a falling body has an initial constant velocity in any direction, it retains that velocity if no other forces are present. If other forces are present, they may change the observed direction and rate of fall of the body, but they do not change the Earth's gravitational pull; therefore a body may still be thought of as freely “failing’’ even though the resultant observed motion is upward.
For a body failing a very large distance from the Earth, the acceleration of gravity can no longer be considered constant. According to Newton's law of gravitation, the force between any two bodies varies inversely with the square of the distance between them; therefore with increasing distance between any body and the Earth, the acceleration of the body toward the Earth decreases rapidly. The final velocity vf, attained when a body falls freely from an infinite distance to the surface of the Earth, is given by Eq. (2),
(2)
where R is the radius of the Earth, which gives a numerical value of 11.3 km/s or 7 mi/s. This is consequently the “escape velocity,’’ the initial upward velocity for a rising body to completely overcome the Earth's attraction.
Because of the independent action of the forces involved, a ball thrown horizontally or a projectile fired horizontally with velocity v will be accelerated downward at the same rate as a body falling from rest, regardless of the horizontal motion.
At a sufficiently large horizontal velocity, a projectile would fall from the horizontal only at the same rate that the surface of the Earth curves away beneath it. The projectile would thus remain at the same elevation above the Earth and in effect become an Earth satellite. See Ballistics, Gravitation
free fall
Motion of a body under the influence of gravity alone, i.e. with no other forces acting. See also weightlessness.
free fall
[′frē ‚fȯl]
(mechanics)
The ideal falling motion of a body acted upon only by the pull of the earth's gravitational field.
(petroleum engineering)
In deep drilling, an arrangement by which the bit is permitted to fall freely to the bottom at each drop or down stroke.
free fall
1. The descent of freshly mixed concrete into forms without dropchutes or other means of confinement.
2. The distance through which such descent occurs.
3. The uncontrolled fall of aggregate.
free fall
A conventional free-fall bomb.
Laser-guided bomb.
i. Any jump in which the parachutist pulls his own ripcord.
ii. The fall or drop of a body, such as a rocket, not guided, not under thrust, and not retarded by a parachute or other braking device. free-fall bomb An unguided bomb that follows the rules of ballistics.
free fall
1. free descent of a body in which the gravitational force is the only force acting on it
2. the part of a parachute descent before the parachute opens
References in periodicals archive ?
However, an object in free fall has no weight, and its speed depends only on gravitational acceleration, not on mass (except for the small effect of air resistance) (5).
Two separate, highly programmable interrupt signals enable immediate notification of free fall, motion or 6D events, giving manufacturers more freedom and flexibility in their design and applications.
* Distance fallen = 1/2 free fall (9.8) x time squared (seconds)
1630 surprisingly led him to see both free fall as "mathematically intractable" and Galileo's solution as extreme in identifying the physical and the mathematical.
One experiment that I will describe tested the equivalence of free fall for different masses in response to local attracting matter (the Earth) rather than to more distant matter (the Sun, which has been used in recent equivalence tests).
Spock drops from the E at South Jordan, Utah for a free-ride free fall (good job, Hubbs).
However, RIMS and Advisen predicted last quarter that underlying economic conditions would keep insurance capacity at levels that would discourage a free fall in pricing.
A large aerospace company designed a structure that was to be dropped from several miles up in the atmosphere and free fall to the ground.
But the plane's maneuvers sent Seth into free fall (the state when gravity isn't balanced by an equal and opposite force).
In the Realms of the Unreal, Yu's Academy Award-short-listed documentary feature (which recently opened in New York and San Francisco), is a cozy, ingratiating introduction to Darger, yet it doesn't shortchange the peculiarities that permitted such a free fall of fascinating aesthetic decisions in his work.
Carnegle introduces three new uplifting textile patterns that support healing environments: Between the Light, Free Fall, and Frameless.
Static line school is the prerequisite to free fall. When making a static line jump, tile student's ripcord is pulled by a line attached to the parachute upon exiting the aircraft.
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Open / Close | 1,329 | 6,330 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2020-24 | longest | en | 0.917996 |
https://forum.freecodecamp.org/t/find-the-longest-word-in-a-string-solved/281146 | 1,657,082,107,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104660626.98/warc/CC-MAIN-20220706030209-20220706060209-00093.warc.gz | 290,816,755 | 6,497 | # Find the Longest Word in a String(solved)
Tell us what’s happening:
So I actually solved it, but I clicked “get a hint” because normally there are better solutions.
In there, there is one similar to mine using reduce instead of map. I can get it and actually maybe it’s a little simpler than mine, but I don’t know, it’s mine quite good enough?
But then there is one using recursiveness. For everything in this world, I can’t ever creatively say “I will use recursivity!”, it just escapes me. Actually in this case I find it quite convulted…
Is recursivity a neccesary trait in the everyday of a coder’s life?
``````
function findLongestWordLength(str) {
let newArr = str.split(' ').map( (element) => element.length );
return newArr.sort((a, b) => b - a)[0];
}
findLongestWordLength("The quick brown fox jumped over the lazy dog");
``````
User Agent is: `Mozilla/5.0 (Macintosh; Intel Mac OS X 10_14_5) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/74.0.3729.169 Safari/537.36`.
Your solution looks really smart to me! It’s great that you made use of map and sort.
I just finished the JavaScript certification and definitely feel your pain… recursive functions are a little mind-bending. You’ll encounter it a few more times as you work through the challenges.
Looking at the variety of solutions in the Hints page is so helpful! It’s nice to be able to compare the solutions as a way of familiarizing yourself with different tools for the same job.
1 Like
The thing about sorting is that it’s computationally expensive compared to iterating over your data. It’s not a big deal for small data sets, but when you want to optimize it’s something you try to avoid.
1 Like
Thanks, yes, I actually feel accompilshed in thinking in those functions, i know it’s quite minimum but well
Got it, so the best solution would be the reduce one?
It is in my opinion. As far as I can tell the three given solutions are all O(n), and the .reduce method is short and clear. Comparison sorts are O(n log n). If you’re not familiar with big O notation, look up a chart and you’ll see why one is preferred to the other.
I am not familiar, I will look into it, thanks!
My fav solution is a functional approach using two functions.
function1 (uppercases first letter of word)
function2 (uses function1 and applies it to each word using map)
Whoops different problem., lol… | 567 | 2,372 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.891531 |
http://mathhelpforum.com/statistics/127619-help-probability-problem-print.html | 1,524,307,650,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945111.79/warc/CC-MAIN-20180421090739-20180421110739-00208.warc.gz | 193,857,513 | 3,675 | # help on a probability problem
• Feb 7th 2010, 09:19 AM
ilc
help on a probability problem
Two cards are drawn from a deck of 52 plating cards. find the probability of each of the following events occurring:
a) both cards are clubs
b) both cards are red
c) both cards are queens
d) both cards are red queens
e) both cards are queens or both cards are red
-----------
a) 13 clubs out of 52 = 13/52
there will be 12 clubs left after picking the first one so: 12/52
so, (13/52)x(12/52) = 156/2704 chances to have both cards as clubs? is this correct?
b) (26/52)x(23/52) = 650/2704
C) (4/52)x(3/52) = 12/2704
d) (26/52)x(23/52) = 2/2704
how would i do e?
• Feb 7th 2010, 09:28 AM
e^(i*pi)
Quote:
Originally Posted by ilc
Two cards are drawn from a deck of 52 plating cards. find the probability of each of the following events occurring:
a) both cards are clubs
b) both cards are red
c) both cards are queens
d) both cards are red queens
e) both cards are queens or both cards are red
-----------
a) 13 clubs out of 52 = 13/52
there will be 12 clubs left after picking the first one so: 12/52
so, (13/52)x(12/52) = 156/2704 chances to have both cards as clubs? is this correct?
b) (26/52)x(23/52) = 650/2704
C) (4/52)x(3/52) = 12/2704
d) (26/52)x(23/52) = 2/2704
how would i do e?
Where did you get 23/52 from in part B? Surely removing 1 red card would mean that 25 remain?
Is the first card replaced? I have assumed it is not
All your second terms should have a denominator of 51.
For example in B:
$\displaystyle \frac{26}{52} \times \frac{25}{51} = \frac{25}{102}$
And in C:
$\displaystyle \frac{4}{52} \times \frac{3}{51} = \frac{1}{221}$
Then part E would be
$\displaystyle \frac{25}{102} + \frac{1}{221}$
• Feb 7th 2010, 09:37 AM
ilc
yes, that should be 25...that was a typo. thanks!
• Feb 7th 2010, 09:42 AM
Quote:
Originally Posted by ilc
Two cards are drawn from a deck of 52 plating cards. find the probability of each of the following events occurring:
a) both cards are clubs
b) both cards are red
c) both cards are queens
d) both cards are red queens
e) both cards are queens or both cards are red
-----------
a) 13 clubs out of 52 = 13/52
there will be 12 clubs left after picking the first one so: 12/52
so, (13/52)x(12/52) = 156/2704 chances to have both cards as clubs? is this correct?
b) (26/52)x(23/52) = 650/2704
C) (4/52)x(3/52) = 12/2704
d) (26/52)x(23/52) = 2/2704
how would i do e?
Hi ilc,
in making these calculations, you are replacing the first card.
This is incorrect as you are choosing 2 from 52,
hence after picking the first card, you have 51 left.
Therefore redo parts a), b) and c).
In part d), you must count the number of red queens.
For part e),
You can calculate the probabilities of getting 2 red cards and add to the probability of getting 2 queens....
but you must subtract the probability of getting 2 red queens as this is a conditional probability question. They will have already been accounted for.
There is overlap since some queens are red.
• Feb 7th 2010, 10:04 AM
ilc
how would i get the probability of the two red queens?
2/53 x 1/51 = 2/2652 correct?
• Feb 7th 2010, 10:13 AM
You only need to know how a pack is organised.
There are 26 red and 26 black.
There are 4 queens, 2 are red and 2 are black.
Therefore 2 of the 26 red cards are red queens.
Therefore there are 24 red cards that are not queens.
There are another 2 queens which are black.
If both cards are queens or both are red, then we must include the probability of also getting 2 black queens.
Also, the queens can be a black and red.
However if we include the 2 red queens, we must be aware that these are already counted,
since the 2 red queens have been already included in 2 red cards total.
Hence we can add the probabilities of 2 red cards to the probability of getting 2 queens,
but we must subtract the probability of getting 2 red queens.
• Feb 7th 2010, 10:33 AM
$\displaystyle \frac{2}{52}\ \frac{1}{51}$
Or choose both red queens at once....$\displaystyle \frac{\binom{2}{2}}{\binom{52}{2}}$ | 1,270 | 4,062 | {"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} | 4.375 | 4 | CC-MAIN-2018-17 | latest | en | 0.956777 |
https://www.daniweb.com/programming/software-development/threads/429958/3d-axis-aligned-bounding-box | 1,709,056,288,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474676.79/warc/CC-MAIN-20240227153053-20240227183053-00550.warc.gz | 733,414,795 | 17,070 | I have two bounding boxes defined by the following values:
``````m1.min.x;//the min x value of model 1
m1.min.y;// " " y " " " "
m1.min.z;// " " z " " " "
m1.max.x;// " max x " " " "
m1.max.y;// " " y " " " "
m1.max.z;// " " z " " " "
m2.min.x;// " min x " " " 2
m2.min.y;// " " y " " " "
m2.min.z;// " " z " " " "
m2.max.x;// " max x " " " "
m2.max.y;// " " y " " " "
m2.max.z;// " " z " " " "
``````
I know that the form is gonna look something like this:
``````if (a&&
b&&
c&&
//...
)
{
//there is a collision
}
``````
I did this in 2d without a problem at all, but now I am in 3d and I am completely out of my element. Any point in the right direction would be helpful.
## All 5 Replies
The 3D case is no different than the 2D case, you just have a lot more cases to worry about. What have you tried?
That is my problem, I have no idea where to start, I just can't figure out how to approach it. So far my idea has been to check the four corners and see if any of the corners is within the other box. I also thought about maybe checking for collision across two dimensions at once... I am just lost in the dealing with the third dimension part.
Does this work... it seems far too simple:
`````` if ((m1.min.x>m2.max.x)||(m2.min.x>m1.max.x)||
(m1.min.y>m2.max.y)||(m2.min.y>m1.max.y)||
(m1.min.z>m2.max.z)||(m2.min.z>m1.max.z))
return true;//there was a collision
else
return false;//there was no collision
``````
I don't think that would work, try this:
``````bool x_in_range = (m1.min.x <= m2.max.x) && (m1.max.x >= m2.min.x);
bool y_in_range = (m1.min.y <= m2.max.y) && (m1.max.y >= m2.min.y);
bool z_in_range = (m1.min.z <= m2.max.z) && (m1.max.z >= m2.min.z);
if( x_in_range && y_in_range && z_in_range )
return true;
else
return false;
``````
I did some boolean logic and came up with what seems like a faster solution (and simpler):
``````bool x_in_range=(m1.min.x<=m2.max.x)&&(m1.max.x>=m2.min.x);
//now using good old boolean logic:
bool x_not_in_range=(m1.min.x>m2.max.x)||(m1.max.x<m2.min.x);//later to be rearranged to use > on both sides
//this can be done to all of the values leading to this:
if (x_not_in_range||y_not_in_range||z_not_in_range)
return false
//since all not_in_range's are composed entirely of || and the final concatenation is done entirely by use of || a statement can be made entirely out of ||
if ((m1.min.x>m2.max.x)||(m2.min.x>m1.max.x)||
(m1.min.y>m2.max.y)||(m2.min.y>m1.max.y)||
(m1.min.z>m2.max.z)||(m2.min.z>m1.max.z))
return false;//no collision
else
return true;//collision
``````
I just want a double-check of my work at this point, but it seems like my original solution works, only backwards :P
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We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge. | 939 | 2,990 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.843408 |
http://blogs.scienceforums.net/swansont/archives/4409 | 1,721,774,832,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518130.6/warc/CC-MAIN-20240723224601-20240724014601-00291.warc.gz | 6,874,881 | 7,965 | How to Kill Mathematicians
Give them pizza, apparently. They’ll starve.
The perfect way to slice a pizza
They can’t think about sharing a pizza, for example, without falling headlong into the mathematics of how to slice it up. “We went to lunch together at least once a week,” says Mabry, recalling the early 1990s when they were both at Louisiana State University, Shreveport. “One of us would bring a notebook, and we’d draw pictures while our food was getting cold.”
The problem that bothered them was this. Suppose the harried waiter cuts the pizza off-centre, but with all the edge-to-edge cuts crossing at a single point, and with the same angle between adjacent cuts. The off-centre cuts mean the slices will not all be the same size, so if two people take turns to take neighbouring slices, will they get equal shares by the time they have gone right round the pizza – and if not, who will get more?
If you really want to mess them up, serve a pizza which isn’t round, and you need to value the crust differently than the rest! That’s like the sheet cake problem — trying to fairness-optimize the volume of cake and surface area of icing. (Gee, I wonder if round food was invented to keep the peace amongst the mathematicians, and allow them to solve other problems)
Or, you could give them a ham sandwich.
(To be fair, I can easily see many scientists falling into a similar analytic trap.) | 313 | 1,406 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-30 | latest | en | 0.969433 |
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# Find the upper limit of the median class from the frequency distribution table.Class$0 - 5$$6 - 11$$12 - 17$$18 - 23$$24 - 29$frequency$3$$10$$15$$8$$11$A. $17$B. $17.5$C. $18$D. $18.5$
Last updated date: 13th Aug 2024
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360.9k+ views
Hint:
For finding the median class we need to calculate the cumulative frequency and convert the data into the continuous format by adding $0.5$ to the upper limit and subtracting the same from the lower limit.
Complete step by step solution:
So here we are given the distribution as:
Class $0 - 5$ $6 - 11$ $12 - 17$ $18 - 23$ $24 - 29$ frequency $3$ $10$ $15$ $8$ $11$
Here the frequency of $5 - 6,11 - 12,17 - 18,23 - 24$ is missing. So here first of all we need to take the average of the upper limit of one class interval and lower limit of the other class interval which is $\dfrac{{5 + 6}}{2} = 5.5$
Similarly the average of $17,18{\text{ is 17}}{\text{.5}}$ and solving for all we will get the averages of all as $5.5,11.5,17.5,23.5$
So now we can write this in the continuous distribution as follows:
Class $0 - 5.5$ $5.5 - 11.5$ $11.5 - 17.5$ $17.5 - 23.5$ $23.5 - 29.5$ frequency $3$ $10$ $15$ $8$ $11$
Now in the continuous for now we can find the cumulative frequency and represent it as follows:
class frequency Cumulative Frequency $0 - 5.5$ $13$ $13$ $5.5 - 11.5$ $10$ $23$ $11.5 - 17.5$ $15$ $38$ $17.5 - 23.5$ $8$ $46$ $23.5 - 29.5$ $11$ $57$
Cumulative frequency is the sum of the preceding all the frequencies to that class interval whose cumulative frequency is to be found.
For the first interval that is $0 - 5.5$ the cumulative frequency is the same as the frequency as there is no preceding interval to that of the interval. Hence in rest of all we keep on adding the preceding all the frequencies to get the cumulative frequencies.
Here $N = 57$
Median is $\dfrac{N}{2} = 28.5$
We need to find the cumulative frequency just after $28.5$ which is $38$ and the interval is here $11.5 - 17.5$
Hence upper limit is $17.5$
Note:
If median class is given as $(a - b)$ then the value of the median would lie between the two values $a{\text{ and }}b$which means between the lower and the higher limit of the median class and the formula is given as:
${\text{median}} = a + \left( {\dfrac{{\dfrac{n}{2} - CF}}{f}} \right)h$
Here $a$is the lower limit of the median class and $n$ is the number of observations, $CF$ is the cumulative frequency of the class preceding the median class, $f$ is the frequency of the median class and $h$ is the class size. | 834 | 2,636 | {"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} | 4.59375 | 5 | CC-MAIN-2024-33 | latest | en | 0.860659 |
https://it.mathworks.com/matlabcentral/profile/authors/870182?detail=cody&page=2 | 1,653,658,033,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662647086.91/warc/CC-MAIN-20220527112418-20220527142418-00153.warc.gz | 383,747,320 | 2,718 | Solved
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https://www.financereference.com/posterior-probability/ | 1,701,308,860,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100164.15/warc/CC-MAIN-20231130000127-20231130030127-00524.warc.gz | 885,694,786 | 21,705 | # Posterior Probability
## What is posterior probability and how is it different from other types of probability
Posterior probability is a type of probability that takes into account both the prior probability of an event and the evidence that has been collected about the event. For example, imagine that you are trying to decide whether or not to attend a concert. The prior probability is the likelihood that you will enjoy the concert without knowing anything about it. The posterior probability is the likelihood that you will enjoy the concert after taking into account all of the evidence that you have collected about it.
This evidence might include reviews from other concertgoers, information about the performers, and your own personal taste in music. In general, the posterior probability will be lower than the prior probability, because the evidence can both confirm and disconfirm our beliefs about an event. However, in some cases, the posterior probability can be higher than the prior probability, if the evidence is particularly strong.
## How to calculate posterior probability
Bayes’ theorem is a formula used to calculate the conditional probability of an event, based on prior knowledge of conditions that might be related to the event. In other words, it allows you to update your estimate of the probability of an event occurring, based on new information. For example, imagine you’re trying to decide whether to buy a particular stock. You know that the stock has a 50% chance of going up and a 50% chance of going down, but you also hear that the company is about to release a new product that is expected to be very popular.
This new information will affect the probability of the stock going up or down, and Bayes’ theorem can be used to calculate the new probabilities. In this example, the posterior probability (the probability after taking into account the new information) would be higher than the prior probability (the probability before taking into account the new information). Bayes’ theorem can be used in many different situations where there is uncertainty about an event occurring, and it can be a valuable tool for making decisions in the face of new information.
## Examples of how to use posterior probability
In statistics, posterior probability is the probability of an event occurring given that another event has already occurred. Bayes’ theorem is a way to calculate posterior probability. It states that the posterior probability of event A occurring is equal to the prior probability of event A multiplied by the likelihood of event A given that event B has occurred, divided by the prior probability of event B. In other words, posterior probability = (prior probability x likelihood) / prior probability of B. An example of how this might be used is if someone wanted to know the likelihood of it raining tomorrow given that the weather forecast says there is a 60% chance of rain.
The prior probability would be the percentage chance of rain without taking into account the forecast, which might be 50%. The likelihood would be the percentage chance of rain if the forecast is accurate, which would be 100%. using Bayes’ theorem, we can calculate that the posterior probability of it raining tomorrow given the forecast is (50% x 100%) / 60% = 83%. This means that there is an 83% chance of it raining tomorrow if the weather forecast is accurate.
## Pros and cons of using posterior probability
The main advantage of using posterior probability is that it allows statisticians to take into account new information as it becomes available. This can be very useful in situations where data is constantly changing, such as in weather forecasting. The main disadvantage of using posterior probability is that it can sometimes be difficult to calculate. Another potential drawback is that, because it relies on data that has already been collected, posterior probability may be less accurate than prior probability in situations where the data is very limited or not representative of the population as a whole.
## When is posterior probability the most useful?
Posterior probability is most useful when making predictions. For example, if you want to predict the likelihood of an event occurring, you can use posterior probability to calculate the chances. This is because posterior probability takes into account all of the available information. This includes both the evidence that supports the event happening and the evidence that goes against it. By considering all of this evidence, posterior probability can give you a more accurate prediction than if you only considered the evidence that supported the event happening.
In addition, posterior probability is also useful for making decisions. For example, if you are trying to decide whether or not to buy a lottery ticket, you can use posterior probability to calculate your chances of winning. This will help you to make a more informed decision about whether or not to purchase the ticket. | 936 | 5,018 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2023-50 | longest | en | 0.953284 |
https://wiki.derivative.ca/index.php?title=Metaball&printable=yes | 1,600,746,632,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400203096.42/warc/CC-MAIN-20200922031902-20200922061902-00175.warc.gz | 723,961,171 | 7,388 | # Metaball
Metaballs can be thought of as force fields whose surface is an implicit function defined at any point where the density of the force field equals a certain threshold. This field can currently be specified as an elliptical or super-quadric shape around a point. When two metaballs overlap in space, their field effects are added together.
## Contents
The field is specified by a weight and a kernel function. The kernel function results in a value of 0 at the outside edge of the metaball and a value of 1 at the center. The kernel function is scaled by the weight to shift the location of the surface closer or further away from the center.
Because the density of the force field can be increased by the proximity of other metaball force fields, metaballs have the unique property that they change their shape to adapt and fuse with surrounding metaballs. This makes them very effective for modeling organic surfaces.
For example, below we have a metaball. The surface of the metaball exists whenever the density of the metaball's field reaches a certain threshold:
When two or more metaball force fields are combined, as in the illustration below, the resulting density of the force fields is added, and the surface extends to include that area where the force fields intersect and create density values with a value of one.
Metaballs are defined by the parameters Center x/y/z, Radius x/y/z, Exponent x/y/z, and a 3×3 rotation matrix which determines the orientation. A metaball is known as a super-quadratic if either exponent is not equal to one.
You can see a metaball's sphere of influence by turning on Display Hulls in a Geometry Viewer's options dialog.
In the SOP editor, a metaball can be selected only by its hull.
## Pusher Metaballs
It is possible for metaballs to have negative Weights (Pusher Metaballs). This allows holes to be created by effectively subtracting from the surface.
## What does an Exponent do?
In the instance of metaballs, the XY and Z exponent determines the inflation towards "squarishness" or contraction towards "starishness" as described below:
• Value > 1 - Results in metaballs that appear more like a "star".
• Value < 1 - Results in metaballs that appear more "squarish".
• Value = 1 - Results in metaballs that appear spherical.
In Touch, metaballs are often used as force fields for particle systems. You can create metaballs with a Metaball SOP, or in the SOP editor.
## Metaball Model Types
• Blinn Kernal - Always puts a sphere at the blob centre, even if the weight is less than 1.0. The Blinn model is the fastest and most stable of all the models.
• Wyvill and Elendt Kernals - These models are very similar; only the weight distribution function is different.
• Links Kernal - This is the slowest method, but provides a good compromise between the Blinn and Wyvill methods in terms of weight distribution.
An Operator Family that reads, creates and modifies 3D polygons, curves, NURBS surfaces, spheres, meatballs and other 3D surface data. | 669 | 3,024 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.915826 |
https://forum.lazarus.freepascal.org/index.php/topic,57287.0/wap2.html?PHPSESSID=unddrode8cmova1oq4soh79cs7 | 1,643,232,856,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304961.89/warc/CC-MAIN-20220126192506-20220126222506-00683.warc.gz | 314,436,454 | 2,306 | Forum > Beginners
If Then else Question
(1/1)
Unkownuser002:
Hey, New here and I got a quick question
I'm trying to figure out the if then else case. I get the error message:
unit1.pas(50,36) Error: Incompatible types: got "Constant String" expected "Real"
unit1.pas(57,4) Fatal: Syntax error, ";" expected but "." found
Sorry if this is relay easy but like I said I'm new and trying to figure this out
So what I'm trying to do is when the amount that is entered in the top is above 200, it will give you a discount of 4% and the extras shipping cost will be clear, but if it's above 200 then there will be a 20 dollar shipping fee.
Sorry for it being in German..
//Eingabe
RBestellBetrag:=strtofloat(EBestellBetrag.text);
//Rechnen
begin
if RBestellBetrag<200 then
RVersandpauschale:=20
else
RVersandpauschale:='';
RRabatt:=4;
end;
RGesamt:=(RBestellBetrag+RVersandpauschale)*0.96;
//Ausgabe
EGesamt.text:=floattostr(RGesamt);
marcov:
Here you assign the integer number 20 to a variable
--- Code: Pascal [+][-]window.onload = function(){var x1 = document.getElementById("main_content_section"); if (x1) { var x = document.getElementsByClassName("geshi");for (var i = 0; i < x.length; i++) { x[i].style.maxHeight='none'; x[i].style.height = Math.min(x[i].clientHeight+15,306)+'px'; x[i].style.resize = "vertical";}};} --- RVersandpauschale:=20
then you assign a string value to the same variable
--- Code: Pascal [+][-]window.onload = function(){var x1 = document.getElementById("main_content_section"); if (x1) { var x = document.getElementsByClassName("geshi");for (var i = 0; i < x.length; i++) { x[i].style.maxHeight='none'; x[i].style.height = Math.min(x[i].clientHeight+15,306)+'px'; x[i].style.resize = "vertical";}};} --- RVersandpauschale:=''
As you don't show types, it is hard to guess which type (string or numeric) is right. But there is definitely a typing logic failure there
Paolo:
Try this
--- Code: Pascal [+][-]window.onload = function(){var x1 = document.getElementById("main_content_section"); if (x1) { var x = document.getElementsByClassName("geshi");for (var i = 0; i < x.length; i++) { x[i].style.maxHeight='none'; x[i].style.height = Math.min(x[i].clientHeight+15,306)+'px'; x[i].style.resize = "vertical";}};} --- //Eingabe RBestellBetrag:=strtofloat(EBestellBetrag.text); //Rechnen if RBestellBetrag<200 then RVersandpauschale:=20 else Begin RVersandpauschale:=0; RRabatt:=4; end; RGesamt:=(RBestellBetrag+RVersandpauschale)*0.96; //Ausgabe EGesamt.text:=floattostr(RGesamt | 801 | 2,651 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2022-05 | latest | en | 0.527276 |
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