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http://blog.lrei.org/math-commons/making-math-relevant/ | 1,600,829,945,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400209665.4/warc/CC-MAIN-20200923015227-20200923045227-00311.warc.gz | 14,657,461 | 23,980 | Making Math Relevant
Simply providing a context to mathematics does not ensure that students will find the task relevant or engaging. Students can be drawn to a problem because it relates to their lives in a real way, or there may be something interesting about a task that intrigues them and challenges them to think deeply. Mathematics that relate to other areas the student is currently studying can also provide relevance. Sometimes it’s not a matter of what specific problem is given, but how it’s posed by the teacher that piques students’ curiosity and results in deep engagement.
Fourth Graders Participated in a Political Art Installation, and then Posed Mathematical Questions About It.
“It’s my first political performance and I just felt I had to do it now. I wanted to show that any wall is dismantlable. We, the public, can tear down walls when society gets together. It could be a mental, physical, or political wall – the point is, it’s ephemeral.”Bosco Sodi, Artist
Here Are the Range of Math Questions They Asked:
Is The Longest Block Area Also The Biggest?
Blog entry by Ariane Stern & Julie Kim
First Graders Explore Area and Perimeter
Earlier in the year, we introduced the concept of area—that you can measure how small or large a space is. After measuring various shapes with non-standard units (e.g. tiles, beans, paperclips), first graders developed an understanding that with bigger units like popsicle sticks, you’d need fewer of them to cover a certain area and that you would need many more small units like beans to fill that same area. Later in the year, we began a unit on linear measurement. Again, we used manipulatives to measure the length of objects. This time, we focused on ways to accurately measure length, such as figuring out which side is the length, and measuring by starting at the edge. After learning how to measure length accurately with one unit, we moved again to the idea of measuring the same object with different sized units. Once again, it takes less popsicle sticks to measure the length of a book, and more cubes, if you were measuring the same length.
This year, to culminate our study on linear measurement, we challenged first graders to measure how long each Lower School classroom’s block area is. In small groups, first graders went to each classroom, calculated which side was the longest side and measured it with string. They brought their strings back to our classroom and proceeded to measure how long they were with popsicle sticks, tiles, and double unit wooden blocks. Afterwards, we collected each group’s data and compared the number of double unit blocks it took to measure the length of each string. We put the lengths in order to see which block area was the longest.
Along the way, we ran into some problems, just as mathematicians do in real life! What would we do if the double units were too long or too short for the end of the string? That led us into a conversation about fractions. We looked at halves, quarters, and thirds of rectangles and also looked at circles to deepen our understanding of fractions. This helped some groups to make more accurate measurements.
Then, Julie showed the class some pictures. Tasha’s block area was the longest, but it was very narrow and skinny. Diane’s block area was the shortest, but it was very wide. We wondered, just because a block area is the longest, does that mean it’s the biggest?
What a big question!
How would we figure that out? We knew what all the lengths of the block areas were, but how would we figure out how big they were? What were we even looking for? One student shouted, “The area! How were we going to figure out the area? We drew a rectangle on the board, showing how we knew the lengths. What else would we need to know to find the area? This was a puzzle. But calling upon what we had already learned about area, we realized that we needed to know the width of each block area. Wow!
Each group then went back to their assigned classroom, this time measuring the width of each block area. When they came back, they measured their new strings again with double unit wooden blocks. Then they used graph paper to replicate their block area. Each square of the graph paper represented one double unit block. Then they counted all the squares to figure out the area. This was hard work! Many of the shapes they drew had over 100 squares! Each group had to count the squares in their shape multiple times to try and get the most accurate count.
Estimating Protest Crowd Size Using the Jacob’s Method of Crowd Counting.
“For many events, especially political rallies or protests, the number of people in a crowd carries political significance and count results are controversial”- Wikipedia
“Almost everyone who has tried to make a crowd estimate has a vested interest in what the outcome of the estimate is”– Charles Seife, Professor of Journalism and Mathematician at NYU.
Fourth graders applied the Jacob’s method of crowd counting to estimate the protest crowd that gathered in Washington Square Park on Wednesday, January 25th, 2017. The emergency rally was organized in response to President Trump’s executive order implementing a ban on immigrants entering the country from large Muslim populations, including Iran, Iraq, Libya, Somalia, Sudan, Syria, and Yemen.
Getting the Numbers Right: The Jacobs Method of Crowd Counting
Herbert Jacobs, a University of California journalism professor in the 1960’s, devised a basic density rule that has been widely accepted. Watching students protesting the Vietnam War from his office window, Jacobs saw that had gathered on a plaza that was arranged in a grid. He counted those in a few squares to get an average number per square and multiplied that by the total number of squares. He also came up with a basic density rule that states a “light crowd” has one person per square meter, and doubled that for a “dense crowd”. A “heavy crowd” would have as many as four people per square meter, according to this method.
Using People Per Square Meter as Density Factors to Determine Crowd Size.
One person per square meter
Three people per square meter
Nine people per square meter
Six people per square meter
Washington Square Park Diagram Showing Density Arrays.
Six people per square meter density computation
Nine people per square meter computation
Kindergartners Use Dramatic Play to Model the Structure of Addition and Subtraction Problems.
Kindergartners related math stories to birds, their current social studies and science topic, by using dramatic play to act out addition and subtraction problems. Through these stories, students explored the action in different structures of problems (see chart below). Some examples of addition structure bird stories that students acted out were result unknown, change unknown, and both addends unknown. “Students who are taught to approach problems by looking at their structures through the use of a visual model are more likely to perform better than students who do not” (NCTM, 2014, pg. 50). This approach helps students make better sense of the action that is happening in problems and rely less on using keywords that can often be interpreted in many different ways (Karp, 2014, pg. 21).
An Example of Student Work Showing Change Unknown Structure Along With a Structure Diagram:
“Six pigeons were pecking on the sidewalk on Bleecker Street. Some more pigeons came to see what they were pecking at. Then there were ten pigeons.” (CHANGE UNKNOWN/ADD TO).
2014 Karp, Karen S., Teaching Children Mathematics. National Council of Teachers of Mathematics
2014 NCTM. Putting Essential Understanding of Addition and Subtraction into Practice
First Graders Cook, Question, and Count
By Julie Kim, First Grade Associate teacher
In the real world, we confront daily math problems through a process of noticing and wondering. After our mind has determined a question about a scenario, whether it is counting how many more blocks you need to walk or how many servings to cook for dinner, we proceed to the next step: plan, search, and gather. We plan for what steps we are going to take in order to answer the question. We search for the separate variables and pieces of information that we need in order to solve the question. We gather these pieces of information then puzzle them up in a way that will help us solve the problem. Will we, as mathematicians, get the answer we are looking for the first time around? Not always. Will we get an answer immediately? Not guaranteed. Will we persevere and try over and over again until we do? We should. Through real world work, first graders develop stamina and perseverance as they attack the challenges and questions that they are eager to solve. This is where the real work happens.
What can a student in the fours do when they run out of a certain sized block?
They can use the “recipe” or conversion chart they made to create the block size they need from other blocks.
Students in Pre-K used classroom blocks to estimate, and then test, how many of each of the same size smaller blocks it would take to cover one double unit block. The class worked in pairs on this investigation, and then the class created posters of their mathematical conversions. They have found this conversion chart, or “recipe” useful to refer to when they run out of a certain block size because they can now create the size they need by combining (composing) other blocks. Continue reading
Do multiplication arrays need to be introduced as a ready-made convention, or can they be “constructed” by children using concrete objects while engaged in an open-ended task?
(An action research collaborative effort by second grade lead teachers, Tasha Hernandez, Bill Miller, and Lower School Math Coordinator, Debra Rawlins)
We began this project as a way to build math relevance into the annual second grade canned food drive. The students visit Saint John’s Food Pantry and spend a morning learning about the needs of the community. Then they pitch in and help out the staff for a few hours. When they return to school, they create signs advertising the canned food drive, and place collection boxes at various locations around the school.
Math with Social Justice Relevance
It was a chilly day in January.
Fourth grade students and teachers went downtown to walk the streets of the busy financial district to meet food cart vendors. They were able to observe the variety of international selections that vendors were selling, and sample the delicious-smelling food.
Over 90% of the food vendors in New York City are first generation, or recent immigrants. This field trip gave students the opportunity to talk directly to the people who stood inside these carts, cooking food that reflected the cuisine of their home countries. Students were curious to hear the stories of where they immigrated from, and how they happened to enter into the business of selling food on the street. Continue reading | 2,236 | 11,006 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2020-40 | latest | en | 0.954568 |
https://www.celebbalah.com/2019/04/04/ohio-u-s-representative-democrat-tim-ryan-seeks-presidency-on-perfect-date/ | 1,652,703,040,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510117.12/warc/CC-MAIN-20220516104933-20220516134933-00631.warc.gz | 824,159,623 | 23,218 | Ohio U.S. Representative Democrat Tim Ryan Seeks Presidency on Perfect Date
As predicted, April 4th, 2019 is a ripe day for Ohioan propaganda.
4/4/2019 => 4+4+20+19 = 47
4/4/19 => 4+4+19 = 27
Democrat = 5+4+5+3+6+9+8+7 = 47 (Reverse Full Reduction)
President = 7+9+5+1+9+4+5+5+2 = 47 (Full Reduction)
Ohio = 15+8+9+15 = 47 (English Ordinal)
Buckeye = 2+3+3+2+5+7+5 = 27 (Full Reduction)
103 is the 27th prime number. Tim Ryan, born July 16, 1973, announces to run on 4/4, 103 days from his upcoming 46th birthday.
Speaking of 4/4, guess what his name equals?
Tim Ryan = 7+9+5 + 9+2+8+4 = 44 (Reverse Full Reduction)
Tim Ryan born 16/7. 167 is the 39th prime number. “Tim” has gematria of 39.
His initials, TR, 20th letter 18th letter, break down to 2 and 9. “Ohio” has gematria of 29.
Next Article
Send this to a friend | 329 | 828 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.870262 |
https://www.physicsforums.com/threads/minimum-velocity-pitched-roof.660140/ | 1,511,215,105,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806225.78/warc/CC-MAIN-20171120203833-20171120223833-00199.warc.gz | 849,725,535 | 20,136 | # Minimum velocity, Pitched roof
1. Dec 20, 2012
### theultimate6
1. The problem statement, all variables and given/known data
What is the minimum velocity required to throw a ball over a house with a pitched roof ( See attachment for the figure) You can choose the throwing point as needed
3. The attempt at a solution
(Choose throwing point standing near $$a$$ and facing left) Assuming b is the hypotenuse of a right triangle and a-c is one side we can find the other side (let's call it d) using the Pythagorean theorem $$b^2 -(a-c)^2 = d$$
$$= -a^2 +2ac +b^2 -c^2$$
a is the maximum height of the projectile
$$a=(v^2sin^2θ)/2g$$
assuming d is the range
$$d=v^2sin(2θ)/g$$
I tried creating an equation for$$v$$from these equations , but it turned out to be too massive to even make a sensible answer.
#### Attached Files:
• ###### Pitched roof.png
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2.9 KB
Views:
59
Last edited: Dec 20, 2012
2. Dec 20, 2012
### haruspex
d2
Not necessarily. You should only assume the trajectory passes through the top of each vertical (a and c).
Different d?
Let x be the distance of the launch point from the first vertical, v the launch velocity and θ the angle. Write down the equations of motion and find what the requirement of just clearing the two verticals tells you. Then use calculus to find the least v. If you're familiar with them, could be a job for Lagrange multipliers.
3. Dec 20, 2012
### Staff: Mentor
If the ball just clears "a" and lands on the roof, but subsequently rolls down the pitch and off at side c, does that count as throwing the ball over the house? Or must it clear the house entirely?
4. Dec 20, 2012
### TSny
You might first consider a related problem. What is the minimum speed VC require to toss a ball from point C to point A in the attached figure? (Or, if you prefer, what's the minimum speed VA required to toss a ball from point A to point C?)
If you could solve that, I think you could use the result to find the answer to the original question. I don't believe you need to worry about what point on the ground the ball should be thrown from.
I'm getting a pretty amazing answer for the original question. But, I've been wrong many times.
#### Attached Files:
• ###### Toss.jpg
File size:
10 KB
Views:
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Last edited: Dec 20, 2012
5. Dec 20, 2012
### ehild
I think the ball can roll on the roof, and that means the smallest kinetic energy required, so the smallest initial speed.
ehild
6. Dec 21, 2012
### haruspex
If you allow it to roll down, and you throw from the peak side of the roof, the answer will clearly be to stand very close and throw almost straight up at speed just over √(2ga).
7. Dec 21, 2012
### theultimate6
Sorry for the late reply, I contacted my professor and he said no rolling is allowed, the ball should aim not to touch the roof at all if only extremely lightly.
$$R=v_ocosθ/g∗(v_osinθ+√(v_o^2sin^2θ)) [tex] [tex] -a^2 +2ac +b^2 -c^2 +x = v_ocosθ/g∗(v_osinθ+√(v_o^2sin^2θ))$$
Actually i have no idea about calculus, :)
8. Dec 21, 2012
### theultimate6
You're right our professor just gave us that hint.
PS: I'm really lost in this problem, i posted a solution but nobody has replied yet, i'm thinking about solving for θ from the maximum height equation then substituting it in the equation i posted, then use calculus (which i have no idea about).
9. Dec 21, 2012
### ehild
I do not understand what you wrote. What are R and x?
ehild
10. Dec 21, 2012
### TSny
For a projectile, fixing the speed at one height will automatically determine the speed at every other height. So, suppose you find the minimum speed the projectile can have at point C, say, such that it barely makes it to point A. That speed at C will fix the speed that the projectile needs to be launched from the ground.
To find the speed at C, you don't need to find the maximum height. Imagine launching the projectile at C at some angle θC. Can you find an expression for the launch speed VC as a function of θC that will just get the projectile to A. Then, as you said, you can use calculus to minimize that function to determine the minimum speed at C that will get the ball to A. Then you can find the launch speed from the ground.
Last edited: Dec 21, 2012
11. Dec 21, 2012
### theultimate6
R is the range and x is the distance of the launch point from the first vertical
I'm not sure if this is correct because time is against me but i'll try.
let y=c
$$c=V_csinθt-0.5gt^2 ?$$
OR
Edit:
$$Sinθ_c = c/v_c$$
$$v_c= c/sinθ_c$$
Last edited: Dec 21, 2012
12. Dec 21, 2012
### theultimate6
Can anybody please type anything my deadline is very near
13. Dec 21, 2012
### theultimate6
$$sinθ=c/v_c$$
$$v_c = c/sinθ$$
$$c=c*sinθ/sinθ*t -0.5gt^2$$
?
14. Dec 21, 2012
### theultimate6
my deadline is in 26 minutes
min = (c^2 csc^2 (theta) sin^2 (alpha))/2g
15. Dec 21, 2012
### ehild
From C to A, the projectile moves along the curve
y=c+xtan(θ)-(g/2) x2/(Vccos(θ))2.
It has to touch the pitch at hight y=a and x=d. You can find Vc2 in terms of theta, and find its minimum.
ehild
16. Dec 21, 2012
### TSny
Sorry, theultimate6.
You mentioned not being familiar with using calculus. I was trying to see if there is a way to solve it without calculus. I don't see another way.
17. Dec 21, 2012
### haruspex
When optimising quadratics, there's often a non-calculus way that consists of expressing the part that becomes zero as a square.
As you said earlier, it reduces to lobbing the ball from one end of the roof just fast enough to clear the far end. Writing down the two equations of motion for a ball thrown at angle theta, speed v, then eliminating time, we get the parabola:
y =x tan(θ) - g x2 sec2(θ)/2v2
To clear the far end:
a-c = d tan(θ) - g d2 sec2(θ)/2v2
where d is the horizontal distance to the far end. Whence
2v2/(gd) = sec2(θ)/(tan(θ) - (a-c)/d)
= (1+t2)/(t - h), where t = tan(θ), h = (a-c)/d
By Pythagoras, 1+h2 = (b/d)2
Whence 1+t2 = 2(h + b/d)(t-h)+(t-h-b/d)2
So 2v2/(gd) = 2(h + b/d)+(t-h-b/d)2/(t-h)
This is minimised wrt t when t-h-b/d = 0, leaving v2/(gd) = h + b/d
v2 = g(a - c + b)
To get the velocity needed from ground level, just need to add the energy to get it up to height c.
18. Dec 21, 2012
### TSny
So far, similar to what I did.
Did not see that trick at all!
Yes, that gets it without calculus. Nice. Did you work out the final answer? I was a little amazed at the form it takes.
19. Dec 22, 2012
### haruspex
Well, of course I solved it with calculus first, then figured that if I subtracted the answer from the non-optimised expression I should get a squared term that vanished at optimum. Wouldn't have found it otherwise.
Yes, it's astonishing. Makes one think there's a much easier way. | 1,969 | 6,740 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2017-47 | longest | en | 0.91802 |
http://www.acmerblog.com/backtracking-3-2937.html | 1,481,457,145,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698544678.42/warc/CC-MAIN-20161202170904-00271-ip-10-31-129-80.ec2.internal.warc.gz | 305,848,541 | 14,330 | 2013
12-27
回溯法（2）
/* 主题：批处理作业调度算法
* 作者：chinazhangjie
* 邮箱：[email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */
* 开发语言：C++
* 开发环境：Mircosoft Virsual Studio 2008
* 时间: 2010.10.24
*/
#include <iostream>
#include <vector>
using namespace std;
class flowshop
{
public:
flowshop(vector<vector<int> >& rhs) {
each_t = rhs ;
machine1_t = 0 ;
total_t = 0 ;
best_total_t = 0 ;
for (int i = 0 ;i < task_count; ++ i) {
current_t[i] = i; // 为了实现全排列
}
}
void backtrack () {
__backtrack (0);
// 显示最佳调度方案和最优完成时间和
cout << "the best flowshop scheme is : ";
copy (best_t.begin(),best_t.end(),ostream_iterator<int> (cout, " "));
cout << endl;
cout << "the best total time is : " << best_total_t << endl;
}
private:
void __backtrack (int i) {
if (total_t < best_total_t || best_total_t == 0) {
// 存储当前最优调度方式
copy (current_t.begin(),current_t.end(),best_t.begin()) ;
best_total_t = total_t;
}
return ;
}
for (int j = i; j < task_count; ++ j) {
// 机器1上结束的时间
machine1_t += each_t[current_t[j]][0] ;
if (i == 0) {
machine2_t[i] = machine1_t + each_t[current_t[j]][1] ;
}
else {
// 机器2上结束的时间
machine2_t[i] =
((machine2_t[i - 1] > machine1_t) ? machine2_t[i - 1] : machine1_t)
+ each_t[current_t[j]][1] ;
}
total_t += machine2_t[i];
// 剪枝
if (total_t < best_total_t || best_total_t == 0) {
// 全排列
swap (current_t[i],current_t[j]) ;
__backtrack (i + 1) ;
swap (current_t[i],current_t[j]) ;
}
machine1_t -= each_t[current_t[j]][0] ;
total_t -= machine2_t[i] ;
}
}
public :
vector<vector<int> > each_t ; // 各作业所需的处理时间
vector<int> current_t ; // 当前作业调度
vector<int> best_t ; // 当前最优时间调度
vector<int> machine2_t ; // 机器2完成处理的时间
int machine1_t ; // 机器1完成处理的时间
int total_t ; // 完成时间和
int best_total_t ; // 当前最优完成时间和
};
int main()
{
// const int task_count = 4;
const int task_count = 3 ;
for (int i = 0;i < task_count; ++ i) {
each_t[i].resize (2) ;
}
each_t[0][0] = 2 ;
each_t[0][1] = 1 ;
each_t[1][0] = 3 ;
each_t[1][1] = 1 ;
each_t[2][0] = 2 ;
each_t[2][1] = 3 ;
// each_t[3][0] = 1 ;
// each_t[3][1] = 1 ;
flowshop fs(each_t) ;
fs.backtrack () ;
}
1)不同列：xi != xj
2)不处于同一正、反对角线：|i-j| != |x(i)-x(j)|
/* 主题：n后问题
* 作者：chinazhangjie
* 邮箱：[email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */
* 开发语言：C++
* 开发环境：Mircosoft Virsual Studio 2008
* 时间: 2010.10.24
*/
#include <iostream>
#include <vector>
using namespace std;
class queen
{
// 皇后在棋盘上的位置
struct q_place {
int x;
int y;
q_place ()
: x(0),y(0)
{}
};
public:
queen(int qc)
: q_count (qc), sum_solution (0) {
curr_solution.resize (q_count);
}
void backtrack () {
__backtrack (0);
}
private:
void __backtrack (int t) {
if (t >= q_count) {
// 找到一个解决方案
++ sum_solution ;
for (size_t i = 0;i < curr_solution.size(); ++ i) {
cout << "x = " << curr_solution[i].x
<< " y = " << curr_solution[i].y << endl;
}
cout << "sum_solution = " << sum_solution << endl;
}
else {
for (int i = 0;i < q_count; ++ i) {
curr_solution[t].x = i;
curr_solution[t].y = t;
if (__place(t)) {
__backtrack (t + 1);
}
}
}
}
// 判断第k个皇后的位置是否与前面的皇后相冲突
bool __place (int k) {
for (int i = 0; i < k; ++ i) {
if ((abs(curr_solution[i].x - curr_solution[k].x)
== abs(curr_solution[i].y - curr_solution[k].y))
|| curr_solution[i].x == curr_solution[k].x) {
return false;
}
}
return true;
}
private:
vector<q_place> curr_solution; // 当前解决方案
const int q_count; // 皇后个数
int sum_solution; // 当前找到的解决方案的个数
};
int main()
{
queen q(5);
q.backtrack ();
return 0;
}
1. bottes vernies blanches
I appreciate the efforts you men and women place in to share blogs on such sort of matters, it was certainly useful. Keep Posting! | 1,631 | 4,436 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2016-50 | longest | en | 0.163239 |
https://mltocups.net/90-ml-to-cups/ | 1,708,470,084,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473347.0/warc/CC-MAIN-20240220211055-20240221001055-00589.warc.gz | 407,392,850 | 23,611 | # Convert 90 ml to cups. How many cups are in 90 ml?
Did you know that there are approximately 90 ml in one cup? Converting between cups and milliliters can be a challenge, but it’s important to understand the conversions when baking or cooking. This guide will walk you through the process of converting 90 ml to cups, so you can always be prepared with the right measurement. Enjoy!
A cup is a unit of measurement typically used to measure volume. It is equal to 8 fluid ounces in the U.S., and to about 250 milliliters in most other countries. There are 0.38 cups in 90 milliliters.
Convert 90 ml to cups. How many cups are in 90 ml?
## Convert 90 milliliters to cups, quick conversion
Convert 90 milliliters to cups, quick conversion
Convert 90 ml to cups by dividing 90 by 236.59. This results in 0.38 cups. It is important to be precise when converting units in cooking and baking. Small variations can greatly affect the outcome of a recipe. Converting units can easily be done using a calculator or an online converter tool. However, it is helpful to have a basic understanding of the conversion process for common kitchen measurements.
See also How many cups is 175 ml of milk? Convert 175 ml to cups
1 cup is equal to 236.59 ml, so to convert from ml to cups, divide the quantity in ml by 236.59. To convert from cups to ml, multiply the number of cups by 236.59.
Additionally, 1 tablespoon is equal to 14.7867648 ml and 1 teaspoon is equal to 4.9289216 ml. Using these conversions, it is possible to convert between teaspoons, tablespoons, and ml. It is also important to note that these conversions may vary slightly in different countries due to variations in the measurement system.
It is important to pay attention to the measurements listed in a recipe and use the proper conversion when necessary. Accurately converting units can ensure the success of your dish.
### Example
• 90 ml water in cups = 90/236.59 = 0.38 cups
• 3 tablespoons in ml = 3 x 14.7867648 = 44.36 ml
• 1 cup sugar in teaspoons = 1 x 236.59/4.9289216 = 48 teaspoons of sugar.
Remember to always double check your conversions and measurements for a successful dish! Happy cooking!
Next, we will have a look at 90 ml to cups conversion table.
## 90 ml to cups conversion table
### 90 ml to cups us
Ml 90 ml to cups us 89,1 89,1 ml = 0,376002 Cups 89,2 89,2 ml = 0,376424 Cups 89,3 89,3 ml = 0,376846 Cups 89,4 89,4 ml = 0,377268 Cups 89,5 89,5 ml = 0,37769 Cups 89,6 89,6 ml = 0,378112 Cups 89,7 89,7 ml = 0,378534 Cups 89,8 89,8 ml = 0,378956 Cups 89,9 89,9 ml = 0,379378 Cups 90,0 89,9999999999999 ml = 0,3798 Cups 90,1 90,0999999999999 ml = 0,380222 Cups 90,2 90,1999999999999 ml = 0,380644 Cups 90,3 90,2999999999999 ml = 0,381066 Cups 90,4 90,3999999999999 ml = 0,381488 Cups 90,5 90,4999999999999 ml = 0,38191 Cups 90,6 90,5999999999999 ml = 0,382332 Cups 90,7 90,6999999999999 ml = 0,382754 Cups 90,8 90,7999999999999 ml = 0,383176 Cups 90,9 90,8999999999999 ml = 0,383598 Cups 91,0 90,9999999999999 ml = 0,38402 Cups 91,1 91,0999999999999 ml = 0,384441999999999 Cups 91,2 91,1999999999999 ml = 0,384863999999999 Cups 91,3 91,2999999999999 ml = 0,385285999999999 Cups 91,4 91,3999999999999 ml = 0,385707999999999 Cups 91,5 91,4999999999999 ml = 0,386129999999999 Cups 91,6 91,5999999999999 ml = 0,386551999999999 Cups 91,7 91,6999999999999 ml = 0,386973999999999 Cups 91,8 91,7999999999998 ml = 0,387395999999999 Cups 91,9 91,8999999999998 ml = 0,387817999999999 Cups 92,0 91,9999999999998 ml = 0,388239999999999 Cups 92,1 92,0999999999998 ml = 0,388661999999999 Cups 92,2 92,1999999999998 ml = 0,389083999999999 Cups 92,3 92,2999999999998 ml = 0,389505999999999 Cups 92,4 92,3999999999998 ml = 0,389927999999999 Cups 92,5 92,4999999999998 ml = 0,390349999999999 Cups 92,6 92,5999999999998 ml = 0,390771999999999 Cups 92,7 92,6999999999998 ml = 0,391193999999999 Cups 92,8 92,7999999999998 ml = 0,391615999999999 Cups 92,9 92,8999999999998 ml = 0,392037999999999 Cups 93,0 92,9999999999998 ml = 0,392459999999999 Cups
### 90 ml to cups canada
Ml 90 ml to cups canada 89,1 0,3919852005 89,2 0,392425139 89,3 0,3928650775 89,4 0,393305016 89,5 0,3937449545 89,6 0,394184893 89,7 0,3946248314 89,8 0,3950647699 89,9 0,3955047084 90,0 0,3959446469 90,1 0,3963845854 90,2 0,3968245239 90,3 0,3972644624 90,4 0,3977044009 90,5 0,3981443394 90,6 0,3985842779 90,7 0,3990242164 90,8 0,3994641549 90,9 0,3999040934 91,0 0,4003440319
See also How much is 460 ml in cups? 460 ml of oil, water to cups
### 90 ml to cups australia
Ml 90 ml to cups australia 89,1 0,4455 89,2 0,446 89,3 0,4465 89,4 0,447 89,5 0,4475 89,6 0,448 89,7 0,4485 89,8 0,449 89,9 0,4495 90,0 0,45 90,1 0,4505 90,2 0,451 90,3 0,4515 90,4 0,452 90,5 0,4525 90,6 0,453 90,7 0,4535 90,8 0,454 90,9 0,4545 91,0 0,455
## F.A.Q 90 ml to cups
### How many cups are in 90 ml?
90 ml is equal to 0.38 cups when converted.
### What cup is 90g?
90g is equal to 0.38 cups in weight when converted. However, it is important to note that 1 cup of a substance does not necessarily equal 1 cup in volume or weight. It all depends on the density of the substance (for example, 1 cup of water will equal both 1 cup in volume and 1 cup in weight, but 1 cup of flour will be lighter and have less volume than 1 cup of water).
### How many ml is a cup of something?
1 cup is equal to 236.59 ml when converted. However, it is important to note that the conversion may vary slightly in different countries due to variations in the measurement system.
### What measure is 100ml?
100ml is equal to 0.42 cups when converted. It is also equal to 3.4 tablespoons or 6.8 teaspoons. Again, it is important to note that these conversions may vary slightly in different countries due to variations in the measurement system.
See more articles in the category: ML to cups | 2,093 | 5,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-10 | latest | en | 0.921562 |
https://www.uio.no/studier/emner/matnat/fys/FYS2140/v14/data/crank-nicolson/cranknicolsonharmonicoscillator.py | 1,603,831,796,000,000,000 | text/plain | crawl-data/CC-MAIN-2020-45/segments/1603107894759.37/warc/CC-MAIN-20201027195832-20201027225832-00283.warc.gz | 882,116,598 | 2,575 | #!/usr/bin/env python """ Created on Sun 27 Oct 2013. Uses the Crank Nicolson scheme to solve the time dependent Schrodinger equation for a harmonic oscillator. Animation is done using the matplotlib.pyplot library. Usage: python CrankNicolsonHarmonicOscillator.py or equivalent: ./CrankNicolsonPotentialHarmonicOscillator.py No commandline arguments are needed. In this script the wave is centered around a point lying away from the center of the potential. It is started without momentum. @author Kristoffer Braekken """ # Tools for sparse matrices import scipy.sparse as sparse import scipy.sparse.linalg # Numerical tools from numpy import * # Plotting library from matplotlib.pyplot import * """Physical constants""" _E0p = 938.27 # Rest energy for a proton [MeV] _hbarc = 0.1973 # [MeV pm] _c = 3.0e2 # Spees of light [pm / as] def Psi0( x ): ''' Initial state for a stationairy gaussian wave packet. ''' x0 = -0.2 # [pm] Starts at center. a = 0.0050 # [pm] A = ( 1. / ( 2 * pi * a**2 ) )**0.25 K1 = exp( - ( x - x0 )**2 / ( 4. * a**2 ) ) return A * K1 def harmonicOscillator( x, k=2*pi*5e1 ): """ The potential for a quantum harmonic oscillator for a proton. @param k The wave number. Default value chosen by eye. """ potential = 0.5*(_E0p/(_c*_c))*k*k*x*x return potential if __name__ == '__main__': nx = 1001 # Number of points in x direction dx = 0.001 # Distance between x points [pm] # Use zero as center, same amount of points each side a = - 0.5 * nx * dx b = 0.5 * nx * dx x = linspace( a, b, nx ) # Time parameters T = 1 # How long to run simulation [as] dt = 1e-5 # The time step [as] t = 0 time_steps = int( T / dt ) # Number of time steps # Constants - save time by calculating outside of loop k1 = - ( 1j * _hbarc * _c) / (2. * _E0p ) k2 = ( 1j * _c ) / _hbarc # Create the initial state Psi Psi = Psi0(x) # Create the matrix containing central differences. It it used to # approximate the second derivative. data = ones((3, nx)) data[1] = -2*data[1] diags = [-1,0,1] D2 = k1 / dx**2 * sparse.spdiags(data,diags,nx,nx) # Identity Matrix I = sparse.identity(nx) # Create the diagonal matrix containing the potential. V_data = harmonicOscillator(x) V_diags = [0] V = k2 * sparse.spdiags(V_data, V_diags, nx, nx) # Put mmatplotlib in interactive mode for animation ion() # Setup the figure before starting animation fig = figure() # Create window ax = fig.add_subplot(111) # Add axes line, = ax.plot( x, abs(Psi)**2, label='\$|\Psi(x,t)|^2\$' ) # Fetch the line object # Also draw a green line illustrating the potential ax.plot( x, V_data, label='\$V(x)\$' ) # Add other properties to the plot to make it elegant fig.suptitle("Solution of Schrodinger's equation with harmonic oscillator") # Title of plot ax.grid('on') # Square grid lines in plot ax.set_xlabel('\$x\$ [pm]') # X label of axes ax.set_ylabel('\$|\Psi(x, t)|^2\$ [1/pm] and \$V(x)\$ [MeV]') # Y label of axes ax.legend(loc='best') # Adds labels of the lines to the window draw() # Draws first window # Time loop while t < T: """ For each iteration: Solve the system of linear equations: (I - k/2*D2) u_new = (I + k/2*D2)*u_old """ # Set the elements of the equation A = (I - dt/2*(D2 + V)) b = (I + dt/2. * (D2 + V)) * Psi # Calculate the new Psi Psi = sparse.linalg.spsolve(A,b) # Update time t += dt # Plot this new state line.set_ydata( abs(Psi)**2 ) # Update the y values of the Psi line draw() # Update the plot # Turn off interactive mode ioff() # Add show so that windows do not automatically close show() | 1,046 | 3,494 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2020-45 | latest | en | 0.661364 |
https://gocardless.com/guides/posts/customer-churn/ | 1,709,245,186,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474853.43/warc/CC-MAIN-20240229202522-20240229232522-00845.warc.gz | 279,549,817 | 58,153 | Pricing
Resources
# Understanding Customer Churn Rate
Written by
Acquiring and losing customers is a natural part of any business that applies to both the largest and smallest organisations in the world. This cycle of gain and loss is referred to in business circles as customer churn.
## .css-g8fzsc{padding:0;margin:0;font-weight:700;}How to calculate customer churn rate
The customer churn rate is a valuable metric that can be used to help businesses understand the efficacy of their marketing and overall customer satisfaction. The easiest way to calculate customer churn rate is to divide the number of customers lost over a set period by the total number of customers from that same period.
This provides you with a percentage of your total customer base. So, for example, if a company had 1000 customers at the start of the month and 800 by the end of the month it would have a customer churn rate of 20%.
## Why is a low customer churn rate important?
Generally speaking, the lower the customer churn, the more customers you retain. A higher churn rate, meanwhile, means you are losing more customers than you are gaining. As a result, low churn rates are a desirable asset. The most desirable outcome is a negative churn rate, which means the business will be gaining more customers than it’s losing.
## Understanding customer churn rate
As a metric, it’s invaluable when it comes to assessing strategy. Because if churn rate is low or negative then you’re obviously doing something right and if it’s high then you must be doing something wrong. To really understand customer churn, however, you need to understand the difference between voluntary and involuntary churn and how to react to each respectively.
### Voluntary churn
This refers to a customer actively deciding to leave. How they achieve this will depend on your business model. If you are a subscription service, for example, if a customer cancels their subscription then this would be voluntary churn.
### Involuntary churn
If a customer loses access to your service unintentionally this is referred to as involuntary churn. This accounts for between 20% and 40% of churn and happens most regularly when customers forget to update their payment information when a card expires.
### How to decrease customer churn rate
There is no catch-all customer churn rate formula and it’s something that all businesses must contend with. However, there are a few ways to prevent high levels of churn.
### Direct debit
The first thing to take care of should be involuntary churn as this is arguably the easiest thing to fix. Bank-to-bank payments such as direct debit are always going to be more reliable than credit cards, as bank accounts don’t just expire or get lost without customers realising it. GoCardless offers a reliable and effective direct debit payment system with failure rates as low as 0.5%.
The best way to fix voluntary customer churn is to understand why your customers are leaving. Market research is always going to be your friend here. Don’t be afraid to ask those who have left why they decided to leave.
### Provide support
Customers are more likely to leave if they don’t feel like they are being listened to or can’t find the answers they are looking for. Provide them with all of the resources and information necessary to ensure they remain happy.
### Target the right audience
Ensure you’re targeting the right people at the right time and that your goals and ideals are aligned with the people you want to be attracting.
### Know when to step in
Have they not logged onto their account for weeks? And when they do, are their sessions decidedly short. Know what warning signs to look out for so you know when to step in and take action to retain their business.
#### We can help
If you’re interested in finding out more about customer churn rate, or any other aspect of your business finances, then get in touch with our financial experts at GoCardless. Find out how GoCardless can help you with ad hoc payments or recurring payments.
Contact sales
Sales
Support | 838 | 4,105 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-10 | longest | en | 0.942895 |
http://www.research.stlouisfed.org/fred2/series/PCDGCC96 | 1,417,112,761,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416931009004.88/warc/CC-MAIN-20141125155649-00171-ip-10-235-23-156.ec2.internal.warc.gz | 803,612,603 | 19,126 | # Real Personal Consumption Expenditures: Durable Goods
2014:Q3: 1,429.9 Billions of Chained 2009 Dollars (+ see more)
Quarterly, Seasonally Adjusted Annual Rate, PCDGCC96, Updated: 2014-11-25 7:51 AM CST
Click and drag in the plot area or select dates: Select date: 1yr | 5yr | 10yr | Max to
BEA Account Code: DDURRX1
A Guide to the National Income and Product Accounts of the United States (NIPA) - (http://www.bea.gov/national/pdf/nipaguid.pdf)
Release: Gross Domestic Product
Restore defaults | Save settings | Apply saved settings
w h
Graph Background: Plot Background: Text:
Color:
(a) Real Personal Consumption Expenditures: Durable Goods, Billions of Chained 2009 Dollars, Seasonally Adjusted Annual Rate (PCDGCC96)
BEA Account Code: DDURRX1
A Guide to the National Income and Product Accounts of the United States (NIPA) - (http://www.bea.gov/national/pdf/nipaguid.pdf)
Real Personal Consumption Expenditures: Durable Goods
Integer Period Range: to copy to all
Create your own data transformation: [+]
Need help? [+]
Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
will be applied to formula result
Create segments for min, max, and average values: [+]
Graph Data
Graph Image
Suggested Citation
``` US. Bureau of Economic Analysis, Real Personal Consumption Expenditures: Durable Goods [PCDGCC96], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/PCDGCC96/, November 27, 2014. ```
Retrieving data.
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Name: Email: | 572 | 2,171 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2014-49 | latest | en | 0.760287 |
https://johnsinclairradio.com/basic-addition-subtraction-worksheets/ | 1,643,460,096,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306181.43/warc/CC-MAIN-20220129122405-20220129152405-00004.warc.gz | 404,409,026 | 8,017 | # First-rate Basic Addition Subtraction Worksheets
Worksheet. September 16th , 2020.
Basic addition subtraction worksheets. Have fun and learn to sum with these addition worksheets. Some effective strategies that will help them recall their basic facts quickly (strategies such as +0, +1, +2, doubles This pack will help students with basic addition and subtraction facts, counting on, fact families, subitizing, and speed/automaticity with math facts.
It is important when learning the basic math operations to develop the skill of looking at the operation itself on each problem. Next to our regular addition, subtraction, divison and multiplication worksheets, we have also added worksheet with mixed operations. It has 77 pages of basic addition and subtraction fact worksheets aimed at helping your students practice and learn their basic facts.
Addition to 11, 12 or 13 generator (lesley richmond) addition to 14, 15 or 16 generator (lesley richmond) addition to 17, 18, 19 or 20 generator (lesley richmond) adding to 10 (philomena shotton) teddy bear subtraction (michelle stone) ways to make 20 (gillian miller) timed addition (katie ward) doc; 5th grade addition and subtraction worksheets including adding and subtracting large numbers, missing addend problems and missing minuend or subtrahend problems. These addition worksheets with pictures pdf will enable kids to perfectly visualize the relationship between numbers and their quantities.
Download and print a subtraction dice game, bingo activity, math mystery pictures, and number line activities. Thankfully, our subtraction worksheets eliminate both of those issues with a variety of lessons that teach this important skill in an entertaining way. Simple addition and subtraction (amalia.
Mixed addition and subtraction word problems. Often, when we focus on only a single type of math fact at a time, progressing our way through addition, subtraction, multiplication and division facts, we can find that students become 'modal' when looking at a worksheet. In the first section, we've included a few addition printables that should help out the beginning student.
Mixing up the operations will enhance the students math skills, as they will be able to reognize the relationshi between the operations, the relationship between addition and subtraction, and the one between division and multiplication. Basic addition and subtraction worksheets high resolution. Free math worksheets for basic operations this worksheet generator allows you to make worksheets for addition, subtraction, division, and multiplication of whole numbers and integers, including both horizontal and vertical forms (long division etc.), and simple equations with variables.
Free addition subtraction worksheets for preschool, kindergarden, 1st grade, 2nd grade, 3rd grade, 4th grade and 5th grade Just grab a bundle of our basic addition worksheets for grade 1 and perfectly understand addition with beautiful models and number lines. Some of the worksheets for this concept are subtraction work 100 vertical subtraction.
Plunge into practice with our addition and subtraction worksheets featuring oodles of exercises to practice performing the two basic arithmetic operations of addition and subtraction. This download contains 217 pages altogether. Addition and subtraction fact family.
Some of the worksheets for this concept are unit 1 easy addition facts, addition, adding with some regrouping a, basic addition scoot, addition work 2 digit plus 2 digit addition with, year 2 maths addition and subtraction workbook, math fact fluency work, domino addition 1. Students are required to figured out which operation to apply given the problem context. Well maybe more than a few, so it's probably a good thing.
This page has a large collection of basic subtraction worksheets. This set of worksheets includes a mix of addition and subtraction word problems.
Pin on TpT Math Lessons
Adding and Subtracting with Facts From 1 to 5 (A) Mixed
Groundhog Subtraction Numbers 05 Subtraction
Simple Addition for winter Kindergarten Pinterest
The Single Digit Addition 100 Horizontal Questions
Summer Addition Worksheet Kindergarten math worksheets
Free Math Worksheets Word Problems For Addition and
First Grade Math Worksheets Math worksheets, Free math
Simple Addition sentences for fall! KinderLand
Love is Key. Math, Math fact fluency, Math facts
subtraction worksheet insects Pinterest Subtraction
Free Math Worksheets and Printouts Math addition
Addition and subtraction activities and worksheets
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Simple Addition up to 10 with built in snowflake
## 1st Grade Language Arts Worksheets Free Printables
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###### Household Budget Worksheet For Retirement
Feb 03, 2021 | Audrey Party | 1,036 | 5,077 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2022-05 | latest | en | 0.922613 |
http://bartleylawoffice.com/faq/what-is-sales-tax-in-florida-on-vehicles-solved.html | 1,680,234,697,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949533.16/warc/CC-MAIN-20230331020535-20230331050535-00230.warc.gz | 6,734,416 | 20,097 | # What Is Sales Tax In Florida On Vehicles? (Solved)
Florida collects a six percent sales tax on the purchase of all new or used vehicles. According to AutoList.com, buyers are required to register and collect vehicles on every vehicle they sell. That said, if you buy a car from someone privately, you are responsible for paying the sales tax.
Florida vehicle sales tax
• Florida collects a 6% state sales tax rate on the purchase of all vehicles. However, the total sales tax can be higher depending on the local tax of the area in which the vehicle is purchased in with a maximum tax rate of 1.5%.
## What is the sales tax when buying a car in Florida?
The estimated cost to register and title a vehicle for the first time is \$420.00, plus any sales tax due. The state of Florida imposes 6% sales tax on the full purchase price less trade-in. Pinellas County residents pay an additional 1% on the first \$5,000.
## How do I avoid paying sales tax on a car in Florida?
Does a trade-in reduce sales tax in Florida? A trade-in does reduce the sales tax in Florida. When you trade-in a vehicle to buy a new one, you will pay sales tax on the price after factoring in the trade-in value.
## How do you calculate sales tax on a car?
To calculate the sales tax on your vehicle, find the total sales tax fee for the city. The minimum is 7.25%. Multiply the vehicle price (before trade-in or incentives) by the sales tax fee. For example, imagine you are purchasing a vehicle for \$20,000 with the state sales tax of 7.25%.
You might be interested: Where To Mail Maryland State Tax Returns? (Solution)
## How much tax will I pay for a used car?
Since it directly impacts their revenue from taxes, they set the sales tax rate based on their own financial conditions and other influencing factors. The national average is around 5.75%. So, if you’re buying a used car for \$10,000, expect to pay around \$575 as sales tax.
## Do I have to pay taxes twice if I buy a car out of state?
Do I have to pay taxes twice if I buy a car out of state? No, you will only pay taxes once to the state where you register the car. If you buy from a dealer they will often collect this for you and remit to your home state.
## Is there a yearly vehicle tax in Florida?
Florida collects a six percent sales tax on the purchase of all new or used vehicles. According to AutoList.com, buyers are required to register and collect vehicles on every vehicle they sell. That said, if you buy a car from someone privately, you are responsible for paying the sales tax.
## Does Florida have a yearly car tax?
New York, New Jersey, Texas, and Florida are a few of the states that don’t charge a vehicle property tax at all. On the other end of the spectrum, it can be rather expensive to own a vehicle in many states. The most expensive vehicle taxes are in Rhode Island, where the average driver pays \$1,133 per year.
## Do you have to pay taxes on a car you buy from a private owner?
When you purchase a vehicle through a private sale, you must pay the associated local and state taxes. In most states, you’ll need to bring your Bill of Sale and signed title to the Department of Motor Vehicles (DMV) or motor vehicle registry agency to pay your taxes and obtain your registration, new title, and plates.
You might be interested: How does a supply curve illustrate the law of supply
## How much tax will I pay on a car?
Stamp duty is calculated at \$3 per \$100, or part thereof, of the vehicle’s value. For passenger vehicles valued over \$45,000 with seating for up to 9 occupants, the rate of stamp duty is \$1,350 plus \$5 per \$100, or part thereof, of the vehicle’s value over \$45,000.
## How do I figure out sales tax?
Multiply the cost of an item or service by the sales tax in order to find out the total cost. The equation looks like this: Item or service cost x sales tax (in decimal form) = total sales tax. Add the total sales tax to the Item or service cost to get your total cost. | 910 | 3,993 | {"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} | 2.5625 | 3 | CC-MAIN-2023-14 | latest | en | 0.960015 |
http://slideplayer.com/slide/2786931/ | 1,526,820,019,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863410.22/warc/CC-MAIN-20180520112233-20180520132233-00372.warc.gz | 278,195,084 | 20,075 | # Section 2.5 Solving Linear Equations in One Variable Using the Multiplication-Division Principle.
## Presentation on theme: "Section 2.5 Solving Linear Equations in One Variable Using the Multiplication-Division Principle."— Presentation transcript:
Section 2.5 Solving Linear Equations in One Variable Using the Multiplication-Division Principle
2.5 Lecture Guide: Solving Linear Equations in One Variable Using the Multiplication-Division Principle Objective: Solve linear equations in one variable using the multiplication-division principle. Just as the addition-subtraction principle allows us to strategically add or subtract the same value on both sides of an equation, the multiplication-division principle allows us to strategically multiply or divide values on both sides of an equation to give the desired coefficient on the variable.
1. Give a verbal description of the correct step to solve each equation below, and then solve the equation: (a) (b) (c) (d)
Multiplication-Division Principle of Equality
Verbally If both sides of an equation are multiplied or divided by the same nonzero number, the result is an equivalent equation. Algebraically Numerical Example If a, b, and c are real numbers and is equivalent to , then is equivalent to and to ; . and is equivalent to .
Strategy for Solving Linear Equations
Step 1: Simplify each side of the equation. (a) If the equation contains fractions, simplify by multiplying both sides of the equation by the least common denominator (LCD) of all the fractions. (b) If the equation contains grouping symbols, simplify by using the distributive property to remove the grouping symbols and then combine like terms. Step 2: Using the addition-subtraction principle of equality, isolate the variable terms on one side of the equation and the constant terms on the other side. Step 3: Using the multiplication-division principle of equality, solve the equation produced in Step 2.
The following examples require using the multiplication-division principle of equality. Some will also require using the addition-subtraction principle of equality. Solve each equation. Note that we can check our solutions of each equation. 2.
The following examples require using the multiplication-division principle of equality. Some will also require using the addition-subtraction principle of equality. Solve each equation. Note that we can check our solutions of each equation. 3.
The following examples require using the multiplication-division principle of equality. Some will also require using the addition-subtraction principle of equality. Solve each equation. Note that we can check our solutions of each equation. 4.
The following examples require using the multiplication-division principle of equality. Some will also require using the addition-subtraction principle of equality. Solve each equation. Note that we can check our solutions of each equation. 5.
The following examples require using the multiplication-division principle of equality. Some will also require using the addition-subtraction principle of equality. Solve each equation. Note that we can check our solutions of each equation. 6.
The following examples require using the multiplication-division principle of equality. Some will also require using the addition-subtraction principle of equality. Solve each equation. Note that we can check our solutions of each equation. 7.
The following examples require using the multiplication-division principle of equality. Some will also require using the addition-subtraction principle of equality. Solve each equation. Note that we can check our solutions of each equation. 8.
The following examples require using the multiplication-division principle of equality. Some will also require using the addition-subtraction principle of equality. Solve each equation. Note that we can check our solutions of each equation. 9.
The following examples require using the multiplication-division principle of equality. Some will also require using the addition-subtraction principle of equality. Solve each equation. Note that we can check our solutions of each equation. 10.
The following examples require using the multiplication-division principle of equality. Some will also require using the addition-subtraction principle of equality. Solve each equation. Note that we can check our solutions of each equation. 11.
The following examples require using the multiplication-division principle of equality. Some will also require using the addition-subtraction principle of equality. Solve each equation. Note that we can check our solutions of each equation. 12.
The following examples require using the multiplication-division principle of equality. Some will also require using the addition-subtraction principle of equality. Solve each equation. Note that we can check our solutions of each equation. 13.
14. Note the difference between simplifying expressions and solving equations:
(a) Simplify (b) Solve
Mentally estimate the solution of each equation and then use your calculator to solve the equation.
15. Mental estimate: _________ Calculator solution: _________
Mentally estimate the solution of each equation and then use your calculator to solve the equation.
16. Mental estimate: _________ Calculator solution: _________
17. Write an algebraic equation for the following statement and then solve the equation.
Verbal Statement: Three times the quantity x plus four is two less than x. Algebraic Equation: Solve this equation:
18. The perimeter of the rectangle shown equals . Find a.
19. Solve the equation by letting equal the left side of the equation and equal the right side of the equation. (a) Create a table on your calculator with the table settings: TblStart = 0; Complete the table below. x Y1 Y2 1 2 3 4 5 6 The x-value at which the two y-values are equal is ______.
19. Solve the equation by letting equal the left side of the equation and equal the right side of the equation. (b) Use your calculator to create a graph of and using a viewing window of . Use the Intersect feature to find the point where these two lines intersect. Draw a rough sketch below. The values in the table will help. The point where the two lines intersect has an x-coordinate of ______.
19. Solve the equation by letting equal the left side of the equation and equal the right side of the equation. (c) Solve the equation algebraically. (d) Check your solution.
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Similar presentations | 1,297 | 6,589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2018-22 | latest | en | 0.838529 |
https://brilliant.org/problems/bamboo-forest/ | 1,669,563,069,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710409.16/warc/CC-MAIN-20221127141808-20221127171808-00610.warc.gz | 178,297,861 | 8,684 | # Bamboo Forest
You are surveying a rectangular area of a bamboo forest of $2 \times 3$ square feet. The four bamboos at the corners are each $12, 27, 56, 59$ feet high, and when you analyze the surface area at the top, you find that it is a partial plain of $f(x,y) = y^2 + x^3 - 2xy +7,$ as shown below.
Assuming the area is densely packed with bamboos, what is the average height of these bamboos in feet?
× | 122 | 413 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 3, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2022-49 | latest | en | 0.921424 |
https://learn.rmotr.com/python/base-python-track/week-1-quiz/all-in-string | 1,558,839,950,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258621.77/warc/CC-MAIN-20190526025014-20190526051014-00544.warc.gz | 530,444,736 | 7,771 | Solution 1
``````def all_in_string(a_string, s1, s2, s3):
total = 0
if s1 in a_string:
total += 1
if s2 in a_string:
total += 1
if s3 in a_string:
total += 1
``````
# All in String
The function `all_in_string` receives a phrase and 3 words, and it should count how many of those words appear in the original phrase at least once.
Example, given the sentence from our previous assignment:
Courage is not the absence of fear, but rather the judgement that something else is more important than fear
(Ambrose Redmoon)
How many of the following words you see included in that sentence: `fear`, `judgement`, `Python`? The answer is `2`, because only `fear` and `judgement` are part of the original sentence.
Examples of `all_in_string`:
``````phrase = "Python is good. Python is Wise. I like Python"
all_in_string(phrase, 'Python', 'good', 'Ruby') # 2
all_in_string(phrase, 'Python', 'Javascript', 'Ruby') # 1
all_in_string(phrase, 'Python', 'Wise', 'good') # 3
all_in_string(phrase, 'Ruby', 'Javascript', 'Perl') # 0
``````
Hint: The `in` operator might be handy.
### Test Cases
test two in string -
``````def test_two_in_string():
assert all_in_string('abcd', 'a', 'b', 'X') == 2
assert all_in_string('abcd', 'a', 'X', 'b') == 2
assert all_in_string('abcd', 'X', 'a', 'b') == 2
``````
test repeated words -
``````def test_repeated_words():
a_string = "Python is good. Python is Wise. I like Python"
assert all_in_string(a_string, 'Python', 'good', 'Ruby') == 2
``````
test only one in string -
``````def test_only_one_in_string():
assert all_in_string('abcd', 'a', 'X', 'Y') == 1
assert all_in_string('abcd', 'X', 'a', 'Y') == 1
assert all_in_string('abcd', 'Y', 'X', 'a') == 1
``````
test three in string -
``````def test_three_in_string():
assert all_in_string('abcd', 'a', 'b', 'c') == 3
``````
def all_in_string(a_string, s1, s2, s3): pass | 585 | 1,866 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-22 | latest | en | 0.629553 |
https://www.smore.com/jpz5 | 1,544,562,930,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823702.46/warc/CC-MAIN-20181211194359-20181211215859-00636.warc.gz | 1,037,006,264 | 12,374 | # What's New in Room 305
## Week of 6/12 - 6/16
Good Morning Room 305 Families!
Summer Birthdays Party! - For all the kids who have birthdays throughout summer, we didn't forget your birthday! On Thursday, June 22nd, we will be getting together with Mr. Christian's Class to do a HUGE summer birthdays party! So if your birthday is between June 24th and August 30th we want to celebrate YOU! I will be sending home an information sheet about this birthday party. Our class alone has 6 birthdays throughout summer so we want to make sure we don't have 6 sets of cupcakes coming into class! I will send the sheet home in Boomerangs today for you to peruse and if you have a child with a summer birthday, begin talking with them about what they may want to bring in (if anything)!
Last Week of School SPIRIT WEEK: On the last week of school (NEXT WEEK?!) we will be having an all school spirit week! The dress up days are listed below:
• Monday - Ridgecrest Pride Day - wear RC shirts or BLUE
• Tuesday - Sports Day!
• Wednesday - Superhero Day!
• Thursday - Crazy Hair Day!
• Friday - Dress up Day!
Math This week, we will finish the Step up to Fourth Grade Lessons (SUTFG)! On Monday, students will use models to add fractions with like denominators. Students will then try to put these sums in simplest form. On Tuesday, students will subtraction fractions with like denominators. Students will then try to put these differences in simplest forms. On Wednesday, we will do our last SUTFG Lesson of the year: decimals.
Reading Groups - This week, students will be meeting in blended reading groups with Mr. Christian’s class Monday through Thursday. The students are reading new books in their reading groups and will be completing a team project for each of their books. This project will require the kids to write chapter summaries as a team for every chapter in the book and illustrate the chapter as well. We will make matchbook foldables to summarize the novels in words and in pictures. We will read these books and take Reading Counts Quizzes on them the last week of school!
Social Studies This week in our Storypath Northwest Coastal Unit, we will engage in a critical event in which they will respond to traders visiting the coast. Students will have to decide how to respond to the traders, trade or not? Students will write about the incident. Next, students will respond to the arrival of settlers. Students will collaborate and discuss how they will respond to either a land dispute or disease. Students will write about this incident. Finally, students will participate in the relocation of the Pacific Northwest Coastal Indians by the settlers and learn about relocation. Students will respond to this critical incident in writing and in character.
Science This week, we will continue our last Science Unit of the year focused on the Physics of Sound. Students will also explore pitch more by using different sized nails and a pencil. One of our last explorations about pitch this week will be using rulers and strumming them over our desks at different lengths to explore pitch. Students will read several nonfiction articles this week about sound, including Thunder and Lightning, Let’s go to the Movies, and The Elephant’s Rumble. We will continue to explore pitch by plucking rulers over desks and by creating a slide whistle and noticing the pitch difference when the air column is long and short. We will also create reed instruments from straws. Last, this week we will begin to learn about the inner workings of the ear and learn about how we hear sounds.
Nightly Homework -
• Read for 20 minutes each night and get Assignment Sheet signed
• Bring a snack for each day of the week to be enjoyed at 10am
This Week's Assignments -
• Math Step Up to 4th Grade Lesson 8 - Due Tuesday
• Math Step Up to 4th Grade Lesson 9 - Due Wednesday
• Math Step Up to 4th Grade Lesson 10 - Due Thursday
Coming up -
- Monday June 12th – Field Trip to the Burke Museum 11:15-2:30pm
- Wednesday June 14th – Visit from the King County Librarian
- Wednesday June 14th – Instrumental Concert
- Thursday June 15th - Buddies with Mrs. Bender 9:45-10:35
- Friday June 16th – Happy Birthday Shannon!
Student of the Week!!!
• Congratulations Eva Wagner for Patience at school!
• We're focusing on Teacher's Choice this week!
Down below, look for your child's LOCKER NUMBER (they should know and if you don't know, to preserve confidentiality, email me and I can let you know what they're number is) and you will see if your child has turned in all of his/her first week's assignments. Find their number, look down and green indicates it was turned in. Red means I don't have it yet. REMEMBER: This could mean that your child finished it, but hasn't put it in the "In-Bin" yet. Check with them if you're sure you completed it together! Email me with questions! | 1,074 | 4,876 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.943997 |
https://oeis.org/A208533 | 1,624,587,034,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488560777.97/warc/CC-MAIN-20210624233218-20210625023218-00403.warc.gz | 389,287,057 | 4,056 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A208533 Number of n-bead necklaces of n colors not allowing reversal, with no adjacent beads having the same color. 2
1, 1, 2, 24, 204, 2635, 39990, 720916, 14913192, 348684381, 9090909090, 261535848376, 8230246567620, 281241174889207, 10371206370593250, 410525522392242720, 17361641481138401520, 781282469565908953017, 37275544492386193492506, 1879498672877604463254424 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 LINKS Andrew Howroyd, Table of n, a(n) for n = 1..80 FORMULA a(n) = (1/n) * Sum_{d | n} totient(n/d) * ((n-1)*(-1)^d + (n-1)^d) for n > 1. - Andrew Howroyd, Mar 12 2017 EXAMPLE All solutions for n=4: ..2....1....1....1....1....1....2....1....1....3....1....1....1....2....1....1 ..3....2....4....4....4....3....4....4....3....4....3....4....2....3....2....2 ..2....4....2....3....2....2....3....1....1....3....4....3....1....4....3....1 ..4....2....4....2....3....3....4....4....3....4....2....4....4....3....2....2 .. ..1....1....2....1....2....1....1....1 ..2....3....3....3....4....2....2....3 ..1....4....2....1....2....4....3....2 ..3....3....3....4....4....3....4....4 MATHEMATICA a[1] = 1; a[n_] = (1/n)*DivisorSum[n, EulerPhi[n/#]*((n-1)*(-1)^# + (n-1)^#)& ]; Array[a, 20] (* Jean-François Alcover, Nov 01 2017, after Andrew Howroyd *) PROG (PARI) a(n) = if (n==1, 1, (1/n) * sumdiv(n, d, eulerphi(n/d) * ((n-1)*(-1)^d + (n-1)^d))); \\ Michel Marcus, Nov 01 2017 CROSSREFS Diagonal of A208535. Sequence in context: A052780 A245019 A189769 * A174668 A302444 A121213 Adjacent sequences: A208530 A208531 A208532 * A208534 A208535 A208536 KEYWORD nonn AUTHOR R. H. Hardin, Feb 27 2012 EXTENSIONS a(14)-a(20) from Andrew Howroyd, Mar 12 2017 STATUS approved
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Last modified June 24 21:50 EDT 2021. Contains 345433 sequences. (Running on oeis4.) | 813 | 2,220 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.506266 |
https://www.studyadda.com/question-bank/sound_q22/3255/264260 | 1,576,452,595,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541310970.85/warc/CC-MAIN-20191215225643-20191216013643-00408.warc.gz | 876,790,693 | 19,831 | • question_answer A hollow sphere is filled with water. It is hung by a long thread. As the water flows out of a hole at the bottom, the period of oscillation will A) first increase and then decrease B) first decrease and then increaseC) go on increasingD) go on decreasing
Time period of simple pendulum $T=2\pi \sqrt{(\frac{l}{g})\propto \sqrt{l}}$where / is effective length. [i.e distance between centre of suspension and centre of gravity of bob] Initially, centre of gravity is at the centre of sphere. When water leaks the centre of gravity goes down until it is half filled; then it begins to go up and finally it again goes at the centre. That is effective length first increases and then decreases. As $T\propto \sqrt{l}$, so time period first increases and then decreases. | 194 | 789 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2019-51 | latest | en | 0.867287 |
http://www.onlineclothingstudy.com/2013/12/answer-post-50-short-questions-for.html | 1,516,431,504,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889473.61/warc/CC-MAIN-20180120063253-20180120083253-00704.warc.gz | 495,489,600 | 81,497 | Answer to merchandising questions on apparel merchandising.
Last week I posted one question set for you titled as '50 Short Questions for Merchandisers: Test Yourself where do You Stand'. I know that many of you tried to give answers of those questions and noted down those answers. As committed to you, here I am publishing answers of those questions.
Question 1.
Initial quantity break-up size wise XS=100, S=200, M=300, L=200, XL=100
Size wise consumption: XS=1.0, S=1.1, M=1.2, L=1.3, XL=1.4
After changes: quantity break-up size wise S=200, M=300, L=300, XL=100
So, Size wise Fabric consumption would be -
S@1.1 * 200=220 mtrs,
M@1.2 * 300=360 mtrs,
L@1.3 * 300 = 390 mtrs
XL@1.4 * 100 =140 mtrs
Therefore average consumption =(220+360+390+140)/900 =1.23333 Meters.
Question 2.
When you break total quantity as per ratio you will size wise quantity as following.
As a general rule size wise consumption XS<S<M<L<XL.
So higher the quantity in higher sizes higher would be the consumption.
XS S M L XL
a. 0 1000 2000 2000 1000
b. 1000 1000 2000 2000 0
c. 1000 2000 2000 1000 0
d. 0 2000 2000 1000 1000
From the above quantity break-up the descending order would be a<d<b<c
Question. 4.
Break total FOB Rs 400 as per cost ratio
12 :1 :3 :1 :3
240 :20 :60 :20 :60
Fabric cost increased by 5% i.e Rs. 12. So margin will be less by Rs. 12.
Current ratio is 252:20:60:20:48
Current margin = Rs. 48.
So reduction in margin = 12/60*100 = 20%
Question 5
Matching left and right hand column
a) – iv);
b) – i);
c) –v);
d) – ii);
e) – iii)
Question 41.
b. is most appropriate. Different agents used different calculation factors. It also depends on importing country. For more reading check here.
a) FPT: Fabric Package Testing
b) GPT: Garment Package Testing
c) CIF: Cost Insurance Freight
d) LC : Letter of Credit
e) SKU: Stock Keeping Unit
f) DTM : Dye to Match
g) CPL: Cut panel Laundry
Friends, if anyone don’t agree with any answer given in the above table please write in the comment box with justification.
Name
5S,3,Activewear,1,Apparel ERP,1,apparel industry,57,apparel news,34,Apparel Production,31,Apparel Software,17,Bamboo Fabric,1,Bangladesh,2,Blogs,1,Books,31,Business Guides,9,Buyer,1,Buying House,4,CAD/CAM,1,Calculation & Formula,1,career,34,Change Management,1,Clothing,2,Clothing Brand,3,Compliance,3,Consumption,4,Costing,11,Cutting,20,Denim,3,Directory,3,Dyeing,1,eCommerce,2,Education,5,Event & Fair,14,Fabric,73,Fashion,5,Fashion Business,41,Fashion Design,3,Fashion Show,2,Fiber,2,Finishing,10,Garment Exporters,1,Garmenting,2,Gartex,1,GSM,1,Guest Post,8,Human Resource,2,Industrial Engineering,214,Infographic,5,Knitting,4,KPI,7,Kurti,1,Lean,27,Line layout,4,Machines,32,Manufacturing,53,Merchandising,105,MIS,24,MISC,22,Operation bulletin,14,Operator,1,Operator Training,1,Performance rating,1,PLM,3,Printing & Embroidery,11,Problem Solving,1,Production Planning,27,Project Report,1,QNA,122,Quality,61,Quality manual,2,Quality system,14,Research project,1,Resources,43,Retailing,4,Salary & Wages,7,Sampling,22,Scheduling,3,Sourcing,8,Spec Sheet,9,Sportswear,4,Stitching,3,Store,2,Stretch fabric,1,Student corner,1,sustainable clothing,7,T-shirt,3,Tailor,1,Tailoring Shop,2,Technology,59,Texprocess,1,Textile processing,11,Textile testing,23,Textiles,42,Threads,1,Towel,1,Trims & Accessories,15,videos,12,Waste,1,Work Study,38,Yoga wear,1,
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Online Clothing Study: Answer Post: 50 Short Questions for Merchandisers
Answer Post: 50 Short Questions for Merchandisers
Answer to merchandising questions on apparel merchandising. | 1,174 | 3,613 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-05 | latest | en | 0.805389 |
https://ask.truemaths.com/question/a-cylindrical-tube-open-at-both-ends-is-made-of-metal-the-internal-diameter-of-the-tube-is-11-2-cm-and-its-length-is-21-cm-the-metal-thickness-is-0-4-cm-calculate-the-volume-of-the-metal/ | 1,675,793,801,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500628.77/warc/CC-MAIN-20230207170138-20230207200138-00520.warc.gz | 125,819,417 | 29,690 | • 0
Newbie
A cylindrical tube open at both ends is made of metal. The internal diameter of the tube is 11.2 cm and its length is 21 cm. The metal thickness is 0.4 cm. Calculate the volume of the metal.
• 0
An Important Question of M.L Aggarwal book of class 10 Based on Mensuration Chapter for ICSE BOARD.
A cylindrical tube open at both ends is made of metal. The internal diameter and length of the tube is given .
And the metal thickness is also given. Calculate the volume of the metal.
This is the Question Number 18, Exercise 17.1 of M.L Aggarwal.
Share
1. Given internal diameter of the tube = 11.2 cm
Internal radius, r = d/2 = 11.2/2 = 5.6 cm
Length of the tube, h = 21 cm
Thickness = 0.4 cm
Outer radius, R= 5.6+0.4 = 6 cm
Volume of the metal = R2h- r2h
= h(R2-r2)
= (22/7)×21×(62-5.62)
= 66×(6+5.6)(6-5.6)
= 66×11.6×0.4
= 306.24 cm3
Hence the volume of the metal is 306.24 cm3.
• 1 | 311 | 909 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2023-06 | longest | en | 0.856154 |
http://morris821028.github.io/2014/07/06/oj/uva/uva-12206/ | 1,560,938,201,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998943.53/warc/CC-MAIN-20190619083757-20190619105757-00393.warc.gz | 112,449,775 | 9,092 | # UVa 12206 - Stammering Aliens
## Problem
Dr. Ellie Arroway has established contact with an extraterrestrial civilization. However, all efforts to decode their messages have failed so far because, as luck would have it, they have stumbled upon a race of stuttering aliens! Her team has found out that, in every long enough message, the most important words appear repeated a certain number of times as a sequence of consecutive characters, even in the middle of other words. Furthermore, sometimes they use contractions in an obscure manner. For example, if they need to say bab twice, they might just send the message babab, which has been abbreviated because the second b of the first word can be reused as the first b of the second one.
Thus, the message contains possibly overlapping repetitions of the same words over and over again. As a result, Ellie turns to you, S.R. Hadden, for help in identifying the gist of the message.
Given an integer m, and a string s, representing the message, your task is to find the longest substring of s that appears at least m times. For example, in the message baaaababababbababbab, the length-5 word babab is contained 3 times, namely at positions 5, 7 and 12 (where indices start at zero). No substring appearing 3 or more times is longer (see the first example from the sample input). On the other hand, no substring appears 11 times or more (see example 2).
In case there are several solutions, the substring with the rightmost occurrence is preferred (see example 3).
## Input
The input contains several test cases. Each test case consists of a line with an integer m (m$\ge$1), the minimum number of repetitions, followed by a line containing a string s of length between m and 40 000, inclusive. All characters in s are lowercase characters from a'' toz’’. The last test case is denoted by m = 0 and must not be processed.
## Output
Print one line of output for each test case. If there is no solution, output none; otherwise, print two integers in a line, separated by a space. The first integer denotes the maximum length of a substring appearing at least m times; the second integer gives the rightmost possible starting position of such a substring. | 492 | 2,212 | {"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} | 2.515625 | 3 | CC-MAIN-2019-26 | longest | en | 0.933387 |
http://darkwing.uoregon.edu/~botvinn/420_F13.html | 1,516,383,363,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888077.41/warc/CC-MAIN-20180119164824-20180119184824-00103.warc.gz | 87,753,860 | 1,828 | # Math 420/520, Ordinary Differential Equations, Fall 2013
• Instructor: Boris Botvinnik
• Class meets MWF 10:00-10:50, Deady 303
• Office: Fenton 304.
• Phone: 346-5636.
• E-mail: botvinn@math.uoregon.edu (Please always include a correct e-mail return address.)
• Office Hours: MW 1:00-1:50 pm,
• Web Page: http://darkwing.uoregon.edu/~botvinn/420_F13.html
• Elementary Differential Equations, by Boyce and DiPrima, 9th edition.
• 1. Background and Goals. Differential equations are used to describe processes that vary continuously
with respect to time. In applications, one often knows some relationship between an unknown function
(or system of functions) and its derivatives, and uses this relation to determine the original function.
This course covers the basic theory of ordinary differential equations. This includes stability theory
and the existence and uniqueness of solutions, and techniques of solutions for linear systems of equations.
• 2. Exams. There will be a midterm in-class exam on Friday, November 1st, 10:00-10:50 a final exam on
Monday, December 9th, 10:15-12:15.
• 3. Homework. Homework problems will be assigned every week and are due in class on Wednesday on
the material of the previous week. No late homework will be accepted.
• 4. Final Exam Review. Here are two hand-outs:
The first one
The second one
• 5. Grading. The grading distribution will be as follows:
Homework: 25%
Midterm Exam: 25%
Final Exam: 50%
• 6. Weekly Schedule:
1. Systems of first order linear differential equations. Read 7.1, 7.2, 7.3.
2. Linear systems with constant coefficients. Read 7.4, 7.5, 7.6.
3. Fundamental matrices and eigenvalues. Read 7.7, 7.8, 7.9.
4. Existence and uniqueness theorems. Read 2.4, 2.8.
5. Stability and the phase plane. Read 9.1, 9.2
6. Almost linear systems, MIDTERM. Read 9.3
7. Applications to population dynamics. Read 9.4, 9.5
8. Liapunov's second method. Read Chapter 9.6.
9. Periodic solutions, limit cycles, and chaos. Read Chapter 9.7, 9.8.
10. Review. | 576 | 1,996 | {"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} | 2.796875 | 3 | CC-MAIN-2018-05 | latest | en | 0.845206 |
https://www.physicsforums.com/threads/if-the-charge-on-an-isolated-conductor-is-negative-is-its.431848/ | 1,726,223,966,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651513.89/warc/CC-MAIN-20240913101949-20240913131949-00424.warc.gz | 842,490,380 | 15,992 | # If the charge on an isolated conductor is negative, is it's
• Ashu2912
In summary, the charge on an isolated conductor can be negative, positive, or neutral, depending on the number of electrons and protons within the conductor. This charge can change from negative to positive if the conductor gains or loses particles. The charge on an isolated conductor affects its behavior by creating an electric field that can attract or repel other charged objects. An isolated conductor can have a neutral charge if it has an equal number of protons and electrons.
Ashu2912
If the charge on an isolated conductor is negative, is it's capacitance negative by the formula C = Q / V??
No because V would also be negative.
## 1. What does it mean if the charge on an isolated conductor is negative?
If the charge on an isolated conductor is negative, it means that the conductor has an excess of electrons. This results in an overall negative charge on the conductor.
## 2. Can the charge on an isolated conductor change from negative to positive?
Yes, the charge on an isolated conductor can change from negative to positive. This can happen if the conductor loses electrons, resulting in a decrease in its negative charge, or if the conductor gains protons, resulting in an increase in its positive charge.
## 3. How does the charge on an isolated conductor affect its behavior?
The charge on an isolated conductor affects its behavior by creating an electric field around the conductor. This electric field can attract or repel other charged objects, depending on their charge and the direction of the field. The strength of the field is determined by the magnitude of the charge on the conductor.
## 4. Is the charge on an isolated conductor always negative?
No, the charge on an isolated conductor is not always negative. It can be positive, negative, or neutral, depending on the number of electrons and protons within the conductor.
## 5. Can an isolated conductor have a neutral charge?
Yes, an isolated conductor can have a neutral charge if it has an equal number of protons and electrons. This means that the positive and negative charges cancel each other out, resulting in a net charge of zero.
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11K | 545 | 2,406 | {"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} | 3.640625 | 4 | CC-MAIN-2024-38 | latest | en | 0.943366 |
https://allnswers.com/mathematics/question3809800 | 1,611,112,957,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519883.54/warc/CC-MAIN-20210120023125-20210120053125-00020.warc.gz | 217,364,575 | 17,866 | , 24.01.2020vaniavidal666
# What are the coordinates of the x-intercept of the equation -5y=4-2x?
Option: A is correct.
A. The x-coordinate of the intersection point of the lines y = 3x + 5 and y = 2x – 7
Step-by-step explanation:
We are asked to find the solution of the equation:
3 x+5=2 x-7.
We can also equate these equation to y so that:
y=3 x+5=2 x-7
i.e. we plot the equation:
y=3 x+5
and y=2 x-7
and see the intersection point and the solution will be the x-value of the coordinate of the intersection point.
The solution is:
x= -12.
As the equation is in a single variable 'x' hence the solution of the equation is the value of x.
Hence, the solution of 3x + 5 = 2x – 7 is:
(A. the x-coordinate of the intersection point of the lines y = 3x + 5 and y = 2x – 7 )
Option: A is correct.
Option: A is correct.
A. The x-coordinate of the intersection point of the lines y = 3x + 5 and y = 2x – 7
Step-by-step explanation:
We are asked to find the solution of the equation:
3 x+5=2 x-7.
We can also equate these equation to y so that:
y=3 x+5=2 x-7
i.e. we plot the equation:
y=3 x+5
and y=2 x-7
and see the intersection point and the solution will be the x-value of the coordinate of the intersection point.
The solution is:
x= -12.
As the equation is in a single variable 'x' hence the solution of the equation is the value of x.
Hence, the solution of 3x + 5 = 2x – 7 is:
(A. the x-coordinate of the intersection point of the lines y = 3x + 5 and y = 2x – 7 )
Option: A is correct.
Keywords:
System of equations, ordered pair, variable, graphic solution
For this case we have the following equation: , which we can rewrite as a system of two equations with two variables of the form:
Algebraic solution:
We equate the equations, obtaining. From here we can clear the value of "x", then substitute in any of the two equations and find the value of "y". Thus, we obtain the ordered pair which is the solution of the system of equations.
Graphic solution:
We graph both equations. The solution of the system of equations is the point of intersection of both lines.
So, we have that the solution of is the x - coordinate of the intersection point of the lines and
The x-coordinate of the intersection point of the lines and
D
Step-by-step explanation:
Just took the Quiz
a
Step-by-step explanation:
4
step-by-step explanation:
2b/3
Step-by-step explanation:
-12
Step-by-step explanation:
3x+5=2x-7
3x+5-5=2x-7-5
3x=2x-12
x=-12
the x-coordinates of the intersection points of the graphs of y = x2 – 2x – 4 and y = –3x + 9
Step-by-step explanation:
The x intercept of -5y=4-2x equals (2,0)
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Medical
1 week, 4 days ago
Which will make basic buffer?
• Option 1)
50 mL of 0.1 M NaOH + 25 mL of 0.1M CH3COOH
• Option 2)
100 mL of 0.1M CH3COOH + 100 mL of 0.1M NaOH
• Option 3)
100 mL of 0.1M HCl + 200 mL of 0.1M NH4OH
• Option 4)
100 mL of 0.1M HCl + 100 mL of 0.1M NaOH
S solutionqc
Answered 1 week, 3 days ago
Basic Buffer is the solution which has salt & weak base. Option 3 is correct.Option 1)50 mL of 0.1 M NaOH + 25 mL of 0.1M CH3COOHOption 2)100 mL of 0.1M CH3COOH + 100 mL of 0.1M NaOHOption 3)100 mL of 0.1M HCl + 200 mL of 0.1M NH4OHOption 4)100 mL of 0.1M HCl + 100 mL of 0.1M NaOH
Medical
1 week, 4 days ago
pH of a saturated solution of $Ca(OH)_2$ is 9. The solubility product $(K_{sp})$ of $Ca(OH)_2$ is:
• Option 1)
$0.5\times 10^{-15}$
• Option 2)
$0.25\times 10^{-10}$
• Option 3)
$0.125\times 10^{-15}$
• Option 4)
$0.5\times 10^{-10}$
Answered 1 week, 3 days ago
Option 1 Given pH = 9 So, So, And will be Option 1)Option 2)Option 3)Option 4)
Medical
1 week, 4 days ago
Conjugate base for Bronsted acids $H_2O$ and $HF$ are :
• Option 1)
$OH^-$ and $H_2F^+$, respectively
• Option 2)
$H_3O^+$ and $F^-$, respectively
• Option 3)
$OH^-$ and $F^-$, respectively
• Option 4)
$H_3O^+$ and $H_2F^+$, respectively
P Plabita
Answered 1 week, 3 days ago
Option 3 Conjugate base of and So, and are the answer.Option 1) and , respectivelyOption 2) and , respectivelyOption 3) and , respectivelyOption 4) and , respectively
Medical
93 Views | 11 months ago
If the value of an equilibrium constant for a particular reaction is 1.6 x 1012, then at equilibrium the system will contain:
• Option 1)
mostly products.
• Option 2)
similar amounts of reactants and products.
• Option 3)
all reactants.
• Option 4)
mostly reactants.
A Ashutosh
Answered 9 months, 1 week ago
1
Medical
126 Views | 11 months, 3 weeks ago
Which one of the following conditions will favour maximum formation of the product in the reaction,
$A_{2}(g) + B_{2}(g) \rightleftharpoons X_{2}(g) \Delta _{r}H = -X kJ$ ?
• Option 1)
High temperature and high pressure
• Option 2)
Low temperature and low pressure
• Option 3)
Low temperature and high pressure
• Option 4)
High temperature and low pressure
S subam
Answered 11 months, 3 weeks ago
As we learnt that Effect of Pressure change - A pressure change obtained by changing the volume can affect the yield of product in case of a gaseous reaction where the total number of moles of gaseous reactant and total number of moles of gaseous product are different. - Effect of temperature change - When a change in temperature occur, the value of equilibrium constant Kc is...
Medical
2062 Views | 11 months, 3 weeks ago
The solubility of BaSO4 in water is 2.42 x 10-3 gL-1 at 298 K. The value of its solubility product (Ksp) will be
(Given molar mass of BaSO4 = 233 g mo1-1)
• Option 1)
1.08 x 10-14 mol2 L-2
• Option 2)
1.08 x 10-12 mol2 L-2
• Option 3)
1.08 x 10-10 mol2 L-2
• Option 4)
1.08 x 10-8 mol2 L-2
Answered 11 months, 3 weeks ago
As we learnt that General expression of solubility product - - wherein Its solubility product is Option 1) 1.08 x 10-14 mol2 L-2 This is incorrect. Option 2) 1.08 x 10-12 mol2 L-2 This is incorrect. Option 3) 1.08 x 10-10 mol2 L-2 This is correct. Option 4) 1.08 x 10-8 mol2 L-2 This is incorrect.
Medical
1115 Views | 11 months, 3 weeks ago
Following solutions were prepared by mixing different volumes of NaOH and HCI of different concentrations :
a. 60 mL —M/10 HC1 + 40 mL —M/10 NaOH
b. 55 mL M/10 HCl + 45 mL M/10 NaOH
c. 75 mL M/5 HCl + 25 mL M/5 NaOH
d. 100 mI—M/10 HCl + 100 mL —M/10 NaOH
pH of which one of them will be equal to 1 ?
• Option 1)
d
• Option 2)
a
• Option 3)
b
• Option 4)
c
V Vakul Arora
Answered 11 months, 3 weeks ago
As we learnt that The p(H) scale - Hydronium ion concentration in molarity is more conveniently expressed on a logarithmic scale known as p(H) scale. - wherein The p(H) of a solution is defined as negative logarithm to base 10 of the activity of hydrogen ion As N1V1 > N2V2 so acid left at the end of reaction Option 1) d This is incorrect. Option 2) a This is incorrect. Option...
Medical
86 Views | 1 year ago
In a buffer solution containing equal concentration of $B^{-}$ and HB, the $k_b \ for \ B^{-}$ is $10^{-10}$ . The pH of buffer soluction is:
• Option 1)
10
• Option 2)
7
• Option 3)
6
• Option 4)
4
As learnt in Value of Kb - - wherein Option 1) 10 This option is incorrect Option 2) 7 This option is incorrect Option 3) 6 This option is incorrect Option 4) 4 This option is correct
Medical
112 Views | 1 year ago
In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains $Ag^{+}$ and $Pb^{2+}$ at a concentration of 0.10M. Aqueous HCl is added to this solution until the $Cl^{-}$ concentration is 0.10M. What will the concentration of $Ag^{+}$ and $Pb^{2+}$ be at equilibrium?
$\left ( K_{sp}\ for AgCl=1.8\times 10^{-10},\ K_{sp}\ for PbCl_{2}=1.7\times 10^{-5} \right )$
• Option 1)
$[Ag^{+}]=1.8\times 10^{-7} M; [Pb^{2+}]=1.7\times 10^{-6}M$
• Option 2)
$[Ag^{+}]=1.8\times 10^{-11} M; [Pb^{2+}]=8.5\times 10^{-5}M$
• Option 3)
$[Ag^{+}]=1.8\times 10^{-9} M; [Pb^{2+}]=1.7\times 10^{-3}M$
• Option 4)
$[Ag^{+}]=1.8\times 10^{-11} M; [Pb^{2+}]=8.5\times 10^{-4}M$
D Divya Saini
As learnt in General expression of solubility product - - wherein Its solubility product is Option 1) This option is incorrect Option 2) This option is incorrect Option 3) This option is correct Option 4) This option is incorrect
Medical
115 Views | 1 year ago
If pH of a saturated solution of $Ba\left ( OH \right )_2$ is 12, the value of its $K_{\left ( sp \right )}$ is
• Option 1)
$4.00\times 10^{-6}M^{3}$
• Option 2)
$4.00\times 10^{-7}M^{3}$
• Option 3)
$5.00\times 10^{-6}M^{3}$
• Option 4)
$5.00\times 10^{-7}M^{3}$
As learnt in General expression of solubility product - - wherein Its solubility product is We know that, S 2S Option 1) This option is incorrect Option 2) This option is incorrect Option 3) This option is incorrect Option 4) This option is correct
Medical
100 Views | 1 year ago
Which one of the following pairs of solution is not an acidic buffer?
• Option 1)
$HClO_{4}\ and\ NaClO_{4}$
• Option 2)
$CH_3COOH\ and\ CH_3COONa$
• Option 3)
$H_2CO_3\ and\ Na_2CO_3$
• Option 4)
$H_3PO_4\ and\ Na_3PO_4$
V Vakul Arora
Law of Chemical equilibrium - At a given temperature, the product of concentration of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentration of the reactants raised to their individual stoichiometric coefficients has a constant value. - wherein are equilibrium concentration Acidic Buffer is a...
Medical
102 Views | 1 year ago
The $K_{sp} \ of \ Ag_{2}CrO_{4}, Agcl, AgBr\ and\ AgI$ are respectively, $1.1\times 10^{-12}, 1.8\times 10^{-10}, 5.0\times 10^{-13}, 8.3\times 10^{-17}$. Which one of the following salts will precipitate last if $AgNO_{3}$ solution is added to the solution containing equal moles of NaCl, NaBr, Nal, and $Na_{2}CrO_{4}$?
• Option 1)
AgCl
• Option 2)
AgBr
• Option 3)
$Ag_{2}CrO_{4}$
• Option 4)
AgI
V Vakul Arora
General expression of solubility product - - wherein Its solubility product is AgCrO4 1.1X10-12= 4 S3 1.8X10-10= S2 AgBr 5X1013 = S2 AgI 8.3 X10-17 = S2 Thus Ag2Cr04 will be precipitated last. Option 1) AgCl incorrect Option 2) AgBr incorrect Option 3) correct Option 4) AgI incorrect
Medical
101 Views | 1 year ago
The $K_{sp} \ of \ Ag_{2}CrO_{4}, Agcl, AgBr\ and\ AgI$ are respectively, $1.1\times 10^{-12}, 1.8\times 10^{-10}, 5.0\times 10^{-13}, 8.3\times 10^{-17}$. Which one of the following salts will precipitate last if $AgNO_{3}$ solution is added to the solution containing equal moles of NaCl, NaBr, Nal, and $Na_{2}CrO_{4}$?
• Option 1)
AgCl
• Option 2)
AgBr
• Option 3)
$Ag_{2}CrO_{4}$
• Option 4)
AgI
V Vakul Arora
General expression of solubility product - - wherein Its solubility product is AgCrO4 1.1X10-12= 4 S3 1.8X10-10= S2 AgBr 5X1013 = S2 AgI 8.3 X10-17 = S2 Thus Ag2Cr04 will be precipitated last. Option 1) AgCl incorrect Option 2) AgBr incorrect Option 3) correct Option 4) AgI incorrect
Medical
95 Views | 1 year ago
Consider the nitration of bezene using mixed conc of $H_{2}SO_{4}$ and $HNO_{3}$. If a large amount of $KHSO_{4}$ is added to the mixture, tne rate of nitration will be
• Option 1)
faster
• Option 2)
slower
• Option 3)
unchanged
• Option 4)
doubled
P Plabita
Effect of concentration change on equilibrium - When the concentration of any of the reactants or products in a reaction at equilibrium is changed, the composition of the equilibrium mixture changes so as to minimize the effect of concentration change. If we add reactant, the equilibrium shift towards right. - In the nitration of benzene in the presence of conc. H2SO4 and HNO3, benzene is...
Medical
136 Views | 1 year ago
MY and $NY_3$, two nearly insoluble salts, have the same $K_{sp}$ value of $6.2\times 10^{-13}$ at room temperature. Which statement would be true in regard to MY and $NY_3$?
• Option 1)
The molar solubilities of MY and $NY_3$ in water are identical.
• Option 2)
The molar solubility of MY in water is less than that of $NY_3$
• Option 3)
The salts MY and $NY_3$ are more soluble in 0.5M KY than in pure water
• Option 4)
The addition of the salt of KY to solution of MY and $NY_3$ will have no effect on their solubilities
General expression of solubility product - - wherein Its solubility product is For MY Ksp= Option 1) The molar solubilities of MY and in water are identical. incorrect Option 2) The molar solubility of MY in water is less than that of correct Option 3) The salts MY and are more soluble in 0.5M KY than in pure water incorrect Option 4) The addition of the salt of KY to...
Medical
111 Views | 1 year ago
What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed?
• Option 1)
12.65
• Option 2)
2.0
• Option 3)
7.0
• Option 4)
1.04
P prateek
Relation between Ka and Kb - - wherein .01 mole of NaOH will be completely neutralised by .01 mole of HCl. Hence remaining part of NaOH is .09 mole. as equal volume is mixed [OH-]=.04514 P(H) = 14- P(OH) = 14-log [OH-] =12.65 Option 1) 12.65 correct Option 2) 2.0 incorrect Option 3) 7.0 incorrect Option 4) 1.04 incorrect
Medical
88 Views | 1 year ago
Which of the following statements is correct for a reversible process in a state of equilibrium?
• Option 1)
$\Delta G=2.30\ RT\ log \ K$
• Option 2)
$\Delta G^{o}=-2.30\ RT\ log \ K$
• Option 3)
$\Delta G^{o}=2.30\ RT\ log \ K$
• Option 4)
$\Delta G=-2.30\ RT\ log \ K$
D Divya Saini
As we discussed in concept Relation between Gibbs energy and reaction Quotient - - wherein is standard Gibbs energy. Option 1) This option is incorrect. Option 2) This option is correct. Option 3) This option is incorrect. Option 4) This option is incorrect.
Medical
88 Views | 1 year ago
The reaction $2A_{\left ( g \right )}+B_{\left ( g \right )}\rightleftharpoons 3C_{\left ( g \right )}+D_{\left ( g \right )}$ is begun with the concentrations of A and B both at an initial value of 1.00M. When equilibrium is reached, the concentrations of D is measured and found to be 0.25M. The value for the equilibrium constant for this reaction is given by the expression
• Option 1)
$[\left ( 0.75 \right )^{3}\left ( 0.25 \right )]\div [\left ( 0.75 \right )^{2}\left ( 0.25 \right )]$
• Option 2)
$[\left ( 0.75 \right )^{3}\left ( 0.25 \right )]\div [\left ( 1.00 \right )^{2}\left ( 1.00 \right )]$
• Option 3)
$[\left ( 0.75 \right )^{3}\left ( 0.25 \right )]\div [\left ( 0.50 \right )^{2}\left ( 0.75 \right )]$
• Option 4)
$[\left ( 0.75 \right )^{3}\left ( 0.25 \right )]\div [\left ( 0.50 \right )^{2}\left ( 0.25 \right )]$
D Divya Saini
As we discussed in concept Law of Chemical equilibrium - At a given temperature, the product of concentration of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentration of the reactants raised to their individual stoichiometric coefficients has a constant value. - wherein are equilibrium...
Medical
87 Views | 1 year ago
In which of the following equilibrium $K_{c}$ and $K_{p}$ are not equal?
• Option 1)
$2NO_{\left ( g \right )}\rightleftharpoons N_{2\left ( g \right )}+O_{2\left ( g \right )}$
• Option 2)
$SO_{2\left ( g \right )}+NO_{2\left ( G \right )}\rightleftharpoons SO_{3\left ( g \right )}+NO_{\left ( g \right )}$
• Option 3)
$H_{2\left ( g \right )}+I_{2\left ( g \right )}\rightleftharpoons 2HI_{\left ( g \right )}$
• Option 4)
$2C_{\left ( s \right )}+O_{2\left ( g \right )}\rightleftharpoons 2CO_{2\left ( g \right )}$
As we discussed in concept Relation between Kp and Kc - while calculating the value of Kp , pressure should be expressed in bar. - wherein = (number of moles of gaseous products) - (number of moles gaseous reaction) In 1), 2) and 3) In 4) Option 1) This option is incorrect. Option 2) This option is incorrect. Option 3) This option is incorrect. Option 4) This option is correct.
Medical
96 Views | 1 year ago
For the reaction $N_{2}\left ( g \right )+O_{2}\left ( g \right )\rightleftharpoons 2NO\left ( g \right )$, the equilibrium constant is $K_{1}$. The equilibrium constant is $K_{2}$ for the reaction
$2NO\left ( g \right )+O_{2}\rightleftharpoons 2NO_{2}\left ( g \right )$
What is K for the reaction
$NO_{2}(g)\rightleftharpoons \frac{1}{2}N_{2}(g)+O_{2}(g)?$
• Option 1)
$1/\left ( 2K_{1}K_{2} \right )$
• Option 2)
$1/\left ( 4K_{1}K_{2} \right )$
• Option 3)
$[1/K_{1}K_{2}]^{\frac{1}{2}}$
• Option 4)
$1/\left ( K_1 K_2 \right )$ | 4,933 | 14,309 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 86, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-22 | latest | en | 0.390607 |
https://www.speedsolving.com/threads/learning-algorithms-backwards-vs-u2.84702/ | 1,627,813,269,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154175.76/warc/CC-MAIN-20210801092716-20210801122716-00032.warc.gz | 1,050,309,536 | 17,302 | # Learning algorithms backwards vs U2
#### jdh3000
##### Member
In cfop 3x3 I know some algorithms right and left. I'm not talking about mirrored algs, but just executing one from the left instead of the right, rather than doing a U2.
I was wondering if anyone felt the need to learn every alg this way, or if doing a U2 was good enough...
Some olls are easy from either side,but others would take a bit of committing to muscle memory to do effectively.
I've just been doing U2 for most but have been pondering learning an opposite.
What are you doing?
#### patricKING
##### Member
In cfop 3x3 I know some algorithms right and left. I'm not talking about mirrored algs, but just executing one from the left instead of the right, rather than doing a U2.
I was wondering if anyone felt the need to learn every alg this way, or if doing a U2 was good enough...
Some olls are easy from either side, but others would take a bit of committing to muscle memory to do effectively.
I've just been doing U2 for most but have been pondering learning an opposite.
What are you doing?
I usually do OLL from only the right side (2 Look), but I can do some F2L cases from both sides, such as sledgehammer.
#### xyzzy
##### Member
Most good RUF algs become mediocre or terrible if you try to execute them unmodified from the U2 angle, since they're now LUB. If you're ambidextrous enough, RU→LU, RUD→LUD, RUL→LUR, RUS→LUS might be reasonable. And of course MU algs remain as MU, just with all the Ms switched with M's and vice versa.
I don't think I use any y2'd algs other than the A perms (y2 x R' U R' D2 R U' R' D2 R2 x' y2 = r' U r' B2 r U' r' B2 r2). For the alg combos involving Sunes, I might do one of the Sunes lefty to avoid a U2 AUF (e.g. R U R' U R U2 R' L' U' L U' L' U2 L), although I very rarely do standalone Sunes lefty.
#### Cubing Forever
##### Member
Most good RUF algs become mediocre or terrible if you try to execute them unmodified from the U2 angle, since they're now LUB.
I'd like to mention a few exceptions:
T perm becomes LUF
N perms remain RUF
V perm is pretty much the same except the first 4 moves are L' U L' U' instead of R' U R' U' and you do y' instead of y.
F perm becomes LUF
There are many more but I'm lazy.
#### abunickabhi
##### Member
I think it boils down to personal preference. Sometimes depending on the hand grip, doing a U2 is effortless, or if we have drilled the backward alg, we may pull it off.
It is all depending on the hand grip, and also personal perference. Tough to give an objective answer for such choices in alg execution.
#### Cubing Forever
##### Member
I wasn't talking about mirroring, and neither was jdh3000.
Well, I wasn't either. I was giving a few exceptions for "most RUF algs become mediocre or terrible if executed unmodified from the U2 angle since they become LUB".
Also, mirroring the alg is somewhat similar to executing it from the U2 angle.
#### xyzzy
##### Member
Well, I wasn't either. I was giving a few exceptions for "most RUF algs become mediocre or terrible if executed unmodified from the U2 angle since they become LUB".
Also, mirroring the alg is somewhat similar to executing it from the U2 angle.
The only one you listed that sorta counts is the V perm, and it's not even an RUF alg to begin with.
When you execute R U R' U' R' F R2 U' R' U' R U R' F' from the U2 angle, it becomes
L U L' U' L' B L2 U' L' U' L U L' B',
not
L' U' L U L F' L2 U L U L' U' L F.
These happen to solve the same case (because it has reflection symmetry), but they're obviously not the same alg. The first one you get by rotating the original alg, the second one you get by mirroring the original alg. The first alg is bad and no one should use it. The second one is good if you're ambidextrous.
AIUI, the point of this thread was to ask about algs that are transformed by rotating 180°, not to find algs specifically optimised to be executed from that angle that are possibly unrelated to the alg for the "standard" angle, or related in ways other than 180° rotation. Of course when you relax the constraints on the algs you choose, you can find better algs. | 1,104 | 4,142 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-31 | latest | en | 0.958519 |
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# TOPIC 7: PRESSURE | PHYSICS FORM 1
1157
0
## TOPIC 7: PRESSURE | PHYSICS FORM 1
#### Concept of Pressure
Pressure is defined as the force per unit area. OR Pressure is the force acting normally (perpendicularly) per unit surface area.
It is calculated by the formula:
Pressure = Force (f)/Area (A)
P = F/A
Where
• P – Pressure
• F – Force
• A – Area
#### The S.I Unit of Pressure
The SI unit of Pressure is Newton per square metre (N/M2). This unit is usually referred to as the Pascal (Pa).
1Pa = 1 N/M2
The other units of pressure are atmosphere, torr bar and mmHg.
• 1 atmosphere = 780mmHg
• 1 atmosphere = 1 105 N/M2 = 1bar (used by meteorologists)
Note: for a given amount of force, the smaller the area of application the greater the pressure exerted.
When a man lifts a bucket of water by its handle that is made with a thin metal, he would experience some discomfort but if the bucket was made with a thicker handle the discomfort will be much less if any.
This is because the area over which the force is applied is larger.
#### Pressure due to Solids
Dependence of Pressure on Surface of Contact
The pressure in solid depends on the surface area of contact. A force (F) applied onto a small area exerts a higher pressure as compared to when it is applied onto a large surface.
Pressure in solid = Force applied/Area of contact.
##### Example 1
A block of wood that weighs 30N and measures 5m by 10m by 4m. If it was placed on a table with the largest possible area (5mx10m) in contact with table, exerts less pressure than it would when placed with its smallest possible area (5mx4m) in contact with table. | 440 | 1,684 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2022-27 | latest | en | 0.921834 |
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during an art contest at your school,you and a classmate each won blue ribbons for 1/3 of the pieces you entered in the contest.you won 2 blue ribbons and your classmate won 3 blue ribbons.explain how this could be.
You must have entered 6 pieces because 1/3 of 6 = 2.
How many pieces did your classmate enter?
nine
## Similar Questions
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I read the meaning of function notation but i still don't understand why are they so important. during an art contest at your school, you and a classmate each won blue ribbons for one third of the pieces you entered in the contest. …
during an art contest at your school, you and a classmate each won blue ribbons for one-third of the pieces you entered in the contest. You won 2 blue ribbons and your classmate won 3 blue ribbons. Explain how this could be
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During an art context at your school, you and a classmate each won blue ribbons for 1/3 of the pieces you entered in the contest. You won 2 blue ribbons and your classmate won 3 blue ribbons. Explain how this could be
More Similar Questions | 677 | 2,627 | {"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} | 2.953125 | 3 | CC-MAIN-2018-05 | latest | en | 0.974101 |
http://www.lertap5.com/Nursing2017/HTML/tosca-(n_z_).html | 1,695,343,153,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506320.28/warc/CC-MAIN-20230922002008-20230922032008-00849.warc.gz | 71,324,736 | 3,871 | Sample studies > TOSCA (N.Z.)
# TOSCA (N.Z.)
Note to myself: workbook saved on Imation USB as: StuIQ_10Sep2017.xlsx
In 1995 the New Zealand government set out to design a new test of "mathematics aptitude".
They assembled a team of expert teachers, and a developed a subject design matrix similar to that used in the FIMS project.
They wrote about 300 test items. About a third of them were supply items; the rest were multiple choice items with 5 options (4 distractors).
They then developed two "parallel" tests, each with 70 items (some supply items, some multiple-choice items).
These tests were referred to as "Form A" and "Form B".
By "parallel" is meant that they selected the same number of items from each of the content categories, and the same proportional number from the performance expectation levels. The 70 items on Form A were different from the 70 items of Form B, but were matched on content and performance levels -- they were judged to be "parallel", having items of similar content and performance difficulty.
They then selected a stratified random sample of 300 junior high-school students to test.
I will display results using Excel and Lertap.
The histograms for the two tests were similar.
The scatterplot of Form A vs Form B is interesting. It basically display the desired pattern of correlation.
The reliability, the correlation between Form A and Form B test scores, was 0.84 -- this would be referred to as "parallel-forms" or "equivalent-forms" reliability. It's a good value.
The csem worksheet can be used to apply the binomial (as in the marbles example) to derive an estimate of a range which is likely to cover "true" percentage-correct score.
The FA35+ worksheet shows what would happen if we applied a cutoff score of 50% (35 items correct).
If we said that students had to get a Form A score of 35 or more to pass and then, if we could test them again with a carefully-developed parallel form (Form B in this example), would we find consistent results? Do the same students pass when tested again?
We will see that not all of them will.
An "item analysis" should be done to see that the items are discriminating as we want them to -- on each item, the top students should do better than the bottom students.
I will show some item response plots.
It can be shown that reliability is related to item discrimination.
And, it can be shown that item discrimination for multiple-choice items requires effective distractors; the weak students will choose the distractors, but the top students will not. | 563 | 2,558 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2023-40 | latest | en | 0.960009 |
https://ctptalk.com/@gungunkrishu/maths-brain-teasers-94-can-you-simplify-the-logic | 1,643,296,633,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305266.34/warc/CC-MAIN-20220127133107-20220127163107-00054.warc.gz | 242,043,248 | 60,267 | # Maths Brain Teasers 94 :: Can You - SIMPLIFY The Logic?
Hey All; @StemGeeks Mathematician;
Without further, ado, let's get to our puzzle for today:: Maths Brain Teasers 94 :: Can You - SIMPLIFY The Logic?
Have a close look at the image given and see if you can decode the logic given in the image to complete the puzzle. Try finding out the right NUMBER that Simplifies or Decodes the logic to the given puzzle. There is logic to resolve the puzzle; try finding out the logic and it should be resolved in seconds.. Guesswork isn't going to help...
With that, I'll leave you all with the Maths Brain Teasers 94 :: Can You - SIMPLIFY The Logic? Good luck solving the puzzle...
##### Maths Brain Teasers 94 :: Can You - Count The Number of LINES? - Solved with Explanation
Let's have a look at the Step by Step instructions to Solve the Maths Brain Teasers 94 :: Can You - Count The Number of LINES?
If you carefully look at the Image in question; You would have noticed the following pattern is being followed::
• We need to count each line segment as a different line
• No need to consider counting the frame as lines & Finally
• We need not consider both i.e. inside and outside of the frame of lines as two it should be counted as ONE
### STEM token GiveAway
I'll be again doing a giveaway of STEM tokens to the lucky random winner with the correct answer. For the last contest, which was Maths Brain Teasers 94 :: Can You - Count The Number of LINES?.
We had 3 entries and I am happy to see the engagement by everyone. At the same time, it's great to see the detailed answers provided by everyone as to how they arrived at the solution of the puzzle.. Well done guys.. Way to GO... Now the results part; out of the three entries ONE entry provided the right answer and hence the WINNER is @yintercept
Congratulations @yintercept You WIN yourself 10 STEM tokens. You should be having the rewards in your STEM Wallet Soon.
### Math Quote for the Day::
Here is the motivation to solve the Maths Equation Puzzle?
If you like my work, then please spread the Word.. that we do have the Math Brain Teasers competition here @StemGeeks platform. Reblog is much appreciated.
Best Regards
Find Me on the Other Social Media Platforms::
PS:: All the Maths Brain Teasers; are made by me using the Pro Canva License Version
0
0
0.000
Should be 68 I guess because 68 - 68 would be 0.
0
0
0.000
68+6-66-8
68+6-74
74-74
0
0
0
0.000 | 594 | 2,442 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2022-05 | latest | en | 0.93852 |
https://www.jiskha.com/users?name=garret | 1,575,910,554,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540519149.79/warc/CC-MAIN-20191209145254-20191209173254-00175.warc.gz | 752,586,715 | 4,263 | # garret
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how does an electric ciircuit work? Books have been written about electrical circuits, what kinds, specilized circuits, etc. You need to be a little more specific with your question.
6. ## Math
The cost in dollars of producing x units of a particular camera is C(x) = x^2 - 10000. (10 points) Find the average rate of change of C with respect to x when the production level is changed from x = 100 to x = 101. Include units in your answer. Find the
7. ## math
how many quarts are in a liter
8. ## Chemistry
How many kcal of heat energy are required or released when 1.73 moles of butanol , C4H9OH, is warmed from -175.8 degrees F to 360.9 degrees F? The following information concerning butanol applies: normal melting point: 12.5 degrees C normal boiling point:
9. ## science
what is the importance of fuses? They are placed in electrical equipment and in houses (although most houses now use circuit breakers) as safety devices. If a short occurs and too much electrical current is going through the circuit, the fuse "blows"
10. ## math
how many miller meaters does it take to make aleader
1. ## math
IT WENT UP 532 %
posted on December 21, 2018
2. ## physics
.935
posted on October 20, 2015
3. ## lll
1.There is no minimum number for a "few." If you're a rancher in Texas, a few acres might be 200. If you're a landowner in New York, a few acres might be 10. A few M & M's is different than a few watermelons. A few students in a class is different than a
posted on November 12, 2012
4. ## chemistry
Does this affect the Calculated molar Mass?
posted on February 4, 2011
100+100+100+100-20
posted on September 3, 2009 | 591 | 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} | 3.140625 | 3 | CC-MAIN-2019-51 | latest | en | 0.932821 |
https://www.peachpit.com/articles/article.aspx?p=711819&seqNum=5 | 1,604,078,369,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107911027.72/warc/CC-MAIN-20201030153002-20201030183002-00201.warc.gz | 839,787,385 | 12,950 | Publishers of technology books, eBooks, and videos for creative people
# Creating the Related Tables
• Print
This chapter is from the book
## Define More Fields
We can now define the final fields in Main; Calculation fields whose formulas rely on fields in the related tables. These new fields are listed in the table below.
Field Name Description Sav_Current Balance Current balance of a savings/money market account Stk_Current Value Value of a stock based on its current price plus earned dividends Stk_Change in Value Change in a stock’s value as of the most recent price update CD_Current Value Original cost of the CD, plus all interest earned to-date
Before creating the fields, choose FileMaker Pro > Preferences (Mac) or Edit > Preferences (PC), click the Layout tab, uncheck Add newly defined fields to current layout, and click OK. If you leave this option checked, FileMaker automatically places new fields on the layout—frequently in the worst possible places.
Sav_Current Balance
• Choose File > Define > Database. In the Define Database dialog box, click the Fields tab and select Main from the Table list.
• Enter Sav_Current Balance in the Field Name box, choose Calculation as the Type, and click Create. The Specify Calculation dialog box appears.
• Double-click Sav_Initial Balance in the field list on the left. Then type a + (or click the + operator button).
• Double-click the Sum function in the list on the right. Its arguments are highlighted.
• Replace the arguments by choosing Savings from the drop-down list above the field list and double-clicking the ::Deposit field. (The :: prefix denotes a related field.)
• Move the cursor to right of the closing parenthesis and then type a - (or click the - operator button).
• Enter another instance of the Sum function by double-clicking it in the function list. Replace the function’s arguments by double-clicking the ::Withdrawal field in the field list. The formula should now read:
Sav_Initial Balance + Sum (Savings::Deposit) - Sum (Savings::Withdrawal)
• To complete the field’s definition, make sure the Calculation result is Number, and then click OK.
Stk_Current Value
• Define a new Calculation field named Stk_Current Value.
• In the Specify Calculation dialog box, enter or create the following formula:
(Stk_Current Share Price * Stk_Number of Shares) +Sum (Stocks::Dividend) +Sum (Stocks::Interest )
This formula multiplies the current share price by the number of shares held, and then adds in any dividends and interest received (by summing these items from the Stocks table). The components Stocks::Dividend and Stocks::Interest are the Dividend and Interest fields from the Stocks table.
• To complete the definition, make sure the Calculation result is Number, and then click OK.
The formula is added to the field’s definition in the Define Database dialog box.
Stk_Change in Value
• Define a new Calculation field named Stk_Change in Value.
• In the Specify Calculation dialog box, type or create the following formula:
Stk_Current Value - Stk_Total Cost
This formula calculates the amount (positive or negative) that the holding has changed by subtracting its original cost (including commission) from the stock’s current value, based on the most recent share price recorded. Both fields in the formula are from the Main table.
• To complete the definition, make sure the Calculation result is Number, and then click OK.
The formula is added to the field’s definition in the Define Database dialog box.
CD _Current Value
• Define a new Calculation field named CD _Current Value.
• In the Specify Calculation dialog box, create the following formula:
CD_Amount + Sum (CD::Interest)
The formula computes each CD’s value by adding its purchase price to the sum of the accrued interest. (CD_Amount is in the Main table; Interest is in the CD table.)
• To complete the definition, make sure the Calculation result is Number, and then click OK. | 830 | 3,960 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-45 | latest | en | 0.782765 |
https://slideplayer.com/slide/5665331/ | 1,618,352,542,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038075074.29/warc/CC-MAIN-20210413213655-20210414003655-00502.warc.gz | 630,364,772 | 18,179 | # EE313 Linear Systems and Signals Fall 2010 Initial conversion of content to PowerPoint by Dr. Wade C. Schwartzkopf Prof. Brian L. Evans Dept. of Electrical.
## Presentation on theme: "EE313 Linear Systems and Signals Fall 2010 Initial conversion of content to PowerPoint by Dr. Wade C. Schwartzkopf Prof. Brian L. Evans Dept. of Electrical."— Presentation transcript:
EE313 Linear Systems and Signals Fall 2010 Initial conversion of content to PowerPoint by Dr. Wade C. Schwartzkopf Prof. Brian L. Evans Dept. of Electrical and Computer Engineering The University of Texas at Austin Fourier Transform Properties
13 - 2 0 1 /2- /2 t f(t)f(t) 0 F()F() Duality Forward/inverse transforms are similar Example: rect(t/ ) sinc( / 2) Apply duality sinc(t /2) 2 rect(- / ) rect(·) is even sinc(t /2) 2 rect( / )
13 - 3 Scaling Given and that a 0 |a| > 1: compress time axis, expand frequency axis |a| < 1: expand time axis, compress frequency axis Extent in time domain is inversely proportional to extent in frequency domain (a.k.a bandwidth) f(t) is wider spectrum is narrower f(t) is narrower spectrum is wider
13 - 4 Shifting in Time Shift in time Does not change magnitude of the Fourier transform Shifts phase of Fourier transform by - t 0 (so t 0 is the slope of the linear phase) Derivation Let u = t – t 0, so du = dt and integration limits stay same
13 - 5 Sinusoidal Amplitude Modulation
13 - 6 Sinusoidal Amplitude Modulation Example: y(t) = f(t) cos( 0 t) f(t) is an ideal lowpass signal Assume 1 << 0 Demodulation (i.e. recovery of f(t) from y(t)) is modulation followed by lowpass filtering Similar derivation for modulation with sin( 0 t) 0 1 -- F()F() 0 Y()Y() - - - + F - + F
13 - 7 Frequency-shifting Property
13 - 8 Time Differentiation Property Conditions f(t) 0 when |t| f(t) is differentiable Derivation of property: Given f(t) F( )
13 - 9 Time Integration Property Example:
13 - 10 Summary Definition of Fourier Transform Two ways to find Fourier Transform Use definition Use transform pairs and properties
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Similar presentations | 946 | 2,442 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2021-17 | latest | en | 0.711205 |
https://thephoenixunderground.com/qa/does-earth-gain-weight.html | 1,611,521,628,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703557462.87/warc/CC-MAIN-20210124204052-20210124234052-00662.warc.gz | 581,332,833 | 8,520 | # Does Earth Gain Weight?
## Is Earth losing its gravity?
“The Earth’s gravity field changes from one month to the next mostly due to the mass of water moving around on the surface,” said Watkins.
“Because water in all its forms has mass and weight, we can actually weigh the ocean moving around..
## Has the Earth gained weight?
Nasa has calculated that the Earth is gaining energy due to rising temperatures. Dr Smith and his colleague Mr Ansell estimate this added energy increases the mass of Earth by a tiny amount – 160 tonnes. This means that in total between 40,000 and 41,000 tonnes is being added to the mass of the planet each year.
## Does Earth’s mass stay the same?
Mass stays the same regardless of location and gravity. You would have the same mass on Mars or Jupiter as you do here on Earth. Your weight is different on other planets due to gravity. However, your mass is the same everywhere!
## How much weight does Earth gain each day?
Earth gains mass through dust and meteorites that are captured by its gravity. If you watched the recent meteor shower you know this can occur on a regular basis. In fact from satellite observations of meteor trails it’s estimated that about 100 – 300 metric tons (tonnes) of material strikes Earth every day.
## Is Earth getting smaller?
The scientists estimated the average change in Earth’s radius to be 0.004 inches (0.1 millimeters) per year, or about the thickness of a human hair, a rate considered statistically insignificant.
## Is ice heavier than water?
In practical terms, density is the weight of a substance for a specific volume. The density of water is roughly 1 gram per milliliter but, this changes with temperature or if there are substances dissolved in it. Ice is less dense than liquid water which is why your ice cubes float in your glass.
## Does the Earth get heavier when a baby is born?
Does the earth get heavier when a baby is born?? No. Matter cannot be created nor destroyed. The mass of the baby was simply moved from another location, into the mother, and became a baby.
## Is Earth getting closer to the sun?
We are not getting closer to the sun, but scientists have shown that the distance between the sun and the Earth is changing. … The rate at which the sun is slowing is also tiny (around 3 milliseconds every 100 years). As the sun loses its momentum and mass, the Earth can slowly slip away from the sun’s pull.
## What will happen if the Earth is bigger than its current size?
First, if the Earth is bigger that its current size, it would also gain a stronger magnetic field and greater gravitational pull because of the additional mass, and this is a bad news for us, because lifting an object would be so much difficult than ever before due to the increased pull of gravity.
## How old is the earth?
4.543 billion yearsEarth/Age
## How does earth gain energy?
Solar power drives Earth’s climate. Energy from the Sun heats the surface, warms the atmosphere, and powers the ocean currents. … This net flow of energy into and out of the Earth system is Earth’s energy budget. The energy that Earth receives from sunlight is balanced by an equal amount of energy radiating into space.
## Does sunlight add mass to Earth?
Since 1 gram is worth 9 x 10^20 ergs, sunlight equals 4 x 10^12 grams/second or 4.4 million metric tons of equivalent mass per second . … During the entire life of the sun…4.5 billion years, the earth has gained 2.7 x 10^17 kilograms, which is only 1/21 millionth of its mass.
## What will Earth weigh in 2020?
You will find that the weight of planet Earth is… 13,170,000,000,000,000,000,000,000 (1.3 x 1025) pounds! That’s thirteen septillion!
## Does Earth weigh more today?
People just weigh more and the Earth therefore weighs more. That’s not true. The Earth is an isolated system where all the of the mass that was on it to start with doesn’t go anywhere. … Debris, dust and other stuff raining in from space contributes a huge amount of weight to the Earth every single year.
## Is Earth losing mass?
According to some calculations, the Earth is losing 50,000 metric tons of mass every single year, even though an extra 40,000 metric tons of space dust converge onto the Earth’s gravity well, it’s still losing weight.
## Why is the earth not getting bigger?
New crust is continually being pushed away from divergent boundaries (where sea-floor spreading occurs), increasing Earth’s surface. But the Earth isn’t getting any bigger. … Deep below the Earth’s surface, subduction causes partial melting of both the ocean crust and mantle as they slide past one another.
## What does the earth weigh?
5.972 × 10^24 kgEarth/Mass
## Is the Earth a closed system?
The Earth is a closed system for matter Because of gravity, matter (comprising all solids, liquids and gases) does not leave the system. | 1,079 | 4,853 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2021-04 | latest | en | 0.954029 |
http://www.efunda.com/glossary/units/units--density--specific_gravity.cfm | 1,563,440,554,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525587.2/warc/CC-MAIN-20190718083839-20190718105839-00282.warc.gz | 196,978,680 | 7,362 | Density Units: Specific Gravity
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Glossary » Units » Density » Specific Gravity
Specific Gravity is a unit in the category of Density. Specific Gravity has a dimension of ML-3 where M is mass, and L is length. It can be converted to the corresponding standard SI unit kg/m3 by multiplying its value by a factor of 1000.
Note that the seven base dimensions are M (Mass), L (Length), T (Time), Q (Temperature), N (Aamount of Substance), I (Electric Current), and J (Luminous Intensity).
Other units in the category of Density include Gram Per Cubic Centimeter (g/cm3), Gram Per Cubic Meter (g/m3), Gram Per Liter (g/l), Kilogram Per Cubic Decimeter (kg/dm3), Kilogram Per Cubic Meter (kg/m3), Kilogram Per Liter (kg/l), Ounce (av.) Per Cubic Foot (oz (av.)/ft3), Ounce (av.) Per Cubic Inch (oz (av.)/in3), Pound Mass Per Cubic Foot (lbm/ft3), Pound Mass Per Cubic Inch (lbm/in3), Pound Mass Per Cubic Yard (lbm/yd3), Pound Mass Per Gallon (UK) (lbm/gal (UK)), Pound Mass Per Gallon (US, Liq.) (lbm/gal (US, liq.)), Slug Per Cubic Foot (slug/ft3), Slug Per Cubic Inch (slug/in3), and Slug Per Cubic Yard (slug/yd3). | 375 | 1,329 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2019-30 | longest | en | 0.665587 |
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#### Resources tagged with Working systematically similar to Making Squares:
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### There are 339 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically
### Geoboards
##### Stage: 2 Challenge Level:
This practical challenge invites you to investigate the different squares you can make on a square geoboard or pegboard.
### Making Squares
##### Stage: 2 Challenge Level:
Investigate all the different squares you can make on this 5 by 5 grid by making your starting side go from the bottom left hand point. Can you find out the areas of all these squares?
### My New Patio
##### Stage: 2 Challenge Level:
What is the smallest number of tiles needed to tile this patio? Can you investigate patios of different sizes?
### Tiles on a Patio
##### Stage: 2 Challenge Level:
How many ways can you find of tiling the square patio, using square tiles of different sizes?
### Ice Cream
##### Stage: 2 Challenge Level:
You cannot choose a selection of ice cream flavours that includes totally what someone has already chosen. Have a go and find all the different ways in which seven children can have ice cream.
### Numerically Equal
##### Stage: 2 Challenge Level:
Can you draw a square in which the perimeter is numerically equal to the area?
### More Transformations on a Pegboard
##### Stage: 2 Challenge Level:
Use the interactivity to find all the different right-angled triangles you can make by just moving one corner of the starting triangle.
### Through the Window
##### Stage: 2 Challenge Level:
My local DIY shop calculates the price of its windows according to the area of glass and the length of frame used. Can you work out how they arrived at these prices?
### Tri.'s
##### Stage: 2 Challenge Level:
How many triangles can you make on the 3 by 3 pegboard?
### Torn Shapes
##### Stage: 2 Challenge Level:
These rectangles have been torn. How many squares did each one have inside it before it was ripped?
### Calcunos
##### Stage: 2 Challenge Level:
If we had 16 light bars which digital numbers could we make? How will you know you've found them all?
### Street Party
##### Stage: 2 Challenge Level:
The challenge here is to find as many routes as you can for a fence to go so that this town is divided up into two halves, each with 8 blocks.
### Halloween Investigation
##### Stage: 2 Challenge Level:
Ana and Ross looked in a trunk in the attic. They found old cloaks and gowns, hats and masks. How many possible costumes could they make?
### Making Boxes
##### Stage: 2 Challenge Level:
Cut differently-sized square corners from a square piece of paper to make boxes without lids. Do they all have the same volume?
### Dicey Perimeter, Dicey Area
##### Stage: 2 Challenge Level:
In this game for two players, you throw two dice and find the product. How many shapes can you draw on the grid which have that area or perimeter?
### Newspapers
##### Stage: 2 Challenge Level:
When newspaper pages get separated at home we have to try to sort them out and get things in the correct order. How many ways can we arrange these pages so that the numbering may be different?
### Ribbon Squares
##### Stage: 2 Challenge Level:
What is the largest 'ribbon square' you can make? And the smallest? How many different squares can you make altogether?
### 3 Rings
##### Stage: 2 Challenge Level:
If you have three circular objects, you could arrange them so that they are separate, touching, overlapping or inside each other. Can you investigate all the different possibilities?
### Cover the Tray
##### Stage: 2 Challenge Level:
These practical challenges are all about making a 'tray' and covering it with paper.
##### Stage: 2 Challenge Level:
How many DIFFERENT quadrilaterals can be made by joining the dots on the 8-point circle?
### Uncanny Triangles
##### Stage: 2 Challenge Level:
Can you help the children find the two triangles which have the lengths of two sides numerically equal to their areas?
### Palindromic Date
##### Stage: 2 Challenge Level:
What is the date in February 2002 where the 8 digits are palindromic if the date is written in the British way?
### Zargon Glasses
##### Stage: 2 Challenge Level:
Zumf makes spectacles for the residents of the planet Zargon, who have either 3 eyes or 4 eyes. How many lenses will Zumf need to make all the different orders for 9 families?
### Egyptian Rope
##### Stage: 2 Challenge Level:
The ancient Egyptians were said to make right-angled triangles using a rope with twelve equal sections divided by knots. What other triangles could you make if you had a rope like this?
### All the Digits
##### Stage: 2 Challenge Level:
This multiplication uses each of the digits 0 - 9 once and once only. Using the information given, can you replace the stars in the calculation with figures?
### Multiples Grid
##### Stage: 2 Challenge Level:
What do the numbers shaded in blue on this hundred square have in common? What do you notice about the pink numbers? How about the shaded numbers in the other squares?
### Ancient Runes
##### Stage: 2 Challenge Level:
The Vikings communicated in writing by making simple scratches on wood or stones called runes. Can you work out how their code works using the table of the alphabet?
### Money Bags
##### Stage: 2 Challenge Level:
Ram divided 15 pennies among four small bags. He could then pay any sum of money from 1p to 15p without opening any bag. How many pennies did Ram put in each bag?
### Seven Pots of Plants
##### Stage: 2 Challenge Level:
There are seven pots of plants in a greenhouse. They have lost their labels. Perhaps you can help re-label them.
### All Seated
##### Stage: 2 Challenge Level:
Look carefully at the numbers. What do you notice? Can you make another square using the numbers 1 to 16, that displays the same properties?
### Family Tree
##### Stage: 2 Challenge Level:
Use the clues to find out who's who in the family, to fill in the family tree and to find out which of the family members are mathematicians and which are not.
### The Pet Graph
##### Stage: 2 Challenge Level:
Tim's class collected data about all their pets. Can you put the animal names under each column in the block graph using the information?
### Counters
##### Stage: 2 Challenge Level:
Hover your mouse over the counters to see which ones will be removed. Click to remover them. The winner is the last one to remove a counter. How you can make sure you win?
### Chocoholics
##### Stage: 2 Challenge Level:
George and Jim want to buy a chocolate bar. George needs 2p more and Jim need 50p more to buy it. How much is the chocolate bar?
### Two by One
##### Stage: 2 Challenge Level:
An activity making various patterns with 2 x 1 rectangular tiles.
### Counting Cards
##### Stage: 2 Challenge Level:
A magician took a suit of thirteen cards and held them in his hand face down. Every card he revealed had the same value as the one he had just finished spelling. How did this work?
### How Much Did it Cost?
##### Stage: 2 Challenge Level:
Use your logical-thinking skills to deduce how much Dan's crisps and ice-cream cost altogether.
##### Stage: 2 Challenge Level:
Can you put plus signs in so this is true? 1 2 3 4 5 6 7 8 9 = 99 How many ways can you do it?
### Plate Spotting
##### Stage: 2 Challenge Level:
I was in my car when I noticed a line of four cars on the lane next to me with number plates starting and ending with J, K, L and M. What order were they in?
### Symmetry Challenge
##### Stage: 2 Challenge Level:
Systematically explore the range of symmetric designs that can be created by shading parts of the motif below. Use normal square lattice paper to record your results.
### Rabbits in the Pen
##### Stage: 2 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
### It Figures
##### Stage: 2 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
### Room Doubling
##### Stage: 2 Challenge Level:
Investigate the different ways you could split up these rooms so that you have double the number.
### Pasta Timing
##### Stage: 2 Challenge Level:
Nina must cook some pasta for 15 minutes but she only has a 7-minute sand-timer and an 11-minute sand-timer. How can she use these timers to measure exactly 15 minutes?
### Bean Bags for Bernard's Bag
##### Stage: 2 Challenge Level:
How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?
### Wonky Watches
##### Stage: 2 Challenge Level:
Stuart's watch loses two minutes every hour. Adam's watch gains one minute every hour. Use the information to work out what time (the real time) they arrived at the airport.
### Two on Five
##### Stage: 1 and 2 Challenge Level:
Take 5 cubes of one colour and 2 of another colour. How many different ways can you join them if the 5 must touch the table and the 2 must not touch the table?
### 1 to 8
##### Stage: 2 Challenge Level:
Place the numbers 1 to 8 in the circles so that no consecutive numbers are joined by a line.
### Pouring the Punch Drink
##### Stage: 2 Challenge Level:
There are 4 jugs which hold 9 litres, 7 litres, 4 litres and 2 litres. Find a way to pour 9 litres of drink from one jug to another until you are left with exactly 3 litres in three of the jugs. | 2,197 | 9,635 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2016-44 | longest | en | 0.875309 |
pmmementos.com | 1,686,351,077,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656833.99/warc/CC-MAIN-20230609201549-20230609231549-00116.warc.gz | 535,314,489 | 21,792 | # How many colors are there in the whole world?
There are innumerable colors in the whole world, there are innumerable colors in the world because the spectrum of colors is infinite so there are innumerable colors in the whole world, and the number of wavelengths of light is also infinite, all the colors we see are different types of waves. There is a combination of wavelengths, so the number of colors in the whole world is infinite.
Contents
## How many colors does a rainbow have?
There are seven colors in the rainbow, which are red, orange, yellow, green, blue, indigo, and violet, the color at the top is red rainbow and the color at the bottom is violet.
## What are the different types of colors?
There are only three types of colors, there are primary, secondary, and tertiary, the primary color is red and also yellow, and blue, secondary is green and orange, and violet color, the tertiary color is yellow and orange, red-orange, There is red and violet, blue and violet, blue and green, yellow and green.
## How do you make different colors?
It is very easy to make different colors, you need some paint and brush to make different colors, now you can easily make any type of shape by dipping the brush in the paint and painting anywhere you want, If you want to make new colors, you can mix different colors together properly, or if you want to make a darker color, you can mix a little bit of black with blue, and if you want to make even lighter colors. If you want to make it, you then have to mix a lot of white paint with a little blue, you can use a variety of colors to make new colors. | 354 | 1,617 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2023-23 | latest | en | 0.942555 |
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Broad Topics > Mathematical Thinking > Working systematically
### Spell by Numbers
##### Age 7 to 11 Challenge Level:
Can you substitute numbers for the letters in these sums?
### X Is 5 Squares
##### Age 7 to 11 Challenge Level:
Can you arrange 5 different digits (from 0 - 9) in the cross in the way described?
### A Mixed-up Clock
##### Age 7 to 11 Challenge Level:
There is a clock-face where the numbers have become all mixed up. Can you find out where all the numbers have got to from these ten statements?
### How Old?
##### Age 7 to 11 Challenge Level:
Cherri, Saxon, Mel and Paul are friends. They are all different ages. Can you find out the age of each friend using the information?
### Arranging the Tables
##### Age 7 to 11 Challenge Level:
There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places.
### ABC
##### Age 7 to 11 Challenge Level:
In the multiplication calculation, some of the digits have been replaced by letters and others by asterisks. Can you reconstruct the original multiplication?
### Trebling
##### Age 7 to 11 Challenge Level:
Can you replace the letters with numbers? Is there only one solution in each case?
### Zargon Glasses
##### Age 7 to 11 Challenge Level:
Zumf makes spectacles for the residents of the planet Zargon, who have either 3 eyes or 4 eyes. How many lenses will Zumf need to make all the different orders for 9 families?
### Being Resourceful - Primary Number
##### Age 5 to 11 Challenge Level:
Number problems at primary level that require careful consideration.
### Rabbits in the Pen
##### Age 7 to 11 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
### A-magical Number Maze
##### Age 7 to 11 Challenge Level:
This magic square has operations written in it, to make it into a maze. Start wherever you like, go through every cell and go out a total of 15!
### Shape Times Shape
##### Age 7 to 11 Challenge Level:
These eleven shapes each stand for a different number. Can you use the number sentences to work out what they are?
### Six Is the Sum
##### Age 7 to 11 Challenge Level:
What do the digits in the number fifteen add up to? How many other numbers have digits with the same total but no zeros?
### Dice and Spinner Numbers
##### Age 7 to 11 Challenge Level:
If you had any number of ordinary dice, what are the possible ways of making their totals 6? What would the product of the dice be each time?
### The Pied Piper of Hamelin
##### Age 7 to 11 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
### How Much Did it Cost?
##### Age 7 to 11 Challenge Level:
Use your logical-thinking skills to deduce how much Dan's crisps and ice-cream cost altogether.
### Today's Date - 01/06/2009
##### Age 5 to 11 Challenge Level:
What do you notice about the date 03.06.09? Or 08.01.09? This challenge invites you to investigate some interesting dates yourself.
### Different Deductions
##### Age 7 to 11 Challenge Level:
There are lots of different methods to find out what the shapes are worth - how many can you find?
### Sums and Differences 1
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This challenge focuses on finding the sum and difference of pairs of two-digit numbers.
### All the Digits
##### Age 7 to 11 Challenge Level:
This multiplication uses each of the digits 0 - 9 once and once only. Using the information given, can you replace the stars in the calculation with figures?
### Multiply Multiples 3
##### Age 7 to 11 Challenge Level:
Have a go at balancing this equation. Can you find different ways of doing it?
##### Age 7 to 11 Challenge Level:
Can you put plus signs in so this is true? 1 2 3 4 5 6 7 8 9 = 99 How many ways can you do it?
### The Dice Train
##### Age 7 to 11 Challenge Level:
This dice train has been made using specific rules. How many different trains can you make?
### Oh! Harry!
##### Age 7 to 11 Challenge Level:
A group of children are using measuring cylinders but they lose the labels. Can you help relabel them?
### Pouring the Punch Drink
##### Age 7 to 11 Challenge Level:
There are 4 jugs which hold 9 litres, 7 litres, 4 litres and 2 litres. Find a way to pour 9 litres of drink from one jug to another until you are left with exactly 3 litres in three of the jugs.
### Forgot the Numbers
##### Age 7 to 11 Challenge Level:
On my calculator I divided one whole number by another whole number and got the answer 3.125. If the numbers are both under 50, what are they?
### Multiply Multiples 2
##### Age 7 to 11 Challenge Level:
Can you work out some different ways to balance this equation?
### Mystery Matrix
##### Age 7 to 11 Challenge Level:
Can you fill in this table square? The numbers 2 -12 were used to generate it with just one number used twice.
### Page Numbers
##### Age 7 to 11 Short Challenge Level:
Exactly 195 digits have been used to number the pages in a book. How many pages does the book have?
### Bean Bags for Bernard's Bag
##### Age 7 to 11 Challenge Level:
How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?
### It Figures
##### Age 7 to 11 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
### Multiply Multiples 1
##### Age 7 to 11 Challenge Level:
Can you complete this calculation by filling in the missing numbers? In how many different ways can you do it?
### Sealed Solution
##### Age 7 to 11 Challenge Level:
Ten cards are put into five envelopes so that there are two cards in each envelope. The sum of the numbers inside it is written on each envelope. What numbers could be inside the envelopes?
### Two Primes Make One Square
##### Age 7 to 11 Challenge Level:
Can you make square numbers by adding two prime numbers together?
### Seven Square Numbers
##### Age 7 to 11 Challenge Level:
Add the sum of the squares of four numbers between 10 and 20 to the sum of the squares of three numbers less than 6 to make the square of another, larger, number.
### Route Product
##### Age 7 to 11 Challenge Level:
Find the product of the numbers on the routes from A to B. Which route has the smallest product? Which the largest?
### The Moons of Vuvv
##### Age 7 to 11 Challenge Level:
The planet of Vuvv has seven moons. Can you work out how long it is between each super-eclipse?
### All Seated
##### Age 7 to 11 Challenge Level:
Look carefully at the numbers. What do you notice? Can you make another square using the numbers 1 to 16, that displays the same properties?
### Sums and Differences 2
##### Age 7 to 11 Challenge Level:
Find the sum and difference between a pair of two-digit numbers. Now find the sum and difference between the sum and difference! What happens?
### Polo Square
##### Age 7 to 11 Challenge Level:
Arrange eight of the numbers between 1 and 9 in the Polo Square below so that each side adds to the same total.
### Journeys in Numberland
##### Age 7 to 11 Challenge Level:
Tom and Ben visited Numberland. Use the maps to work out the number of points each of their routes scores.
##### Age 5 to 11 Challenge Level:
How could you put these three beads into bags? How many different ways can you do it? How could you record what you've done?
### Dart Target
##### Age 7 to 11 Challenge Level:
This task, written for the National Young Mathematicians' Award 2016, invites you to explore the different combinations of scores that you might get on these dart boards.
### Build it up More
##### Age 7 to 11 Challenge Level:
This task follows on from Build it Up and takes the ideas into three dimensions!
### Open Squares
##### Age 7 to 11 Challenge Level:
This task, written for the National Young Mathematicians' Award 2016, focuses on 'open squares'. What would the next five open squares look like?
### Two Egg Timers
##### Age 7 to 11 Challenge Level:
You have two egg timers. One takes 4 minutes exactly to empty and the other takes 7 minutes. What times in whole minutes can you measure and how?
### On Target
##### Age 7 to 11 Challenge Level:
You have 5 darts and your target score is 44. How many different ways could you score 44?
### Worms
##### Age 7 to 11 Challenge Level:
Place this "worm" on the 100 square and find the total of the four squares it covers. Keeping its head in the same place, what other totals can you make?
### Number Detective
##### Age 5 to 11 Challenge Level:
Follow the clues to find the mystery number.
### Hubble, Bubble
##### Age 7 to 11 Challenge Level:
Winifred Wytsh bought a box each of jelly babies, milk jelly bears, yellow jelly bees and jelly belly beans. In how many different ways could she make a jolly jelly feast with 32 legs? | 2,205 | 9,287 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2020-10 | latest | en | 0.863948 |
http://slideplayer.com/slide/3942995/ | 1,527,275,381,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867173.31/warc/CC-MAIN-20180525180646-20180525200646-00215.warc.gz | 268,753,593 | 26,657 | # Digital System Ch5-1 Chapter 5 Synchronous Sequential Logic Ping-Liang Lai ( 賴秉樑 ) Digital System 數位系統.
## Presentation on theme: "Digital System Ch5-1 Chapter 5 Synchronous Sequential Logic Ping-Liang Lai ( 賴秉樑 ) Digital System 數位系統."— Presentation transcript:
Digital System Ch5-1 Chapter 5 Synchronous Sequential Logic Ping-Liang Lai ( 賴秉樑 ) Digital System 數位系統
Digital System Ch5-2 Outline of Chapter 5 5.1Introduction 5.2Sequential Circuits 5.3 Storage Element: Latches 5.4Storage Element: Flip-Flops 5.5Analysis of Clocked Sequential Circuits 5.7State Reduction and Assignment 5.8Design Procedure
Digital System Ch5-3 5-1Introduction Combinational circuits Contains no memory elements The outputs depends on the inputs
Digital System Ch5-4 5-2Sequential Circuits Sequential circuits A feedback path The state of the sequential circuit (inputs, current state) (outputs, next state) Synchronous: the transition happens at discrete instants of time Asynchronous: at any instant of time
Digital System Ch5-5 Synchronous sequential circuits A master-clock generator to generate a periodic train of clock pulses The clock pulses are distributed throughout the system Clocked sequential circuits Most commonly used No instability problems The memory elements: flip-flops » Binary cells capable of storing one bit of information. » Two outputs: one for the normal value and one for the complement value. » Maintain a binary state indefinitely until directed by an input signal to switch states.
Digital System Ch5-6 Fig. 5.2 Synchronous clocked sequential circuit
Digital System Ch5-7 5-3Latches ( 栓鎖器 ) Basic flip-flop circuit Two NOR gates More complicated types can be built upon it Directed-coupled RS flip-flop: the cross-coupled connection An asynchronous sequential circuit (S, R)= (0, 0): no operation (S, R)=(0, 1) : reset (Q=0, the clear state) (S, R)=(1, 0) : set (Q=1, the set state) (S, R)=(1, 1) : indeterminate state (Q=Q'=0) Consider (S, R) = (1, 1) (0, 0)
Digital System Ch5-8 SR latch with NAND gates Fig. 5.4 SR latch with NAND gates
Digital System Ch5-9 SR latch with control input En=0, no change En=1, 0/1 1/S' 1/R' Fig. 5.5 SR latch with control input
Digital System Ch5-10 D Latch Eliminate the undesirable conditions of the indeterminate state in the RS flip-flop D: data Gated D-latch D Q when En=1; no change when En=0 Fig. 5.6 D latch S R 0/1 1/D' 1/D
Digital System Ch5-11 Fig. 5.7 Graphic symbols for latches
Digital System Ch5-12 5-4Flip-Flops ( 正反器 ) A trigger The state of a latch or flip-flop is switched by a change of the control input. Level triggered – Latches Edge triggered – Flip-Flops Fig. 5.8 Clock response in latch and flip-flop
Digital System Ch5-13 If level-triggered flip-flops are used ( 準位觸發 ) The feedback path may cause instability problem Edge-triggered flip-flops ( 邊緣觸發 ) The state transition happens only at the edge Eliminate the multiple-transition problem
Digital System Ch5-14 Edge-triggered D flip-flop Master-slave D flip-flop Two separate flip-flops A master flip-flop (positive-level triggered) A slave flip-flop (negative-level triggered) Fig. 5.9 Master-slave D flip-flop
Digital System Ch5-15 CP = 1: (S, R) (Y, Y'); (Q, Q') holds. CP = 0: (Y, Y') holds; (Y, Y') (Q, Q'). (S, R) could not affect (Q, Q') directly. The state changes coincide with the negative-edge transition of CP. 第三版內容，參考用 !
Digital System Ch5-16 Edge-triggered flip-flops The state changes during a clock-pulse transition A D-type positive-edge-triggered flip-flop Fig. 5.10 D-type positive-edge-triggered flip-flop
Digital System Ch5-17 Three basic flip-flops (S, R) = (0, 1): Q = 1 (S, R) = (1, 0): Q = 0 (S, R) = (1, 1): no operation (S, R) = (0, 0): should be avoided Fig. 5.10 D-type positive-edge-triggered flip-flop
Digital System Ch5-18 1 0 0 1 0 0 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 0 0 1 Holding data Set Reset Clk=0 Clk=1 positive-edge-triggered
Digital System Ch5-19 Setup and Hold Time for FF The setup time D input must be maintained at a constant value prior to the application of the positive CP pulse. Data to the internal latches. The hold time D input must not changes after the application of the positive CP pulse. Clock to the internal latch. Q D CK D T setup T hold D Q T propagation delay
Digital System Ch5-20 Summary CP=0: (S, R) = (1, 1), no state change CP= ↑ : state change once CP=1 and ↓ : state holds Eliminate the feedback problems in sequential circuits All flip-flops must make their transition at the same time
Digital System Ch5-21 Other Flip-Flops The edge-triggered D flip-flops The most economical and efficient Positive-edge and negative-edge Fig. 5.11 Graphic symbols for edge-triggered D flip-flop
Digital System Ch5-22 JK FF JK flip-flop D=JQ ’ +K ’ Q J=0, K=0: D=Q, no change J=0, K=1: D=0 Q=0 J=1, K=0: D=1 Q=1 J=1, K=1: D=Q' Q=Q' Fig. 5.12 JK flip-flop JKQ(t+1) 00Q(t) (No change) 010 (Reset) 101 (Set) 11Q’(t) (Complement)
Digital System Ch5-23 T FF T (toggle) flip-flop D = T ⊕ Q = TQ'+T'Q » T=0: D=Q, no change » T=1: D=Q' Q=Q' Fig. 5.13 T flip-flop T Q ( t +1) 0 Q ( t ) (No change) 1 Q’ ( t ) (Complement)
Digital System Ch5-24 Summary Characteristic tables
Digital System Ch5-25 Characteristic equations D flip-flop » Q(t+1) = D JK flip-flop » Q(t+1) = JQ'+K'Q T flop-flop » Q(t+1) = T ⊕ Q
Digital System Ch5-26 Direct inputs Asynchronous set and/or asynchronous reset Fig. 5.14 D flip-flop with asynchronous reset
Digital System Ch5-27 5-5Analysis of Clocked Sequential Circuits A sequential circuit (inputs, current state) (output, next state). A state equation (also called transition equation) specified the next state as a function of the present state and inputs. A state transition table or state transition diagram.
Digital System Ch5-28 State Equations A(t+1) = A(t)x(t) + B(t)x(t) B(t+1) = A'(t)x(t) A compact form A(t+1) = Ax + Bx B(t+1) = A'x The output equation y(t) = (A(t)+B(t))x'(t) y = (A+B)x' Fig. 5.15 Example of sequential circuit
Digital System Ch5-29 State Table State transition table = State equations A ( t + 1) = Ax + Bx B ( t + 1) = Ax y = Ax + Bx
Digital System Ch5-30 State Equation A ( t + 1) = Ax + Bx B ( t + 1) = Ax y = Ax + Bx
Digital System Ch5-31 State Diagram State transition diagram A circle: a state. A directed lines connecting the circles: the transition between the states » Each directed line is labeled “ inputs/outputs ”. A logic diagram a state table a state diagram. Fig. 5.16 State diagram of the circuit of Fig. 5.15
Digital System Ch5-32 Flip-Flop Input Equations The part of circuit that generates the inputs to flip-flops Also called excitation functions D A = Ax +Bx D B = A'x The output equations To fully describe the sequential circuit y = (A+B)x' Note Q(t+1)=D Q DADA DBDB y
Digital System Ch5-33 Analysis with D FFs The input equation D A =A ⊕ x ⊕ y The state equation A(t+1)=A ⊕ x ⊕ y Fig. 5.17 Sequential circuit with D flip-flop
Digital System Ch5-34 Analysis with JK flip-flops The next-state can be derived by following procedure 1.Determine the flip-flop equations in terms of the present state and input variables ( 用目前狀態和輸入變數的觀點決定正反器的輸入方程式 ); 2.List the binary values of each input equation ( 列出每一個輸入方程式的 二進位值 ); 3.Use the corresponding flip-flop characteristic table to determine the next state values in the state table ( 使用相對應的的正反器特性表，決定狀 態表中的次一狀態值 ).
Digital System Ch5-35 Analysis with JK Flip-Flops An sequential circuit with JK FFs Determine the flip-flop input function in terms of the present state and input variables; Used the corresponding flip-flop characteristic table to determine the next state. Fig. 5.18 Sequential circuit with JK flip-flop J A =B, K A =Bx’ J B =x’, K B =A’x+Ax’
Digital System Ch5-36 J A = B, K A = Bx' J B = x', K B = A'x + Ax' Derive the state table Or, derive the state equations using characteristic equation. JKQ(t+1) 00Q(t) (No change) 010 (Reset) 101 (Set) 11Q’(t) (Complement)
Digital System Ch5-37 State transition diagram State equation for A and B: Fig. 5.19 State diagram of the circuit of Fig. 5.18 J A =B, K A =Bx’ J B =x’, K B =A’x+Ax’
Digital System Ch5-38 Analysis with T Flip-Flops The characteristic equation Q(t+1)= T ⊕ Q = TQ'+T'Q Fig. 5.20 Sequential circuit with T flip-flop
Digital System Ch5-39 The input and output functions T A =Bx T B = x y = AB The state equations A(t+1) = (Bx)'A+(Bx)A' =AB'+Ax'+A'Bx B(t+1) = x ⊕ B
Digital System Ch5-40 State Table T Q ( t +1) 0 Q ( t ) (No change) 1 Q’ ( t ) (Complement)
Digital System Ch5-41 Mealy and Moore Models The Mealy model ( 密利模型 ): the outputs are functions of both the present state and inputs (Fig. 5-15). The outputs may change if the inputs change during the clock pulse period. » The outputs may have momentary false values unless the inputs are synchronized with the clocks. The Moore model ( 莫爾模型 ): the outputs are functions of the present state only (Fig. 5-20). The outputs are synchronous with the clocks.
Digital System Ch5-42 Mealy and Moore Models Fig. 5.21 Block diagram of Mealy and Moore state machine
Digital System Ch5-43 5-7State Reduction and Assignment State Reduction ( 狀態簡化 ) Reductions on the number of flip- flops and the number of gates. A reduction in the number of states may result in a reduction in the number of flip-flops. An example state diagram showing in Fig. 5.25. Fig. 5.25State diagram
Digital System Ch5-44 State Reduction Only the input-output sequences are important. Two circuits are equivalent » Have identical outputs for all input sequences; » The number of states is not important. Fig. 5.25State diagram State:aabcdeffgfga Input:01010110100 Output:00000110100
Digital System Ch5-45 Equivalent states Two states are said to be equivalent » For each member of the set of inputs, they give exactly the same output and send the circuit to the same state or to an equivalent state. » One of them can be removed.
Digital System Ch5-46 Reducing the state table e = g (remove g); d = f (remove f);
Digital System Ch5-47 The reduced finite state machine State:aabcdeddedea Input:01010110100 Output:00000110100
Digital System Ch5-48 The checking of each pair of states for possible equivalence can be done systematically (section 9-5). The unused states are treated as don't-care condition fewer combinational gates. Fig. 5.26 Reduced State diagram
Digital System Ch5-49 State Assignment State Assignment ( 狀態指定 ) To minimize the cost of the combinational circuits. Three possible binary state assignments. (m states need n-bits, where 2 n > m)
Digital System Ch5-50 Any binary number assignment is satisfactory as long as each state is assigned a unique number. Use binary assignment 1.
Digital System Ch5-51 5-8Design Procedure Design Procedure for sequential circuit ( 設計程序 ) The word description of the circuit behavior to get a state diagram; ( 從文 字敘述及所需的操作規格，獲得電路的狀態圖 ) State reduction if necessary; ( 如果需要，簡化狀態數量 ) Assign binary values to the states; ( 指定狀態的二進位值 ) Obtain the binary-coded state table; ( 獲得二進位編碼的狀態表 ) Choose the type of flip-flops; ( 選擇欲使用的正反器形式 ) Derive the simplified flip-flop input equations and output equations; ( 推導 出已化簡的輸入方程式，及輸出方程式 ) Draw the logic diagram; ( 繪製邏輯圖 )
Digital System Ch5-52 Synthesis using D flip-flops An example state diagram and state table: design a circuit to detect a sequence of three or more consecutive 1’s. Fig. 5.27 State diagram for sequence detector
Digital System Ch5-53 The flip-flop input equations A(t+1) = D A (A, B, x) = (3, 5, 7) B(t+1) = D B (A, B, x) = (1, 5, 7) The output equation y(A, B, x) = (6, 7) Logic minimization using the K map D A = Ax + Bx D B = Ax + B'x y = AB
Digital System Ch5-54 Fig. 5.28 Maps for sequence detector
Digital System Ch5-55 Sequence Detector The logic diagram Fig. 5.29 Logic diagram of sequence detector
Digital System Ch5-56 Excitation Tables ( 激勵表 ) A state diagram flip-flop input functions Straightforward for D flip-flops. We need excitation tables for JK and T flip- flops. JKQ(t+1) 00 Q(t) (No change) 01 0 (Reset) 10 1 (Set) 11 Q’(t) (Complement) TQ ( t +1) 0 Q ( t ) (No change) 1 Q’ ( t ) (Complement)
Digital System Ch5-57 Synthesis using JK FFs The same procedure with D FFs, but a extra excitation table must be used. The state table and JK flip-flop inputs.
Digital System Ch5-58 J A = Bx'; K A = Bx J B = x; K B = (A ⊕ x) ‘ y = ? Fig. 5.30 Maps for J and K input equations
Digital System Ch5-59 Fig. 5.31 Logic diagram for sequential circuit with JK flip-flops
Digital System Ch5-60 Synthesis using T flip-flops A n-bit binary counter The state diagram. No inputs (except for the clock input). Fig. 5.32 State diagram of three-bit binary counter
Digital System Ch5-61 The state table and the flip-flop inputs
Digital System Ch5-62 Fig. 5.33 Maps of three-bit binary counter
Digital System Ch5-63 Logic simplification using the K map T A2 = A 1 A 2 T A1 = A 0 T A0 = 1 The logic diagram Fig. 5.34 Logic diagram of three-bit binary counter
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http://www.physicsforums.com/showthread.php?t=542431 | 1,386,417,306,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163054353/warc/CC-MAIN-20131204131734-00031-ip-10-33-133-15.ec2.internal.warc.gz | 502,776,820 | 7,079 | Ramp problem with friction
by sakebu
Tags: friction, ramp
P: 6 1. The problem statement, all variables and given/known data μs = .300 Mass = 3.00kg Incline = 35° What is the minimum normal force applied to prevent the crate from sliding down the incline? 2. Relevant equations μs = Fk / n 3. The attempt at a solution The book solution is 32.1N but I have no idea how they got that.
HW Helper
P: 2,315
Quote by sakebu 1. The problem statement, all variables and given/known data μs = .300 Mass = 3.00kg Incline = 35° What is the minimum normal force applied to prevent the crate from sliding down the incline? 2. Relevant equations μs = Fk / n 3. The attempt at a solution The book solution is 32.1N but I have no idea how they got that.
If this crate is left alone, it will slip down as friction is not big enough.
One way to stop it would be to apply an appropriate force parallel to the slope.
What is planned here is to apply a force perpendicular to the slope, which will increase the friction force so that it is strong enough to prevent motion.
Related Discussions Classical Physics 6 Introductory Physics Homework 7 Introductory Physics Homework 5 Introductory Physics Homework 1 Introductory Physics Homework 7 | 308 | 1,226 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-48 | longest | en | 0.903165 |
https://homework.cpm.org/category/CC/textbook/cc2/chapter/4/lesson/4.2.4/problem/4-69 | 1,726,458,540,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.29/warc/CC-MAIN-20240916012328-20240916042328-00814.warc.gz | 270,378,072 | 14,952 | ### Home > CC2 > Chapter 4 > Lesson 4.2.4 > Problem4-69
4-69.
At your first job, you may be amazed to learn that one fourth of your paycheck will go to pay taxes.
Suppose the amounts listed in parts (a) through (c) below are the earnings for three employees. Determine how much of each paycheck will go to pay taxes.
1. $\84$
1. $\128$
$\32$
1. $\210$
$\52.50$ | 108 | 368 | {"found_math": true, "script_math_tex": 5, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-38 | latest | en | 0.902557 |
https://gamedev.stackexchange.com/questions/109056/how-to-implement-view-bobbing/109061 | 1,718,539,827,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861659.47/warc/CC-MAIN-20240616105959-20240616135959-00690.warc.gz | 237,263,541 | 37,387 | # How to implement view bobbing?
I'm developing a First-Person shooter game. Everything is just fine but I've a little problem with my camera moving.
My FPS camera is moving smoothly! But the camera should move like a human is walking and his head is doing up and down.
Now I'm stuck! So what kind of algorithm can I use for that purpose?
Any kind of help is appreciated.
• Any particular engine? Also, why not just walk and see how your head bobs and come up with a height (y) versus time graph, turn it into a mathematical approximation, and then implement it in-game? Similarly you could look at other games with head bobbing and copy their style of motion. Commented Oct 1, 2015 at 7:11
• No particular engine. Just developing my own game engine with direct x-9. Sorry, I'm little weak in mathematics. But if you post the algorithm as an answer, it would be a great help for me. Thanks for your comment. Commented Oct 1, 2015 at 7:18
• There really is no standard head bob algorithm. I gave the basic components of what you'd need to create head bobbing but playing with the math is what gives you differing end results (more/less bob, faster/slower, more jerkiness, etc). The differing end results are all based on taste and not suited for a stack exchange question. Commented Oct 1, 2015 at 7:56
Head bobbing consists of transformations to the camera to imply human movements of the player. The player would be using his/her feet to step from one foot to the other. This causes all sorts of changes to the viewpoint of the player but we're just going to look at a few.
Let's define a function to call with some time value which we can then plug into an oscillator to then apply the bobbing. I'm using Unity3D and C# as an example but the concepts apply to anything. For those unfamiliar Transform holds object transform data (position, rotation) and Rigidbody holds physics information (velocity).
void ApplyBob(float time, GameObject playerObj, GameObject camObj)
{
Transform t = camObj.GetComponent<Transform>(); //Position and rotation of camera
Rigidbody rb = playerObj.GetComponent<RigidBody>(); //Velocity of player
We want the head bob to be based both on time and the player's velocity so let's calculate the velocity as well.
Vector3 vel = rb.velocity;
vel.y = 0; //0 out y (up) component for xz velocity only
float velocity = vel.magnitude;
Obviously if the player isn't moving, we don't need to apply any head bob (We check it against a threshold instead of 0)
if(Mathf.Abs(velocity) <= 0.0001f)
return;
Now we need to plug it into some sort of oscillating function, we'll stick with sine.
float bobOscillate = Mathf.Sin(time * velocity * (2 * Mathf.PI));
So now we have an oscillating function from -1 to 1 which completes one cycle every at the same rate as the velocity (1 unit/second in velocity would give 1 Hz in bobbing).
What we do now is take this and apply it to all the values we want to bob.
t.position.y += bobOscillate; //Oscillate the position in y (up) axis)
t.rotation.x += bobOscillate * 5; //Oscillates 5 degrees in each direction around x axis, presented this way for clarity (if using Unity3D, use Transform.Rotate)
}
This is the most basic setup you'd need for head bobbing. To perfect it for your specific game you'd want to do with following.
• Replace the sine function with a different or transformed function. Sine is a very smooth, ideal function which probably doesn't best represent the jerkiness and subtleties of human motion. Using a different oscillating function (or even one that draws on a custom data set and interpolates between the values similar to what might happen with motion capture data) would be much better.
• Changing the oscillator frequency to better match your players velocity for your given engine and player walking animation. This would require adding some constant multiplier into the time * velocity * 2*pi portion of the line of code with the sine function.
• Adding more values to oscillate instead of just the height of the camera and the rotation to better simulate head bobbing.
All of these things would require more research into how humans actually walk and move or just playing around with things until they seem realistic enough / desirable. | 964 | 4,267 | {"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} | 3.484375 | 3 | CC-MAIN-2024-26 | latest | en | 0.948356 |
https://www.coursehero.com/file/1151651/Study-Guides/ | 1,493,202,164,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121267.21/warc/CC-MAIN-20170423031201-00034-ip-10-145-167-34.ec2.internal.warc.gz | 897,871,602 | 25,060 | Study Guides
Study Guides - STUDY GUIDE/SUGGESTED PROBLEMS Chapter 1:...
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Unformatted text preview: STUDY GUIDE/SUGGESTED PROBLEMS Chapter 1: Typical Test Questions: 1. Perform the following calculations, reporting the answer to the correct number of significant figures and reporting the answer in both scientific notation and decimal notation. a. 2.64E4 x 1.15E-3 d. 32.5 x 1.0E-3 b. 205.1 + 1.386 e. 1.3E4 – 5.41E3 c. 14.69 / 1.1E5 f. 14.3 – (12.12 x 0.650) 2. How many feet are there in 0.50 km? 3. How many ml are there in 1.25E-2 L 4. If the students in a particular class are 40% male and 60% female and there are 16 males, what is the number in the total class? How many females in the class? 5. What is the volume occupied by 15.2g of a material with a density of 2.41 g/ml? 6. Convert 87 ° F into ° C and ° K. Convert -40 ° C into ° F and ° K. Chapter 2: Typical test questions: 1. An atom has an atomic number of 17 and an atomic weight of 35. How many protons, electron, and neutrons. 2. A fluoride ion has an atomic number of 9, and atomic mass of 19, and a charge of – 1. How many electrons, protons, and neutrons. 3. How many protons, electrons, and neutrons does a typical calcium cation have? 4. In the following, supply the name or the formula, depending on what is missing. Hydrogen peroxide NaF Nitric acid FeSO 3 Ammonia HClO 4 Benzene H 2 CO 3 Magnesium nitrate Sn(OH) 2 Aluminum carbonate KHSO 4 Copper (I) phosphate Mg(CN) 2 Potasium acetate AgF Silver oxide NH 4 NO 3 Zinc hydroxide LiCH 3 COO Ch 3: Typical test questions: Consider the following reactions in answering the questions below: a. 2 NaOH(aq) + H 2 SO 4(aq) 2 H 2 O (l) + Na 2 SO 4(aq) b. Ag (S) + H 2 S (g) Ag 2 S (s) + H 2(g) c. CaCO 3(s) + HCl (aq) CaCl 2(aq) + CO 2(g) + H 2 O (l) d. Fe(OH) 3(s) + H 3 PO 4(aq) FePO 4(aq) + H 2 O (l) 1. The first reaction is balanced. Balance the rest of the reactions. 2. How many moles of Na 2 SO 4 could be produced from 0.23 moles of NaOH in a? 3. How many moles of sulfuric acid would be consumed in the above reaction? 4. How many grams of Ag 2 S would be produced by the complete reaction of 0.25 moles of hydrogen sulfide in b? 5. If, in c, the reaction produced 2.42 g of CO 2 , how many grams of CaCO 3 reacted (assuming 100% yield)? 6. Assume that you started with 15.1g of Fe(OH) 3 and 25.2 g of phosphoric acid, what is the potential yield of FePO 4 ? Which reagent was limiting? How much of the excess reagent is left after the reaction? If the yield of Iron(III) phosphate were 17.9g, what was the % yield? 7. If I dissolve 25.3 g of HCl in 0.360 L of total solution, what is the molarity of the solution? How many moles of HCl would there be in 450 ml of this solution? What volume of solution would I need to take if I needed 15.1 g of HCl?...
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Ask a homework question - tutors are online | 1,086 | 3,404 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2017-17 | longest | en | 0.835161 |
https://community.tableau.com/thread/130632 | 1,561,373,403,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999482.38/warc/CC-MAIN-20190624104413-20190624130413-00397.warc.gz | 380,667,732 | 24,449 | 1 Reply Latest reply on Oct 8, 2013 12:18 PM by Alex Kerin
# How can I get the minimum of two measures? e.j. The minimum value between inventory and sales.
How can I get the minimum of two measures? e.j. The minimum value between inventory and sales.
I tried with formula MIN, IF and variants but seems not to work.
thank you,
RQ
Additional information: What I am trying to achieve is to perform a calculation where Tableau can identify which measure has the lower value between Sales and Inventory, and use the result for a calculation.
e.g. Sales 100, Avg Inv (50) then use the result to calculate the fill rate as a % of sales, 50/100 = 50%
• ###### 1. Re: How can I get the minimum of two measures? e.j. The minimum value between inventory and sales.
Can you provide a few lines of what your data looks like? | 202 | 821 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2019-26 | latest | en | 0.938357 |
http://www.sophia.org/tutorials/calculating-forces--2 | 1,440,800,063,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644064019.39/warc/CC-MAIN-20150827025424-00234-ip-10-171-96-226.ec2.internal.warc.gz | 716,667,120 | 15,428 | ### Free Educational Resources
+
Calculating Forces
4 Tutorials that teach Calculating Forces
Next Generation: MS.PS2.2
# Calculating Forces
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Author: Nathan Lampson
##### Objective:
This lesson will explain how to calculate net forces on an object being pushed or pulled.
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##### Test Yourself
Got it
Tutorial
Net force is the combinations of all forces acting on an object. When forces act in the same direction, net force can be calculated by adding the forces together. When forces act in opposite directions, the net force is in the direction of the greater force. The amount of force is found by combining opposing forces with subtraction. The strength of a force is measured in Newtons (N).
Example:
Bob and George are having a contest and are trying to push a crate in opposing directions. George's force (known as F1) is 8 Newtons, while Bob's force in the opposite direction is 12 Newtons. In order to find the net force, George's lesser force is subtracted from Bob's greater force.
Forces in opposing directions can be calculated by finding the difference of the forces. The net force is in the direction of the greater force.
12 Newtons (The force of Bob's push) - 8 Newtons (The force of George's push) = 4 Newtons (Net force)
A net force of 4 newtons is produced in the direction of Bob's force.
## Calculating Forces
### No bones about it. Our Human Biology course is only \$329.
Sophia's online courses not only save you money, but credits are also eligible for transfer to over 2,000 colleges and universities.* | 387 | 1,630 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2015-35 | latest | en | 0.8957 |
https://www.includehelp.com/dot-net/calculate-the-standard-deviation-of-a-set-of-given-numbers.aspx?ref=rp | 1,675,608,724,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500255.78/warc/CC-MAIN-20230205130241-20230205160241-00743.warc.gz | 851,803,450 | 148,188 | # C# program to calculate the Standard Deviation of a set of given numbers
Here, we are going to learn how to calculate the Standard Deviation of a set of given numbers in C#?
Submitted by Nidhi, on September 21, 2020
Here we will calculate the Standard Deviation of a set of given numbers and then print the calculated Standard Deviation on the console screen.
Program:
The source code to calculate the Standard Deviation of a set of given numbers is given below. The given program is compiled and executed successfully on Microsoft Visual Studio.
```//C# Program to Find the Standard Deviation of a
//set of given numbers.
using System;
using System.Collections.Generic;
class Program
{
private static void Main()
{
List<double> list = new List<double> { 1, 2, 3, 4, 5, 6};
double mean = 0;
double variance = 0;
double standard_deviation = 0;
double sum = 0;
double temp = 0;
int loop = 0;
for (loop = 0; loop < list.Count; loop++)
{
sum += list[loop];
}
mean = sum / (list.Count - 0);
for (loop = 0; loop < list.Count; loop++)
{
temp += Math.Pow((list[loop] - mean), 2);
}
variance=temp / (list.Count - 0);
standard_deviation = Math.Sqrt(variance);
Console.WriteLine("Mean : " + mean );
Console.WriteLine("Variance : " + variance );
Console.WriteLine("Standard deviation: " + standard_deviation);
}
}
```
Output:
```Mean : 3.5
Variance : 2.91666666666667
Standard deviation: 1.70782512765993
Press any key to continue . . .
```
Explanation:
Here, we created a class Demo that contains a static method Main(). The Main() method is the entry point of the program. Here we created the list of numbers, and then calculate the MEAN, VARIANCE, and STANDARD DEVIATION of the given list of numbers. After that, we printed the calculated values on the console screen.
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More: » Articles » Puzzles » News/Updates | 635 | 2,486 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | longest | en | 0.710496 |
https://learnhool.in/rs-aggarwal-class-6-solutions-chapter-10-ratio-proportion-and-unitary-method-ex-10c/ | 1,701,757,388,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100545.7/warc/CC-MAIN-20231205041842-20231205071842-00051.warc.gz | 403,772,258 | 10,242 | # RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10C - Learn Hool
## RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10C
These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10C.
Other Exercises
Question 1.
Solution:
Cost of 14 m of cloth = Rs. 1890
Cost of 1 m = Rs. (\ frac { 1890 }{ 14 })
and cost of 6 m = Rs. (\ frac { 1890times 6 }{ 14 })
= Rs. 135 x 6
= Rs. 810
Question 2.
Solution:
Cost of 1 dozen or 12 soaps = Rs. 285.60
Cost of 1 soap = Rs. (\ frac { 285.60 }{ 12 })
Cost of 15 soaps = Rs. (\ frac { 285.60times 15 }{ 12 })
= Rs. 357.00
Question 3.
Solution:
Cost of 9 kg of rice = Rs. 327.60
Cost of 1 kg = Rs. (\ frac { 327.60 }{ 9 })
and cost of 50 kg = Rs. (\ frac { 327.60times 50 }{ 9 })
= Rs. 36.40 x 50
= Rs. 1820
Question 4.
Solution:
Weight of 22.5 metres of the iron rod: = 85.5 kg
Weight of 1 metre of the iron rod
Question 5.
Solution:
Quantity of oil in 15 tins = 234 kg
Quantity of oil in 1 tin = (\ frac { 234 }{ 15 }) kg
Quantity of oil in 10 tins
Question 6.
Solution:
Distance covered by the car in 12 litres of diesel = 222 kms
Distance covered by the car in 1 litre of diesel = (\ frac { 222 }{ 12 }) km
Question 7.
Solution:
Charges of 25 tonnes of weight = Rs. 540
charges of 1 ton = Rs.(\ frac { 540 }{ 25 })
Question 8.
Solution:
Weight of copper in 4.5 g of alloy = 3.5g
Weight of copper in 1 g of alloy
Question 9.
Solution:
In Rs. 87.50, the inland letter are purchased = 35
In Re. 1, letters can be purchased
= (\ frac { 35 }{ 87.50 })
and in Rs. 315, letters can be purchased
Question 10.
Solution:
4 dozen = 4 x 12 = 48 bananas
In Rs. 104, banana are purchased = 48
Question 11.
Solution:
In Rs. 22770, chairs are purchased =18
In Re. 1, chairs will be purchased
= (\ frac { 18 }{ 22770 })
and in Rs. 10120, chairs will be
purchased = (\ frac { 18times 10120 }{ 22770 })
= 8
Question 12.
Solution:
(i) A car travels 195 km distance in = 3 hours
It will travel 1 km distance in = (\ frac { 3 }{ 195 }) hr.
and it will travel 520 km distance in
= (\ frac { 3times 520 }{ 195 })
= 8 hr
(ii) A car travels in 3hr = 195 km
Question 13.
Solution:
(i) A laborer earn in 12 days = Rs. 1980
He will earn in 1 day = Rs. (\ frac { 1980 }{ 12 })
and he will earn in 7 days
Question 14.
Solution:
(i) Weight of 65 books = 13 kg
Then weight of 1 book = (\ frac { 13 }{ 65 }) kg
Question 15.
Solution:
Number of boxes needed for 6000 pens = 48
Number of boxes needed for 1 pen 48 = (\ frac { 48 }{ 6000 })
Question 16.
Solution:
Clearly, less workers will build the wall in more days.
And, more workers will build the wall in less days.
24 workers can build the wall in 15 days
1 worker can build the wall in (15 x 24) days
(less worker, more days)
9 workers will build the wall in
Question 17.
Solution:
Men needed to finish a piece of work in 26 days = 40
Men needed to finish a piece of work in 1 day = 40 x 26 (less days, more men)
Question 18.
Solution:
Clearly, less men will take more days to consume the food.
And, more men will take less days to consume the food.
550 men have provisions for 28 days
1 men has provisions for (28 x 550) days [less men, more days]
700 men will have provisions for
Question 19.
Solution:
Clearly, less persons will consume the rice in more days.
And more persons will consume the rice in less days.
60 persons consume the bag of rice in 3 days.
1 person will consume the bag of rice in
(3 x 60) days (less persons, more days)
18 persons will consume the bag of rice
Hope given RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10C are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
@ Designed By : Vikas Copyright 2023 @ LearnHool.In | 1,252 | 3,926 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.852946 |
https://quantumcomputing.stackexchange.com/questions/6519/does-the-choi-jamiolkowski-isomorphism-really-establish-a-connection-between-kin | 1,657,212,035,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104495692.77/warc/CC-MAIN-20220707154329-20220707184329-00640.warc.gz | 511,902,958 | 67,493 | # Does the Choi-Jamiolkowski isomorphism really establish a connection between kinematics and dynamics?
I understand the mathematical construction of the Choi-Jamiolkowski isomorphism aka channel-state duality. It all makes sense formally, yet I still struggle to grasp its physical (or quantum-informational) meaning. Does the isomorphism between quantum states and quantum channels in any sense establish some kind of connection relating the constitution (state) of systems to evolution (channel) of other systems?
Cross-posted on physics.SE
• If you give me the state dual of a Clifford channel, I can teleport a new state through it in order to apply the channel to the new state. Jun 19, 2019 at 23:12
Mathematically it is a relationship between a bipartite linear operator vector space $$L(X\otimes Y)$$ and a superoperator vector space $$C(X): L(X)\to L(Y)$$ (maps of linear operators to linear operators). Bipartite density matrices are contained in the former, and quantum channels in the latter. The real "physical" meaning of the isomorphism for quantum information theory is that a superoperator is a valid quantum channel if and only if its Choi-state is a valid density matrix. You can read more about properties of the Choi-representation (and other representations of quantum channels) in my review article [1].
This also has several quantum computing applications outside of quantum information theory. The most common two are probably
1. Quantum process tomography
2. Gate teleportation (especially in measurement based quantum computing)
Matthew Leiffer has a good blog post [2] which describes point 2 (and talks more generally about the Choi-Jamiolkowski isomorphism in the context of some of his research on conditional states).
For point 1. If if one wants to experimentally reconstruct a description of a quantum channel they basically need to reconstruct the Choi-matrix representation of the channel. This is typically done by preparing a tomographically complete set of input states (states that span the input state space), and perform a tomographyically complete set of measurements (measurements with outcomes projectors that span the output state space) and use the resulting probabilities to reconstruct the description of the channel.
From the point of view of measurement probabilities, the "channel" is indistinguishable from a bipartite "state" that we would have performed state tomography on, and so we can reconstruct the description as if we were simply reconstructing a bipartite density matrix using state tomography (there are some subtleties concerning the trace preserving property which I won't get into). You can see an example of process tomography in this Qiskit Ignis tutorial notebook [3].
The is also another way to do process tomography called Ancilla assisted process tomography [4] which is a physical implementation of the Choi-Jamiolkowski isomorphism. You could prepare a maximally entangled bell state, send half of it through the channel to be investigated (and do nothing to the other half), and then perform state tomography of the full output to directly reconstruct the Choi-state for the unknown channel. This is rarely used in practice as it is typically less accurate than the standard method due to errors in preparing the maximally entangled input state.
[1] Quant. Inf. Comp. 15, 0579-0811 (2015) arXiv:1111.6950 [quant-ph]
[4] Physical Review Letters 90, 193601 (2003), arXiv:quant-ph/0303038
• Yes, thank you very much Christopher. Btw, I had actually read your publication [1] when I first studied superoperators and their representations - it was very helpful. I understand the mathematical procedure; guess I am still hoping for some further insight as to why this construction works or what it implies beyond the mere encoding of the evolution law in a state. Sometimes there is some additional/different operational perspective on the same math - that is what I was angling for... Oct 10, 2019 at 18:19
• Hi @cjwood , the reference 3(github link) appears to be broken. Jun 20, 2020 at 15:18 | 880 | 4,080 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-27 | latest | en | 0.911404 |
https://www.hindawi.com/journals/ijmms/1982/319374/ | 1,606,585,301,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195687.51/warc/CC-MAIN-20201128155305-20201128185305-00294.warc.gz | 683,879,196 | 50,061 | Open Access
Volume 5 |Article ID 319374 | https://doi.org/10.1155/S0161171282000532
Richard H. Hudson, Kenneth S. Williams, "A new formulation of the law of octic reciprocity for primes ±3(mod8) and its consequences", International Journal of Mathematics and Mathematical Sciences, vol. 5, Article ID 319374, 20 pages, 1982. https://doi.org/10.1155/S0161171282000532
# A new formulation of the law of octic reciprocity for primes ≡±3(mod8) and its consequences
Revised01 Sep 1981
#### Abstract
Let p and q be odd primes with q±3(mod8), p1(mod8)=a2+b2=c2+d2 and with the signs of a and c chosen so that ac1(mod4). In this paper we show step-by-step how to easily obtain for large q necessary and sufficient criteria to have (1(q1)/2q(p1)/8(ab)d/ac)j(modp) for j=1,,8 (the cases with j odd have been treated only recently [3] in connection with the sign ambiguity in Jacobsthal sums of order 4. This is accomplished by breaking the formula of A.E. Western into three distinct parts involving two polynomials and a Legendre symbol; the latter condition restricts the validity of the method presented in section 2 to primes q3(mod8) and significant modification is needed to obtain similar results for q±1(mod8). Only recently the author has completely resolved the case q5(mod8), j=1,,8 and a sketch of the method appears in the closing section of this paper.Our formulation of the law of octic reciprocity makes possible a considerable extension of the results for q±3(mod8) of earlier authors. In particular, the largest prime 3(mod8) treated to date is q=19, by von Lienen [6] when j=4 or 8 and by Hudson and Williams [3] when j=1,2,3,5,6, or 7. For q=19 there are 200 distinct choices relating a,b,c,d which are equivalent to (q)(p1)/8((ab)d/ac)j(modp) for one of j=1,,8. We give explicit results in this paper for primes as large as q=83 where there are 3528 distinct choices.This paper makes several other minor contributions including a computationally efficient version of Gosset's [2] formulation of Gauss' law of quartic reciprocity, observations on sums γi,j where the γi,j's are the defining parameters for the distinct choices mentioned above, and proof that the results of von Lienen [6] may not only be appreciably abbreviated, but may be put into a form remarkably similar to the case in which q is a quadratic residue but a quartic non-residue of p.An important contribution of the paper consists in showing how to use Theorems 1 and 3 of [3], in conjunction with Theorem 4 of this paper, to reduce from (q+1/4)2 to (q1)/2 the number of cases which must be considered to obtain the criteria in Theorems 2 and 3.
Copyright © 1982 Hindawi Publishing Corporation. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
#### More related articles
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Views77 | 748 | 2,959 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2020-50 | latest | en | 0.888803 |
https://mathematicalenquiries.blogspot.com/2016/02/investigating-relationships-and.html | 1,534,401,248,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210463.30/warc/CC-MAIN-20180816054453-20180816074453-00217.warc.gz | 739,874,595 | 19,799 | ## Wednesday, 17 February 2016
### Investigating Relationships and Connections of Angles
To help enquire further into our central idea:
When angles co-exist, connections and relationships form.
I placed provocations along the hallway. The children chose ones they found interesting to enquire into and using protractors etc used the taped shapes or angles to investigate connections or relationships between them.
Children discovered how all the angles of all quadrilaterals add up to 360°!!!
Why? Hmmmmm.......good question!
How are the angles of a triangle connected with the angles of a quadrilateral?
Some children discovered that all the interior angles added up to 180°!!
"But why is that?" A student overhearing this asked.
We often discuss how understanding the 'whys' in maths is actually more important than the 'hows' as it deepens our understandings, so it's great when I hear the children now asking themselves and others these questions.
What relationships exist when angles add to a straight angle of 180°?
How does knowing that a circle is 360° help us to find connections or relationships with these angles?
When two sets of parallel lines intercept, what connections or relationships amongst the angles exist and why is that?
What connections exist with the interior or exterior angles of pentagons?
A lot of excited discussions took place as the children investigated. Theories were formed and tested and they recorded their discoveries and wonderings to share later in small group discussions.
Whilst doing this towards the end of our unit, in retrospect I think this could have been a successful lead in provocation engagement at the beginning of our unit to get the children thinking about angle relationships.
As it is, they felt it was great to help deepen their understandings of their central idea and always appreciate being able to choose what interests them rather than being told what to do on a worksheet.
Fun, engaging and deep thinking took place with this simple enquiry-based experience :)
#### 1 comment:
1. Great idea! I gave my students a roll of masking tape and let them stick and write on the tables with white board markers. I took photos and recorded some of their quotes/thinking on an ipad using Book Creator. Their discussion and discoveries were amazing. Really deepened their understanding. Thanks for the inspiration!
What do you think? ........... | 482 | 2,422 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2018-34 | longest | en | 0.963009 |
https://homework.study.com/explanation/princeton-company-manufactures-only-two-products-l1-and-l2-the-company-expects-to-produce-and-sell-1-000-units-of-product-l1-and-1-000-units-of-product-l2-during-the-current-year-the-company-used-t.html | 1,680,069,417,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948951.4/warc/CC-MAIN-20230329054547-20230329084547-00516.warc.gz | 343,209,910 | 24,069 | # Princeton Company manufactures ONLY two products: L1 and L2. The company expects to produce and...
## Question:
Princeton Company manufactures ONLY two products: L1 and L2. The company expects to produce and sell 1,000 units of Product L1 and 1,000 units of Product L2 during the current year. The company used to allocate overhead costs to products on the basis of direct labor hours. Recently, the company implemented activity-based costing to compute unit product costs. Data on total estimated costs relating to the company's production department are given below for the current year:
Activity Center Estimated Expected Activity
(Cost Driver) OH costs Product L1 Product L2
Machine setups (setups) $30,000 100 200 Purchase orders (orders)$60,000 1,000 2,000
General factory (MH) 15,000 $60,000 2,000 4,000 Total Direct Material Cost$35,000 $60,000 Total Direct Labor Cost$20,000 $40,000 Direct Labor Hours per unit 1 2 a. If the company had used the old method of overhead allocation, what would be the unit cost for each product? b. Using the activity-based costing approach, determine the cost per unit for each product c. The company's controller is puzzled by the results in parts a and b. Can you offer an explanation? Support your arguments with numbers. Is ABC useful for this company? ## Activity-Based Costing (ABC) Manufacturing costs consist of direct materials, direct labor and manufacturing overhead. While direct materials and direct labor are traceable costs, manufacturing overhead is an indirect cost and has to be allocated to products. The two common allocation methods are the traditional overhead allocation method and the activity-based costing (ABC) method. As the traditional method applies overhead cost using one volume-based cost driver, it is generally less accurate as compared to the ABC system. The ABC system requires the initial classification of overhead costs into various cost pools with a cost driver selected for each pool. A cost driver is a common activity shared among overhead costs in each cost pool. An activity rate will subsequently be calculated for each cost pool and costs from each cost pool will be applied to products using the rate. ## Answer and Explanation: 1 Become a Study.com member to unlock this answer! a. Total budgeted manufacturing overhead =$30,000 + $60,000 +$60,000 = \$150,000
Total budgeted direct labor hours = (1,000 x 1) + (1,000 x 2) =... | 530 | 2,423 | {"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} | 2.828125 | 3 | CC-MAIN-2023-14 | latest | en | 0.923077 |
http://mathhelpforum.com/trigonometry/47106-trigonometric-identity-print.html | 1,527,366,494,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867859.88/warc/CC-MAIN-20180526190648-20180526210648-00110.warc.gz | 193,236,941 | 3,294 | # Trigonometric Identity
• Aug 29th 2008, 09:56 AM
PersaGell
Trigonometric Identity
Hi, I need help on this problem:
Use the identity $\displaystyle 2\sin{\frac{x}{2}}\cos{kx} = \sin{(2k+1)}\frac{x}{2} - \sin{(2k-1)}\frac{x}{2}$ and the telescoping property of finite sums to prove that if $\displaystyle x \neq 2m\pi$ (m an integer), we have $\displaystyle \sum_{k=1}^{n} \cos{kx} = \frac{\sin{\frac{nx}{2}}\cos{\frac{1}{2}(n+1)x}}{\ sin{\frac{1}{2}x}}$
Here is what I have so far:
Taking $\displaystyle k= 1,2,...,n$
$\displaystyle 2\sin{\frac{x}{2}}\cos{kx}= 2\sin{\frac{x}{2}}\sum_{k=1}^{n}\cos{kx}$ =
$\displaystyle [\sin{\frac{3x}{2}}-\sin{\frac{x}{2}} ] +$ $\displaystyle [\sin{\frac{5x}{2}}-\sin{\frac{3x}{2}}]+ ... + [\sin{\frac{(2n-1)x}{2}}-\sin{\frac{(2n-3)x}{2}}] +$ $\displaystyle [\sin{\frac{(2n+1)x}{2}} -\sin{\frac{(2n-1)x}{2}}]$
$\displaystyle = \sin{\frac{(2n+1)x}{2}} - \sin{\frac{x}{2}}$ = $\displaystyle \sin{(n+ \frac{1}{2})x } - \sin{\frac{x}{2}}$
This reduces to $\displaystyle \sin{nx}\cos{\frac{x}{2}} + \cos{nx}\sin{\frac{x}{2}} -\sin{\frac{x}{2}}$
Now I'm stuck.
• Aug 29th 2008, 10:07 AM
Moo
Hi !
Quote:
Originally Posted by PersaGell
Hi, I need help on this problem:
Use the identity $\displaystyle 2\sin{\frac{x}{2}}\cos{kx} = \sin{(2k+1)}\frac{x}{2} - \sin{(2k-1)}\frac{x}{2}$ and the telescoping property of finite sums to prove that if $\displaystyle x \neq 2m\pi$ (m an integer), we have $\displaystyle \sum_{k=1}^{n} \cos{kx} = \frac{\sin{\frac{nx}{2}}\cos{\frac{1}{2}(n+1)x}}{\ sin{\frac{1}{2}x}}$
Here is what I have so far:
Taking $\displaystyle k= 1,2,...,n$
$\displaystyle {\color{red}\sum_{k=1}^n} 2\sin{\frac{x}{2}}\cos{kx}= 2\sin{\frac{x}{2}}\sum_{k=1}^{n}\cos{kx}$ =
$\displaystyle [\sin{\frac{3x}{2}}-\sin{\frac{x}{2}} ] +$ $\displaystyle [\sin{\frac{5x}{2}}-\sin{\frac{3x}{2}}]+ ... + [\sin{\frac{(2n-1)x}{2}}-\sin{\frac{(2n-3)x}{2}}] +$ $\displaystyle [\sin{\frac{(2n+1)x}{2}} -\sin{\frac{(2n-1)x}{2}}]$
$\displaystyle = \sin{\frac{(2n+1)x}{2}} - \sin{\frac{x}{2}}$
(There's the red part that was missing :))
Once you're here, try to think "how can I transform a difference of sines into a product of a sine and a cosine ?" and then use this formula (which is quite the same as the identity you're given at the very beginning) :
$\displaystyle \sin(a)-\sin(b)=2 \cos \left(\frac{a+b}{2} \right) \cdot \sin \left(\frac{a-b}{2}\right)$
And you're done :)
(this formula is available, as well as many others, here : Trigonometry)
• Aug 29th 2008, 01:35 PM
ThePerfectHacker
It is even easier.
Use complex numbers and geometric series.
• Aug 29th 2008, 05:01 PM
Shyam
Your last line is wrong, which is
$\displaystyle \sin{nx}\cos{\frac{x}{2}} + \cos{nx}\sin{\frac{x}{2}} -\sin{\frac{x}{2}}$
See here,
$\displaystyle =\sin{(n+ \frac{1}{2})x } - \sin{\frac{x}{2}}$
$\displaystyle =2\cos{(\frac{n+1}{2})x }\sin{\frac{nx}{2}}$ | 1,137 | 2,884 | {"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} | 4.34375 | 4 | CC-MAIN-2018-22 | latest | en | 0.44051 |
https://lonewolfonline.net/shutter-speed/ | 1,660,482,664,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572033.91/warc/CC-MAIN-20220814113403-20220814143403-00089.warc.gz | 358,621,600 | 16,328 | # Shutter Speed Explained for the Beginning Photographer
Last Updated January 15, 2021 by . First Published in 2015.
Introduction to Photography Series
Accurate control of shutter speed can make or break a shot. Take control of your shutter speed and master sharp photos and creative effects.
## What Does Shutter Speed Do?
When your camera takes a photo, it opens the shutter to expose the imaging sensor (or film) to light. How long it opens the shutter for, and therefore how much light the sensor receives, is determined by the shutter speed. The term shutter dates back to film cameras, where there was a physical shutter covering the film. Nowadays it refers to when the sensor starts recording light. The longer the shutter is open, the more light is captured.
## How is Shutter Speed Measured?
Shutter speed is measured in fractions of seconds to seconds. Typical shutter speeds range from 30 seconds through to 1/8000 seconds. Some cameras also feature a "bulb" setting which keeps the shutter open as long as the shutter button is depressed. Traditional SLR cameras often used a shutter release cable which locks the shutter open indefinitely. This is typically used in very dark situations, such as astrophotography, or for when you need to capture a lot of motion in a timelapse.
Shutter speed is written as a fraction of a second, so when we write 1/4 that is quarter of a second and 1/200 is one two-hundredth of a second. The higher the number the shorter the duration the shutter is open for.
Shutter speeds are given in specific increments which are double or half the previous. For example, 1/125 is half the speed of 1/250 and 1/500 is twice as fast as 1/250.
## What are Fast and Slow Shutter Speeds?
You may come across shutter speeds being described as fast or slow. When we say a shutter speed is fast we are saying that the shutter is open for a very short amount of time. Slow shutter speeds on the other hand are when the shutter is open for a long time. This can have a drastic affect on the final image recorded. This is particularly true for the way we capture motion within a photo.
## How Shutter Speed Affects Movement
The longer the shutter speed (slow shutter speed), the more likely you are to introduce vibration and motion blur to an image, however, you can capture the motion of objects. The faster the shutter speed, the less blur and motion blur will be captured. With a fast shutter setting, you will "freeze" an object's motion. An example of this can be seen below.
You can control the way in which movement is captured in your pictures by getting to grips with your cameras full range of shutter speeds. Shutter speed works hand in hand with aperture and ISO to balance the perfect exposure. You can create dramatic effects by either freezing action or blurring motion by controlling shutter speed.
This effect is used quite often in advertisements of cars and motorbikes, where a sense of speed and motion is communicated to the viewer by intentionally blurring the moving wheels.
Shutter speed can also be used to freeze motion. If you use especially fast shutter speeds, you can eliminate motion even from fast-moving objects, like birds in flight, water drops, helicopters in flight, cars and motorbikes.
Controlling exposure is a fine balance between shutter speed, aperture and ISO. This balance is called the exposure triangle which we will come to later on.
If you are shooting handheld without a tripod you'll need a fast enough shutter speed to make sure camera shake doesn't cause blurred photos.
🛈︎
A good rule of thumb for the slowest shutter speed for handheld is to use a shutter speed that's faster than the focal length of the lens, so for example, if you were using a 200mm lens, the minimum shutter speed you should use is 1/250 seconds.
Keep an eye on the shutter speed in the viewfinder and widen the aperture if necessary. In low light, you may need to increase the ISO or use a tripod.
## How to Set Shutter Speed
Almost all cameras handle shutter speeds automatically by default. When the camera is set to "Auto" or "A" mode. The shutter speed is selected by the camera without your input (along with every other setting). To change the shutter speed yourself you will need to select a semi-automatic or manual camera mode. These are typically called Shutter Priority (Tv for Time value) or just simply "S" on some cameras. You can also use the semi-automatic program mode, P, where the camera will allow you to change the shutter and it will adjust the rest of the settings. You will then get the option to select a shutter speed.
Shutter speed is generally shown in the LCD display if you have one, the viewfinder and on the live view window.
The final setting you can use is the full manual. In this mode, you have to set everything, shutter speed, aperture, ISO. I would suggest for now that you not use this until you fully understand the exposure triangle.
### How to Set Shutter Speed in Canon dSLR
On Canon dSLR cameras shutter speed can be selected using the adjustment wheel depending on the camera shooting mode selected.
In Shutter Priority mode, the adjustment wheel directly changes the shutter speed. In full manual shutter speed is changed using the adjustment wheel whilst holding the AE-Lock button. In program mode shutter speed changes along with aperture as you move the adjustment wheel. It will increase one and decrease the other to maintain the correct exposure. This relation is explained in the exposure triangle relationship.
### How to Set Shutter Speed on Android
On Android phones you can set the shutter speed only in Pro mode. To access the camera pro mode tap on "more" on the mode section list. From the menu that pops up select Pro. This mode will put the Android camera into full manual mode, however you can set each exposure and focus setting to auto and only adjust the shutter speed. Tap the icon for aperture and shutter speed and the camera app will show a sliding control which you can slide to set shutter speed from auto to any value you wish.
### How to Set Shutter Speed on iPhone
Unfortunately, iOS doesn't offer manual controls in the default Camera app so you'll need to install a third-party app. There are many camera apps available on the App Store with some of the popular ones being VSCO, Halide Mark II and ProCamera+.
## What is Bulb Mode?
Bulb mode is a special shutter speed mode which allows for super-slow shutter speeds. The shutter will be open for as long as the shutter is depressed. Bulb mode can usually be accessed by selecting "B" on the mode dial, or on "M" and selecting bulb on the shutter speed selection. On modern DSLRs, this may cause camera shake as you will need to hold the button in. For this reason you should use the self-timer mode, or you can use a remote shutter release cable which is either a manual switch or a more advanced timer device.
Bulb mode is much simpler than it sounds, but the small matter of knowing how long to hold the shutter open for puts people off using it. If you were shooting a landscape with a 10-stop ND filter, then the exposure chart supplied with the filter, or a downloadable smartphone app will give you an idea. Otherwise, it's a bit of trial and error.
For photographing star trails or time-lapse night scenes you may be interested in the article "How to take long exposures on a Canon dSLR".
## Examples of Shutter Speeds
Shutter SpeedUsage
1/4000Freezing really fast moving objects such as a tennis ball, football.
1/2000Freezing the flight of birds without blurring the wings.
1/1000Freezing very fast moving objects, such as moving vehicles.
1/500Freezing fast moving people, such as runners and cyclists.
1/250A great speed for freezing your still subject, without having to think too much about focal length and how that affects the motion blur. Great for portrait photography.
1/125You won't typically want to go much slower than this if you're shooting hand-held, otherwise, you will likely capture motion blur from your hands.
1/60Again, this is a great speed for panning photography, and handheld photography in low light.
1/30This is about as slow as you will want to go while capturing panning photography, as much slower and your photo will become too much of a blur.
1/15Mounted on a tripod, at this speed you can capture sight movement from moving objects. Think people walking, cars moving in traffic, water blurring slightly.
1/8Capturing motion blur in water.
1/4Blurred movement in a scene. Not so little that it appears accidental, but not so much that it's hard to tell what's going on.
1/2More motion blur, only much stronger than before. Think of water starting to appear like mist.
1 SecondTwilight photography. The sun may not be completely gone, but there's not enough light to make up the exposure you're looking for. You may incorporate a flash, and you're more than likely using a tripod.
> 1 SecondThis is where night photography starts to come into play. You can play with different speeds and capture awesome night time photos.
BulbThis is used for exposures longer than 30 seconds, where you can manually control the exposure time with the shutter release. This is used for astrophotography where you may want to capture some stars. You may also use this mode for slow sync flash where you want to have immediate control of the shutter speed.
Tutorial Series
This post is part of the series Introduction to Photography. Use the links below to advance to the next tutorial in the couse, or go back and see the previous in the tutorial series. | 2,055 | 9,601 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-33 | latest | en | 0.960312 |
https://www.media4math.com/MA.2.M.2.2 | 1,685,717,328,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648695.4/warc/CC-MAIN-20230602140602-20230602170602-00028.warc.gz | 957,044,141 | 10,911 | ## MA.2.M.2.2: Solve one- and two-step addition and subtraction real-world problems involving either dollar bills within $100 or coins within 100¢ using$ and ¢ symbols appropriately.
There are 70 resources.
Title Description Thumbnail Image Curriculum Topics
## Quizlet Flash Card Collection: Counting
Overview This is a collection of Quizlet Flash Cards on the topic of Counting. There are a total of 5 Quizlet Flash Card sets.
This is a collection of Quizlet Flash Cards on the topic of Counting. There are a total of 5 Quizlet Flash Card sets.
Units of Measure, Counting, Addition Facts to 25, Subtraction Facts to 10 and Working with Money
## Math Examples Collection: Piggy Bank Math
Overview This collection aggregates all the math examples around the topic of Piggy Bank Math. There are a total of 12 Math Examples.This collection aggregates all the math examples around the topic of Piggy Bank Math. There are a total of 12 Math Examples. This collection of resources is made up of downloadable JPG images that you can easily incorporate into your lesson plans. Working with Money
## Counting and Cardinality Collection: Worksheets
This is a collection of worksheets on the topic of Counting and Cardinality.
Counting, Money and Addition Facts to 100
## Counting and Cardinality Collection: Quizlet
This is a collection of Quizlet Flash Card sets on the topic of Counting and Cardinality.
Addition Facts to 25, Subtraction Facts to 10, Units of Measure, Working with Money and Counting
## Math Examples Collection: Analyzing Money
Overview This collection aggregates all the math examples around the topic of Analyzing Money. There are a total of 9 Math Examples.This collection aggregates all the math examples around the topic of Analyzing Money. There are a total of 9 Math Examples. Working with Money
## Math Examples Collection: Labor, Income, and Expenses
Overview This collection aggregates all the math examples around the topic of Labor, Income, and Expenses. There are a total of 8 Math Examples.This collection aggregates all the math examples around the topic of Labor, Income, and Expenses. There are a total of 8 Math Examples. Working with Money
## Math Examples Collection: Saving Money (K-2)
Overview This collection aggregates all the math examples around the topic of Saving Money (K-2). There are a total of 12 Math Examples.This collection aggregates all the math examples around the topic of Saving Money (K-2). There are a total of 12 Math Examples. Working with Money
## Math Worksheet Collection: Adding Money (K-2)
Overview This collection aggregates all the math worksheets around the topic of Adding Money. There are a total of 5 worksheets. Each worksheet includes an answer key.This collection aggregates all the math worksheets around the topic of Adding Money. There are a total of 5 worksheets. Money
## Math Clip Art Collection: Money
Overview This collection aggregates all the math clip art around the topic of Money. There are a total of 4 images.This collection aggregates all the math clip art around the topic of Money. There are a total of 4 images. Working with Money
## Math Clip Art--Money---Penny
Math Clip Art--Money---Penny
This is part of a collection of math clip art images that show money amounts.
Working with Money
## Math Clip Art--Money--Dime
Math Clip Art--Money--Dime
This is part of a collection of math clip art images that show money amounts.
Working with Money
## Math Clip Art--Money--Nickel
Math Clip Art--Money--Nickel
This is part of a collection of math clip art images that show money amounts.
Working with Money
## Math Clip Art--Money--Quarter
Math Clip Art--Money--Quarter
This is part of a collection of math clip art images that show money amounts.
Working with Money
## Math Example--Math of Money--Analyzing Money--Example 1
Math Example--Math of Money--Analyzing Money--Example 1
This is part of a collection of math examples that focus on money.
Working with Money
## Math Example--Math of Money--Analyzing Money--Example 10
Math Example--Math of Money--Analyzing Money--Example 10
This is part of a collection of math examples that focus on money.
Working with Money
## Math Example--Math of Money--Analyzing Money--Example 2
Math Example--Math of Money--Analyzing Money--Example 2
This is part of a collection of math examples that focus on money.
Working with Money
## Math Example--Math of Money--Analyzing Money--Example 3
Math Example--Math of Money--Analyzing Money--Example 3
This is part of a collection of math examples that focus on money.
Working with Money
## Math Example--Math of Money--Analyzing Money--Example 4
Math Example--Math of Money--Analyzing Money--Example 4
This is part of a collection of math examples that focus on money.
Working with Money
## Math Example--Math of Money--Analyzing Money--Example 5
Math Example--Math of Money--Analyzing Money--Example 5
This is part of a collection of math examples that focus on money.
Working with Money
## Math Example--Math of Money--Analyzing Money--Example 6
Math Example--Math of Money--Analyzing Money--Example 6
This is part of a collection of math examples that focus on money.
Working with Money
## Math Example--Math of Money--Analyzing Money--Example 7
Math Example--Math of Money--Analyzing Money--Example 7
This is part of a collection of math examples that focus on money.
Working with Money
## Math Example--Math of Money--Analyzing Money--Example 8
Math Example--Math of Money--Analyzing Money--Example 8
This is part of a collection of math examples that focus on money.
Working with Money
## Math Example--Math of Money--Analyzing Money--Example 9
Math Example--Math of Money--Analyzing Money--Example 9
This is part of a collection of math examples that focus on money.
Working with Money
## Math Example--Math of Money--Labor, Income, and Expenses--Example 1
Math Example--Math of Money--Labor, Income, and Expenses--Example 1
This is part of a collection of math examples that focus on money.
Working with Money
## Math Example--Math of Money--Labor, Income, and Expenses--Example 2
Math Example--Math of Money--Labor, Income, and Expenses--Example 2
This is part of a collection of math examples that focus on money.
Working with Money | 1,415 | 6,296 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2023-23 | latest | en | 0.86728 |
http://coq.inria.fr/pylons/contribs/files/CoLoR/v8.3/CoLoR.Term.Varyadic.VTerm.html | 1,371,727,290,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368711441609/warc/CC-MAIN-20130516133721-00072-ip-10-60-113-184.ec2.internal.warc.gz | 59,693,551 | 5,475 | CoLoR, a Coq library on rewriting and termination. See the COPYRIGHTS and LICENSE files.
• Frederic Blanqui, 2005-06-10
algebraic terms with no arity
Set Implicit Arguments.
Require Import LogicUtil.
Require Import BoolUtil.
Require Export List.
Require Import ListUtil.
Require Import EqUtil.
Require Export VSignature.
Require Import Peano_dec.
Section S.
Variable Sig : Signature.
Inductive term : Type :=
| Var : variable -> term
| Fun : forall f : Sig, list term -> term.
Reset term_rect.
Notation terms := (list term).
induction principle
Section term_rect.
Variables
(P : term -> Type)
(Q : terms -> Type)
(H1 : forall x, P (Var x))
(H2 : forall f v, Q v -> P (Fun f v))
(H3 : Q nil)
(H4 : forall t v, P t -> Q v -> Q (t :: v)).
Fixpoint term_rect t : P t :=
match t as t return P t with
| Var x => H1 x
| Fun f v => H2 f
((fix vt_rect (v : terms) : Q v :=
match v as v return Q v with
| nil => H3
| cons t' v' => H4 (term_rect t') (vt_rect v')
end) v)
end.
End term_rect.
Definition term_ind (P : term -> Prop) (Q : terms -> Prop) := term_rect P Q.
Require Import ListForall.
Lemma term_ind_forall : forall (P : term -> Prop)
(H1 : forall x, P (Var x))
(H2 : forall f v, lforall P v -> P (Fun f v)),
forall t, P t.
Proof.
intros. apply term_ind with (Q := fun v => lforall P v).
assumption. assumption. constructor.
intros. apply lforall_intro. intros.
destruct H3. subst t0. assumption.
apply lforall_in with term v; assumption.
Qed.
Lemma term_ind_forall2 : forall (P : term -> Prop)
(H1 : forall x, P (Var x))
(H2 : forall f v, (forall t, In t v -> P t) -> P (Fun f v)),
forall t, P t.
Proof.
intros. apply term_ind with (Q := fun v => forall t, In t v -> P t); simpl.
intros. destruct H3. subst. assumption. apply H0. assumption.
Qed.
Section term_rec_forall.
Variable term_eq_dec : forall t u : term, {t=u} + {t<>u}.
Lemma term_rect_forall : forall (P : term -> Type)
(H1 : forall x, P (Var x))
(H2 : forall f v, (forall t, Inb term_eq_dec t v = true -> P t) ->
P (Fun f v)),
forall t, P t.
Proof.
intros. apply term_rect with
(Q := fun v => forall t, Inb term_eq_dec t v = true -> P t); simpl.
assumption. assumption. intros. discriminate.
intros. destruct (term_eq_dec t1 t0). subst t1. assumption.
apply X0. assumption.
Qed.
End term_rec_forall.
equality
Lemma term_eq : forall f f' v v', f = f' -> v = v' -> Fun f v = Fun f' v'.
Proof.
intros. rewrite H. rewrite H0. refl.
Qed.
Lemma fun_eq : forall f f' v, f = f' -> Fun f v = Fun f' v.
Proof.
intros. rewrite H. refl.
Qed.
Lemma args_eq : forall f v v', v = v' -> Fun f v = Fun f v'.
Proof.
intros. rewrite H. refl.
Qed.
Lemma term_eq_dec : forall t u : term, {t = u} + {t <> u}.
Proof.
intro. pattern t. apply term_rect with
(Q := fun ts : terms => forall us, {ts=us} + {~ts=us}); clear t; intros.
destruct u.
destruct (eq_nat_dec x n).
intuition. right. congruence.
right. discriminate.
destruct u.
right. discriminate.
destruct (eq_symbol_dec f f0). destruct (X l).
left. congruence.
right. intro diff. apply n. congruence.
right. intro diff. apply n. congruence.
destruct us.
left. trivial.
right. discriminate.
destruct us.
right. discriminate.
destruct (X t0).
destruct (X0 us).
left. congruence.
right. intro diff. apply n. congruence.
right. intro diff. apply n. congruence.
Defined.
Section beq.
Variable beq_var : variable -> variable -> bool.
Variable beq_var_ok : forall x y, beq_var x y = true <-> x = y.
Variable beq_symb : Sig -> Sig -> bool.
Variable beq_symb_ok : forall f g, beq_symb f g = true <-> f = g.
Fixpoint beq (t u : term) {struct t} :=
match t with
| Var x =>
match u with
| Var y => beq_var x y
| _ => false
end
| Fun f ts =>
match u with
| Fun g us =>
let fix beq_terms (ts us : terms) {struct ts} :=
match ts with
| nil =>
match us with
| nil => true
| _ => false
end
| t :: ts' =>
match us with
| u :: us' => beq t u && beq_terms ts' us'
| _ => false
end
end
in beq_symb f g && beq_terms ts us
| _ => false
end
end.
Lemma beq_terms : forall ts us,
(fix beq_terms (ts us : terms) {struct ts} :=
match ts with
| nil =>
match us with
| nil => true
| _ => false
end
| t :: ts' =>
match us with
| u :: us' => beq t u && beq_terms ts' us'
| _ => false
end
end) ts us = beq_list beq ts us.
Proof.
induction ts; destruct us; refl.
Qed.
Lemma beq_fun : forall f ts g us,
beq (Fun f ts) (Fun g us) = beq_symb f g && beq_list beq ts us.
Proof.
intros. rewrite <- beq_terms. refl.
Qed.
Lemma beq_ok : forall t u, beq t u = true <-> t = u.
Proof.
intro t. pattern t. apply term_ind_forall2; destruct u.
simpl. rewrite beq_var_ok. intuition. inversion H. refl.
intuition; discriminate. intuition; discriminate.
rewrite beq_fun. split; intro. destruct (andb_elim H0).
rewrite beq_symb_ok in H1. subst f0.
rewrite beq_list_ok_in in H2. subst l. refl. exact H.
inversion H0. apply andb_intro. apply (beq_refl beq_symb_ok).
ded (beq_list_ok_in H). subst v. rewrite H1. refl.
Qed.
End beq.
maximal index of a variable
Require Import ListMax.
Fixpoint maxvar (t : term) : nat :=
match t with
| Var x => x
| Fun f v =>
let fix maxvars (v : terms) : nats :=
match v with
| nil => nil
| cons t' v' => cons (maxvar t') (maxvars v')
end
in lmax (maxvars v)
end.
Lemma maxvar_fun : forall f ts, maxvar (Fun f ts) = lmax (map maxvar ts).
Proof.
intros. simpl. apply (f_equal lmax).
induction ts. auto. rewrite IHts. auto.
Qed.
Lemma maxvar_var : forall k x, maxvar (Var x) <= k -> x <= k.
Proof.
intros. simpl. intuition.
Qed.
Definition maxvar_le k t := maxvar t <= k.
End S.
Implicit Arguments Var [Sig].
Implicit Arguments maxvar_var [Sig k x]. | 1,766 | 5,574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2013-20 | latest | en | 0.599158 |
https://www.ecybertechdesigns.com/the-way-to-get-lottery-through-the-use-of-examination-algorithms-for-lottery-prediction/ | 1,716,403,426,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058560.36/warc/CC-MAIN-20240522163251-20240522193251-00540.warc.gz | 676,262,514 | 26,147 | # The way to Get Lottery through the use of Examination Algorithms for Lottery Prediction
Lottery predictions is kind of well-known in recent times. Persons was skeptical Along with the predictions since they thought the profitable figures absolutely are a matter of luck and fortunes. Not Many of us feel that lottery might be won by using some type of a complicated science based mostly predictions. It was not till the late 90s when lottery gamers started utilizing lottery predictions to help them to get lottery or at least catch up with on the profitable figures. When Gonzalo Garcia-Pelayo, a Spanish man who managed to check and assess several games in two unique countries, Spain as well as the US and get a lot of cash by making use of various strategies. Following him people today started to believe that lottery effects may be predicted.
Lottery gamers commence thinking about tips on how to earn the lotteries utilizing predictions. They use quite a few kinds of predictions: from mechanical predictions on mechanical lotteries to technological predictions applying Togel dingdong Personal computer software package. A lot of folks use algorithm to research and forecast lottery results.
Lotto Investigation algorithm is tested to get incredibly helpful to help lottery players get closer into the winning quantities and even make the lottery properties go bankrupt! There are plenty of types of lottery Evaluation utilized by lottery predictors and Here are a few of them:
During this Investigation, the predictors use some comprehensive scientific tests which document the frequency of each adjoining pairs of figures inside the similar lottery earn in the timeframe then put quite possibly the most Regular figures along with the ranks and do it consecutively.
Evaluation of Equilibrium
By Examination of harmony, lottery players consider to analyze if particular combinations will provide them with opportunities to win including combos of modest and massive quantities, odd and also quantities and also the array of the total sum quantities.
Analysis of Digits
When examining lottery profitable quantities working with digit Investigation, lottery predictors should be able to know the exact figures in particular range may be drawn in a specific stretch of time. To be able to make the profitable likelihood more substantial, the gamers really need to limit the array of figures every time they find every single digit of their combination.
Investigation of Elapse Time
This Assessment operates by Understanding and noting the period every time a quantity is in its waiting time to be drawn again following its final profitable time. Gamers may also know the chance or perhaps the winning probability of specified numbers determined by the elapse time. Should the elapse time is extended, the possibility to gain is larger. This Evaluation is taken into account extra exact when compared to the Many others as it gives additional facts about inclination of some numbers to win or not so that it’s simpler to know the following successful numbers in a few lotteries for example Powerball, Mega million, California Tremendous Lotto Plus and a few Other individuals. | 577 | 3,194 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.957315 |
https://oneclass.com/class-notes/us/west-chester/che-crl/che-104/1352300-che-104-lecture-11.en.html | 1,624,614,544,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487630081.36/warc/CC-MAIN-20210625085140-20210625115140-00224.warc.gz | 385,858,976 | 69,168 | # CHE 104 Lecture 11: Lecture Notes 11
4 views2 pages
6 Feb 2017
School
Department
Course
Professor
February 3rd, 2017
Dr. Townsend
Lecture #11
Bonding in Metals and Semiconductors: Band Theory
Molecular orbital theory (MO) was introduced in chapter 9 to rationalize
covalent bonding in molecules
Dictionary Definition: molecular orbital (MO) theory is a method for
determining molecular structure in which electrons are not assigned
to individual bonds between atoms, but are treated as moving
under the influence of the nuclei in the whole molecule
Molecular orbital theory can also be used to describe metallic bonding
Metals can be thought of as a “supermolecule”
Metallic bonding is described as delocalized
The electrons are associated with all the atoms in the crystal and
not with specific bonded atoms
Valence electrons are free to move around the entire atom because
they are no localized (aka delocalized)
The theory of the metallic bonding is call BAND THEORY
Recall that if we combine a 2s orbital on one atom with 2s orbital on
another atom, we get two molecular orbitals, a 𝝈*2s bonding orbital
𝝈*2s antibonding orbital
If we combine a large number (n) of atomic orbitals we get n
molecular orbitals
n/2 bonding orbitals and n/2 antibonding orbitals
These will form a continuous band of energy levels lower in energy
and a continuous band higher in energy than the original orbitals
The lowest energy for a system occurs with all electrons in orbitals
with the lowest possible energy
● The lower
the energy the more stable
The Fermi band is the highest filled level at 0 K
Look at diagram and the top level that is filled with atoms is
the fermi band
In order for a material to conduct electricity, electrons must move
from the lower band (called the valence band) into the higher band
(called the conduction band)
Once conduction band, the electrons can move toward the positive
pole of the electric field and the resulting hole in the valence band
can shift towards the negative pole
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Textbook Notes | 530 | 2,220 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-25 | latest | en | 0.906041 |
https://www.statsmodels.org/devel/_modules/statsmodels/sandbox/regression/try_ols_anova.html | 1,721,522,894,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517544.48/warc/CC-MAIN-20240720235600-20240721025600-00652.warc.gz | 838,512,934 | 10,928 | # Source code for statsmodels.sandbox.regression.try_ols_anova
``````''' convenience functions for ANOVA type analysis with OLS
Note: statistical results of ANOVA are not checked, OLS is
checked but not whether the reported results are the ones used
in ANOVA
includes form2design for creating dummy variables
TODO:
* ...
*
'''
from statsmodels.compat.python import lmap
import numpy as np
#from scipy import stats
import statsmodels.api as sm
[docs]
def data2dummy(x, returnall=False):
'''convert array of categories to dummy variables
by default drops dummy variable for last category
uses ravel, 1d only'''
x = x.ravel()
groups = np.unique(x)
if returnall:
return (x[:, None] == groups).astype(int)
else:
return (x[:, None] == groups).astype(int)[:,:-1]
[docs]
def data2proddummy(x):
'''creates product dummy variables from 2 columns of 2d array
drops last dummy variable, but not from each category
singular with simple dummy variable but not with constant
quickly written, no safeguards
'''
#brute force, assumes x is 2d
#replace with encoding if possible
groups = np.unique(lmap(tuple, x.tolist()))
return (x==groups[:,None,:]).all(-1).T.astype(int)[:,:-1]
[docs]
def data2groupcont(x1,x2):
'''create dummy continuous variable
Parameters
----------
x1 : 1d array
label or group array
x2 : 1d array (float)
continuous variable
Notes
-----
useful for group specific slope coefficients in regression
'''
if x2.ndim == 1:
x2 = x2[:,None]
dummy = data2dummy(x1, returnall=True)
return dummy * x2
# Result strings
#the second leaves the constant in, not with NIST regression
#but something fishy with res.ess negative in examples ?
#not checked if these are all the right ones
anova_str0 = '''
ANOVA statistics (model sum of squares excludes constant)
Source DF Sum Squares Mean Square F Value Pr > F
Model %(df_model)i %(ess)f %(mse_model)f %(fvalue)f %(f_pvalue)f
Error %(df_resid)i %(ssr)f %(mse_resid)f
CTotal %(nobs)i %(uncentered_tss)f %(mse_total)f
R squared %(rsquared)f
'''
anova_str = '''
ANOVA statistics (model sum of squares includes constant)
Source DF Sum Squares Mean Square F Value Pr > F
Model %(df_model)i %(ssmwithmean)f %(mse_model)f %(fvalue)f %(f_pvalue)f
Error %(df_resid)i %(ssr)f %(mse_resid)f
CTotal %(nobs)i %(uncentered_tss)f %(mse_total)f
R squared %(rsquared)f
'''
'''update regression results dictionary with ANOVA specific statistics
not checked for completeness
'''
ad.update(res.__dict__) #dict does not work with cached attributes
anova_attr = ['df_model', 'df_resid', 'ess', 'ssr','uncentered_tss',
'mse_model', 'mse_resid', 'mse_total', 'fvalue', 'f_pvalue',
'rsquared']
for key in anova_attr:
[docs]
def form2design(ss, data):
'''convert string formula to data dictionary
ss : str
* varname : for simple varnames data is used as is
* F:varname : create dummy variables for factor varname
* P:varname1*varname2 : create product dummy variables for
varnames
* G:varname1*varname2 : create product between factor and
continuous variable
data : dict or structured array
data set, access of variables by name as in dictionaries
Returns
-------
vars : dictionary
dictionary of variables with converted dummy variables
names : list
list of names, product (P:) and grouped continuous
variables (G:) have name by joining individual names
sorted according to input
Examples
--------
>>> xx, n = form2design('I a F:b P:c*d G:c*f', testdata)
>>> xx.keys()
['a', 'b', 'const', 'cf', 'cd']
>>> n
['const', 'a', 'b', 'cd', 'cf']
Notes
-----
with sorted dict, separate name list would not be necessary
'''
vars = {}
names = []
for item in ss.split():
if item == 'I':
vars['const'] = np.ones(data.shape[0])
names.append('const')
elif ':' not in item:
vars[item] = data[item]
names.append(item)
elif item[:2] == 'F:':
v = item.split(':')[1]
vars[v] = data2dummy(data[v])
names.append(v)
elif item[:2] == 'P:':
v = item.split(':')[1].split('*')
vars[''.join(v)] = data2proddummy(np.c_[data[v[0]],data[v[1]]])
names.append(''.join(v))
elif item[:2] == 'G:':
v = item.split(':')[1].split('*')
vars[''.join(v)] = data2groupcont(data[v[0]], data[v[1]])
names.append(''.join(v))
else:
raise ValueError('unknown expression in formula')
return vars, names
[docs]
def dropname(ss, li):
'''drop names from a list of strings,
names to drop are in space delimited list
does not change original list
'''
newli = li[:]
for item in ss.split():
newli.remove(item)
return newli
if __name__ == '__main__':
# Test Example with created data
# ------------------------------
nobs = 1000
testdataint = np.random.randint(3, size=(nobs,4)).view([('a',int),('b',int),('c',int),('d',int)])
testdatacont = np.random.normal( size=(nobs,2)).view([('e',float), ('f',float)])
import numpy.lib.recfunctions
dt2 = numpy.lib.recfunctions.zip_descr((testdataint, testdatacont),flatten=True)
# concatenate structured arrays
testdata = np.empty((nobs,1), dt2)
for name in testdataint.dtype.names:
testdata[name] = testdataint[name]
for name in testdatacont.dtype.names:
testdata[name] = testdatacont[name]
#print(form2design('a',testdata)
if 0: # print(only when nobs is small, e.g. nobs=10
xx, n = form2design('F:a',testdata)
print(xx)
print(form2design('P:a*b',testdata))
print(data2proddummy(np.c_[testdata['a'],testdata['b']]))
xx, names = form2design('a F:b P:c*d',testdata)
#xx, names = form2design('I a F:b F:c F:d P:c*d',testdata)
xx, names = form2design('I a F:b P:c*d', testdata)
xx, names = form2design('I a F:b P:c*d G:a*e f', testdata)
X = np.column_stack([xx[nn] for nn in names])
# simple test version: all coefficients equal to one
y = X.sum(1) + 0.01*np.random.normal(size=(nobs))
rest1 = sm.OLS(y,X).fit() #results
print(rest1.params)
X = np.column_stack([xx[nn] for nn in dropname('ae f', names)])
# simple test version: all coefficients equal to one
y = X.sum(1) + 0.01*np.random.normal(size=(nobs))
rest1 = sm.OLS(y,X).fit()
print(rest1.params)
# Example: from Bruce
# -------------------
#get data and clean it
#^^^^^^^^^^^^^^^^^^^^^
# requires file 'dftest3.data' posted by Bruce
# read data set and drop rows with missing data
dt_b = np.dtype([('breed', int), ('sex', int), ('litter', int),
('pen', int), ('pig', int), ('age', float),
('bage', float), ('y', float)])
print('missing', [dta.mask[k].sum() for k in dta.dtype.names])
droprows = m.reshape(-1,len(dta.dtype.names)).any(1)
# get complete data as plain structured array
# maybe does not work with masked arrays
dta_use_b1 = dta[~droprows,:].data
print(dta_use_b1.shape)
print(dta_use_b1.dtype)
#Example b1: variables from Bruce's glm
#^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
# prepare data and dummy variables
xx_b1, names_b1 = form2design('I F:sex age', dta_use_b1)
# create design matrix
X_b1 = np.column_stack([xx_b1[nn] for nn in dropname('', names_b1)])
y_b1 = dta_use_b1['y']
# estimate using OLS
rest_b1 = sm.OLS(y_b1, X_b1).fit()
# print(results)
print(rest_b1.params)
#compare with original version only in original version
# Example: use all variables except pig identifier
#^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
allexog = ' '.join(dta.dtype.names[:-1])
#'breed sex litter pen pig age bage'
xx_b1a, names_b1a = form2design('I F:breed F:sex F:litter F:pen age bage', dta_use_b1)
X_b1a = np.column_stack([xx_b1a[nn] for nn in dropname('', names_b1a)])
y_b1a = dta_use_b1['y']
rest_b1a = sm.OLS(y_b1a, X_b1a).fit()
print(rest_b1a.params) | 2,173 | 7,478 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-30 | latest | en | 0.63231 |
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Making conclusions from graphs doesn't need to seem like magic. Learners explore an interactive lesson on reading graphs to strengthen their skills. The content discusses how to read the axes of a graph and draw conclusions based on the... | 1,081 | 5,168 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2021-25 | latest | en | 0.892183 |
https://www.sarthaks.com/2692482/in-control-systems-if-g-is-the-gain-then-damping-varies-with | 1,685,662,612,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648209.30/warc/CC-MAIN-20230601211701-20230602001701-00494.warc.gz | 1,069,445,028 | 15,246 | # In control systems, if 'G' is the gain then Damping varies with:
57 views
closed
In control systems, if 'G' is the gain then Damping varies with:
1. √G
2. 1/G
3. 1/√G
4. G
by (30.1k points)
selected by
Correct Answer - Option 3 : 1/√G
Concept:
Standard second-order closed-loop transfer function is given by:
$\frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{ω _n^2}}{{{s^2} + 2\xi {ω _n}s + ω _n^2}}$ .... (1)
$\frac{C(s)}{R(s)}=\frac{{G}}{{{s^2} + 2\xi {ω _n}s + ω _n^2}}$ ... (2)
ξ = damping ratio
ωn = undamped natural frequency
G = Gain
From equation (1) and (2),
G = ωn2
∴ $\omega _n=\sqrt{G}$ .... (3)
Now we have to find the pole of the system, and it will be located at,
$s=-\zeta \omega_n\pm\omega_n \sqrt{1-\zeta^2}$
The decaying exponential has a time constant equal to,
$1=\frac{1}{\zeta \omega_n}$
∴ $\zeta =\frac{1}{\omega_n}$ .... (4)
From equation (3) and (4),
$\large{\zeta=\frac{1}{\sqrt{G}}}$ | 367 | 943 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2023-23 | latest | en | 0.747732 |
https://engineerscommunity.com/t/if-a-629-8-and-b-131-3-a-b-with-range-of-doubt-is/14556 | 1,657,121,525,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104675818.94/warc/CC-MAIN-20220706151618-20220706181618-00461.warc.gz | 273,536,989 | 4,566 | # If A = 629 ± 8 and B = 131 ± 3, A - B with range of doubt is ___
If A = 629 ± 8 and B = 131 ± 3, A - B with range of doubt is
A. 498 ± 8
B. 498 ± 4
C. 498 ± 11
D. 498 ± 13 | 86 | 175 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-27 | latest | en | 0.87388 |
http://rogerpielkejr.blogspot.com/2010/11/did-paul-rip-have-skill.html?showComment=1288624326763 | 1,529,541,060,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863980.55/warc/CC-MAIN-20180621001211-20180621021211-00299.warc.gz | 282,125,625 | 21,788 | ## 01 November 2010
### Did Paul (RIP) Have Skill?
Paul the Octopus died last week. For those unaware Paul, who spent his days at the Sea Life Centre in Oberhausen, Germany, gained fame by correctly picking the winners of all seven of Germany's World Cup matches last summer, as well as the final. In this post I ask a few questions about the Paul phenomena and then distill some broader lessons for making sense of predictions.
First, did Paul have predictive skill? That is to ask, were his picks better than those that would have been made at the time by a naive forecasting methodology?
The answer is yes. Paul had skill. The naive forecasting methodology that I employed last summer in my World Cup pool was the estimated market value of each team in the transfer market, under the assumption that the higher-valued squad wins. For the 8 games that Paul picked in the World Cup the naive forecasting methodology would have gone 5-3, missing out on Germany's wins over Argentina and England, and Serbia's victory over Germany in the Group stage. Paul's picks easily bested the skill threshold.
What can we learn from the Paul phenomena?
One lesson might be that his results indicate that some octopi have Delphic capabilities and can see the future. Call me a cephalopod skeptic, but I don't think that Paul could actually pick the winners of World Cup matches. If so, then skill, by itself, is not a sufficient basis for evaluating a forecast. Rather than being about Paul, his fame and predictive successes say something about us, and how we act in response to forecasts made in many practical venues such as finance and science.
Consider the math of Paul's feat. If each team has an equal chance of winning a World Cup match, then the odds of picking 8 of 8 winners is 1 in 256 (2^8). The are long odds to be sure, but not impossibly odd. Once Paul attracted attention, he had already picked several games correctly, thus increasing the odds that he'd be viewed as an oracle (for instance, if you only heard of Paul before the World Cup final, he then had a 50% chance of "proving" his predictive capabilities to you.)
In fact, I would go so far as to argue that the odds of a Paul -- some predictive oracle -- emerging were in fact 100%. Wikipedia's recounting illustrates why this is so:
Some other oracles did not fare so well in the World Cup. The animals at the Chemnitz Zoo in Germany were wrong on all of Germany's group-stage games, with Leon the porcupine picking Australia, Petty the pygmy hippopotamus spurning Serbia's apple-topped pile of hay, Jimmy the Peruvian guinea-pig and Anton the tamarin eating a raisin representing Ghana. Mani the Parakeet of Singapore,[55][56] Octopus Pauline of Holland,[57] Octopus Xiaoge of Qingdao China,[58] Chimpanzee Pino and Red River Hog Apelsin in Tallinn zoo Estonia[59] picked the Netherlands to win the final.[60] Crocodile Harry of Australia picked Spain to win.[61]
If the solution space is covered by a range of predictions, then it is a guarantee that one of those predictions (or sets of predictions) will prove correct. We to selectively forget about the bad predictions (who among us knows of Petty the pygmy hippo?) and focus on the successes. To put a judgment of skill into context, we have to know something about the universe of competing predictions and methods.
This sets up a rather up-is-down situation in which we allow reality to select our oracles, rather than our oracles selecting our futures. We thus very easily risk being fooled by randomness into thinking that forecasters have enhanced chances for future skill based on past performance, when in fact those success may just be a combination of (a) coverage of the solution space by a range of forecasts, and (b) our selective memories and focus of attention on forecast successes. Anyone who has invested in last year's hot mutual fund, will likely have learned this lesson.
Scholars who study judgment and decision making are well aware of these sorts of cognitive biases. One is called the "hot hand fallacy" which is based on the assumption that a recent pattern will continue. For instance, the assumption that because Paul got 8 of 8 right in the World Cup, that he'd have good chances to do well in predicting the next competition. In my view the "hot hand fallacy" is very poorly named because it is based on studies of basketball players (who in making a streak of baskets show a "hot hand") who actually do have "hot hands" at times. They also get lucky. So the phenomena actually combines true skill and illusory skill.
Another implication of the Paul phenomena is that in some instances, it may not be possible to rigorously evaluate a forecast methodology. With an octopus, it is easy to assert that whatever methods he employed, they probably were not very rigorous. But what if it had been JP Morgan or Goldman Sachs with the remarkable record based on their lengthy quantitative analyses? Then it would be more difficult to assess whether their results were the consequence of a true forecasting ability, or just luck.
If you look around, on a daily basis you'll see all sorts of examples of the potential challenges presented by predictions in important settings. Are the economists who anticipated the financial downturn in the past few years actually smarter than others? Or were they just the few outliers in a fully covered distribution? Or both?
You can see a discussion of these subjects in a bit more depth in the following articles and book chapters:
Pielke, Jr., R.A. (2009), United States hurricane landfalls and damages: Can one-to five-year predictions beat climatology?. Environmental Hazards 8 187-200, issn: 1747-7891, doi: 10.3763/ehaz.2009.0017
Pielke, R.A. Jr, (2003), The role of models in prediction for decision. Models in Ecosystem Science 111-135, Princeton University Press.
Pielke, Jr., R. A., D. Sarewitz, and R. Byerly, (2000), Decision making and the future of nature: Understanding, using, and producing predictions. Prediction: Decision Making and the Future of Nature 361-387, Island Press (D. Sarewitz, D., R. A. Pielke, Jr., and R. Byerly, editors). | 1,393 | 6,177 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2018-26 | latest | en | 0.971034 |
https://www.sapling.com/6472599/measure-stock-performance | 1,726,004,931,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651318.34/warc/CC-MAIN-20240910192923-20240910222923-00144.warc.gz | 921,054,061 | 47,621 | # How to Measure Stock Performance
Measure stock performance with ROI.
There are two main ways to invest in a company: through debt or equity. Debt represents a claim against the future earnings of the company, whereas equity represents ownership and is purchased with shares of stock. The most common way to calculate stock performance is with the measure ROI (return on investment). ROI looks at investment earnings compared with the original cost of the investment.
## Step 1
Determine the original stock price. This is the price of the stock when you purchased it. Let's say you purchased the stock for \$50 per share.
Video of the Day
## Step 2
Determine the current or ending stock price. The ending stock price is its price when sold, say, at the end of the year for tax purposes. Let's say you are considering the sale of your stock, but want to know its performance first. The current value of the stock is \$60.
## Step 3
Determine the stock's earnings. This is the difference between the ending (or current) price and the original purchase price. The calculation is: \$60 - \$50 = \$10. | 237 | 1,106 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-38 | latest | en | 0.952141 |
https://stats.stackexchange.com/questions/196769/multiple-comparison-correction-for-temporally-correlated-tests | 1,708,596,256,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473738.92/warc/CC-MAIN-20240222093910-20240222123910-00268.warc.gz | 554,121,940 | 39,519 | # Multiple Comparison correction for temporally correlated tests
My data consists of N~200 cells. For each cell I am seeing the response to a stimulus for the next 300ms, in bins of 5ms (150bins total). In order to get confidence bounds for each bin on whether a response is significant, I perform a shuffling/bootstrapping test. This gives me graphs which look like those in figure below for each cell:
[Figure shows 4 cell responces (blue lines) w/ 95% confidence bounds (grey).]
If cells responses go above bounds for even 1 bin, they are considered significant and have a blue background. I understand that essentially I am doing 150 comparisons here, but a Bonferroni correction here seems incorrect due to the tests being temporally correlated.
Any suggestions on how to correctly ascertain whether the the cells have a significant response?
According to http://www.fil.ion.ucl.ac.uk/spm/doc/papers/NicholsHolmes.pdf, you can correct the p-values for multiple comparisons by comparing the lengths of the regions out of the confidence interval to ones obtained by permutation. This is presented for 3D signal. In your case, it is about 1D signal but I see no restriction to the dimension of the space. More precisely, the method is the following:
1. fix a suprathreshold value $\alpha$ typically 0.05,
2. build the permutation distribution of the maximal suprathreshold group size (shuffling the p-values over the line and computing the size of the largest group of points with p-values greater than $\alpha$ for $N$ generated permutation)
3. set the critical suprathreshold group size as the [$\alpha$ × $N$]+1 th largest value over the sampling distribution.
4. then a response is significant if a region of $\alpha$ significant point of size larger than the suprathreshold group size exists.
• thank you! which section of the paper did you get this from? Feb 16, 2016 at 18:10
• @DankMasterDan page 7 subsection Suprathreshold cluster tests. This is presented for 3D signal. In your case, it is about 1D signal but I see no restriction to the dimension of the space. Hope it helps. Feb 17, 2016 at 8:56
• Thnx, just the course I was looking for ! Feb 17, 2016 at 18:07
• Great, this method is pretty classic in functional mri (the paper is almost cited 3000 times). Happy to help Feb 17, 2016 at 19:05 | 559 | 2,319 | {"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} | 2.859375 | 3 | CC-MAIN-2024-10 | latest | en | 0.881002 |
https://scan.stockcharts.com/discussion/1110/intraday-swing-scans | 1,702,013,516,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100724.48/warc/CC-MAIN-20231208045320-20231208075320-00563.warc.gz | 543,320,438 | 11,858 | #### Howdy, Stranger!
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Hello, Sorry I have recently started toying around with stockcharts and was wondering how can I scan for stocks which are up by 90 cents or down by 90 cents and having a volume above 10000. I think its fairly easy but just that I am not able to write it properly. Can someone help?
• I wrote this
[type = stock] AND [exchange is TSX]
AND [close > 0.90*[high - low] + low] AND
[volume > 10000]
• I think this is the proper one
[type = stock] AND [exchange is TSX]
AND [[high-low > .9] OR [high-low < .9] ]AND
[volume > 10000]
• mod
edited March 2017
and
[
[close > 1 day ago close +.90]
or
[close < 1 day ago close - .90]
]
That says, find stocks that closed up MORE THAN 90 cents from yesterday's close, or closed down MORE THAN 90 cents.
If you mean that you want today's close to be WITHIN 90 cents of yesterday's close, reverse the > and < signs.
Notice this is an "or" statement. Click on the "Instructions" link at the bottom of the Advanced Scan page to read about how they have to be constructed (they need a set of brackets to group the choices) and when to use them. (I noticed you did this correctly in the last version, so this note is for other readers.)
You can put an "or" statement all on one line, but it's easier to see what you are doing if each statement within the or condition gets its own line, as shown above.
Also, it's easier to keep track of things if you start a new line each time you come to an "and".
So
[type = stock]
AND [exchange is TSX]
AND [volume > 10000]
AND [[high-low > .9] OR [high-low < .9] ]
For a simple scan like this, it seems unimportant, but as your scans get more complicated, it helps.
Also, the scan will run a little faster if you put the characteristics of the stocks you want (the "universe") at the top of the scan (as I've re-arranged yours) - things that don't have to be calculated, like exchange, volume, sector or industry group, etc. If you do that, the scan engine only has to do the calculations you request for those selected stocks.
• You are the best @markd . Hats off to your knowledge.. | 565 | 2,162 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-50 | latest | en | 0.963547 |
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What Is Gravity? Explain the equation for determining the gravitational force between two objects and the rate at which objects accelerate toward Earth.
The Description Of Gravity: Equation For Determining The Gravitational Force
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Tellus molestie nunc non blandit massa enim nec dui. Tellus molestie nunc non blandit massa enim nec dui. Ac tortor vitae purus faucibus ornare suspendisse sed nisi. Pharetra et ultrices neque ornare aenean euismod. Pretium viverra suspendisse potenti nullam ac tortor vitae. Morbi quis commodo odio aenean sed. At consectetur lorem donec massa sapien faucibus et. Nisi quis eleifend quam adipiscing vitae proin sagittis nisl rhoncus. Duis at tellus at urna condimentum mattis pellentesque. Vivamus at augue eget arcu dictum varius duis at. Justo donec enim diam vulputate ut. Blandit libero volutpat sed cras ornare arcu. Ac felis donec et odio pellentesque diam volutpat commodo. Convallis a cras semper auctor neque. Tempus iaculis urna id volutpat lacus. Tortor consequat id porta nibh.
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On APP - grab it while it lasts! | 4,745 | 18,274 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.681265 |
https://www.doubtnut.com/question-answer/find-the-mean-deviation-about-median-for-the-following-data-marks-0-10-10-20-20-30-30-40-40-50-50-60-1085 | 1,627,241,069,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151760.94/warc/CC-MAIN-20210725174608-20210725204608-00405.warc.gz | 752,669,252 | 79,875 | Home
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# Find the mean deviation about median for the following data :Marks 0-10 10-20 20-30 30-40 40-50 50-60Number of 6 8 14 16 4 2Girls
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2:53 | 396 | 1,033 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-31 | latest | en | 0.356722 |
https://www.mathworks.com/matlabcentral/fileexchange/72825-solving-a-system-of-non-linear-equations?s_tid=blogs_rc_6 | 1,725,892,794,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651103.13/warc/CC-MAIN-20240909134831-20240909164831-00358.warc.gz | 829,607,955 | 21,794 | # Solving a System of Non - Linear Equations
This script demonstrates the use of "The Newton - Raphson Method" to solve a "System of Non-Linear Equations" in 3 Independent Variables.
Updated 22 Sep 2019
We have learnt about solving the systems of linear equations. Different techniques like Cramer's rule, Gauss Elimination, etc exist for them.
But a large number of equations in real life are Non-linear in nature.We know various numerical methods like Newton-Raphson method, Regula Falsi method, etc in order to find a numerical solutions to such equations.
But what if the equations have more than one independent variable? Then as per rule, no. of equations must be equal to the number of equations ! These equations the form a System. How do we solve such "System of Non-Linear Equations" ?
Here comes the generalization of the iteration methods mentioned above applied to a
"System of Non-Linear Equations". The system of non-linear equations may consist of "Transcendental Equations" or "Nth order equations" or "Polynomial Equations" or Combinations of them.
This script demonstrates the use of "The Newton - Raphson Method" to solve a "System of Non-Linear Equations" in 3 Independent Variables.
The method proceeds as follows. Let f = f (x,y,z) ; g = g (x,y,z) ; h = h (x,y,z) be 3 non-linear equations. Let (fx , fy , fz) ; (gx , gy , gz) ; (hx , hy , hz) be the partial derivatives w.r.t x,y,z of f,g,h respectively. Let x0 , y0 , z0 be the initial approximations to the exact root.
Denote xk , yk ,zk = kth approximation to the root
fk , gk , hk = Value of f,g,h at (xk , yk ,zk)
fxk , gxk , hxk = Value of fx , gx , hx at (xk , yk ,zk)
fyk , gyk , hyk = Value of fy , gy , hy at (xk , yk ,zk)
fzk , gzk , hzk = Value of fx , gx , hx at (xk , yk ,zk)
Tran(A) = Transpose of matrix "A"
Inv(A) = Inverse of Matrix "A"
Eig(A) = Eigenvalues of matrix "A"
We create 3 matrices as follows :-
J = [fxk fyk fzk ; gxk gyk gzk ; hxk hyk hzk] = The jacobian matrix , ";"(semicolon) denotes a new row
xyzk = Tran([xk yk zk]) = The kth values of x,y,z
Fk = Tran([fk gk hk]) = Matrix of function values
The formula to find (k+1)th approximation to root is : - xyzkplus1 = (xyzk - inv(J)*Fk)
"The Newton - Raphson Method" uses one initial approximation to solve a given equation y = f(x).In this method the function f(x) , is approximated by a tangent line, whose equation is found from the value of f(x) and its first derivative at the initial approximation.
The tangent line then intersects the X - Axis at second point. This second point is again used as next approximation to find the third point.
In the similar way, we find (k+1)th "Matrix" of approximation at each Iteration. Instead of a "Derivative" we take a "Partial derivative".
IMPORTANT NOTE : There exists a "Sufficient Condition" that need to be fulfilled so that the iterations converge.
The condition is as follows :- || Max( eig ( inv ( J ) ) ) || >= 1 ,that is, the magnitude of the largest eigenvalue of inverse of the jacobian must be greater than unity.
The script proceeds in the same way and performs upto 5 iterations. The Accuracy required (required no. of decimal places) is taken as input from the user. The error between solutions of each iteration is checked every time and if found less than required accuracy, the iterations are stopped.
Screenshot Source : Wikipedia
### Cite As
अंबरीश प्रशांत चांदूरकर Ambarish Prashant Chandurkar (2024). Solving a System of Non - Linear Equations (https://www.mathworks.com/matlabcentral/fileexchange/72825-solving-a-system-of-non-linear-equations), MATLAB Central File Exchange. Retrieved .
##### MATLAB Release Compatibility
Created with R2018a
Compatible with any release
##### Platform Compatibility
Windows macOS Linux
##### Categories
Find more on Newton-Raphson Method in Help Center and MATLAB Answers | 1,005 | 3,845 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2024-38 | latest | en | 0.808642 |
https://stackoverflow.com/questions/5667888/counting-the-occurrences-frequency-of-array-elements/57028486#57028486 | 1,660,362,184,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571869.23/warc/CC-MAIN-20220813021048-20220813051048-00224.warc.gz | 506,259,136 | 90,488 | # Counting the occurrences / frequency of array elements
In Javascript, I'm trying to take an initial array of number values and count the elements inside it. Ideally, the result would be two new arrays, the first specifying each unique element, and the second containing the number of times each element occurs. However, I'm open to suggestions on the format of the output.
For example, if the initial array was:
``````5, 5, 5, 2, 2, 2, 2, 2, 9, 4
``````
Then two new arrays would be created. The first would contain the name of each unique element:
``````5, 2, 9, 4
``````
The second would contain the number of times that element occurred in the initial array:
``````3, 5, 1, 1
``````
Because the number 5 occurs three times in the initial array, the number 2 occurs five times and 9 and 4 both appear once.
I've searched a lot for a solution, but nothing seems to work, and everything I've tried myself has wound up being ridiculously complex. Any help would be appreciated!
Thanks :)
• If all you needed was to see if a value appears only once (instead of two or more times), you could use `if (arr.indexOf(value) == arr.lastIndexOf(value))` Mar 6, 2016 at 12:58
• We can use `ramda.js` to achieve this the easy way. `const ary = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4]; R.countBy(r=> r)(ary)` Jan 4, 2017 at 5:50
• `arr.filter(x => x===5).length` would return `3` to indicate that there are '3' fives in the array. Jun 15, 2020 at 21:21
• Let us assume My response is array of object
– Ajay
May 12, 2021 at 15:04
You can use an object to hold the results:
``````const arr = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
const counts = {};
for (const num of arr) {
counts[num] = counts[num] ? counts[num] + 1 : 1;
}
console.log(counts[5], counts[2], counts[9], counts[4]);``````
So, now your `counts` object can tell you what the count is for a particular number:
``````console.log(counts[5]); // logs '3'
``````
If you want to get an array of members, just use the `keys()` functions
``````keys(counts); // returns ["5", "2", "9", "4"]
``````
• It should be pointed out, that `Object.keys()` function is only supported in IE9+, FF4+, SF5+, CH6+ but Opera doesn't support it. I think the biggest show stopper here is IE9+. Jul 28, 2011 at 14:45
• Similarly, I also like `counts[num] = (counts[num] || 0) + 1`. That way you only have to write `counts[num]` twice instead of three times on that one line there. Jul 21, 2014 at 1:18
• This is a nice answer. This is easily abstracted into a function that accepts an array and returns a 'counts' object. Feb 23, 2017 at 0:08
• This is true for the specific example in the question, but for the sake of googlers it's worth pointing out that this is not always a safe technique for wider usage. Storing the values as object keys to count them means you're casting those values to strings and then counting that value. `[5, "5"]` will simply say you've got `"5"` two times. Or counting instances some different objects is just gonna tell you there's a lot of `[object Object]`. Etc. etc. Feb 5, 2020 at 9:51
• How could I then filter the returned object to show me highest to lowest, or lowest to highest count on a numbers Feb 11, 2020 at 22:00
``````const occurrences = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4].reduce(function (acc, curr) {
return acc[curr] ? ++acc[curr] : acc[curr] = 1, acc
}, {});
console.log(occurrences) // => {2: 5, 4: 1, 5: 3, 9: 1}``````
• Thanks, very nice solution ;) ... and to get the "key" and "value" arrays: `const keys = Object.keys(a);` `const values = Object.values(a);` Jun 7, 2020 at 8:15
• Short hand: `acc[curr] = (acc[curr] || 0) + 1` instead of using `if/else`. You can check the answer below Apr 28, 2021 at 5:55
• Clean and neat! What I was looking for :) Mar 24 at 4:44
• Note that elements in the array are always converted to a string when used as an object key. So when you would pass `[1, "1", { toString: () => "1" }]` you would give the result `{ 1: 3 }` May 3 at 9:24
``````const arr = [2, 2, 5, 2, 2, 2, 4, 5, 5, 9];
function foo (array) {
let a = [],
b = [],
arr = [...array], // clone array so we don't change the original when using .sort()
prev;
arr.sort();
for (let element of arr) {
if (element !== prev) {
a.push(element);
b.push(1);
}
else ++b[b.length - 1];
prev = element;
}
return [a, b];
}
const result = foo(arr);
console.log('[' + result[0] + ']','[' + result[1] + ']')
console.log(arr)``````
• has side-effect of sorting the array (side effects are bad), also sorting is `O(N log(N))` and the elegance gain isn't worth it May 25, 2011 at 17:05
• In absence of a nice high-level primitive from a third-party library, I would normally implement this like the `reduce` answer. I was about to submit such an answer before I saw it already existed. Nevertheless the `counts[num] = counts[num] ? counts[num]+1 : 1` answer also works (equivalent to the `if(!result[a[i]])result[a[i]]=0` answer, which is more elegant but less easy to read); this answers can be modified to use a "nicer" version of the for loop, perhaps a third-party for-loop, but I sort of ignored that since the standard index-based for-loops are sadly the default. May 25, 2011 at 18:51
• For small arrays sorting it in-place can be faster than creating an associative array. Sep 29, 2017 at 20:24
• @ŠimeVidas I added a disclaimer for `Array.sort`, because missing this fact has tripped me up in real code. (It's easy to naively treat it as if it makes a copy, since it returns the sorted array.) Jan 10, 2019 at 22:59
• Agree with @ninjagecko. `dictionary` would be better. Here is my answer to another approach. Feb 2, 2021 at 2:09
If using underscore or lodash, this is the simplest thing to do:
``````_.countBy(array);
``````
Such that:
``````_.countBy([5, 5, 5, 2, 2, 2, 2, 2, 9, 4])
=> Object {2: 5, 4: 1, 5: 3, 9: 1}
``````
As pointed out by others, you can then execute the `_.keys()` and `_.values()` functions on the result to get just the unique numbers, and their occurrences, respectively. But in my experience, the original object is much easier to deal with.
• Just worth noting that with countBy it only includes items that exist in the list, so if you wanted to count items that may not exist in the list then you would need to handle the exception. Or use lodash filter and length like this: `filter([true, true, true, false], function(m){return m==true}).length`. This would just return 0 if no values exist.
– Doug
Feb 3, 2021 at 16:24
• Worth adding that you need: const _ = require("lodash")
– Alex
May 20, 2021 at 23:30
One line ES6 solution. So many answers using object as a map but I can't see anyone using an actual Map
``````const map = arr.reduce((acc, e) => acc.set(e, (acc.get(e) || 0) + 1), new Map());
``````
Use `map.keys()` to get unique elements
Use `map.values()` to get the occurrences
Use `map.entries()` to get the pairs [element, frequency]
``````var arr = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4]
const map = arr.reduce((acc, e) => acc.set(e, (acc.get(e) || 0) + 1), new Map());
console.info([...map.keys()])
console.info([...map.values()])
console.info([...map.entries()])``````
• I used this and It worked pretty fine! May 31 at 22:13
Don't use two arrays for the result, use an object:
``````a = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
result = { };
for(var i = 0; i < a.length; ++i) {
if(!result[a[i]])
result[a[i]] = 0;
++result[a[i]];
}
``````
Then `result` will look like:
``````{
2: 5,
4: 1,
5: 3,
9: 1
}
``````
``````const a = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
const aCount = new Map([...new Set(a)].map(
x => [x, a.filter(y => y === x).length]
));
``````
``````aCount.get(5) // 3
aCount.get(2) // 5
aCount.get(9) // 1
aCount.get(4) // 1
``````
This example passes the input array to the `Set` constructor creating a collection of unique values. The spread syntax then expands these values into a new array so we can call `map` and translate this into a two-dimensional array of `[value, count]` pairs - i.e. the following structure:
``````Array [
[5, 3],
[2, 5],
[9, 1],
[4, 1]
]
``````
The new array is then passed to the `Map` constructor resulting in an iterable object:
``````Map {
5 => 3,
2 => 5,
9 => 1,
4 => 1
}
``````
The great thing about a `Map` object is that it preserves data-types - that is to say `aCount.get(5)` will return `3` but `aCount.get("5")` will return `undefined`. It also allows for any value / type to act as a key meaning this solution will also work with an array of objects.
``````function frequencies(/* {Array} */ a){
return new Map([...new Set(a)].map(
x => [x, a.filter(y => y === x).length]
));
}
let foo = { value: 'foo' },
bar = { value: 'bar' },
baz = { value: 'baz' };
let aNumbers = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4],
aObjects = [foo, bar, foo, foo, baz, bar];
frequencies(aNumbers).forEach((val, key) => console.log(key + ': ' + val));
frequencies(aObjects).forEach((val, key) => console.log(key.value + ': ' + val));``````
• do you have by any chance an improved answer of this just for an object array? im having trouble trying to modify it for an object array, where you just create a new array/map/set in which you remove duplicates, and add a new value for the object, let say called "duplicatedCount: value". i managed to remove duplicates in my nested objects array from this answer stackoverflow.com/a/36744732 Apr 4, 2017 at 14:32
• `Set` uses object references for uniqueness and offers no API for comparison of "similar" objects. If you want to use this approach for such a task you'd need some intermediate reduction function that guarantees an array of unique instances. It's not the most efficient but I put together a quick example here. Apr 6, 2017 at 9:25
• Thanks for the answer! but i actually solved it a lilttle bit differently. if you can see the answer i added here stackoverflow.com/a/43211561/4474900 i gave example of what i did. it works well, my case had a complex object needed comparing. dont know about the efficiency of my solution though Apr 6, 2017 at 15:26
• This might use nice new data structures but has runtime in O() while there are plenty of simple algorithms here that solve it in O(n). Jan 30, 2018 at 20:16
I think this is the simplest way how to count occurrences with same value in array.
``````var a = [true, false, false, false];
a.filter(function(value){
return value === false;
}).length
``````
• or `a.filter(value => !value).length` with the new js syntax Jun 1, 2018 at 8:03
– Ry-
Mar 19, 2020 at 2:44
``````const data = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4]
function count(arr) {
return arr.reduce((prev, curr) => (prev[curr] = ++prev[curr] || 1, prev), {})
}
console.log(count(data))``````
• Would anyone care to explain this (prev[curr] = ++prev[curr] || 1, prev) ? May 14, 2017 at 9:34
• The comma operator “evaluates each of its operands (from left to right) and returns the value of the last operand”, so this increments the value of prev[curr] (or initialises it to 1), then returns prev. Jun 25, 2017 at 16:37
• but is the output an array? Oct 8, 2017 at 23:47
## 2021's version
### The more elegant way is using `Logical nullish assignment (x ??= y) ` combined with `Array#reduce()` with O(n) time complexity.
The main idea is still using `Array#reduce()` to aggregate with output as `object` to get the highest performance (both time and space complexity) in terms of `searching` & `construct bunches of intermediate arrays` like other answers.
``````const arr = [2, 2, 2, 2, 2, 4, 5, 5, 5, 9];
const result = arr.reduce((acc, curr) => {
acc[curr] ??= {[curr]: 0};
acc[curr][curr]++;
return acc;
}, {});
console.log(Object.values(result));``````
### Clean & Refactor code
Using Comma operator (,) syntax.
`The comma operator (,)` evaluates each of its operands (from left to right) and returns the value of the last operand.
``````const arr = [2, 2, 2, 2, 2, 4, 5, 5, 5, 9];
const result = arr.reduce((acc, curr) => (acc[curr] = (acc[curr] || 0) + 1, acc), {});
console.log(result);``````
Output
``````{
"2": 5,
"4": 1,
"5": 3,
"9": 1
}
``````
• How can I get the highest tally from this object? Aug 5, 2021 at 10:44
• Could you pls give me your expected result? @kontenurban Aug 5, 2021 at 10:49
• This is a very nice answer. you could make it more concise with: `const result = arr.reduce((acc, curr) => (acc[curr] = -~(acc[curr]), acc), {});` See stackoverflow.com/a/47546846/1659476 for an explanation. Apr 27 at 18:55
• Thanks @YoniRabinovitch. `Bitwise` of your answer looks elegant & concise as well. Apr 28 at 4:45
If you favour a single liner.
`arr.reduce(function(countMap, word) {countMap[word] = ++countMap[word] || 1;return countMap}, {});`
Edit (6/12/2015): The Explanation from the inside out. countMap is a map that maps a word with its frequency, which we can see the anonymous function. What reduce does is apply the function with arguments as all the array elements and countMap being passed as the return value of the last function call. The last parameter ({}) is the default value of countMap for the first function call.
• You should explain this. that would make it a much better answer so people can learn how to use it in other use cases. Apr 25, 2015 at 23:38
• A single liner that just removes the linebreak that would usually follow `;`, `{` and `}`. ... OK. I think with that definition of a one liner we can write Conway's Game of Life as a "oneliner". Nov 30, 2018 at 21:42
ES6 version should be much simplifier (another one line solution)
``````let arr = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
let acc = arr.reduce((acc, val) => acc.set(val, 1 + (acc.get(val) || 0)), new Map());
console.log(acc);
// output: Map { 5 => 3, 2 => 5, 9 => 1, 4 => 1 }
``````
A Map instead of plain Object helping us to distinguish different type of elements, or else all counting are base on strings
Edit 2020: this is a pretty old answer (nine years). Extending the native `prototype` will always generate discussion. Although I think the programmer is free to choose her own programming style, here's a (more modern) approach to the problem without extending `Array.prototype`:
``````{
// create array with some pseudo random values (1 - 5)
const arr = Array.from({length: 100})
.map( () => Math.floor(1 + Math.random() * 5) );
// frequencies using a reducer
const arrFrequencies = arr.reduce((acc, value) =>
({ ...acc, [value]: acc[value] + 1 || 1}), {} )
console.log(arrFrequencies);
console.log(`Value 4 occurs \${arrFrequencies[4]} times in arrFrequencies`);
// bonus: restore Array from frequencies
const arrRestored = Object.entries(arrFrequencies)
.reduce( (acc, [key, value]) => acc.concat(Array(value).fill(+key)), [] );
console.log(arrRestored.join());
}``````
``.as-console-wrapper { top: 0; max-height: 100% !important; }``
The old (2011) answer: you could extend `Array.prototype`, like this:
``````{
Array.prototype.frequencies = function() {
var l = this.length,
result = {
all: []
};
while (l--) {
result[this[l]] = result[this[l]] ? ++result[this[l]] : 1;
}
// all pairs (label, frequencies) to an array of arrays(2)
for (var l in result) {
if (result.hasOwnProperty(l) && l !== 'all') {
result.all.push([l, result[l]]);
}
}
return result;
};
var freqs = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4].frequencies();
console.log(`freqs[2]: \${freqs[2]}`); //=> 5
// or
var freqs = '1,1,2,one,one,2,2,22,three,four,five,three,three,five'
.split(',')
.frequencies();
console.log(`freqs.three: \${freqs.three}`); //=> 3
// Alternatively you can utilize Array.map:
Array.prototype.frequencies = function() {
var freqs = {
sum: 0
};
this.map(function(a) {
if (!(a in this)) {
this[a] = 1;
} else {
this[a] += 1;
}
this.sum += 1;
return a;
}, freqs);
return freqs;
}
}``````
``.as-console-wrapper { top: 0; max-height: 100% !important; }``
Based on answer of @adamse and @pmandell (which I upvote), in ES6 you can do it in one line:
• 2017 edit: I use `||` to reduce code size and make it more readable.
``````var a=[7,1,7,2,2,7,3,3,3,7,,7,7,7];
a.reduce((r,k)=>{r[k]=1+r[k]||1;return r},{})
));``````
It can be used to count characters:
``````var s="ABRACADABRA";
s.split('').reduce((a, c)=>{a[c]++?0:a[c]=1;return a},{})
));``````
• It would be more readable if you used `|| 0`: `(r,k)=>{r[k]=(r[k]||0)+1;return r}` May 10, 2018 at 17:29
• You can do anything in one line in JavaScript.
– Ry-
Apr 15, 2019 at 7:28
• And why is it a bad thing, @Ry-?
– ESL
Apr 15, 2019 at 13:49
• Sometimes it's more clear in several lines, other is clearer in one line. Althoug it's a matter of "taste".
– ESL
Apr 15, 2019 at 13:50
• I mean “in ES6 you can do it in one line” applies to every answer, and you could also do this in ES5 in one line.
– Ry-
Apr 15, 2019 at 18:01
A shorter version using `reduce` and tilde (`~`) operator.
``````const data = [2, 2, 2, 2, 2, 4, 5, 5, 5, 9];
function freq(nums) {
return nums.reduce((acc, curr) => {
acc[curr] = -~acc[curr];
return acc;
}, {});
}
console.log(freq(data));``````
If you are using underscore you can go the functional route
``````a = ['foo', 'foo', 'bar'];
var results = _.reduce(a,function(counts,key){ counts[key]++; return counts },
_.object( _.map( _.uniq(a), function(key) { return [key, 0] })))
``````
``````_.keys(results)
``````
and the second array is
``````_.values(results)
``````
most of this will default to native javascript functions if they are available
``````var array = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
function countDuplicates(obj, num){
obj[num] = (++obj[num] || 1);
return obj;
}
// answer => {2:5, 4:1, 5:3, 9:1};
``````
If you still want two arrays, then you could use answer like this...
``````var uniqueNums = Object.keys(answer);
// uniqueNums => ["2", "4", "5", "9"];
// countOfNums => [5, 1, 3, 1];
``````
Or if you want uniqueNums to be numbers
``````var uniqueNums = Object.keys(answer).map(key => +key);
// uniqueNums => [2, 4, 5, 9];
``````
• es6/7 makes this all much nicer. You may also want to reduce to a `Map` instead, since it will avoid the typecasting that using a number as an object key (casting as string) does. `const answer = array.reduce((a, e) => a.set(e, (a.get(e) || 0) + 1), new Map())` .You can get `answer.keys()` for the keys, and `answer.values()` for the values as arrays. `[...answer]` will give you a big array with all the key/values as 2d arrays. Jul 9, 2017 at 13:08
Here's just something light and easy for the eyes...
``````function count(a,i){
var result = 0;
for(var o in a)
if(a[o] == i)
result++;
return result;
}
``````
Edit: And since you want all the occurences...
``````function count(a){
var result = {};
for(var i in a){
if(result[a[i]] == undefined) result[a[i]] = 0;
result[a[i]]++;
}
return result;
}
``````
• Ok, missed that. Should be fixed now. Jul 10, 2017 at 19:38
So here's how I'd do it with some of the newest javascript features:
First, reduce the array to a `Map` of the counts:
``````let countMap = array.reduce(
(map, value) => {map.set(value, (map.get(value) || 0) + 1); return map},
new Map()
)
``````
By using a `Map`, your starting array can contain any type of object, and the counts will be correct. Without a `Map`, some types of objects will give you strange counts. See the `Map` docs for more info on the differences.
This could also be done with an object if all your values are symbols, numbers, or strings:
``````let countObject = array.reduce(
(map, value) => { map[value] = (map[value] || 0) + 1; return map },
{}
)
``````
Or slightly fancier in a functional way without mutation, using destructuring and object spread syntax:
``````let countObject = array.reduce(
(value, {[value]: count = 0, ...rest}) => ({ [value]: count + 1, ...rest }),
{}
)
``````
At this point, you can use the `Map` or object for your counts (and the map is directly iterable, unlike an object), or convert it to two arrays.
For the `Map`:
``````countMap.forEach((count, value) => console.log(`value: \${value}, count: \${count}`)
let values = countMap.keys()
let counts = countMap.values()
``````
Or for the object:
``````Object
.entries(countObject) // convert to array of [key, valueAtKey] pairs
.forEach(([value, count]) => console.log(`value: \${value}, count: \${count}`)
let values = Object.keys(countObject)
let counts = Object.values(countObject)
``````
• "Without a Map, some types of objects will give you strange counts" +1 for explaining how a map is better than an Object. Feb 27 at 22:03
Solution using a map with O(n) time complexity.
``````var arr = [2, 2, 2, 2, 2, 4, 5, 5, 5, 9];
const countOccurrences = (arr) => {
const map = {};
for ( var i = 0; i < arr.length; i++ ) {
map[arr[i]] = ~~map[arr[i]] + 1;
}
return map;
}
``````
• My upvote to you, this works like butter with O(n) time complexity May 7, 2020 at 10:46
• I timed many answers from this thread (link jsbench.me/2wkzzagdu3/1 and this was the fastest). I imagine even without the bit twiddling, it would be similarly fast, just given the plain for loop. The lodash solution is 9% slower, and without the tilde tilde and replace with (map[arr[i]] || 0) it is also 9% slower. Other answers are much slower than this one. Feb 23 at 8:32
There is a much better and easy way that we can do this using `ramda.js`. Code sample here
```const ary = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4]; R.countBy(r=> r)(ary) ``` countBy documentation is at documentation
My solution with ramda:
``````const testArray = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4]
const counfFrequency = R.compose(
R.map(R.length),
R.groupBy(R.identity),
)
counfFrequency(testArray)
``````
Using MAP you can have 2 arrays in the output: One containing the occurrences & the other one is containing the number of occurrences.
``````const dataset = [2,2,4,2,6,4,7,8,5,6,7,10,10,10,15];
let values = [];
let keys = [];
var mapWithOccurences = dataset.reduce((a,c) => {
if(a.has(c)) a.set(c,a.get(c)+1);
else a.set(c,1);
return a;
}, new Map())
.forEach((value, key, map) => {
keys.push(key);
values.push(value);
});
console.log(keys)
console.log(values)``````
Using Lodash
``````const values = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
const frequency = _.map(_.groupBy(values), val => ({ value: val[0], frequency: val.length }));
console.log(frequency);``````
``<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.15/lodash.min.js"></script>``
I know this question is old but I realized there are too few solutions where you get the count array as asked with a minimal code so here is mine
``````// The initial array we want to count occurences
var initial = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
// The count array asked for
var count = Array.from(new Set(initial)).map(val => initial.filter(v => v === val).length);
// Outputs [ 3, 5, 1, 1 ]
``````
Beside you can get the set from that initial array with
``````var set = Array.from(new Set(initial));
//set = [5, 2, 9, 4]
``````
• This code is terribly inefficient as it iterates on the initial array length^2 times. Dec 24, 2020 at 5:31
• Yes this code is `length²` time complex, that's why I insisted its aim is to provide a minimal code that solves that problem! Dec 25, 2020 at 19:39
Check out the code below.
``````<html>
<script>
// array with values
var ar = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
var Unique = []; // we'll store a list of unique values in here
var Counts = []; // we'll store the number of occurances in here
for(var i in ar)
{
var Index = ar[i];
Unique[Index] = ar[i];
if(typeof(Counts[Index])=='undefined')
Counts[Index]=1;
else
Counts[Index]++;
}
// remove empty items
Unique = Unique.filter(function(){ return true});
Counts = Counts.filter(function(){ return true});
var a=[];
for(var i=0; i<Unique.length; i++)
{
a.push(Unique[i] + ':' + Counts[i] + 'x');
}
</script>
<body>
</body>
</html>
``````
You can simplify this a bit by extending your arrays with a `count` function. It works similar to Ruby’s `Array#count`, if you’re familiar with it.
``````Array.prototype.count = function(obj){
var count = this.length;
if(typeof(obj) !== "undefined"){
var array = this.slice(0), count = 0; // clone array and reset count
for(i = 0; i < array.length; i++){
if(array[i] == obj){ count++ }
}
}
return count;
}
``````
Usage:
``````let array = ['a', 'b', 'd', 'a', 'c'];
array.count('a'); // => 2
array.count('b'); // => 1
array.count('e'); // => 0
array.count(); // => 5
``````
Gist
### Edit
You can then get your first array, with each occurred item, using `Array#filter`:
``````let occurred = [];
array.filter(function(item) {
if (!occurred.includes(item)) {
occurred.push(item);
return true;
}
}); // => ["a", "b", "d", "c"]
``````
And your second array, with the number of occurrences, using `Array#count` into `Array#map`:
``````occurred.map(array.count.bind(array)); // => [2, 1, 1, 1]
``````
Alternatively, if order is irrelevant, you can just return it as a key-value pair:
``````let occurrences = {}
occurred.forEach(function(item) { occurrences[item] = array.count(item) });
occurences; // => {2: 5, 4: 1, 5: 3, 9: 1}
``````
This question is more than 8 years old and many, many answers do not really take ES6 and its numerous advantages into account.
Perhaps is even more important to think about the consequences of our code for garbage collection/memory management whenever we create additional arrays, make double or triple copies of arrays or even convert arrays into objects. These are trivial observations for small applications but if scale is a long term objective then think about these, thoroughly.
If you just need a "counter" for specific data types and the starting point is an array (I assume you want therefore an ordered list and take advantage of the many properties and methods arrays offer), you can just simply iterate through array1 and populate array2 with the values and number of occurrences for these values found in array1.
As simple as that.
Example of simple class SimpleCounter (ES6) for Object Oriented Programming and Object Oriented Design
``````class SimpleCounter {
constructor(rawList){ // input array type
this.rawList = rawList;
this.finalList = [];
}
mapValues(){ // returns a new array
this.rawList.forEach(value => {
this.finalList[value] ? this.finalList[value]++ : this.finalList[value] = 1;
});
this.rawList = null; // remove array1 for garbage collection
return this.finalList;
}
}
module.exports = SimpleCounter;
``````
• Sticking a function in a class for no reason doesn’t make it object-oriented, `finalList` has no reason to be an array, and this has no advantages over doing it properly.
– Ry-
Mar 19, 2020 at 2:41
Given the array supplied below:
``````const array = [ 'a', 'b', 'b', 'c', 'c', 'c' ];
``````
You can use this simple one-liner to generate a hash map which links a key to the number of times it appears in the array:
``````const hash = Object.fromEntries([ ...array.reduce((map, key) => map.set(key, (map.get(key) || 0) + 1), new Map()) ]);
// { a: 1, b: 2, c: 3 }
``````
Expanded & Explained:
``````// first, we use reduce to generate a map with values and the amount of times they appear
const map = array.reduce((map, key) => map.set(key, (map.get(key) || 0) + 1), new Map())
// next, we spread this map into an array
const table = [ ...map ];
// finally, we use Object.fromEntries to generate an object based on this entry table
const result = Object.fromEntries(table);
``````
credit to @corashina for the `array.reduce` code
To return an array which is then sortable:
``````let array = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4]
let reducedArray = array.reduce( (acc, curr, _, arr) => {
if (acc.length == 0) acc.push({item: curr, count: 1})
else if (acc.findIndex(f => f.item === curr ) === -1) acc.push({item: curr, count: 1})
else ++acc[acc.findIndex(f => f.item === curr)].count
return acc
}, []);
console.log(reducedArray.sort((a,b) => b.count - a.count ))
/*
Output:
[
{
"item": 2,
"count": 5
},
{
"item": 5,
"count": 3
},
{
"item": 9,
"count": 1
},
{
"item": 4,
"count": 1
}
]
*/`````` | 8,564 | 27,867 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.891739 |
http://www.ck12.org/book/CK-12-Middle-School-Math---Grade-6/r4/section/3.8/ | 1,490,362,912,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218187945.85/warc/CC-MAIN-20170322212947-00232-ip-10-233-31-227.ec2.internal.warc.gz | 464,254,959 | 35,855 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 3.8: Use Estimation
Difficulty Level: At Grade Created by: CK-12
## Introduction
Summer Job Benefits
Jose has enjoyed working all summer. He loved helping Mr. Harris and his recycling idea ended up being very profitable.
Jose began the summer with an estimate of how much money he thought he would make.
Jose earned $7.00 per hour and he worked ten 30 hour weeks. Jose ended up earning$2100.00 for the summer, and he is very pleased with his accomplishment.
Now that the summer is over, Jose wishes to spend part of his money on new clothes for school.
He has selected the following items.
$19.95$32.95
$46.75 Jose brought$100.00 with him to purchase the items.
If he estimates the total cost, what would it be?
How much change will Jose receive from the $100.00? Using estimation will help Jose with his purchases. Let’s look at some situations where estimation makes the most sense, then we will come back to this problem to help Jose with his shopping. What You Will Learn In this lesson, you will learn to use the following skills: • Read and understand given problem situations • Develop and use the strategy: Use Estimation • Plan and compare alternative approaches to solving problems • Solve real-world problems using selected strategies as part of a plan Teaching Time I. Read and Understand Given Problem Situations We can use estimation in several different problem situations. To use estimation, we need to read and understand the problem. There will be clues in the problem to let us know if estimation is a good option for solving that specific problem. Let’s review what it means to estimate. Estimating means that we are going to be finding an answer that is an approximate answer. When estimating, our answer must make sense, but it does not need to be exact. We can find an answer that is reasonable to provide us information for our problem. When looking at a problem, we need to read the problem to see if estimating is a good option in the problem. We can look for key words to help us with this. Here are some of the key words that we use when estimating: • Close to • Approximate • Estimate • An answer that makes sense • About If you see these words in a word problem, you can use estimating to find the answer. Let’s look at an example. Example Kelly wanted to get an idea how much she was spending at the store. On the way to the checkout she looked at the items in her cart. Here are the prices of the food in her cart:$.50, $2.50,$ 3.45 and $6.79. About how much will Kelly spend when she checks out? Are there any key words in this problem? Yes, the word ABOUT lets us know that we can estimate to find our answer. Now that we know that we can estimate, how do we use estimation to solve this problem? II. Develop and Use the Strategy: Use Estimation Once you have figured out that you can estimate to solve the problem, you will need to apply the estimation strategy. We can do this in one of two ways. 1. Rounding 2. Front – end Estimation. For the problem that we just looked at, let’s use rounding. Here is the problem once again. Example Kelly wanted to get an idea how much she was spending at the store. On the way to the checkout she looked at the items in her cart. Here are the prices of the food in her cart,$.50, $2.50,$ 3.45 and $6.79. About how much will Kelly spend when she checks out? Next, let’s round each price. .50 becomes 1 2.50 becomes 3.00 3.45 becomes 3.00 6.79 becomes 7 Now we can add up the rounded answers: 1 + 3 + 3 + 7 = 14 Our answer is$14.00. Kelly will spend approximately 14.00 at the store. III. Plan and Compare Alternative Approaches to Solving Problems There are many different ways to approach solving a problem. In the last example, we used rounding and estimation. We know that this is an approach that works when we are looking for an approximate answer. If we had been working with large numbers in the thousands, we would have been using estimation and front – end estimation. Sometimes, we will need to draw a picture to solve a problem. That is what will make the most sense. Let’s look at an example where we would draw a picture to solve an estimation problem. Example Carl is working on building a small cd rack out of wood. He can buy material in a 6' \begin{align*}\times\end{align*} 8' rectangular piece of plywood. Carl needs to build two sides from one piece of wood. The sides have the dimensions 2' \begin{align*}\times\end{align*} 4'. If Carl buys one sheet of plywood, will he have enough wood for the two sides of the cd rack? Hmmmm…. How can we work on this problem? We don’t need an exact measurement we just need to know the rough dimensions to figure out if the two sides of the cd rack will fit on piece of plywood. We can use estimation to do this. First, let’s underline the important information in the problem. Carl is working on building a small cd rack out of wood. He can buy material in a \begin{align*}\underline{6’ \times 8’}\end{align*} rectangular piece of plywood. Carl needs to build two sides from one piece of wood. The sides have the dimensions \begin{align*}\underline{2’ \times 4’}\end{align*}. If Carl buys one sheet of plywood, will he have enough wood for the two sides of the cd rack? Notice here that we show three pictures. The first one is of the rectangular piece of wood that is 6 \begin{align*}\times\end{align*} 8. The second two are the two rectangles that will make up the side of the cd rack. This is a visual way to estimate whether the two pieces will fit on the one piece of plywood. Visually it looks like it will work. Visual estimation is one strategy. What about if we want to be sure our estimate is accurate? We can estimate the dimensions of the two sides of the cd rack combined. 2 \begin{align*}\times\end{align*} 4 + 2 \begin{align*}\times\end{align*} 4 = 4 \begin{align*}\times\end{align*} 4 We need a piece of wood that is 4 \begin{align*}\times\end{align*} 4 to build the sides of the cd rack. Since our piece is 6 \begin{align*}\times\end{align*} 8 it will work for us. Our visual estimation is accurate. ## Real Life Example Completed Summer Job Benefits Now we can help Jose with his shopping. Shopping is a great real life example where estimation is very useful. We can get an idea of how much we are spending as well as about how much change we can receive when estimation. Let’s take another look at the problem. Jose has enjoyed working all summer. He loved helping Mr. Harris and his recycling idea ended up being very profitable. Jose began the summer with an estimate of how much money he thought he would make. Jose earned7.00 per hour and he worked ten 30 hour weeks.
Jose ended up earning $2100.00 for the summer, and he is very pleased with his accomplishment. Now that the summer is over, Jose wishes to spend part of his money on new clothes for school. He has selected the following items.$19.95
$32.95$46.75
Jose brought $100.00 with him to purchase the items. If he estimates the total cost, what would it be? How much change will Jose receive from the$100.00?
We could use a couple of different strategies to estimate the total of Jose’s purchases.
We could use rounding or front – end estimation.
Let’s use rounding first.
$19.95 rounds to$20.00
$32.95 rounds to$33.00
$46.75 rounds to$47.00
Our estimate is 100.00. Hmmm. Ordinarily, rounding would give us an excellent estimate, but in this case our estimate is the amount of money Jose wishes to pay with. Because of this, let’s try another strategy. Let’s use front – end estimation and see if we can get a more accurate estimate. \begin{align*}19 + 32 + 46 & = 97 \\ 1 + 1 + 80 & = 2.80\end{align*} Our estimate is99.80.
With front – end estimation, we can estimate the Jose will receive .20 change from his $100.00. While he isn’t going to get a lot of change back, he is going to receive some change so he does have enough money to make his purchases. ## Technology Integration ## Time to Practice Directions: Look at each problem and use what you have learned about estimation to solve each problem. 1. Susan is shopping. She has purchased two hats at$5.95 each and two sets of gloves at $2.25 each. If she rounds each purchase price, how much can she estimate spending? 2. If she uses front – end estimation, how does this change her answer? 3. Which method of estimation gives us a more precise estimate of Susan’s spending? 4. If she brings$20.00 with her to the store, about how much change can she expect to receive?
5. If she decided to purchase one more pair of gloves, would she have enough money to make this purchase?
6. Would she receive any change back? If yes, about how much?
7. Mario is working at a fruit stand for the summer. If a customer buys 3 oranges at $.99 a piece and two apples for$.75 a piece, about how much money will the customer spend at the fruit stand? Use rounding to find your answer.
8. What is the estimate if you use front – end estimation?
9. Why do you think you get the same answer with both methods?
10. If the customer gives Mario a $10.00 bill, about how much change should the customer receive back? 11. Christina is keeping track of the number of students that have graduated from her middle school over the past five years. Here are her results. 2004 – 334 2005 – 367 2006 – 429 2007 – 430 2008 – 450 Estimate the number of students who graduated in the past five years. 12. Did you use rounding or front – end estimation? 13. Why couldn’t you use front – end estimation for this problem? 14. Carlos has been collecting change for the past few weeks. He has 5 nickels, 10 dimes, 6 quarters and four dollar bills. Write out each money amount. 15. Use rounding to estimate the sum of Carlos’ money. 16. Use front – end estimation to estimate the sum of Carlos’ money. 17. Which method gives you a more accurate estimate? Why? 18. Tina is working to buy presents for her family for the holidays. She has picked out a cd for her brother for$14.69, a vase for her Mother at $32.25 and a picture frame for her father at$23.12. Use rounding to estimate the sum of Tina’s purchases.
19. Use front – end estimation to find an estimate for the purchases.
20. Which estimate is more accurate?
21. Why?
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Subjects: | 2,551 | 10,457 | {"found_math": true, "script_math_tex": 11, "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} | 4.3125 | 4 | CC-MAIN-2017-13 | longest | en | 0.950958 |
http://mathoverflow.net/revisions/75794/list | 1,368,971,129,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368697552127/warc/CC-MAIN-20130516094552-00081-ip-10-60-113-184.ec2.internal.warc.gz | 164,321,781 | 7,269 | 4 correction
As many other have just said you cannot think to study just some particular subjects ignoring some other areas, expecially if you want to do research. Most of math was born from the observation of some similar phenomena in many different areas: for instance the concept of category itself was born from the observation that in math we deal every time with collections of structures and morphisms preserving those structures, that led to the abstraction of category, similarly I strongly doubt that Grothendieck could invent the concept of (generalized) sheaf if first he hadn't known the many concrete sheaves that appear in topology, differential geometry and algebraic geometry, so it couldn't get to the concept of (Grothendieck's) topos, and without that I'm not so sure that Lawvere could get to the concept of elementary topos while doing his research in logic. This are just some example of as math have evolved thanks to interaction of different areas (for instance, as you can see in the example above, from interaction of geometry and logic).
Just to answer to your comment about analysis there's a professor in Italy who studies higher dimensional category theory for his research in analysis, so analysis need higher category theory.
Of course the best place where you can get a lot of intuition of higher category theory is algebraic topology where higher categories are used to model homotopy types for topological spaces, via $\infty$-groupoids, and directed space, via $(n,r)$-categories where $n,r \in \omega \cup {\infty}$ but you can find a lot of higher dimensional category theory in logic and computer science too, I've seen some application in calculability theory and model theory where (higher) category theory is used to model the semantic of theories, in particular type theory (if you're interested in application of higher categorical logic-model theory you can take a look to Makkai's work and also Mike Shulman's work on homotopy type theory). Also in mathematical physics there are a lot of higher category theory as John Baez's work prove.
I suppose above you were referring to Cheng-Lauda "Illustrated guide book", that's a good book if you want to learn many approaches to $n$-categories, but in higher category theory there's a lot of more then just $(n,r)$-categories (like usually Mr.Shulman says), Leinster's "Higher operads, Higher categories" is more complete from this point of view because it presents a lot of stuff like generalized multicategories/operads or $fc$-multicategories. Anyway if you want some references on higher category theory you can find some here.
(Edit: I've improved a little the answer now that I've found some other references.)
3 fixed grammar, added other references
As many other have just said you cannot think to study just some particular subject subjects ignoring some other areas, expecially if you want to do research. Most of math was born from the observation of some similar phenomena in many different areas: for instance the concept of category itself was born from the observation that in math we deal every time with collections of structures and morphisms preserving those structures, that led to the abstraction of category, similarly I strongly doubt that Grothendieck could invent the concept of sheaf if first he hadn't known the many concrete sheaves that appear in topology, differential geometry and algebraic geometry, so it couldn't get to the concept of (Grothendieck's) topos, and without that I'm not so sure that Lawvere could get to the concept of elementary topos while doing his research in logic. This are just some example of as math have evolved thanks to interaction of different areas (for instance, as you can see in the example above, from interaction of geometry and logic).
Just to answer to your comment about analysis there's a professor in Italy with study who studies higher dimensional category theory and is an analystfor his research in analysis, so analysis to need higher category theory.
Of course the best place where you can get a lot of intuition of higher category theory is algebraic topology where higher categories are used to model homotopy types for topological spaces, via $\infty$-groupoids, and directed space, via $(n,r)$-categories where $n,r \in \omega \cup {\infty}$ but you can find a lot of higher dimensional category theory in logic and computer science too, I've seen some application in calculability theory and model theory where (higher) category theory is used to model the semantic of theories, but in particular type theory (if you're interested in application of higher categorical logic-model theory you can take a look to Makkai's work and also Mike Shulman's work on homotopy type theory). Also in mathematical physics there are a lot of higher category theory as John Baez's work prove.
I suppose you're above you were referring to Cheng-Lauda illustrated "Illustrated guide bookbook", that's a good book if you want to learn many approaches to higher categories, $n$-categories, but in higher category theory there's a lot of more then just $(n,r)$-categories (like usually say Mike Shulman)Mr.Shulman says), Leinster's book "Higher operads, Higher categories" is more complete from this point of view because it presents a lot of stuff like generalized multicategories/operads or $fc$-multicategories. Anyway if you want some references on higher category theory you can find some here.
(Edit: I've improved a little the answer now that I've found some other references.)
2 deleted 4 characters in body
As many other have just said you cannot think to study just some particular subject and just to answer to your comment about analysis there's a professor in Italy with study higher dimensional category theory and is an analyst, so analysis to need higher category theory. Of course the best place where you can get a lot of intuition of higher category theory is algebraic topology where higher categories are used to model homotopy types for topological spaces, via $\infty$-groupoids, and directed space, via $(n,r)$-categories where $n,r \in \omega \cup {\infty}$ but you can find a lot of higher dimensional category theory in logic and computer science, I've seen some application in calculability theory, but also in mathematical physics there are a lot of higher category theory as John Baez's work prove. I suppose you're referring to Cheng-Lauda illustrated guide book, that's a good book if you want to learn many approaches to higher categories, but in higher category theory there's a lot of more then just $(n,r)$-categories (like usually say Mike Shulman), Leinster's book is more complete from this point of view because it presents a lot of stuff like $fc$-multicategories. Anyway if you want some references on higher category theory you can find some here in a my previous answer on MO. | 1,448 | 6,886 | {"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} | 2.796875 | 3 | CC-MAIN-2013-20 | latest | en | 0.962908 |
https://www.jacksonvilleillinois-homes.com/2019/07/11/how-much-loan-can-you-afford/ | 1,571,771,110,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987823061.83/warc/CC-MAIN-20191022182744-20191022210244-00558.warc.gz | 938,925,650 | 9,751 | # How Much Loan Can You Afford
Salary And House Price Top 10 Mortgage Lenders For First Time Buyers Top 10 95% LTV First time buyer mortgages | money.co.uk – Compare first time buyer mortgages appropriate for those with a deposit of 5%. As much as 95% of the property’s value could be offered to customers who are accepted for the mortgages listed here.What Price of House Can I Buy If My Salary is \$50,000. – What Price of House Can I Buy If My Salary is \$50,000? By: Karina C. Hernandez.. Once you know how much you can borrow add to that your down payment to calculate the maximum house price you can afford. In this example, the maximum loan amount is calculated at \$203,000. If you have a \$20,000.
Now that you know how much of a car loan you might qualify for, your next step is to determine if you can truly afford it. Just because you might qualify for a high-value loan doesn’t mean you can.
Calculator Use. How much house can you afford to be looking for? This calculator will help you calculate how much you can afford. Shopping for a new home? Calculate the home price you can pay and the mortgage schedule you will need based on the payment, down payment, taxes and insurance you can afford.
Methodology. In general, that means your total debt payments should be no more than 36% of your gross income. Once you enter your monthly debt (including credit cards, student loan and car payments), we come up with a maximum monthly home payment you could handle while staying under that threshold.
What Price Of A House Can I Afford How Much A Month Can I Afford in House Payments? Formula For. – The reason these formulas talk about the size of the mortgage you can afford rather than the actual cost of the house is because everybody brings a different down payment to the table. If you just sold a house or have saved or inherited a big bucket of money, maybe you can put down 50% or more of the house price.
How Much Can I Afford? fha mortgage calculator. Use the following calculator to help you determine an affordable monthly payment so that you know what you can afford before you make an offer on the home you want to purchase.
When you are looking into getting a loan, it is easier to estimate the amount you can pay monthly, based on your current financial situation, as opposed to the.
This borrowing power calculator can help you find out how much you. out how much you could afford to borrow and how much lenders might.
You can plug in your combined annual income, along with any monthly payment obligations, and then experiment with new loan assumptions. Try different interest rates, down payment amounts, property taxes, and mortgage terms to see how they impact how much house you can afford.
· If you want to do the calculation manually, let’s look at five ways to calculate how much house you can afford, beginning with a standard rule of thumb. 1. Multiply Your Annual Income By 2.5 or 3
If you earn \$56,516, the average household income, you can afford \$1,695 in total monthly payments, according to the 36% rule. The rule, which measures your debt relative to your income, is used by lenders to evaluate how much you can afford.
^ | 689 | 3,178 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2019-43 | longest | en | 0.94593 |
https://unmethours.com/question/64375/numerical-equations-with-ems/ | 1,638,218,617,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358842.4/warc/CC-MAIN-20211129194957-20211129224957-00491.warc.gz | 670,363,920 | 15,080 | Question-and-Answer Resource for the Building Energy Modeling Community
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# Numerical equations with EMS
Hello,
I am trying to write a subroutine that calculates the temperature of my storage medium at each timestep. I am trying to apply a finite difference model that solves N ODEs for each node, the model is very similar to that of WaterHeater:Stratified where the temperature update equation for each node is:
Tfinal= (Tinitial +b/a).exp(a.deltaT)-b/a
Does EMS allow this sort of formulations ?
My idea is to create a trend variable for the temperature at each node, but I dont understand how initialization works in EMS Any insights? Thank you
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If I'm understanding correctly, you do not need to use a trend variable just to retrieve a variable's value from a previous timestep. Rather, just use an EnergyManagementSystem:GlobalVariable. For example:
EnergyManagementSystem:Program,
TemperatureCalculation,
SET Tfinal = Tinitial + 10,
SET Tinitial = Tfinal;
And then set Tinitial as a global EMS variable so that its value at the end of the previous timestep can be used in the program in the subsequent timestep.
If you need to set the initial value for Tinitial for the beginning of the simulation, you can do this by creating a second EMS program that sets it value (e.g., SET Tinitial = 50) and uses the BeginNewEnvironment and/or AfterNewEnvironmentWarmUpIsComplete` calling points (I forget which offhand).
more
Thank you very much this is really helpful!
( 2021-11-11 14:34:35 -0600 )edit | 365 | 1,611 | {"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} | 2.546875 | 3 | CC-MAIN-2021-49 | longest | en | 0.890889 |
https://forum.amibroker.com/t/how-to-change-the-time-frame-to-2-days-back/11008/3 | 1,716,016,319,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057327.58/warc/CC-MAIN-20240518054725-20240518084725-00211.warc.gz | 228,228,599 | 10,585 | How to change the time frame to 2 days back
Hi,
How to find price of a stock at 2 days back particular time.
For example i want to compare today 10 AM stock price with 2 days back same 10 AM stock price(If i am using 5 minute interval). If any pseudo code is available that is much helpful.
Thanks
There are several ways. One way is using ValueWhen.
``````/// @link https://forum.amibroker.com/t/how-to-change-the-time-frame-to-2-days-back/11008/2
daysback = 2;
SetBarsRequired((daysback+1)*inDaily/Max(1,Interval()));
time_cond = tn == time;// equality check so time bar has to exists
today_price = ValueWhen(time_cond, C);
n_days_back_price = ValueWhen(time_cond, C, daysback+1);
printf("today's price: %g\nprice of %g days back: %g", today_price, daysback, n_days_back_price );
``````
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@TRDK, getting "ready to use" code is always nice, but we will learn more if we also spend time studying the corresponding AFL functions reference pages and additional examples.
For instance, the above posted answer uses the following functions:
So even a short snippet of code can be a good starting point to increase our knowledge of the wide range of functions that are ready for use in AFL.
In particular, I also suggest reviewing this previous thread to better understand the use of ValueWhen(), which is quite a useful function, but seems to be a little difficult for many AmiBroker users to understand.
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Hi @fxshrat
Can you pls explain your thinking behind:
, I understand that you're limiting the number of prior bars to the condition set by the OP, but the logic of it escapes me at the moment.
Ordinarily I'd just use `SetBarsRequired(sbrAll,0)` to ensure that all the bars needed are available, leaving it up to the logic in code to manage the actual number of bars used, and not rely on/ be restricted by limits set by AB's internal decision-making.
No, setting SetBarsRequired(sbrall) would be insane idea. If you would have a symbol with one million 5-min bars then it would load all 1 million bars.
The reason for my version is that I am not restricting but just making sure that minimum number of bars is activated.
Why that way? It's not difficult.
He is using 5-minute. Maximum possible bars per day on 5-minute (== 300 seconds) time frame are 86400/300 = 288 bars. 86400 == inDaily. A single day has maximum 86400 seconds or 1440 minutes or 24 hours. Since he wants to get two days back from today it is today+yesterday+day before yesterday. So ensuring three days (since daysback is a variable it is not just 3 days being ensured but n-days look back are getting ensured depending on which lookback day you want to compare with). 3 days x 288 bars = 864 bars minimum on 5-minute TF to ensure proper output on large zoom in (e.g if zooming in so far that only todays data being visible on chart, or even further zoom in). daysback with "+1" within SetBarsRequired in case someone would set zero daysback. Anyway just try yourself, zoom in to just few bars being visible (let's say 20 bars) on 5 minute and then comment SetBarsRequired. You will get Null (-1e+10) output then.
Now, Interval() ensures that it works on any intraday time frame but not just 5-minute. Interval() returns seconds. And using max(1, Interval()) ensures that if applying indicator from editor it does not return error because Interval function returns zero on formula verification.
Mysteries solved?
7 Likes
Thanks @fxshrat,
Yes "Mysteries solved"!
I tend to analyse/ trade on daily timeframe, so wasn't thinking about the explosion of bars on intra-day - your code and logic makes sense.
Thanks @fxshrat. Now i am getting the desired output for 10 AM.
But if i want to compare all the 5 minute intervals i am removing the timenum condition .
But not getting the desired output.. Can you please let me know what modification do i need to do?
``````daysback = 2;
SetBarsRequired((daysback+1)*inDaily/Max(1,Interval()));
//time_cond = tn == time;// equality check so time bar has to exists
today_price = ValueWhen(C,1);
n_days_back_price = ValueWhen(C, daysback+1);
``````
When you set all the bars required, then you are telling AB to use all bars so "leaving it up to the logic" will be discounted by AB.
Its either "you tell me how many bars" OR "i'll calculate it myself", but not both.
You are effectively disabling QuickAFL for that code, and all bars will be forced into computation.
Since I didn't design, cant commit, but Array processing works on the whole array, so you cant then have partial arrays being computed, and all Arrays are of size BarCount (excluding Matrices)
In other words, last half of Barcount in Array won't(cannot) be processed alone is what I mean.
If you have a chart, say updating every 1 second, the moment you change code and save the AFL, you will see Barcount using all the Bars (or a preety large number), and from the 2nd run onwards, QuickAFL(TM) kicking in and reducing the Bars drastically.
1 Like
You cannot remove parts of a code and then expect it to work like that.
When you ask for something, be sure its what you want, a small change in your thinking could change the whole approach.
If you want a bar by bar comparison, you might as well Ref() the whole array forward by a certain bars instead.
have a look at this:
Post you have withdrawn:
Listen, the code of 2nd post works exactly as you described in first post!
You do not believe it? (Besides in your quote there is not a single exploration code to be seen).
So just adding exploration output code without changing any previous code.
``````/// Do NOT remove link from code you did NOT create yourself!
/// If you do changes then add changes and add your name below and describe changes done
/// by fxshrat@gmail.com
daysback = 2;
time = 100000; // timenum -> so e.g. it means "10:00:00 AM"
SetBarsRequired((daysback+1)*inDaily/Max(1,Interval()));
tn = TimeNum();// array of time numbers
time_cond = tn == time;// equality check so time bar has to exists
today_price = ValueWhen(time_cond, C);// gets "today's" price of specified time
n_days_back_price = ValueWhen(time_cond, C, daysback+1);// gets price of specified time of a past day
/// Now printing result
printf("today's price: %g\nprice of %g days back: %g", today_price, daysback, n_days_back_price );
/// Now exploration output code
Filter = time_cond;
AddColumn( today_price, "Today's price", 1.2 );
AddColumn( n_days_back_price, StrFormat("Price of %g days back", daysback), 1.2 );
``````
And what do we get then?
Right, we get what is described in 1st post (once again).
Do you understand term "days back"? Do you understand meaning of term "2 days back"? Two days back from today is not yesterday.
I am not sure whether you are serious?
So what do you think does 100000 mean? Is it really such a one billion \$ brain twister?
No, it is not. It is a time number. A time number consists of max. 6 digits.
Since it is a time number then e.g. 100000 stands for -> "10:00:00 AM". Your mentioned time. Wow...amazing, isn't it. What a "Sixth Sense" revelation. Not!
We specify time number and then compare it with TimeNum() array via equality check ( `==` ) since we want to get values if true.
So the code is exactly doing what you described in first post.
Since it is exactly doing what you have described the issue is solved from my POV. It is even so much solved that it works on any intraday time frame. If something is solved we hit thanks button as token of appreciation here and to distinguish contributors from non contributors. And in addition there is solution forum button below of a post so that other users can easily recognize solved threads.
That change is incorrect! Please read formula reference of ValueWhen. It clearly and unmistakably says that
when the EXPRESSION was true on the n -th most recent occurrence
So first argument has to be true/false expression.... an array condition returning true(1) / false(0). If you add C then it is always returning true since price is greater than zero (false) (so output of your change will be "1" for today_price and "3" for n_days_back_price from start to end. So clearly nonsense output). time_cond of my posted code is a true/false array. It checks for equality ( via `==` operator).
You have to iterate to precisely shift entire array forward for each time number (precisely means it has to work for cases where number of intraday bars differ). And what you mean is not interval but times. And your first post mentions single time only! Do you understand? Now please do your homework.
Now another thing, if you copy code you did not create yourself then you leave everything in which includes double-clickable reference link to source but you do not remove (as you did). It is exactly not you who came up with idea. Absolutely not you. This is disrespectful and will put you quickly on ignore list. Do we understand each other? You may wail as much as you want now but having zero clue in the first place, being absolute beginner and begging for copy&paste code and not showing any own effort in first place and then removing on top of that... is no go!
2 Likes
Let's 1st learn how to encourage the ones who are HELPing the learners. Otherwise....... The Learners will remain learners because the Help/Guide will remain Silent.
Edit:
Please understand that when i disrespect someone who is selflessly Helping/Guiding me and he goes silent , it's not just my loss but a huge loss to all the genuine learners.
1 Like
First learn how to show some appreciation and respect! Maybe this forum is not for you!
Statements from both @TRDK and @fxshrat went into "inappropriate" zone. Closing this thread now. | 2,295 | 9,676 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-22 | latest | en | 0.902123 |
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## Section3.3Linear Maps and Matrices
### Subsection3.3.1Matrices as Maps
###### Investigation3.3.1.
Let
\begin{equation*} A= \left( \begin{array}{rr} 1 \amp 0 \\ -3 \amp 1 \end{array} \right),\ \vec{w}= \left( \begin{array}{rr} 5\\ 1 \end{array} \right),\ \mbox{ and } \vec{v}= \left( \begin{array}{rr} -1\\ 10 \end{array} \right).\ \end{equation*}
We can "quickly" verify that $A(2\vec{v}+\vec{w})=2(A\vec{v})+A\vec{w}\text{:}$
\begin{align*} \left( \begin{array}{rr} 1 \amp 0 \\ -3 \amp 1 \end{array} \right) \left( 2\left( \begin{array}{rr} -1\\ 10 \end{array} \right)+ \left( \begin{array}{rr} 5\\ 1 \end{array} \right) \right)\amp= \left( \begin{array}{rr} 1 \amp 0 \\ -3 \amp 1 \end{array} \right) \left( \begin{array}{rr} -2+5\\ 20+1 \end{array} \right)\\ \amp= \left( \begin{array}{rr} 1(-2+5)+0(20+1)\\ -3(-2+5)+1(20+1)\\ \end{array} \right)\\ \amp= \left( \begin{array}{rr} 1(-2)+0(20)\\ -3(-2)+1(20)\\ \end{array} \right)+ \left( \begin{array}{rr} 1(5)+0(1)\\ -3(5)+1(1)\\ \end{array} \right)\\ \amp= 2\left( \begin{array}{rr} 1(-1)+0(10)\\ -3(-1)+1(10)\\ \end{array} \right)+ \left( \begin{array}{rr} 1 \amp 0 \\ -3 \amp 1 \end{array} \right) \left( \begin{array}{rr} 5\\ 1\\ \end{array} \right)\\ \amp= 2\left( \left( \begin{array}{rr} 1 \amp 0 \\ -3 \amp 1 \end{array} \right) \left( \begin{array}{rr} -1\\ 10\\ \end{array} \right) \right)+ \left( \begin{array}{rr} 1 \amp 0 \\ -3 \amp 1 \end{array} \right) \left( \begin{array}{rr} 5\\ 1\\ \end{array} \right) \end{align*}
What is true, though perhaps not immediately obvious, is that this doesn't depend on the values of $A\text{,}$ $\vec{w}\text{,}$ or $\vec{v}\text{.}$
###### Investigation3.3.2.
The previous relationship works both ways. Define $T:\mathbb{R}^3\rightarrow\mathbb{R}^2$ by
\begin{equation*} T(\left\lt x_1,x_2,x_3\right\gt)=\left\lt x_1-2x_2, 2x_2-3x_3\right\gt. \end{equation*}
If we apply this to each element of the elementary basis $\mathcal{E}_3$ for $\mathbb{R}^3$ we get:
\begin{align*} T(\left\lt 1,0,0\right\gt)\amp=\left\lt 1, 0\right\gt,\\ T(\left\lt 0,1,0\right\gt)\amp=\left\lt -2, 2\right\gt,\ \mbox{ and }\\ T(\left\lt 0,0,1\right\gt)\amp=\left\lt 0, -3\right\gt. \end{align*}
But, this is the same as multiplying each of the vectors by the matrix
\begin{equation*} \left( \begin{array}{rrr} 1 \amp -2 \amp 0\\ 0 \amp 2 \amp -3 \end{array} \right) \end{equation*}
in which each column is the image of one of the elementary vectors.
### Subsection3.3.2Row, Column, and Null Spaces (again)
The row, column, and null spaces for the matrix
\begin{equation*} A=\left[ \begin{array}{rrr} 1 \amp 0 \amp -1\\ 0 \amp 1 \amp 0\\ 0 \amp 1 \amp 0\\ \end{array} \right] \end{equation*}
are
\begin{align*} Row\ A\amp=\left\{\left\lt 1,0,-1\right\gt, \left\lt 0,1,0\right\gt \right\}, \\ Col\ A\amp=\left\{\left\lt 1,0,0\right\gt, \left\lt 0,1,1\right\gt \right\}, and\\ Nul\ A\amp=\left\{\left\lt 1,0,1\right\gt\right\}. \end{align*}
We can see these visually in Figure 3.3.3 where you can see the plain of the row space transition to the plain of the column space while the vector of the null space transitions to zero.
If we multiply each of the standard basis elements by the matrix $A$ we get
\begin{gather*} \vec{e}_1=\left\lt 1,0,0 \right\gt \leadsto \left\lt 1,0,0 \right\gt\\ \vec{e}_2=\left\lt 0,1,0 \right\gt \leadsto \left\lt 0,1,1 \right\gt\\ \vec{e}_3=\left\lt 0,0,1 \right\gt \leadsto \left\lt -1,0,0 \right\gt \end{gather*}
which are the columns of the matrix. The domain of the transformation defined by $A$ is therefore $\mathbb{R}^3\text{,}$ Figure 3.3.4. Meanwhile, the image (or range) of the transformation is the column space of $A$ which is a copy of $\mathbb{R}^2\text{,}$ Figure 3.3.5. So the basis for $\mathbb{R}^3$ has been turned into a basis for a plain that is a copy of $\mathbb{R}^2\text{.}$ 1 This is the perspective that we will spend time on in Section 3.4.
We can also define a linear transformation $T_A$ algebraically by
\begin{equation*} \left( \begin{array}{c} x\\ y \\ z \end{array} \right) \stackrel{T_A}{\leadsto} \left( \begin{array}{c} x-z\\ y \\ y \end{array} \right) \end{equation*}
which has the same effect as multiplying the vector $\left\lt x,y,z\right\gt$ by $A\text{.}$
We have three ways to think of each transformation. Each transformation is a manipulation of a physical space, a change from one basis to another, or a series of algebraic instructions. All of these are reasonable ways to view a transformation and which perspective we use depends on what we wish to accomplish. | 1,805 | 4,572 | {"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} | 4.75 | 5 | CC-MAIN-2021-04 | latest | en | 0.364897 |
https://www.physicsforums.com/threads/valid-proof-of-cauchy-schwarz-inequality.684566/ | 1,518,980,173,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812247.50/warc/CC-MAIN-20180218173208-20180218193208-00324.warc.gz | 932,929,466 | 16,403 | Valid proof of Cauchy-Schwarz inequality?
1. Apr 10, 2013
phosgene
1. The problem statement, all variables and given/known data
I was discussing the proof for the Cauchy-Schwarz inequality used in our lectures, and another student suggested an easier way of doing it. It's really, really simple. But I haven't seen it anywhere online or in textbooks, so I'm wondering if it's either wrong or is only valid in certain scenarios.
The quick version is this:
$u \bullet v = ||u|| ||v|| cosθ$ where θ is between 0 and $\pi$.
The upper and lower bounds of |cosθ| are 1 and 0, respectively, so $|u \bullet v | \leq ||u|| ||v||$
2. Apr 10, 2013
Ray Vickson
You need to tell us what is the definition of inner product that you start with. If it is $u \cdot v = \sum_i u_i v_i,$ then you first need to prove that $u \cdot v = |u| |v| \cos \theta.$ Actually, it is sometimes done in the opposite way: from $u \cdot v = \sum_i u_i v_i,$, one proves the C_S inequality, and then notes that one can *define* $\theta$ from
$$\cos \theta =\frac{u \cdot v}{|u| |v|}.$$
3. Apr 12, 2013
phosgene
Thanks for the reply, so it seems that the proof in my original post would probably not be a good one to use in an exam? The course in question is multivariable and complex calculus. The proof that I've used myself relies on algebraically manipulating the square of the length of the difference of two vectors ( $||a \widehat{x} - b \widehat{y} ||^2 \geq 0$ ) using the properties of the dot product, rather than the cosine argument. It's the proof I've seen in the textbooks I have, but it's a bit longer to write out. | 459 | 1,609 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2018-09 | longest | en | 0.909938 |
http://users.ipfw.edu/CoffmanA/pov/lsoc.html | 1,501,108,191,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426639.7/warc/CC-MAIN-20170726222036-20170727002036-00562.warc.gz | 350,978,105 | 5,812 | Linear Systems of Conics
Conics.
A "conic curve" (or a "conic section," or just a "conic") in the two-dimensional plane is the solution set of an equation of the form:
Ax2 + By2 + Cxy + Dx + Ey + F = 0
for real constant coefficients A, B, C, D, E, F, which are not all zero. There are different types of conic curves in the real plane. The following chart gives a representative for each type of quadratic equation, and a name for the corresponding conics. For our purposes, they are divided into five cases:
Real Nondegenerate Imaginary Nondegenerate Ellipse x2+y2-1=0 Imaginary Ellipse x2+y2+1=0 Hyperbola x2-y2-1=0 Parabola x2-y=0
Real Degenerate Imaginary Degenerate Intersecting Pair of Lines x2-y2=0 Imaginary Pair of Lines, Intersecting at a Real Point x2+y2=0 Parallel Lines x2-1=0 Imaginary Parallel Lines x2+1=0 Pair of Lines, one at Infinity x=0
Square Degenerate Coincident Lines x2=0 Coincident Lines at Infinity 1=0
If you know the coefficients of a quadratic equation, you can find its type in the above chart, by using the table from a Geometry Center web page. We can abbreviate an equation as E1(x,y) = 0, and multiplying all the coefficients of E1 by the same nonzero constant, c, gives
(c . E1)(x,y) = cAx2 + cBy2 + cCxy + cDx + cEy + cF.
Then E1(x,y) = 0 if and only if (c . E1)(x,y) = 0, so E1 = 0 and c . E1 = 0 have the same solution set, and they define the same conic curve.
Linear systems.
A "linear system of conics" (also called a "pencil of conics") is the set of linear combinations of polynomials E1 and E2:
t1 . E1 + t2 . E2.
We assume E1 and E2 are linearly independent: one is not a constant multiple of the other. Plugging specific numbers into these coefficients, (t1,t2), gives a polynomial in (x,y), so each ordered pair (t1,t2) corresponds to a conic curve (t1 . E1 + t2 . E2)(x,y) = 0. Multiplying both t1 and t2 by the same nonzero constant c results in the same curve:
((c . t1) . E1 + (c . t2) . E2)(x,y) = c . (t1 . E1 + t2 . E2)(x,y) = 0.
So, a linear system is really just a one-parameter family of different conic curves, and the non-zero pairs (t1,t2) could be considered as homogeneous coordinates [t1 : t2] for a projective line. (See my Steiner surfaces page for some details on homogeneous coordinates.)
• Theorem: In a real linear system of conics, there is at least one conic which contains a real line (so that conic is either "real degenerate" or "square degenerate").
• Theorem: If there is one nondegenerate conic in a linear system, then there are at most three degenerate conics. However, there exist linear systems where all the conics are degenerate.
Intersections.
A point (x,y) is on both conics if E1(x,y) = 0 and E2(x,y) = 0. Any point on this intersection will also be a point on every conic in the linear system: for any t1, t2,
(t1 . E1 + t2 . E2)(x,y) = t1 . 0 + t2 . 0 = 0.
In fact, if any two conics in a linear system meet at a point, then that point is on every member of the linear system. Such points of intersection are called base points. If two conics in a linear system are disjoint, then every pair of conics in the linear system will also be disjoint, and there are no base points.
• Theorem: If a real linear system of conics has finitely many base points, then there can be 0, 1, 2, 3, or 4 of them, but no more than four. However, there exist linear systems with infinitely many base points.
• Theorem: Every point in the plane which is not a base point lies on exactly one conic in the linear system.
The Classification
The quadratic equation of a conic could also be written as a homogeneous expression in three variables, Ax2 + By2 + Cxy + Dxz + Eyz + Fz2 = 0, which gives the original x,y equation when z = 1. This corresponds to the geometric definition of a conic section as the intersection of a cone (defined by the homogeneous equation in xyz-space) with a plane (defined by z = 1).
The following examples list all the types of linear systems of conics, up to "real projective equivalence" (real linear transformations of the homogeneous coordinates [x : y : z]). It turns out there are only finitely many types of linear systems; roughly speaking, they can be categorized by the number of base points and degenerate conics, but there are the usual issues of "multiplicity," and whether the base points have real coordinates.
The numbering system for types I, Ia, ..., V follows H. Levy's book. The Roman numerals I, ..., VIII represent eight complex projective equivalence classes, some of which are further subdivided into real projective equivalence classes, for a total of 13 types.
Click on the picture to see an animation.
Type I t1 . (x2-y2) + t2 . (x2-1). These conics meet at four real base points. There are three degenerate conics: [t1:t2]=[1:0] The intersecting lines x2-y2=0, [t1:t2]=[0:1] The vertical parallel lines x2-1=0, [t1:t2]=[-1:1] The horizontal parallel lines y2-1=0. To see how I rendered the conic sections in the picture to the left, Click Here to see a "side view," showing the cones in xyz-space, whose intersection with the plane z=1 (the top of the box) is shown in the two-dimensional figure.
Type Ia t1 . (x2+y2+1) + t2 . (x). The type Ia linear system has no real base points. In this representative, most of the conics you see in the picture are circles, but there are also "imaginary ellipses" with no visible points: for some values of [t1:t2], the polynomial equation (t1 . E1 + t2 . E2)(x,y) = 0 has no real solutions. There are three degenerate conics: [t1:t2]=[0:1] The vertical line x=0, [t1:t2]=[1:2] The point (-1,0) is the only real solution of x2+2x+1+y2=0, [t1:t2]=[1:-2] The point (1,0) is the only real solution of x2-2x+1+y2=0. In complex homogeneous coordinates, the linear system t1 . (x2+y2+z2) + t2 . (xz) = 0 has four base points: [0:i:1], [0:-i:1], [1:i:0], [1:-i:0].
Type Ib t1 . (x2+y2-1) + t2 . (x). The type Ib linear system has two real base points. Since all the conics in the system must meet both real base points, there are no "imaginary ellipses." There is one degenerate conic: [t1:t2]=[0:1] The vertical line x=0. In the complex linear system, where x, y, t1, and t2 are allowed to be complex, there are two more degenerate conics, at [t1:t2]=[1:2i] and [1:-2i]. In complex homogeneous coordinates, there are two more base points, [1:i:0], [1:-i:0].
Type II t1 . y(x+y-1) + t2 . x(x+2y-1). The type II linear system has three real base points, and at one of these points, all the nondegenerate conics have a common tangent line. There are two degenerate conics: [t1:t2]=[1:0] The pair of lines y(x+y-1)=0. [t1:t2]=[0:1] The pair of lines x(x+2y-1)=0.
Type IIa t1 . (y2+(x-y-1)2) + t2 . x(x-y-1). The type IIa linear system has one real base point, at which all the nondegenerate conics have a common tangent line. There are two degenerate conics: [t1:t2]=[1:0] The pair of imaginary lines y2+(x-y-1)2=0, meeting at the real base point (1,0). [t1:t2]=[0:1] The pair of lines x(x-y-1)=0. In complex coordinates, there are two more base points, (0,(-1+i)/2), (0,(-1-i)/2).
Type III t1 . (x2-1/4) + t2 . y2. The type III linear system has two real base points. At each of the two base points, all the nondegenerate conics have a common tangent line. There are two degenerate conics: [t1:t2]=[1:0] The vertical parallel lines x2-1/4=0, [t1:t2]=[0:1] The horizontal coincident lines y2=0.
Type IIIa t1 . (x2+y2) + t2 . (y+1)2. The type IIIa linear system has no real base points. Like Type Ia, there are some imaginary ellipses. There are two degenerate conics: [t1:t2]=[1:0] The point (0,0) is the only real solution of x2+y2=0, [t1:t2]=[0:1] The horizontal coincident lines (y+1)2=0. In complex coordinates, there are two base points, (i,-1), (-i,-1).
Type IV t1 . (x2+x+y2) + t2 . (xy). The type IV linear system has two real base points. At one of the base points, any two conics meet each other transversely, and at the other, they meet at an intersection with multiplicity 3. There is only one degenerate conic: [t1:t2]=[0:1] The pair of lines xy=0.
Type V t1 . (x+y2) + t2 . (x2). The type V linear system has one real base point, at which any two conics meet each other at an intersection with multiplicity 4. There is only one degenerate conic: [t1:t2]=[0:1] The vertical coincident lines x2=0.
Type VI t1 . (x2-y2) + t2 . (2xy). The type VI linear system has only real degenerate conics, which are pairs of lines that meet at the one real base point. In the complex linear system, where x, y, t1, and t2 are allowed to be complex, there are two square degenerate conics, at [t1:t2]=[1:i] and [1:-i].
Type VIa t1 . (x2+y2) + t2 . (2xy). The type VIa linear system has real and imaginary degenerate conics, which are pairs of lines that meet at the one real base point. There are also two square degenerate conics, at [t1:t2]=[1:1] and [1:-1]
Type VII t1 . (x2-y2) + t2 . (x+y)(y-1). The type VII linear system has only real degenerate conics, which are pairs of lines, one of which is a line of base points. There is also one real base point not on that line. There are no square degenerate conics.
Type VIII t1 . (x2) + t2 . (xy). The type VIII linear system has base points which form exactly one line. There are real degenerate conics, and one square degenerate conic at [t1:t2]=[1:0]. This case is related to the previous one, by moving the isolated base point so that it is on the line of base points. | 2,825 | 9,396 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | longest | en | 0.881166 |
https://www.studypool.com/discuss/495355/-79?free | 1,508,812,787,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187827853.86/warc/CC-MAIN-20171024014937-20171024034937-00091.warc.gz | 1,015,185,716 | 13,853 | -----------------------------------------------------------------
label Algebra
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
Student tickets for the football game cost \$12 each. An adult ticket costs \$20. \$1720 was collected for the 120 tickets sold at the last game. Which system of equations can be used to solve for the number of each kind of ticket sold?
Apr 25th, 2015
assume there are x student tickets, y adult tickets
thus, we can set up equations:
12x+20y=1720
x+y=120
thus you can solve for x and y
Apr 25th, 2015
...
Apr 25th, 2015
...
Apr 25th, 2015
Oct 24th, 2017
check_circle | 166 | 625 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2017-43 | latest | en | 0.872959 |
http://edx.readthedocs.io/projects/open-edx-building-and-running-a-course/en/open-release-ginkgo.master/exercises_tools/circuit_schematic_builder.html | 1,508,758,739,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825900.44/warc/CC-MAIN-20171023111450-20171023131450-00391.warc.gz | 105,713,293 | 9,456 | # 9.7. Circuit Schematic Builder Problem¶
Note
EdX does not support this problem type.
In circuit schematic builder problems, students can arrange circuit elements such as voltage sources, capacitors, resistors, and MOSFETs on an interactive grid. They then submit a DC, AC, or transient analysis of their circuit to the system for grading.
## 9.7.1. Create a Circuit Schematic Builder Problem¶
1. In the unit where you want to create the problem, under Add New Component select Problem, and then select Advanced.
2. Select Circuit Schematic Builder.
3. In the component that appears, select Edit.
4. In the component editor, replace the example code with your own code.
5. Select Save.
Problem Code
The illustration above shows a condensed version of an actual problem from MITx’s 6.002.1x. To create the entire problem, paste the following code into the advanced editor.
<problem>
<p>A circuit that combines two or more signals is called a <i>mixer</i>. In
this lab, your goal is to build a mixer that combines the signals generated
by two voltage sources, V1 and V2, where:</p>
<ul style="margin-left:2em;">
<li>
<p>V1 is a 1 kHz square wave that varies between 0V and +1V, and</p>
</li>
<li>
<p>V2 is a 5 kHz sine wave that varies between -1V and +1V.</p>
</li>
</ul>
<p>Please design a circuit that mixes V1 and V2 to produce Vout such that</p>
<center>$V_\mathrm{out} \approx \frac{1}{2}V_1 + \frac{1}{6}V_2.$</center>
<p>The resulting output should be similar to that shown in Figure 1. The
maximum value of the output is approximately $$667mV$$ and the minimum value
is approximately $$-167mV$$.</p>
<center><img src="/static/images/circuits/Lab2B_1.png"/><br/>Figure 1. Desired output waveform</center>
<p>Hint: Figure 2 shows a simple resistive mixer for combining two signals.</p>
<center><img src="/static/images/circuits/Lab2B_2.png"/><br/>Figure 2. Simple resistive mixer</center>
<p>Enter your circuit below, using the appropriate configuration of
resistors. Please do not modify the wiring or parameters of the voltage
sources -- your goal is to take the signals they generate and combine them,
not to change what is generated. Run a 5ms transient analysis to verify the
correct operation of your circuit. We will be checking for the transient
waveform at the "output" node.</p>
<schematicresponse>
<center>
<schematic height="500" width="650" parts="g,r,s" analyses="dc,tran" submit_analyses="{"tran":[["output",0.00025,0.00035,0.00065,0.00075]]}" initial_value="[["v",[56,48,0],{"name":"V1","value":"square(0,1,1k)","_json_":0},["2","0"]],["g",[56,96,0],{"_json_":1},["0"]],["v",[56,128,0],{"name":"V2","value":"sin(0,1,5k,0,0)","_json_":2},["1","0"]],["g",[56,176,0],{"_json_":3},["0"]],["w",[56,48,88,48]],["w",[56,128,88,128]],["L",[224,48,3],{"label":"output","_json_":6},["output"]],["w",[224,48,200,48]],["w",[224,48,224,128]],["w",[224,128,200,128]],["s",[224,48,0],{"color":"magenta","_json_":10},["output"]],["view",0,0,2,"5","10","10MEG",null,"100","5ms"]]"/>
</center>
# for a schematic response, submission[i] is the json representation
# of the diagram and analysis results for the i-th schematic tag
def get_tran(json,signal):
for element in json:
if element[0] == 'transient':
return element[1].get(signal,[])
return []
output = get_tran(submission[0],'output')
[0.00035, 0.333],
[0.00065, 0.166],
[0.00075, -0.166]]
okay = True
if not output or output[0][1] == 'undefined': # No transient or output node floating
okay = False
else:
for (t,v) in output:
if at==t and abs(av - v) < 0.05*abs(av):
# found a good match for this answer, on to the next one
break
else:
print 'check',at,av
# no submission matched answer, complain
okay = False;
break;
correct = ['correct' if okay else 'incorrect']
</schematicresponse>
<p>When you're done or if you wish to save your work, please click CHECK.
The checker will be verifying the voltage of the output node at several
different times, so you'll earn a point only <i>after</i> you've performed
the transient simulation so that the checker will have a waveform to check!</p>
<solution>
<div class="detailed-solution"><p>Explanation:</p>
<p>The goal is to design a mixer circuit with characteristics of
$$V_{out}=\frac{1}{2}\cdot V_1+\frac{1}{6}\cdot V_2$$
You might have started to design your mixer with two resistors only as the example suggests.
But working through the math, soon you'll realize that the equations return no non-zero value for the resistor components.
Thus you have to change the design. The next simplest design will be to add a resistor $$R_3$$ that connects the node Vout to ground.
See the schematic below:</p>
<img src="/static/images/circuits/lab2fmt.png"/>
<p>Since we are going to use only linear elements in this circuit
(resistors are linear), superposition will hold
and thus one can look at the effect of each source $$V_1$$ and $$V_2$$
one at the time:</p>
[mathjax] V_{out1} = V_1 \cdot \frac{\left(R_2 \parallel R_3\right)}
{\left(R_2 \parallel R_3+R_1\right)}\\ V_{out2} = V_2 \cdot
\frac{\left(R_1 \parallel R_3\right)}{\left(R_1 \parallel R_3+R_2\right)}
\\ V_{out} = V_{out 1} + V_{out 2} \\ V_{out} = V_1 \cdot
\frac{\left(R_2 \parallel R_3\right)}{\left(R_2\parallel R_3+R_1\right)} +
V_2 \cdot \frac{\left(R_1 \parallel R_3\right)}{\left(R_1 \parallel R_3 +
R_2\right)} = \frac{1}{2} \cdot V_1+\frac{1}{6} \cdot V_2 [/mathjax]
<p>Therefore:</p>
[mathjax] \frac{ \left(R_2 \parallel R_3 \right) }
{ \left( R_2 \parallel R_3 + R_1 \right)} =
\frac{1}{2} \\\frac{\left( R_1 \parallel R_3 \right) }
{ \left(R_1 \parallel R_3 + R_2 \right)} = \frac{1}{6} [/mathjax]
<p>So we have to solve for the resistors given these two equations. You
might notice that we have 2 equations and 3 unknowns, and that there is
therefore not a unique solution. That is okay, though. We only have to
worry if there is no solution, not if there are too many solutions. We will
simply find one of the many possible correct answers by arbitrarily
choosing a value for one of the variables later.</p>
<p>The first equation simplifies to $$R_1 = R_2\parallel R_3$$ and the
second simplifies to $$R_2 = 5 \cdot R_1\parallel R_3$$
Expanding the notation gives: </p>
[mathjax]\frac{1}{R_1}=\frac{1}{R_2}+\frac{1}{R_3}
\tag{*} \\\frac{1}{R_1}+\frac{1}{R_3}=\frac{5}{R_2} [/mathjax]
<p>Subtracting these two equations will yield $$R_2 = 2 \cdot R_3$$
And putting this back to the starred equation , will result in
$$R_1 = \frac{2}{3} \cdot R_3$$
So now we have $$R_2$$ and $$R_1$$ in terms of $$R_3$$ with the following
ratios:</p>
[mathjax]R_2 = 2 \cdot R_3 \\ R_1 = \frac{2}{3} \cdot R_3 \\[/mathjax]
use a simple value of $$R_3= 3Ω$$ and find the rest accordingly:</p>
<p>With these resistor values, doing a transient analysis shows a result which meets the required specs of $$V_{out}$$.</p> | 2,333 | 7,249 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2017-43 | longest | en | 0.758014 |
https://matholympiad.org.bd/forum/viewtopic.php?f=8&t=5423&p=20530&sid=e963d435a5173f3b356ae3bc6d4f5c43 | 1,621,008,143,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991428.43/warc/CC-MAIN-20210514152803-20210514182803-00411.warc.gz | 409,395,647 | 6,614 | ## BdMO regional 2018 set 4 Secondary P 06
Forum rules
Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
samiul_samin
Posts: 1007
Joined: Sat Dec 09, 2017 1:32 pm
### BdMO regional 2018 set 4 Secondary P 06
\$ABCD\$ is a quadrilateral, where \$BC=CD=2\$ and \$\angle DCA=\angle DBA\$,
\$\angle
BAC=60^{\circ}\$ . What is the length of \$BD\$?
Tasnood
Posts: 73
Joined: Tue Jan 06, 2015 1:46 pm
### Re: BdMO regional 2018 set 4 Secondary P 06
\$\angle DCA=\angle DBA\$
\$\angle BAC=\angle BDC=60^{\circ}\$
So, \$\triangle BDC\$ is isosceles and \$BD=2\$
samiul_samin
Posts: 1007
Joined: Sat Dec 09, 2017 1:32 pm
### Re: BdMO regional 2018 set 4 Secondary P 06
Tasnood wrote:
Thu Jan 17, 2019 5:32 pm
\$\angle DCA=\angle DBA\$ | 346 | 1,009 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.742916 |
https://smart-answers.com/mathematics/question20785769 | 1,660,041,277,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570921.9/warc/CC-MAIN-20220809094531-20220809124531-00318.warc.gz | 519,334,475 | 35,616 | , 21.01.2021 02:50, mendezmarco2004
# Are the triangles congruent? If so, state the theorem that can be used to prove that they are, and what additional information would need to be shown in the proof.
### Other questions on the subject: Mathematics
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The width of a rectangle is half as long as the length. the rectangle has an area of 18 square feet. what are the length and width of the rectangle?
Mathematics, 22.06.2019 00:30, gthif13211
I've been working on this for a few days and i just don't understand, it's due in a few hours. you. the direction of a vector is defined as the angle of the vector in relation to a horizontal line. as a standard, this angle is measured counterclockwise from the positive x-axis. the direction or angle of v in the diagram is α. part a: how can you use trigonometric ratios to calculate the direction α of a general vector v = < x, y> similar to the diagram? part b suppose that vector v lies in quadrant ii, quadrant iii, or quadrant iv. how can you use trigonometric ratios to calculate the direction (i. e., angle) of the vector in each of these quadrants with respect to the positive x-axis? the angle between the vector and the positive x-axis will be greater than 90 degrees in each case. part c now try a numerical problem. what is the direction of the vector w = < -1, 6 > ? | 429 | 1,673 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.93014 |
https://discourse.julialang.org/t/values-from-dictionary/32996 | 1,660,645,495,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572286.44/warc/CC-MAIN-20220816090541-20220816120541-00184.warc.gz | 214,678,439 | 4,855 | # Values from Dictionary
hi,
I have the following dictionary called “states”:
``````states=Dict{Any,Any}((0, 0) => Array{Any,1}[[[0], (0, 0), Any[0, 0]], [[1], (1, 0), Any[1, 1]]],(1, 0) => Array{Any,1}[[[0], (0, 1), Any[1, 0]], [[1], (1, 1), Any[0, 1]]],(0, 1) => Array{Any,1}[[[0], (0, 0), Any[1, 1]], [[1], (1, 0), Any[0, 0]]],(1, 1) => Array{Any,1}[[[0], (0, 1), Any[0, 1]], [[1], (1, 1), Any[1, 0]]])
``````
and I want to take the values of states and define them as “val_0” and “val_1” so that I get the following output:
``````val_0=[[0], (0, 0), Any[0, 0]]
val_1=[[1], (1, 0), Any[1, 1]]
``````
(so val_0 and val_1 are from (0, 0) => …).
(In the next step val_0 shall be:
`val_0=[[0], (0, 1), Any[1, 0]] `
`val_1= [[1], (1, 1), Any[0, 1]]`
from key (1, 0) and so on).
Hey @ar12y
Is this what you want?
``````for v in values(states)
val_0, val_1 = v
# do something
end
``````
2 Likes
@ar12y, I think you want:
``````for (val_0, val_1) in values(states)
println(val_0)
println(val_1)
println()
end
``````
Hope this helps.
1 Like
this is exactly the solution I was looking for. | 487 | 1,096 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.679131 |
http://functions.wolfram.com/GammaBetaErf/Erf/26/02/01/0004/ | 1,519,522,971,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816083.98/warc/CC-MAIN-20180225011315-20180225031315-00622.warc.gz | 138,058,015 | 7,816 | html, body, form { margin: 0; padding: 0; width: 100%; } #calculate { position: relative; width: 177px; height: 110px; background: transparent url(/images/alphabox/embed_functions_inside.gif) no-repeat scroll 0 0; } #i { position: relative; left: 18px; top: 44px; width: 133px; border: 0 none; outline: 0; font-size: 11px; } #eq { width: 9px; height: 10px; background: transparent; position: absolute; top: 47px; right: 18px; cursor: pointer; }
Erf
http://functions.wolfram.com/06.25.26.0006.01
Input Form
Erf[Sqrt[z]] == (1/Sqrt[Pi]) MeijerG[{{1}, {}}, {{1/2}, {0}}, z]
Standard Form
Cell[BoxData[RowBox[List[RowBox[List["Erf", "[", SqrtBox["z"], "]"]], "\[Equal]", RowBox[List[FractionBox[RowBox[List["1", " "]], SqrtBox["\[Pi]"]], RowBox[List["MeijerG", "[", RowBox[List[RowBox[List["{", RowBox[List[RowBox[List["{", "1", "}"]], ",", RowBox[List["{", "}"]]]], "}"]], ",", RowBox[List["{", RowBox[List[RowBox[List["{", FractionBox["1", "2"], "}"]], ",", RowBox[List["{", "0", "}"]]]], "}"]], ",", "z"]], "]"]]]]]]]]
MathML Form
erf ( z ) 1 π G 1 , 2 1 , 1 ( z 1 1 2 , 0 ) TagBox[RowBox[List[SubsuperscriptBox[TagBox["G", MeijerG], RowBox[List["1", ",", "2"]], RowBox[List["1", ",", "1"]]], "\[InvisibleApplication]", RowBox[List["(", RowBox[List[TagBox["z", MeijerG, Rule[Editable, True]], "\[VerticalSeparator]", GridBox[List[List[TagBox["1", MeijerG, Rule[Editable, True]]], List[RowBox[List[TagBox[FractionBox["1", "2"], MeijerG, Rule[Editable, True]], ",", TagBox["0", MeijerG, Rule[Editable, True]]]]]]]]], ")"]]]], MeijerG, Rule[Editable, False]] Erf z 1 2 1 1 2 -1 MeijerG 1 1 2 0 z [/itex]
Rule Form
Cell[BoxData[RowBox[List[RowBox[List["HoldPattern", "[", RowBox[List["Erf", "[", SqrtBox["z_"], "]"]], "]"]], "\[RuleDelayed]", FractionBox[RowBox[List["MeijerG", "[", RowBox[List[RowBox[List["{", RowBox[List[RowBox[List["{", "1", "}"]], ",", RowBox[List["{", "}"]]]], "}"]], ",", RowBox[List["{", RowBox[List[RowBox[List["{", FractionBox["1", "2"], "}"]], ",", RowBox[List["{", "0", "}"]]]], "}"]], ",", "z"]], "]"]], SqrtBox["\[Pi]"]]]]]]
Date Added to functions.wolfram.com (modification date)
2001-10-29 | 770 | 2,141 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-09 | latest | en | 0.22926 |
http://www.markedbyteachers.com/gcse/maths/broadsheets-use-longer-words-than-tabloids.html | 1,545,195,699,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376831334.97/warc/CC-MAIN-20181219045716-20181219071716-00371.warc.gz | 414,762,240 | 24,414 | • Join over 1.2 million students every month
• Accelerate your learning by 29%
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• Level: GCSE
• Subject: Maths
• Word count: 2113
# Broadsheets use longer words than tabloids.
Extracts from this document...
Introduction
Katie Sanderson 11L1
Maths Statistics Course Work
Introduction
There are many different newspapers; they range from tabloid papers to the broadsheet papers.
Hypothesis One
Broadsheets use longer words than tabloids
Prediction
I predict that broadsheets will tend to use longer words than a tabloid paper because the tabloids are a lighter read to the more involving descriptive broadsheet papers. Different newspapers are written to suit these preferences.
In the tabloid papers the wording used is less profound and therefore is more easily understood.
Method
The data will be collected in samples of 100 words per article. One tabloid and one broadsheet paper will be chosen with similar stories to make this a fair test. The first and last 50 words of each article will be counted and recorded. The reason for using a 100-word sample is because this should give a good overall view to the word length in a section. The samples are political samples based on the Iraq War. The samples are taken from the papers the Daily Telegraph (broadsheet) and the Mirror (tabloid) I am going to work out the Mean Median Mode and the range for each paper so that the data is easier to compare.
Results
Number of letters per word Tally Total Frequency Frequency x No of letters per word 1 3 3 2 13 26 3 17 51 4 13 52 5 21 105 6 7 42 7 8 56 8 5 40 9 5 45 10 3 30 11 4 44 12 1 12 Total 100 506
Mirror 6th January (Tabloid)
Number of letters per word Tally Total Frequency Frequency x No of letters per word 1 3 3 2 11 22 3 19 57 4 20 86 5 19 95 6 9 54 7 11 77 8 3 24 9 1 9 10 3 30 11 1 11 Total 100 462
Daily Telegraph
Mean: 506/100 = 5.06
Median: 5
Mode: 5
Range: 11
## Mirror
Mean: 462/100 = 4.62
Median: 4
Mode: 4
Range: 11
Looking at my tables and results above these show that the broadsheet paper has a larger range of longer words compared with the tabloid. The broadsheets most common word length is 5 compared that to that of the tabloid which is 4.
I aim going to show my data in a bar chart as it shows it more clearly and makes the data easier to see and read it also shows it in a more interesting way.
Looking at my bar chart (see graph one) I can clearly see that the broadsheet paper has a larger spread of longer words compared to that of the tabloid, which has most of its data down at the lower end of the scale. I decided to show the trends of my data by using a frequency polygon (graph two)
Looking at graph two you can see that the trend of data for each paper is almost identical which shows that there is not much difference between each paper. The graph does show however that the broadsheet does contain more longer words than the tabloid.
I have come to the question are the words used in a political article longer than those used in a sport article so I have decided to extend my original investigation.
### Extended Hypothesis
I expect that the word length in a political article to be longer than that of a sport article.
I repeated the process that I had done for my political article and my results were:
#### Daily Telegraph (broadsheet Birmingham Vs Arsenal
Number of letters per word Total Frequency Frequency x No of letters per word 1 0 0 2 6 12 3 22 66 4 22 88 5 13 65 6 10 60 7 14 98 8 2 16 9 6 54 10 2 20 11 1 11 12 1 12 13 1 13 Total 100 515
Daily Mail (Tabloid Birmingham Vs Arsenal)
Number of letters per word Total Frequency Frequency x No of letters per word 1 1 1 2 17 34 3 22 66 4 23 27 5 11 55 6 9 54 7 6 42 8 7 56 9 1 9 10 2 20 11 0 0 12 0 0 13 1 13 Total 100 377
Middle
To compare all the data I have decided to show it in the form of cumulative frequency charts. The cumulative frequency was worked out for all 4 articles used and here are the results.
Cumulative Frequency for Daily Telegraph article on the Iraq War (Broadsheet Political)
Word Length Frequency • Length Cumulative Frequency 1 0 1 3 2 3 2 13 3 16 3 17 4 33 4 13 5 46 5 21 6 67 6 7 7 74 7 8 8 82 8 5 9 87 9 5 10 92 10 3 11 95 11 4 12 99 12 1 13 100
Cumulative Frequency for Mirror Article on the Iraq War (Tabloid Political)
Word Length Frequency • Length Cumulative Frequency 1 0 1 3 2 3 2 11 3 14 3 19 4 33 4 20 5 53 5 19 6 72 6 9 7 81 7 11 8 92 8 3 9 95 9 1 10 96 10 3 11 99 11 1 12 100
Cumulative Frequency for Daily Telegraph on the match Birmingham Vs Arsenal (Broadsheet Sport)
Word Length Frequency • Length Cumulative Frequency 1 0 1 0 2 0 2 6 3 6 3 22 4 28 4 22 5 50 5 13 6 63 6 10 7 73 7 14 8 87 8 2 9 89 9 6 10 95 10 2 11 97 11 1 12 98 12 1 13 99 13 1 14 100
Conclusion
Count the number of polysyllabic words in the chain of 30 sentences and look up the approximate grade level on the chart.
If Article contains less than 30 sentences:
• Count all of the polysyllabic words in the article
• Count the number of sentences
• Find the average number of polysyllabic words per sentence as follows:
• Multiply that average by the number of sentences short of 30
• Add that figure to total number of polysyllabic words and look up the approximate grade level on the chart.
Smog Conversion Table
Total Polysyllabic word counts Approximate Grade Level(+/-1.5 grades) 0-2 4 3-6 5 7-12 6 13-20 7 21-30 8 31-42 9 43-56 10 57-72 11 73-90 12 91-110 Collage level 111-132 Collage level 133-156 Collage level 157-182 Collage level 183-210 Beyond Collage 211-240 Beyond Collage
My Results were as follows
Had 31 sentences and 50 polysyllabic words which gives a reading level of 10
Tabloid (Political) the Iraq War
Had 15 sentences and 37 polysyllabic words
Had 30 sentences and 64 polysyllabic words and so gives a reading level of 11
Tabloid (Sports) Birmingham Vs Arsenal
Had 32 sentences and 68 polysyllabic words which gives a reading level of 11
Looking at my results I can see that this proves my hypothesis wrong because the tabloid for the sports had a higher reading level than that for the political but this could have been due to the long words such as Birmingham that are within the sports articles.
This student written piece of work is one of many that can be found in our GCSE Comparing length of words in newspapers section.
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# Related GCSE Comparing length of words in newspapers essays
1. ## &quot;Broadsheet newspapers have a longer average word length than tabloid newspapers&quot;
35% of the total word sample would be words from the news pages. This method ensures that the sample is representative of the whole newspaper rather than randomly selected samples. Investigation 2 Hypothesis 2 "Tabloid newspapers have a higher reading age than broadsheet newspapers."
2. ## Assesment of Reading Difficulties in Patient AM Following the Development of Vascular Dementia.
AM was required to read these words aloud. In a non-word reading task AM was presented a list of non-words which again had to be read aloud. Image ability and frequency spelling was the next test given. The first in a series of tests looking at spelling.
1. ## Are High Imagery Words Easier To Retrieve From The Short Term Memory Than Abstract ...
You don't have to take part and you can stop at any time you like" De-brief: The participant was thanked and told their score. It was carefully explained that their results were normal and to be expected. It was explained that if they wished to withdraw their result they might do so at any time.
2. ## Maths Statistics Coursework on the Readability of a Tabloid Newspaper Compared to a Broadsheet
7 to 12 letter words than the Tabloid article. Also again similar to that of the first article the means might be close but again the Broadsheet's is higher meaning on average that the Broadsheet had larger words, which is evident from the graphs.
1. ## For my Coursework I will use the following newspapers: ...
1 8 8 2 2 10 3 1 11 I will again calculate all the numbers. Paragraph length Paragraph length The Sun The times Daily Mail Mean 7.00 1.60 1.36 Mode 2 1 1 Median 1.5 5 2 Range 15 1 2 But: The paragraph length does not really help
2. ## A recognised 'High Quality' Magazine will have longer words than a recognised 'Low Quality' ...
We can also state that the highest frequency value of LPW occurs for words with 2, 3, and 5 letters, with a frequency of 17 occurrences. The most commonly used diagrams are the various forms of bar charts. 5.2 Bar Chart Analysis The bar chart shown above allows us to
1. ## THE STROOP EFFECT: FURTHER TESTS OF THE ATTENTION-CAPTURE HYPOTHESIS
Like Kahneman and Henik (1981) and Kahneman and Chajcek (1983), they conclude that attention is involved at an early stage in processing and that 'the first word processed captures attention, thereby screening any others out'. The present study constitutes a further investigation of this attention-capture hypothesis.
2. ## The hypotheses are: 1. Broadsheet newspapers have longer words ...
In hypothesis 2, I will do many of the things I did in hypothesis 1 and I'll do them for many of the same reasons. I will choose to use 100 words to make analysis simpler and I also choose an article that had the same subject to make it fair.
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• Ideas and feedback to | 2,755 | 9,805 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2018-51 | longest | en | 0.843885 |
https://www.physicsforums.com/threads/what-is-the-frequency-of-the-harmonic-potential.791494/ | 1,695,417,119,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506423.70/warc/CC-MAIN-20230922202444-20230922232444-00618.warc.gz | 1,043,407,165 | 15,731 | # What is the frequency of the harmonic potential?
• KFC
#### KFC
Hi there,
I am reading an introduction on trapping atoms in space with magnetic potential. The article said the lab usually use a harmonic potential to trap the atoms and the potentials is in the form
##\dfrac{m}{2}(\omega_x^2x^2 + \omega_y^2y^2 + \omega_z^2z^2)##
and ##\omega_{x,y,z}## has the unit of frequency. I wonder how do you understand the frequency in the potential from physical point of view. Why there is frequency?
Let's see what force that potential gives. For the x component, we have $F_x=-\frac{\partial V}{\partial x}=-m \omega_x^2 x$. But hey, that's Hooke's law!(With the spring constant $k_x=m \omega_x^2$.) So this potential is actually a non-isotropic harmonic oscillator potential and this is the reason you have frequencies in it.
Let's see what force that potential gives. For the x component, we have $F_x=-\frac{\partial V}{\partial x}=-m \omega_x^2 x$. But hey, that's hooks law!(With the spring constant $k_x=m \omega_x^2$.) So this potential is actually a non-isotropic harmonic oscillator and this is the reason you have frequencies in it.
Thanks. So if can I say under that potential, it just like 3 harmonic oscillators along x, y and z each is oscillating at the frequency ##\omega_x##, ##\omega_y## and ##\omega_z## independently?
Thanks. So if can I say under that potential, it just like 3 harmonic oscillators along x, y and z each is oscillating at the frequency ##\omega_x##, ##\omega_y## and ##\omega_z## independently?
No, its just one harmonic oscillator having three independent oscillations with different frequencies in different dimensions. But yes, in terms of degrees of freedom, its no different than having three independent harmonic oscillators with different frequencies. But you should note this resemblance may not be usable in the context you're considering.
No, its just one harmonic oscillator having three independent oscillations with different frequencies in different dimensions. But yes, in terms of degrees of freedom, its no different than having three independent harmonic oscillators with different frequencies. But you should note this resemblance may not be usable in the context you're considering.
Got it.
One more question. Usually, if you solve the harmonic oscillator with 3 oscillating frequencies along 3 different frequency, we will get a solution in 3-dimensional also. But if the frequency along x and y are way larger than the ##\omega_z##. In some articles, I saw that people simply approximate the solution along the z direction only. I stuck on the explaining this approximation in physics.
The first thing come to my mind is if the oscillator oscillating along x and y much faster than z, can we consider the system may see the average motion along x and y instead because of high frequency? So we could consider the amplitude of the solution along x and y just like a constant? Only the z direction depends on time?
But before I find the explanation, I am also thing that if ##\omega_z## is way smaller than the
##\omega_{x,t}##, can we consider the profile on the z direction is changing slowly in time, so we could consider the solution in z direction is a constant, the effective solution is along x and y direction.
I know those two statements are contradictory. But I cannot tell which one (or all) is wrong. and why?
Got it.
One more question. Usually, if you solve the harmonic oscillator with 3 oscillating frequencies along 3 different frequency, we will get a solution in 3-dimensional also. But if the frequency along x and y are way larger than the ##\omega_z##. In some articles, I saw that people simply approximate the solution along the z direction only. I stuck on the explaining this approximation in physics.
The first thing come to my mind is if the oscillator oscillating along x and y much faster than z, can we consider the system may see the average motion along x and y instead because of high frequency? So we could consider the amplitude of the solution along x and y just like a constant? Only the z direction depends on time?
But before I find the explanation, I am also thing that if ##\omega_z## is way smaller than the
##\omega_{x,t}##, can we consider the profile on the z direction is changing slowly in time, so we could consider the solution in z direction is a constant, the effective solution is along x and y direction.
I know those two statements are contradictory. But I cannot tell which one (or all) is wrong. and why?
Can you point me to one of those "articles"? | 1,033 | 4,580 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2023-40 | latest | en | 0.898646 |
https://wiki.zcubes.com/index.php?title=Manuals/calci/NE&oldid=208258 | 1,652,736,382,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662512249.16/warc/CC-MAIN-20220516204516-20220516234516-00278.warc.gz | 679,850,714 | 6,320 | # Manuals/calci/NE
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
NE(A,B)
• and are any real numbers.
## Descreiption
• This function checks if given two numbers are not equal.
• In , and are any real and Imaginary number.
• NE will return the result as "TRUE" when A is not equal to B.
• NE will return the result as "FALSE" when A is equal to B.
• NE compares both real and imaginary parts A and B.
• This function will check for alphabets and special characters also.
## Examples
1. NE(55,55) = false
2. NE(-123,123) = true
3. NE(5.43, 5.43) = false
4. NE(2/3,6/9) = false
5. NE("june","june") = false | 197 | 638 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.801115 |
https://certifiedcalculator.com/raid-value-calculator/ | 1,721,765,125,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518115.82/warc/CC-MAIN-20240723194208-20240723224208-00121.warc.gz | 145,307,344 | 13,596 | # Raid Value Calculator
The Raid Value Calculator is a handy tool for gamers looking to calculate raid values efficiently. Whether you’re involved in gaming or any other field where raid values play a crucial role, this calculator simplifies the process.
Formula: To calculate the raid value, use the formula: Result = Base Value x Multiplier.
How to Use:
1. Enter the base value in the designated field.
2. Input the multiplier value.
3. Click the “Calculate” button to get the result.
Example: Suppose you have a base value of 100 and a multiplier of 1.5. The calculated raid value would be 150 (100 x 1.5).
FAQs:
1. Q: What is a raid value?
• A: A raid value is a calculated value obtained by multiplying the base value with a specific multiplier.
2. Q: How accurate is the Raid Value Calculator?
• A: The calculator provides accurate results based on the entered values.
3. Q: Can I use decimals for base and multiplier values?
• A: Yes, the calculator supports decimal values for precise calculations.
4. Q: Is this calculator specific to gaming raids only?
• A: While designed with gaming raids in mind, it can be used for any scenario where a similar calculation is needed.
5. Q: Can I use negative values?
• A: The calculator only accepts positive values for base and multiplier.
Conclusion: The Raid Value Calculator simplifies the process of calculating raid values, making it a valuable tool for gamers and professionals alike. Easily obtain accurate results by entering the base value and multiplier, streamlining your raid value calculations. | 334 | 1,563 | {"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} | 2.9375 | 3 | CC-MAIN-2024-30 | latest | en | 0.808208 |
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# Partial Differential Equations : Eigenvalue Expansion
Solve the Boundary value problem
PDF: Ut = Uxx + 7sin(5x/2)
b. C. U(0,t) = 3
Ux(Pi,t) = 0
i. c. U(x,0) = 3 + 6sin(7x/2)
#### Solution Summary
A boundary value problem is solved using an eigenvalue expansion. The solution is detailed and well presented. The response received a rating of "5/5" from the student who originally posted the question.
\$2.19 | 133 | 441 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2016-50 | latest | en | 0.861301 |
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# After more than four decades of research and development, a
Author Message
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Joined: 19 Jun 2009
Posts: 6
### Show Tags
29 Jun 2009, 06:25
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Question Stats:
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### HideShow timer Statistics
After more than four decades of research and development, a new type of jet engine is being tested that could eventually propel aircraft anywhere in the world within two hours or help boost cargoes into space at significantly lower costs than current methods permit.
A. tested that could eventually propel aircraft anywhere in the world within two hours or help
B. tested that could eventually have the capability of propelling aircraft anywhere in the world within two hours or to help
C. tested, eventually able to propel aircraft anywhere in the world within two hours, or helping
D. tested, and it eventually could propel aircraft anywhere in the world within two hours or helping
E. tested, and it could eventually have the capability to propel aircraft anywhere in the world within two hours or help
I was told that the correct answer is A, but I think E is correct. I think that the relative pronoun in A is wrong.
Would anyone explain what is the correct answer?
Director
Joined: 05 Jun 2009
Posts: 805
WE 1: 7years (Financial Services - Consultant, BA)
Re: Sentence Correction Question #3 [#permalink]
### Show Tags
29 Jun 2009, 09:13
IMO A
In E, 'could' and 'have the capability to' I think create redundancy.
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Joined: 25 Mar 2009
Posts: 300
Re: Sentence Correction Question #3 [#permalink]
### Show Tags
29 Jun 2009, 10:35
The word "that" refers directly to the new jet engine. The meaning is clear.
sudeep's reasoning sounds right. Also, "capability to" might be wrong? I'm not sure, but when someone says, "I am capable of learning," the correct idiom is "capable of." I'm not sure if "capability of propelling" would have been correct, but it's just a thought.
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Joined: 14 May 2009
Posts: 141
Schools: AGSM '16
Re: Sentence Correction Question #3 [#permalink]
### Show Tags
29 Jun 2009, 11:33
I'd go with A.
I think E is too wordy and "have the capability" slightly alters the meaning of the sentence.
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Joined: 18 Jun 2009
Posts: 165
Location: Tbilisi, Georgia
Re: Sentence Correction Question #3 [#permalink]
### Show Tags
29 Jun 2009, 11:40
A 'that' can't refer to anything else other than 'jet engine'
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Las cualidades del agua...porque el agua no olvida que su destino es el mar, y que tarde o temprano deberá llegar a él.
Manager
Joined: 15 Jun 2009
Posts: 151
Re: Sentence Correction Question #3 [#permalink]
### Show Tags
29 Jun 2009, 23:33
IMO A...
In E "have the capability to" changes meaning...
Manager
Joined: 16 Apr 2009
Posts: 147
Re: Sentence Correction Question #3 [#permalink]
### Show Tags
08 Jul 2009, 23:38
OA is (A)
Re: Sentence Correction Question #3 [#permalink] 08 Jul 2009, 23:38
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https://www.reddit.com/user/DTKahn | 1,493,458,715,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123484.45/warc/CC-MAIN-20170423031203-00202-ip-10-145-167-34.ec2.internal.warc.gz | 938,475,669 | 20,265 | [–] 0 points1 point (0 children)
The serenity to accept the things I cannot change, Courage to change the things I can, And wisdom to know the difference.
[–][S] 2 points3 points (0 children)
That's quite the fragment you have there.
[–][S] 1 point2 points (0 children)
How to count to 1000 on your fingers.
tl;dr first – use binary.
Longer: L = Left hand, R = Right hand L00000 R00000 = 0, L00000 R00001 = 1, L00000 R00010 = 2, L00000 R00011 = 3, L00000 R00100 = 4, L00000 R00101 = 5, L00000 R00111 = 6, L00000 R01000 = 7, etc., L11111 R11111 = 1023
[–] 0 points1 point (0 children)
This was downvoted, but this is good advice. Not telling you to stop. Keep rocking it! Just that medical input with major life changes is never a bad idea.
[–] 12 points13 points (0 children)
[–][S] 0 points1 point (0 children)
I'm obviously perfectly happy to let you make the connection. Nothing to hide here :)
An upvote for you internet user!
[–][S] 0 points1 point (0 children)
Ha! It seemed like a good idea, and I'm always happy to steal a good idea!!
[–][S] 1 point2 points (0 children)
PMing you now. Thanks!
[–][S] 2 points3 points (0 children)
Hey,
That's great! Yeah I understand 30min can be an obstacle. If you'd be willing I'd be happy to do it over the phone and could probably cut it down to about 15 if we move through the questions quickly.
I've got availability between 4:00 - 5:00 today, or anytime after 12:00 tomorrow if that works for your schedule.
[–][S] 3 points4 points (0 children)
I appreciate the feedback. I suspect yours and everyone's time is worth a lot more than half an hour per coffee!
I'm more asking for volunteers out of an interest in helping a stranger/ goodness of your heart perspective. The coffee is just a bonus/friendly gesture!
[–][S] 0 points1 point (0 children)
I absolutely would be!
What's your availability like? I'd be happy to talk on Facetime, Google Hangouts, or Skype.
[–][S] 0 points1 point (0 children)
That would be great! I can be free any time before 5:00 on Wednesday, any time before 5:00 on Thursday, and any time before 4:00 on Friday.
[–][S] 0 points1 point (0 children)
Looking to do about a 20min-30min video interview to ask about things like buying and selling experience, what websites were used in the process, where you found you had the most stress in the process, motivation for buying or selling a used system, etc.
Let me know if you're interested. I'd love to chat!
[–][S] 0 points1 point (0 children)
That would be great! I'm pretty flexible with timing during the day before about 4:00pm and operate out of Toronto. (Eastern Time Zone UTC-5)
When would be good for you?
[–][S] 0 points1 point (0 children)
Hi! Thanks for the response! We've found that we often get as much information at this point in the process from the way people react to questions as much as we do from the content of their answers. Ideally we're actually trying to meet people in person for coffee to do these interviews, but we thought Reddit might be a cool community to hit up.
I'll let you know if we decide to start collecting information online! Thanks again for the response! | 888 | 3,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2017-17 | latest | en | 0.903823 |
http://www.carrawaydavie.com/lowes-5-gallon-bucket/ | 1,563,553,849,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526324.57/warc/CC-MAIN-20190719161034-20190719183034-00431.warc.gz | 184,202,285 | 8,560 | ## Lowes 5 gallon bucket
lets to see lowes 5 gallon bucket
## So lowes 5 gallon bucket
A gallon has 4 quarts. Each quart has 2 pints, so a gallon additionally contains 8 pints. Each pint is 2 cups, so a gallon has 16 cups, lowes 5 gallon bucket.
## Lets to see lowes 5 gallon bucket
Each mug holds 8 liquid ounces. Therefore, there are 128 fluid ounces in 1 gallon. The basic recommendations are to consume 8 8-ounce glasses of water a day, which is equivalent to consuming half a gallon of water daily, due to the fact that a gallon amounts 64 ounces. So, lowes 5 gallon bucket.
## How to know lowes 5 gallon bucket
So, lowes 5 gallon bucket. When it comes to the U.K. standard is concerned, 1 gallon amounts 160 liquid oz, which is 4.546 liters of water. Significance, you would require 9.46 bottles of water to load a gallon. Consider this to be 9 containers or a little much less than 9 as well as a half containers. Individuals utilize gallons in order to carry a significant amount of liquid, lowes 5 gallon bucket!
## United States gallon to mugs, lowes 5 gallon bucket
1 Gallon (United States, fluid) = 16 Mugs (US).
1 Gallon (Dry, US) = 18.618355 Mugs (US).
lowes 5 gallon bucket | 318 | 1,197 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2019-30 | latest | en | 0.874918 |
https://community.qlik.com/t5/QlikView-App-Dev/Comma-seperated-field-evaluation/m-p/1122086 | 1,627,894,325,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154310.16/warc/CC-MAIN-20210802075003-20210802105003-00356.warc.gz | 190,843,922 | 51,372 | # QlikView App Dev
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## Comma seperated field evaluation
I have a field like below.
ProductValue
23,5,34,55,90,106,200
9,8,88,92,7,55
22,44,99,82
Question1: How can i find max value for Product Value Field.
Ans:
ProductValue Max
23,5,34,55,90,106,200 200
9,8,88,92,7,55 92
22,44,99,82 99
Question2: How can i find counts for field that is greater then 30.
Ans:
ProductValue Count > 30
23,5,34,55,90,106,200 5
9,8,88,92,7,55 3
22,44,99,82 3
15 Replies
MVP
What is your expected output for:
ProductValue Output
23,5,34,55,90,106,200 ?
9,8,88,92,7,55 ?
22,44,99,82 ?
Creator
Author
Indicator
Over 100 -- Selected
Under 100
Results:
Product Value
23,5,34,55,90,106,200
Indicator
Over 100
Under 100 -- Selected
Results:
9,8,88,92,7,55
22,44,99,82
Employee
Max(SingleValue)>100 As Over100
Add this to the Max: load from above
Max:
RecId,
Max(SingleValue) as MaxValue,
sum(If(SingleValue>30,1)) as Over30Count,
Max(SingleValue)>100 As Over100
Group By RecId
;
Hi,
try below script
test:
LOAD RowNo() as row,ProductValue inline [
ProductValue
"23,5,34,55,90,106,200"
"9,8,88,92,7,55"
"22,44,99,82"
];
t1:
LOAD row,if(SubField(ProductValue,',')<100,'Less than 100','higher') as t Resident test;
t2:
Load row, Concat(distinct t,',') as t2 Resident t1 group by row;
test_final:
LOAD row,if(wildmatch(t2,'*higher*'),'higher','Less than 100') as testfield Resident t2;
Right join
Load * Resident test;
DROP Table t2,test,t1;
It is working for me and I think you can optimize it.
Regards
Great dreamer's dreams never fulfilled, they are always transcended.
Specialist
Hi,
Then, add another LOAD based on Rob's solution, also works.
Max:
if (Over100Count>0, 'Over 100',
if (Over30Count>0, '30 -100', 'Under 30')) as Indicator,
*
;
RecId,
Max(SingleValue) as MaxValue,
sum(If(SingleValue>30,1)) as Over30Count,
sum(If(SingleValue>100,1)) as Over100Count
Group By RecId
;
RecId,
SubField(ProductValue,',') as SingleValue
Resident Data
;
Zhihong
Luminary Alumni
A simple way would be to add another LOAD on top. Example attached.
Max:
*,
if(MaxValue > 100, 'Over 100', 'Under 100') as Indicator
;
RecId,
Max(SingleValue) as MaxValue,
sum(If(SingleValue>30,1)) as Over30Count
Group By RecId
;
RecId,
SubField(ProductValue,',') as SingleValue
Resident Data
;
-Rob | 861 | 2,828 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-31 | latest | en | 0.709973 |
http://moleseyhill.com/blog/2009/09/22/bayesian-probability-of-success/ | 1,519,523,780,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816083.98/warc/CC-MAIN-20180225011315-20180225031315-00559.warc.gz | 238,242,971 | 5,413 | # Bayesian Probability of Success
One of the tricky things with probability is what you do when your sample size isn’t very big. For example, say a new rocket has had one successful launch and no launch failures. How safe is it to ride on the next flight? A naive frequency based approach would say, “1 launch, 1 success, therefore probability of success is 1”, completely discounting the possibility of failure all together.
You know that this is wrong. Say you have a choice of on a different rocket which has had 99 successful launches and one failure. The probability for success is 0.99. What would you choose? I’d go for the one with the 100 launches, you just know a lot more about it.
Simple statistics don’t work properly with small sample sizes. What we need is a formula which takes account of the fact we don’t know anything at the beginning. Bayesian statistics to the rescue! On space launch report I found this ace formula for estimating the probability properly:
Here P is the probability of success next time, k is the number of successful events so far, and n is the total number of trials so far. It’s called “the first level Bayesian estimate of mean predicted probability of success”. Therefore:
• New rocket: k=1, n=1 therefore P=0.67
• Established rocket: k=99, n=100 therefore P=0.99
For the new rocket it predicts a 2/3 chance that the next flight will be successful. This captures your state of ignorance about the rocket. It’s had one successful flight, and that’s good, but you still don’t know that much about it. For the established rocket it predicts a success probability of 0.99, which is pretty much what normal probability would tell you.
What about if no launches have yet occurred? Well the formula says probability of success is 0.5. A fifty-fifty chance of failure, kind of makes sense; you’ve not tried something yet, and you have no other information, so what else can you say?
This last point does raise a tricky problem with Bayesian reasoning. You have to make an initial guess at the probability, called the prior probability. In this case the choice was a 50-50 chance, which does sound reasonable, but it’s still a guess.
Of course the formula doesn’t just apply to rockets. It works fine for anything which is success/failure, true/false. | 504 | 2,295 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2018-09 | latest | en | 0.960839 |
http://www.commercialrealtorsofma.com/difference-between-fixed-rate-and-apr/ | 1,601,513,468,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402130531.89/warc/CC-MAIN-20200930235415-20201001025415-00102.warc.gz | 147,274,711 | 8,342 | Fixed rate (or fixed APR) definition – Glossary – CreditCards.com – Fixed rate (or fixed APR). An annual percentage rate that does not change throughout the year, unlike an introductory APR that changes after a specific period of.
The biggest difference between APR and APY lies in how they relate to your savings or investment growth, or the cost of borrowing. With savings or investments, APY factors in how often the interest is applied to the balance, which can range anywhere from daily to annually.
The interest rate is the amount a lender charges for the use of assets expressed as a percentage of the principal. The interest rate is typically noted on an annual basis known as the annual.
Fixed APR vs. Variable APR: What's the Difference? | SuperMoney! – A loan with a fixed APR has an interest rate that typically stays the same throughout the. So what are the major differences between the two?
When evaluating the cost of a loan or line of credit, it is important to understand the difference between the advertised interest rate and the annual percentage rate, or APR. Interest Rate
Difference Between Interest Rate and APR (with Comparison. – The basic difference between interest rate and APR is that, while interest rate shows current borrowing cost, APR is used to present the true picture of total cost of financing, where the interest rate and the lender fees needed to finance the loan are taken into consideration.
APR Calculator – Calculator.net – The real APR, or annual percentage rate, considers these costs as well as the interest rate of a loan.. Use the calculator below for mortgage loan in the United States.. For these, if the rate is fixed, the interest rate and APR should be the same.. The main difference between these and APR is that the former considers .
30-Year vs. 5/1 ARM Mortgage: Which Should I Pick? – As of this writing, a buyer with this credit profile can expect an APR of about 5.46% on a 30-year. What is the difference in interest rates and monthly payments? The actual difference between.
What is the difference between flat interest rate and. – Consider a loan of Rs. 100000 at 12% per year (1% per month) interest for 3 years. flat interest for 3 years would be Rs. 36000 (1000000 X 12/100 X 3). Total amount to be repaid Rs. 136000. The monthly installment would be 136000/36 = 3777 Now let.
Usda Loan Eligibility Requirements Aiken County eligible for USDA assistance – Aiken County is eligible for USDA. emergency loans from USDA’s Farm Service Agency. The disaster areas were designated on March 12. All qualified farm operators in the designated counties are.Refinance House Loan Calculator Some Colleges Are Failing to Comply With cost calculator requirements – Some institutions did not differentiate the difference between grants and loans in the cost calculator. And two used a definition. financial aid, books and housing, and gauge if a particular.
APR and Flat Rate Interest | The Car Loan Warehouse – APR and Flat Rate Interest: What’s The Difference? Whether you’re new to the world of car finance or experienced, some still get confused about interest rates. We have tried to simplify the differences between APR and flat interest rates, they are very different and you need to understand. | 680 | 3,267 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.966089 |
https://joemendola.com/5th-grade-math-worksheets-long-division-with-decimals/ | 1,620,372,632,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988775.25/warc/CC-MAIN-20210507060253-20210507090253-00168.warc.gz | 321,493,794 | 6,704 | # 45++ 5th grade math worksheets long division with decimals Images
Written by Wayne May 30, 2021 ยท 5 min read
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5th Grade Math Worksheets Long Division With Decimals. We can use. Dividing with decimals worksheets for grade 5 Our grade 5 decimal division worksheets start with simple mental math. Long division decimals 3rd grade math free printable. These long division worksheets have problem sets with varying levels of difficulty all with decimal division problems.
Division Worksheets Printable Division Worksheets For Teachers Clase De Matematicas Divisiones Matematicas Matematicas De Sexto Grado From pinterest.com
Get the full course at. Below are six versions of our grade 5 math worksheet on long division with decimals. 5th grade multiplying decimals worksheets including multiplying decimals by decimals multiplying decimals by whole numbers missing factor problems multiplying by 10 100 or 1000 and multiplication in columns with decimals. Below are six versions of our grade 5 math worksheet on dividing 1-3 digit decimals by whole numbers 1-9 using long division. We can use. These worksheets are pdf files.
### Sample Grade 5 Decimal Division Worksheet What is K5.
These fifth grade decimals worksheets including. 5th grade multiplying decimals worksheets including multiplying decimals by decimals multiplying decimals by whole numbers missing factor problems multiplying by 10 100 or 1000 and multiplication in columns with decimals. Long division with decimals is a subtle extension of regular long division problems and a skill worth practicing. May 28 2018 - Grade 5 Decimals Worksheet long division with decimals. Whether youre just starting out need a quick refresher or here to master your math skills this is. Then we divide as we would with whole numbers.
Source: pinterest.com
Children apply their prior knowledge to divide fractions and decimals and practice division of four-digit numbers with two-digit numbers. These long division worksheets have problem sets with varying levels of difficulty all with decimal division problems. The worksheets provide calculation practice for decimal division topics. Decimal Division using a Number Line Worksheets 50 Worksheets Dividing Decimals by Powers of Ten Practice this collection of printable worksheets and make headway dividing decimal numbers involving digits in the tenths hundredths and. These worksheets are pdf files.
Source: pinterest.com
Grade Adding And Subtracting Decimals Worksheets 6th Pdf Printable. Grade 5 Decimals Worksheet Keywords. We can use. Grade Adding And Subtracting Decimals Worksheets 6th Pdf Printable. Long division decimals 3rd grade math free printable.
Source: pinterest.com
07102020 Fifth grade word problems include multiplication division fractions averages and a variety of other math concepts. Both dividends and divisors are decimal numbers. Grade 5 Decimals Worksheet - Long division with decimals math practice printable elementary school. Below are six versions of our grade 5 math worksheet on long division with decimals. Dividing Decimals Vertically Version 1 Worksheet.
Source: pinterest.com
6th grade dividing decimals worksheets including decimals divided by whole numbers decimals divided by decimals decimal division with missing divisors or dividends dividing by 10 100 1000 or 10000 and long division with decimals. The worksheets provide calculation practice for decimal division topics. Dividing with decimals worksheets for grade 5 Our grade 5 decimal division worksheets start with simple mental math. 6th grade dividing decimals worksheets including decimals divided by whole numbers decimals divided by decimals decimal division with missing divisors or dividends dividing by 10 100 1000 or 10000 and long division with decimals. Grade 5 Decimals Worksheet Keywords.
Source: pinterest.com
While the earlier sets of worksheets have long division problems with varying numbers of digits the later sets focus on decimal long division. 6th grade dividing decimals worksheets including decimals divided by whole numbers decimals divided by decimals decimal division with missing divisors or dividends dividing by 10 100 1000 or 10000 and long division with decimals. Long division with decimals is a subtle extension of regular long division problems and a skill worth practicing. SplashLearns online division games for Grade 5 teach children how to estimate quotients identify patterns when dividing decimal numbers and use visual models to simplify division problems. We can use.
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Apr 06 . 2 min read | 1,325 | 6,736 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2021-21 | latest | en | 0.852426 |
https://www.studymode.com/essays/Arbitrage-Pricing-Theory-1793260.html | 1,539,642,647,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583509845.17/warc/CC-MAIN-20181015205152-20181015230652-00152.warc.gz | 1,077,129,080 | 24,770 | # Arbitrage Pricing Theory
Topics: Capital asset pricing model, Risk, Arbitrage pricing theory Pages: 2 (607 words) Published: June 24, 2013
Summary of the Arbitrage Pricing Theory (APT)
The APT model was developed as an alternative to the CAPM. Like the CAPM, this model provides implications for the relationship between expected returns and risk on securities. However, the model differs from CAPM in its assumptions, its implications, and in the way that equilibrium prices are reached. • Assumptions: The CAPM model assumes that all investors are risk-averse utility maximizers. In other words, all investors solve the investment problem in the way we described in portfolio theory (Lecture 3). The APT model makes a less restrictive assumption about the way investors behave. Here we just assume that there are at least some investors out there who would like to take extremely large positions in any risk-free arbitrage opportunities that arise. So, if securities are priced correctly in equilibrium, there cannot be any remaining risk-free arbitrage opportunities. Equilibrium: Both the APT model and the CAPM describe what expected returns should look like in equilibrium. However, equilibrium is reached in different ways. In the CAPM, all investors solve the portfolio theory problem we described in Lecture 3 using mean-variance optimization and maximizing their utility. This leads to an equilibrium where all investors hold the same optimal risky portfolio (the “market” portfolio) and also results in the expected return – beta equation that we typically refer to as the CAPM equation. In the APT model, any mispriced securities will create arbitrage opportunities that will immediately be traded on by investors. Equilibrium results when the trades of these investors push prices back to their correct values. Implications: Both the CAPM and the APT lead to equations for expected returns. These equations are shown below. The equations are very similar, but have several important differences. First, the CAPM has only one risk factor (market risk), while the APT model can have multiple sources of risk. Second, the... | 428 | 2,138 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-43 | latest | en | 0.941146 |
https://questioncove.com/updates/528cfdcce4b0fd0888873dea | 1,503,031,852,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104565.76/warc/CC-MAIN-20170818043915-20170818063915-00565.warc.gz | 801,936,456 | 2,640 | OpenStudy (anonymous):
Prove: C-(AUB)=(C-A)N(C-B)
OpenStudy (anonymous):
By N I mean intersection
OpenStudy (amistre64):
truth tables help if you dont have the rules for the tautologies
OpenStudy (anonymous):
I do realize this is one of De Morgran's Laws. I need proof of it though.
OpenStudy (anonymous):
$C - (A \cup B ) = (C - A) n (C - B)$ to do this you want to prove the left is equal to the right vice versa let $x \in C - (A \cup B)$ then $x \in C$and $x \notin (A \cup B)$ so because$x \notin (A \cup B)$ then $x \notin A, x \notin B$ since $x \in C$ and $x \notin A$ then $x \in (C - A )$ and since $x \in C$and$x \notin B$ then $x \in (C - B )$ so since $x \in (C - A )$ and $x \in (C - B )$ then $x \in (C - A ) n (C - B )\$ now you prove the RHS to the LHS
OpenStudy (anonymous):
Thanks!
OpenStudy (anonymous): | 288 | 835 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2017-34 | latest | en | 0.69181 |
https://betterlesson.com/lesson/resource/2904501/snowmans-coat1-jpg | 1,542,338,193,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742970.12/warc/CC-MAIN-20181116025123-20181116051123-00071.warc.gz | 570,760,255 | 24,351 | # A Jacket for Frosty
9 teachers like this lesson
Print Lesson
## Objective
Students will be able to investigate and describe the relationship between heat energy and the temperature of materials.
#### Big Idea
Would putting a jacket on a snowman keep it from melting? Find out in this chilling experiment about insulation and heating and cooling!
## Introduction
5 minutes
For this lesson, students will need the attached worksheets in the resources menu, two frozen water bottles per group, one glove per group, and two digital thermometers per group.
CCSS-CCSS.Math.Content.3.MD.A.1-Tell and write time to the nearest minute and measure time intervals in minutes. Solve word problems involving addition and subtraction of time intervals in minutes.
NGSS-3-5-ETS1-3.-Plan and carry out fair tests in which variables are controlledand failure points are considered to identify aspects of a model or prototype that can be improved.
Begin the lesson by building background knowledge and making connections. Ask the students to think back to a time that they were really cold or really hot. Ask them, "what were you doing or where were you that caused you to feel this way? What did you do to cool off or warm up?" Expect answers like, "I put on my jacket or I turned on a fan, etc." Ask the students, "what happens to your body when you are cold? What happens to your body when you are hot?" Expect answers such as you shiver or sweat. Explain to students that as our body heats up, the particles inside of us move faster. When this happens, heat is released through our skin causing us to sweat. The heat leaving our body, causes us to cool off. The opposite occurs when we are cold. The particles slow down, causing us to shiver to regain warmth. Explain to the students, "today we will solve a chilling problem, regarding heating and cooling."
## Whole Group
10 minutes
Display the Frozen Snowman video. Have students think-pair-share with their elbow partner. The students will discuss their point of view of the video and what they believe the problem is. As a group, discuss possible ways to preserve a snowman outside for as long as possible without the use of a freezer. Ask the students is it possible at all? Display the Snowman Coat picture. Have the students analyze the picture and ask them, "what is the problem that the students are trying to solve? Discuss this with your partner." Have the students form a hypothesis and share with an elbow partner. Walk around and encourage students to explain their reasoning. Encourage students to ask and answer questions while defining the problem as stated in NGSS 3-5-ETS1-1. Give each student a recording sheet (snowmanblresource sheet)to write down their hypothesis. Allow students to share their predictions.
Explain to students, that today they will be able to investigate to determine if placing a coat on a snowman will melt it or keep it frozen longer. Inform the students that each group will receive two frozen water bottles, which will represent the snowman, a digital thermometer for each, and a glove. Introduce the vocabulary word, "insulator." Explain to students that an insulator is a substance that does not readily allow the passage of heat.
Therefore, the glove (which serves as the jacket) is our insulator. Will the insulator (the glove) keep the heat in our snowman (the bottle) and make it melt faster or will the insulator keep the bottle cool causing it to remain frozen longer? Ask the students, "why do we need to have two frozen waterbottles?" Introduce the word control to the students and explain that the control is the part of an experiment that does not change. The control in this experiment is the frozen water bottle without the glove. Lead a discussion on why having a control is important in an experiment.
## The Experiment
20 minutes
Provide students with the materials to complete the experiment. It is important to instruct the students to not continuously touch the water bottles, as their body heat assists with the melting process. Allow students time to monitor and record the temperatures of each bottle and answer the reflection questions. Be sure to remind the students to keep track of the time they are recording the temperatures. You may provide timers if available.
## Conclusion
15 minutes
As students conclude their experiment, allow students an opportunity to share their findings. Discuss as a class, what they discovered from their experiment; was their hypothesis correct? Discuss how their results compared to their original thinking when they were first introduced to the picture. Make a connection back to the students by asking, "how do the results compare to how you feel when you wear your coat?" Ask the students, "what did this investigation teach you about how coat's keep you warm?" Explain to students that the coat is one example of an insulator, however there are many more.
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# Yogi Avayogi technique
Introduction: Among the most brilliant extensions and applications of the principles of classical jyotish were those researched and published by Seshadri Iyer of Bangalore, India. Iyers work has been extensively taught in the West by Hart deFouw who has verified the validity of these principles through years of applying them in his own practice. After studying these principles, I too, have used them with my clients and have come to rely on their validity and accuracy. Among many wide-ranging and useful topics, Iyer put forward a methodology for examining prosperity in a chart that is quick and easy to pinpoint and evaluate. Although embedded in a larger context that requires more study and experience, the points detailed below should give an astrologer good insight into the prosperity promised in a chart and when the promised prosperity may be more likely to deliver the goods. Because these principles are taken from a more holistic and complex context, however, there may be instances when the simple applications described below do not fully reflect the true experience. Therefore please enjoy applying these techniques with the knowledge that it is not yet the total story. Calculation of the Yogi Point: Iyer was inspired by some Nadi methods that were available to him to define and calculate specific planetary yogas that were correlated with prosperity. The basis of these yogas involved identifying the yogi point. Although the calculation of this point is available in many computer programs, it is important that every jyotishi understands the basis for the computer calculation and is capable of calculating it independently. To find this point one simply adds (for a particular chart) the longitude of the Sun, the Moon and a constant of 93 degrees 20 minutes. The resultant sum will be the longitude known as the yogi point. It has been my experience that many aspiring students need practice at calculations, and I, therefore, will provide an example in great detail, including how one reduces the resulting sums to find the correct longitudes. (For those who find these operations simple and obvious, this section can be quickly skimmed.) Suppose in a given chart, the Sun is at 14 degrees 22 minutes of Pisces and the Moon is at 5 degrees and 45 minutes of Gemini. One could proceed with the calculation in two different ways. My preferred method is the Indian way of expressing the longitude of the Sun as 11 signs 14 degrees 22 minutes (11 signs have been completed when a planet is placed in Pisces and the Sun in this case is 14 degrees 22 minutes into the sign of Pisces.) This can be a point of confusion as there will be some that will think of Pisces as the 12th sign but in fact, only 11 signs have actually been completed). The longitude of the Moon placed in Gemini can be expressed as 2 signs 5 degrees and 45 minutes (here again, placement in Gemini means 2 signs are completed plus the 5 degrees and 45 minutes into Gemini that the Moon has traversed). To this one must also add the constant of 93 degrees 20 minutes which in this way of handling the calculation is most easily expressed as 3 signs (30 degrees x 3 = 90 degrees) 3 degrees and 20 minutes. Adding these three longitudes we arrive at 16 signs 22 degrees and 87 minutes. 11s 14 degrees 22 minutes 02s 05 degrees 45 minutes 03s 03 degrees 20 minutes 16 s 22 degrees 87 minutes Expunging 60 minutes from the last column and adding it to the degree column as one degree (60 minutes = one degree) , we get 16 signs 23 degrees and 27 minutes.
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Whenever the sign portion of the answer exceeds 12, we subtract 12 (360 degrees) and our final figure is 4 signs 23 degrees and 27 minutes. 16s 22 degrees 87 minutes +1 degree -60 minutes
## 16 s 23 degrees 27 minutes -12s 04s 23 degrees 27 minutes = Leo 23 degrees 27 minutes
This places the yogi point in the sign of Leo at 23 degrees and 27 minutes. An alternate method is to convert everything to degrees with 0 degrees of Aries as the starting point and add the longitudes directly, divide the sum by 30 degrees and arrive at the same answer. Calculation of the Avayogi Point: As relative creation exists in duality, the existence of the yogi point implies that an opposite energy must exist as well. Iyer called this the avayogi point and this calculation does not seem to be generally available in computer programs. It is a simple operation of adding 6 signs 6 degrees and 40 minutes to the yogi point. For the above example, the avayogi point would be calculated by adding 4 signs 23 degrees and 27 minutes to the constant of 6 signs 6 degrees and 40 minutes. 04s 23 degrees 27 minutes 06s 06 degrees 40 minutes 10s 29 degrees 67 minutes The resultant sum is 10 signs 29 degrees and 67 minutes which must be reduced by expunging 60 minutes from the minutes column leaving 7 minutes and then expunging 30 degrees from the degree column (29 degrees plus the 1 degree derived from expunging 60 minutes from the 67 minute total) leaving 0 degrees and a total of 11 signs (10 signs plus the extra sign derived from expunging 30 degrees). This gives an avayogi point located in Pisces at 0 degrees 7 minutes. 10s 29 degrees 67 minutes +1 degree -60 minutes 10s 30 degrees 07 minutes +1s -30 degrees 07 minutes 11s minutes Finding the Yogi, Duplicate Yogi and Avayogi: Having the yogi and avayogi points available, the method for finding the yogi, duplicate yogi and avayogi is as follows: The yogi will be the ruler of the nakshatra in which the yogi point is placed. The duplicate yogi will be the ruler of the sign in which the yogi point is placed. The avayogi will be the ruler of the nakshatra in which the avayogi point is placed. All of these are based on 0 degrees 07 minutes = Pisces, 0 degrees 07
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rulerships according to the vimshottari dasha system only. For the example above, the yogi point falls in the sign of Leo at 23 degrees 27 minutes putting it into the nakshatra of Purva Phalguni ruled by Venus. Therefore, Venus is the yogi of the chart. The Sun would be the duplicate yogi as it is the ruler of the sign in which the yogi point is placed. Jupiter is the ruler of Purva Bhadrapada which spans from 20 degrees of Aquarius through 3 degrees 20 of Pisces. The avayogi point is located at 0 degrees 07 minutes of Pisces making Jupiter the avayogi of the chart. In five instances, the duplicate yogi and the avayogi are the same planet. This fact will color some of the principles detailed below. In two instances, the yogi and the duplicate yogi are the same planet. A shortcut method of finding the avayogi once you know the yogi is by counting 6 inclusively from the yogi planet in vimshottari dasha sequence order. For the above example, Venus would be the starting point as the yogi and counting inclusively in the order of Venus, Sun, Moon, Mars, Rahu and Jupiter, you would arrive at Jupiter as sixth from Venus. Trinal nakshatras of the yogi and avayogi nakshatra: In this methodology, it is also important to be aware that each planet in the vimshottari dasha scheme rules three nakshatras. For a given chart, the nakshatra containing the actual yogi or avayogi point is the most important. However, results can also be seen when considering the two other trinal nakshatras for each of the respective points. In the example above, the nakshatras of Bharani and Purvashadra can also be indicative of prosperity and the nakshatras of Purnavasu and Vishaka can be obstructive in this regard. Principles of Prosperity: Once these calculations have been accomplished, the intriguing process of applying these points to a given chart can begin. Each principle will be stated below with an illustration so that the methodology is clear. There will not be an attempt at any further analysis of the chart other than to illustrate the principle and comment where necessary on the life experience with respect to prosperity. As with any other set of jyotish principles, good common astrological sense needs to be applied. There is a vast difference between a yogi planet that is in its own sign, vargottama, in a good house and unafflicted and one that is debilitated, in dusthana, combust and afflicted. Another factor to be aware of is that it is often the tendency when introduced to a new principle to immediately examine charts for example in this case, of wealthy individuals and see if these principles apply. A true test of the validity of these principles would be to examine charts where these combinations exist and see if they manifest in prosperity or poverty; however, one should note that there are many significations that can support prosperity and it does not necessarily follow that all people who have wealth will have these particular indicators. Still, it SHOULD be the case that people with the combinations listed below manifest the indicated results in a comfortable majority of the cases (although this is subject to the greater context in which this work is embedded which will be unfolded in future writings.) Finally, it is important to remember that the yogi and avayogi will give results according to their condition or strength. Jyotishis evaluate strength in different ways and in accordance with their own experience and training. It is not my purpose to elaborate on this topic in the context of this paper but it is important to bear in mind that for maximum expression of the prosperity indicators, the yogi and duplicate yogi should be strong and well located. Likewise, for maximum expression of lack of prosperity, the avayogi should be strong and well placed. In this system, the weaker the avayogi, the less it will obstruct prosperity. The charts of private parties listed below have some or all of the birth details withheld to maintain confidentiality. Principle 1: If the ascending degree of the horoscope is placed in the nakshatra of the yogi point, the person will be predisposed towards material prosperity. This will be all the more marked if the yogi is strong and favorably placed for good results.
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Above is the birth chart of Tony Blair ( 6th of May in 1953, at 6.13 AM in Edinburgh, Scotland, 3degreesW13, 55degrees N 57, daylight). The ascending degree is 12 degrees 43 Taurus. The yogi point is 13 degrees 48 Taurus. The Moon is the yogi and it is noteworthy that it is well placed in the 9th house in its own nakshatra (and therefore one of the trinal nakshatras of the yogi point) of Shravana. Venus is the duplicate yogi exalted in the 11th house and in a parivartana yoga with Jupiter who is placed in the ascendant with dig bala. Principle 2: If the ascending degree of the horoscope is placed in the nakshatra of the avayogi point, the person will tend toward material deprivation. This will be more marked if the avayogi is strong and able to give its obstructive results. Please note that in the cases where the avayogi and the duplicate yogi are the same planet, this principle does not apply.
In the chart of this individual, the avayogi nakshatra is Ardra and the ascendant degree falls into this nakshatra. This person though capable of bringing in some good income, has shown a marked propensity towards wild spending and financial irresponsibility, and is currently without resources. Principle 3: If the yogi planet is placed in the ascendant, there is a marked tendency towards material prosperity. This will be more marked if the yogi is strong and well disposed.
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This is the nativity of William Randolph Hearst born April 29th, 1863 at 6:07 AM in San Francisco 122W25, 37N47 standard time. The lagna is 2 degrees 5 minutes Taurus. The yogi planet for the chart is Venus in its own sign in the ascendant. Principle 4: If the avayogi is in the ascendant, there is a marked tendency toward material deprivation. This will be more marked if the avayogi is strong and able to give its obstructive results. The chart given for principle 2 also has the configuration listed in the above principle. The avayogi of the chart is Rahu which is placed in the ascendant. . Principle 5: If the yogi planet is the lord of the ascendant and well placed, the native is pre-disposed toward material prosperity. This principle is somewhat less powerful than the principles involving the yogi listed above.
The birth chart of Jack Nicklaus is shown above. He was born January 21st, 1940 at 3.10 AM in Columbus Ohio 83W00, 39N58, Standard Time. The lagna degree is 1 degree 40 Scorpio. The lagna lord Mars is the yogi of the chart well placed in the 5th house in the company of the 2nd and 5th lord and friend, Jupiter. Principle 6: If the avayogi is the lord of the ascendant, the native will tend toward material deprivation. This will be more marked if the avayogi is strong and able to give its obstructive results. This principle is somewhat less powerful than the avayogi principles listed above.
This person was born into a relatively comfortable family but has had a series of menial jobs and an income level well below the standard of his family and friends. Saturn is the avayogi of the chart. Principle 7: A connection between the yogi and the duplicate yogi by conjunction, mutual aspect etc. creates a powerful yoga for prosperity. If these two planets are also forming other yogas indicating material gains, this association greatly augments the indications of those yogas. The extent of the effect is again
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dependent on the strength of the planets involved and their ability to realize their potential in the chart.
The yogi of the chart is Venus and the duplicate yogi is Mars. They are conjoined in the ascendant, enlivening principle 3 as well as principle 7. Venus is the yogakaraka for this Capricorn ascendant and the duplicate yogi Mars is exalted. Both of these planets form several other yogas for success and prosperity. This person has the experience of abundant financial resources. Conclusion and food for thought: This has been an introduction to some of the basic principles of Iyers methods that incline a chart towards prosperity or poverty. Many other factors play into these dynamics including the influence of the yogi, duplicate yogi and avayogi as participants in yoga combinations, their influence by transits, the effects of their dasha periods, their role in divisional chart analysis, etc. For example, consider the situation where the yogi planet is transiting the yogi nakshatra during its dasha. This can be a powerful indication for timing a prosperous period. Consider also the times when, by transit, the duplicate yogi and the yogi reform a yoga combination present in the natal chart. This too may lead to striking material success. These are topics to be presented in more detail for future study and consideration. Initially, the most effective way to gain skills in applying these principles is to find those charts in which they operate cleanly and elegantly. In these cases, the results should be self evident. When you find charts that are not working, put them aside for future study. As more principles are introduced and assimilated, you may find that these problem charts become excellent examples of more subtle and interesting dynamics.
Source: http://www.jupitersweb.com/iyer_point_of_prosperity.htm
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# Edgeworth series
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### Edgeworth series
The Gram–Charlier A series (named in honor of Jørgen Pedersen Gram and Carl Charlier), and the Edgeworth series (named in honor of Francis Ysidro Edgeworth) are series that approximate a probability distribution in terms of its cumulants.[1] The series are the same; but, the arrangement of terms (and thus the accuracy of truncating the series) differ.[2]
## Contents
• Gram–Charlier A series 1
• The Edgeworth series 2
• Illustration: density of the sample mean of 3 Χ² 3
• Disadvantages of the Edgeworth expansion 4
• References 6
## Gram–Charlier A series
The key idea of these expansions is to write the characteristic function of the distribution whose probability density function F is to be approximated in terms of the characteristic function of a distribution with known and suitable properties, and to recover F through the inverse Fourier transform.
We examine a continuous random variable. Let f be the characteristic function of its distribution whose density function is F, and \kappa_r its cumulants. We expand in terms of a known distribution with probability density function Ψ, characteristic function ψ, and cumulants \gamma_r. The density Ψ is generally chosen to be that of the normal distribution, but other choices are possible as well. By the definition of the cumulants, we have (see Wallace, 1958)[3]
f(t)= \exp\left[\sum_{r=1}^\infty\kappa_r\frac{(it)^r}{r!}\right] and
\psi(t)=\exp\left[\sum_{r=1}^\infty\gamma_r\frac{(it)^r}{r!}\right],
which gives the following formal identity:
f(t)=\exp\left[\sum_{r=1}^\infty(\kappa_r-\gamma_r)\frac{(it)^r}{r!}\right]\psi(t)\,.
By the properties of the Fourier transform, (it)^r\psi(t) is the Fourier transform of (-1)^r[D^r\Psi](-x), where D is the differential operator with respect to x. Thus, after changing x with -x on both sides of the equation, we find for F the formal expansion
F(x) = \exp\left[\sum_{r=1}^\infty(\kappa_r - \gamma_r)\frac{(-D)^r}{r!}\right]\Psi(x)\,.
If Ψ is chosen as the normal density with mean and variance as given by F, that is, mean \mu = \kappa_1 and variance \sigma^2 = \kappa_2, then the expansion becomes
F(x) = \exp\left[\sum_{r=3}^\infty\kappa_r\frac{(-D)^r}{r!}\right]\frac{1}{\sqrt{2\pi}\sigma}\exp\left[-\frac{(x-\mu)^2}{2\sigma^2}\right].
since \gamma_r=0 for all r >2 as higher cumulants of the normal distribution are 0. By expanding the exponential and collecting terms according to the order of the derivatives, we arrive at the Gram–Charlier A series. If we include only the first two correction terms to the normal distribution, we obtain
F(x) \approx \frac{1}{\sqrt{2\pi}\sigma}\exp\left[-\frac{(x-\mu)^2}{2\sigma^2}\right]\left[1+\frac{\kappa_3}{3!\sigma^3}H_3\left(\frac{x-\mu}{\sigma}\right)+\frac{\kappa_4}{4!\sigma^4}H_4\left(\frac{x-\mu}{\sigma}\right)\right]\,,
with H_3(x)=x^3-3x and H_4(x)=x^4 - 6x^2 + 3 (these are Hermite polynomials).
Note that this expression is not guaranteed to be positive, and is therefore not a valid probability distribution. The Gram–Charlier A series diverges in many cases of interest—it converges only if F(x) falls off faster than \exp(-(x^2)/4) at infinity (Cramér 1957). When it does not converge, the series is also not a true asymptotic expansion, because it is not possible to estimate the error of the expansion. For this reason, the Edgeworth series (see next section) is generally preferred over the Gram–Charlier A series.
## The Edgeworth series
Edgeworth developed a similar expansion as an improvement to the central limit theorem.[4] The advantage of the Edgeworth series is that the error is controlled, so that it is a true asymptotic expansion.
Let {Xi} be a sequence of independent and identically distributed random variables with mean μ and variance σ2, and let Yn be their standardized sums:
Y_n = \frac{1}{\sqrt{n}} \sum_{i=1}^n \frac{X_i - \mu}{\sigma}.
Let Fn denote the cumulative distribution functions of the variables Yn. Then by the central limit theorem,
\lim_{n\to\infty} F_n(x) = \Phi(x) \equiv \int_{-\infty}^x \tfrac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}q^2}dq
for every x, as long as the mean and variance are finite.
Now assume that the random variables Xi have mean μ, variance σ2, and higher cumulants κrrλr. If we expand in terms of the standard normal distribution, that is, if we set
\Psi(x)=\frac{1}{\sqrt{2\pi}}\exp(-\tfrac{1}{2}x^2)
then the cumulant differences in the formal expression of the characteristic function fn(t) of Fn are
\kappa^{F(n)}_1-\gamma_1 = 0,
\kappa^{F(n)}_2-\gamma_2 = 0,
\kappa^{F(n)}_r-\gamma_r = \frac{\kappa_r}{\sigma^rn^{r/2-1}} = \frac{\lambda_r}{n^{r/2-1}}; \qquad r\geq 3.
The Edgeworth series is developed similarly to the Gram–Charlier A series, only that now terms are collected according to powers of n. Thus, we have
f_n(t)=\left[1+\sum_{j=1}^\infty \frac{P_j(it)}{n^{j/2}}\right] \exp(-t^2/2)\,,
where Pj(x) is a polynomial of degree 3j. Again, after inverse Fourier transform, the density function Fn follows as
F_n(x) = \Phi(x) + \sum_{j=1}^\infty \frac{P_j(-D)}{n^{j/2}} \Phi(x)\,.
The first five terms of the expansion are[5]
\begin{align} F_n(x) &= \Phi(x) \\ &\quad -\frac{1}{n^{\frac{1}{2}}}\left(\tfrac{1}{6}\lambda_3\,\Phi^{(3)}(x) \right) \\ &\quad +\frac{1}{n}\left(\tfrac{1}{24}\lambda_4\,\Phi^{(4)}(x) + \tfrac{1}{72}\lambda_3^2\,\Phi^{(6)}(x) \right) \\ &\quad -\frac{1}{n^{\frac{3}{2}}}\left(\tfrac{1}{120}\lambda_5\,\Phi^{(5)}(x) + \tfrac{1}{144}\lambda_3\lambda_4\,\Phi^{(7)}(x) + \tfrac{1}{1296}\lambda_3^3\,\Phi^{(9)}(x)\right) \\ &\quad + \frac{1}{n^2}\left(\tfrac{1}{720}\lambda_6\,\Phi^{(6)}(x) + \left(\tfrac{1}{1152}\lambda_4^2 + \tfrac{1}{720}\lambda_3\lambda_5\right)\Phi^{(8)}(x) + \tfrac{1}{1728}\lambda_3^2\lambda_4\,\Phi^{(10)}(x) + \tfrac{1}{31104}\lambda_3^4\,\Phi^{(12)}(x) \right)\\ &\quad + O \left (n^{-\frac{5}{2}} \right ). \end{align}
Here, Φ(j)(x) is the j-th derivative of Φ(·) at point x. Remembering that the derivatives of the density of the normal distribution are related to the normal density by ϕ(n)(x)=(-1)nHn(x)ϕ(x), (where Hn is the Hermite polynomial of order n), this explains the alternative representations in terms of the density function. Blinnikov and Moessner (1998) have given a simple algorithm to calculate higher-order terms of the expansion.
Note that in case of a lattice distributions (which have discrete values), the Edgeworth expansion must be adjusted to account for the discontinuous jumps between lattice points.[6]
## Illustration: density of the sample mean of 3 Χ²
Density of the sample mean of three chi2 variables. The chart compares the true density, the normal approximation, and two edgeworth expansions
Take X_i \sim \chi^2(k=2) \qquad i=1, 2, 3 and the sample mean \bar X = \frac{1}{3} \sum_{i=1}^{3} X_i .
We can use several distributions for \bar X :
• The exact distribution, which follows a gamma distribution: \bar X \sim \mathrm{Gamma}\left(\alpha=n\cdot k /2, \theta= 2/n \right) = \mathrm{Gamma}\left(\alpha=3, \theta= 2/3 \right)
• The asymptotic normal distribution: \bar X \xrightarrow{n \to \infty} N(k, 2\cdot k /n ) = N(2, 4/3 )
• Two Edgeworth expansion, of degree 2 and 3
## Disadvantages of the Edgeworth expansion
Edgeworth expansions can suffer from a few issues:
• They are not guaranteed to be a proper probability distribution as:
• The integral of the density needs not integrate to 1
• Probabilities can be negative
• They can be inaccurate, especially in the tails, due to mainly two reasons:
• They are obtained under a Taylor series around the mean
• They guarantee (asymptotically) an absolute error, not a relative one. This is an issue when one wants to approximate very small quantities, for which the absolute error might be small, but the relative error important.
## References
1. ^ Stuart, A., & Kendall, M. G. (1968). The advanced theory of statistics. Hafner Publishing Company.
2. ^ Kolassa, J. E. (2006). Series approximation methods in statistics (Vol. 88). Springer Science & Business Media.
3. ^ Wallace, D. L. (1958). Asymptotic approximations to distributions. The Annals of Mathematical Statistics, 635-654.
4. ^ Hall, P. (2013). The bootstrap and Edgeworth expansion. Springer Science & Business Media.
5. ^ Weisstein, Eric W., "Edgeworth Series", MathWorld.
6. ^ Kolassa, John E.; McCullagh, Peter (1990). "Edgeworth series for lattice distributions".
• H. Cramér. (1957). Mathematical Methods of Statistics. Princeton University Press, Princeton.
• D. L. Wallace. (1958). "Asymptotic approximations to distributions". Annals of Mathematical Statistics, 29: 635–654.
• M. Kendall & A. Stuart. (1977), The advanced theory of statistics, Vol 1: Distribution theory, 4th Edition, Macmillan, New York.
• P. McCullagh (1987). Tensor Methods in Statistics. Chapman and Hall, London.
• D. R. Cox and O. E. Barndorff-Nielsen (1989). Asymptotic Techniques for Use in Statistics. Chapman and Hall, London.
• P. Hall (1992). The Bootstrap and Edgeworth Expansion. Springer, New York.
• Hazewinkel, Michiel, ed. (2001), "Edgeworth series",
• S. Blinnikov and R. Moessner (1998). Expansions for nearly Gaussian distributions. Astronomy and astrophysics Supplement series, 130: 193–205.
• J. E. Kolassa (2006). Series Approximation Methods in Statistics (3rd ed.). (Lecture Notes in Statistics #88). Springer, New York.
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Crowd sourced content that is contributed to World Heritage Encyclopedia is peer reviewed and edited by our editorial staff to ensure quality scholarly research articles. | 3,122 | 10,671 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2021-04 | latest | en | 0.818281 |
https://mathoverflow.net/questions/314835/characterizing-bounded-distributivity-in-terms-of-dense-open-sets | 1,624,373,566,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488517820.68/warc/CC-MAIN-20210622124548-20210622154548-00480.warc.gz | 361,133,243 | 39,005 | # Characterizing “bounded” distributivity in terms of dense open sets
Prikry forcing is a well-known example of a forcing which adds an $$\omega$$-sequence to some cardinal $$\kappa$$, but does not add bounded subsets to $$\kappa$$.
There are other examples of forcings which add $$\omega$$-sequences, but do not add reals (for example Namba forcing in the presence of $$\sf CH$$).
In these situations the forcing is not $$\sigma$$-distributive, namely there is a countable sequence of dense open sets whose intersection is empty; but the added $$\omega$$-sequences live relatively "far up stairs" in terms of the von Neumann hierarchy.
Question. Is there a well-known characterizations of "$$\Bbb P$$ does not add new sets of rank $$\alpha$$" in terms of dense open sets and the intersection thereof?
• I guess that using properties of the Boolean completion of the forcing poset is not useful for this question, right? – Yair Hayut Nov 8 '18 at 13:55
• What do you mean? – Asaf Karagila Nov 8 '18 at 14:37
• Are you sure you mean "rank"? Because Namba adds new set of rank $\omega+8$. – Not Mike Dec 9 '18 at 8:59
• @NotMike: Yes, I'm pretty sure I mean rank. Also, since reals have rank $\omega$, I don't see how $\omega+8$ contradicts my statement. – Asaf Karagila Dec 9 '18 at 9:02
• @NotMike: For crying out loud, don't use $\Bbb N$ to denote stuff. It's enough we have $\Bbb Q$ and $\Bbb R$ outside their standard mathematical notation (and even $\Bbb C$ sometimes). But $\Bbb N$ is a step too far, mate! – Asaf Karagila Dec 9 '18 at 17:52
This is just a comment on the question (but too long for a comment box), not an answer.
Let $$\kappa$$ and $$\lambda$$ be cardinals. A complete Boolean algebra $$\mathbb B$$ is called $$(\kappa,\lambda)$$-distributive if $$\prod_{\alpha < \kappa} \ \sum_{\beta < \lambda} u_{\alpha,\,\beta} \,=\, \sum_{f: \kappa \rightarrow \lambda} \ \prod_{\alpha < \kappa} u_{\alpha,\,f(\alpha)}$$ for any $$u_{\alpha,\,\beta}$$ in $$\mathbb B$$. In Jech's set theory book (third edition, Theorem 15.38, p. 246) it is proved that
$$\mathbb B$$ is $$(\kappa,\lambda)$$-distributive if and only if every $$f: \kappa \rightarrow \lambda$$ in the generic extension by $$\mathbb B$$ is already in the ground model.
So for example, $$\mathbb B$$ does not add new reals if and only if it is $$(\aleph_0,2)$$-distributive, and (the Boolean completion of) the Prikry forcing is a good example of an algebra that is $$(\aleph_0,2)$$-distributive but not $$\aleph_0$$-distributive. (It adds new $$\omega$$-sequences, but not new reals.)
Using the highlighted theorem above, we can deduce that
$$\mathbb B$$ adds a set of rank $$\leq\!\alpha$$ if and only if it fails to be $$(|V_\alpha|,2)$$-distributive.
The reason is that a set of rank $$\leq\!\alpha$$ is just a subset of $$V_\alpha$$, and can be identified with a characteristic function $$V_\alpha \rightarrow 2$$. So getting new sets of rank $$\leq\!\alpha$$ is equivalent to getting new functions of this kind.
Of course, I'm not sure this directly answers your question (because $$(\kappa,\lambda)$$-distributivity is not defined in terms of dense open sets). But here is something a little closer: $$\mathbb B$$ is $$(\kappa,\lambda)$$-distributive if and only if every collection of $$\kappa$$ partitions of $$\mathbb B$$, each containing at most $$\lambda$$ members of $$\mathbb B$$, has a common refinement. (This is Lemma 15.37 from Jech.)
• Right. I am aware of these things that you mention. But they are not given in terms of open sets, and ultimately I want to apply them (also) in ZF, where chain conditions are meaningless (so the common refinement argument is somewhat irrelevant, or at least much harder to appeal to). – Asaf Karagila Nov 8 '18 at 14:37
• OK, that's fair enough -- I'll leave this here in case anyone finds it helpful, but you're right that it's not really an answer. – Will Brian Nov 8 '18 at 14:38
• I wonder if you can get a version that makes sense with $\neg\text{AC}$ by using pre-dense sets, instead of antichains. Something like: the intersection of $\kappa$ many dense sets, each having a pre-dense subset of size $\lambda$, is dense. – Joel David Hamkins Jan 10 '19 at 14:11
This is just a translation of @WillBrian's answer into a statement about dense sets.
For each $$\delta \in \mathsf{On}$$, fix $$\dot{R}_\delta \in V^{\mathbb{P}}$$ of minimal rank, such that $$1 \Vdash_\mathbb{P} \dot{R}_\delta = \{ x : \mathsf{rank}(x) < \check{\delta}\}$$. Then, because of how the rank function works, we always have that the least $$\delta \in \mathsf{On}$$, such that $$1 \Vdash_{\mathbb{P}} \dot{R}_\delta \neq \check{V}_\delta$$ is a successor; so the thing to do is characterize when $$1 \Vdash_{\mathbb{P}} \dot{R}_{\gamma + 1} = \check{V}_{\gamma + 1}$$. To this end,
Theorem: For every $$\delta \in \mathsf{On}$$ such that $$1 \Vdash_{\mathbb{P}} \dot{R}_\delta = \check{V}_\delta$$, the following are equivalent,
1. $$1 \Vdash_{\mathbb{P}} \dot{R}_{\delta+1} = \check{V}_{\delta+1}$$,
2. For every $$p \in \mathbb{P}$$ and every $$\{ \mathcal{B}_x: x \in V_{\delta} \} \subset \mathcal{P}(\mathbb{P})$$, there is some $$p_0 \le p$$, such that, for any $$x \in V_{\delta}$$, either $$\{q_0 \in \mathbb{P}: (\exists s \in \mathcal{B}_x)( q_0 \le s)\}\text{, or }\{q_0 \in \mathbb{P}: (\forall s \in \mathcal{B}_x)( q_0 \perp s)\},$$ is dense below $$p_0$$.
Proof: Fix $$\delta \in \mathsf{On}$$, and assume that $$1 \Vdash_{\mathbb{P}} \dot{R}_{\delta+1} = \check{V}_{\delta+1}$$. Then, for every $$p\in \mathbb{P}$$ and $$\dot{g} \in V^{\mathbb{P}}$$ such that $$p \Vdash_{\mathbb{P}} \dot{g}:\check{V}_\delta \rightarrow \check{2}$$, the set $$\{ q \le p : (\exists f: V_\delta\rightarrow 2)(q \Vdash_{\mathbb{P}} \dot{g} = \check{f} ) \}$$ is dense. So let $$\{ \mathcal{B}_x : x \in V_\delta\} \subset \mathcal{P}(\mathbb{P})$$ be arbitrary, and define $$\dot{g},\sigma_x, \tau_x \in V^{\mathbb{P}}$$ ($$x\in V_\delta$$) as follows $$\dot{g} = \{ \langle\sigma_x, 1\rangle: x\in V_\delta \}$$, $$\sigma_x = \mathsf{pair}_\mathbb{P}(\check{x}, \tau_x)$$, and $$\tau_x = \{ \langle \emptyset, p \rangle: p \in \mathcal{B}_x \}$$. Then, $$1 \Vdash_{\mathbb{P}} \dot{g}: \check{V}_\delta \rightarrow \check{2}$$ and so by assumption the set $$\{ q \in \mathbb{P}: (\exists f: V_\delta\rightarrow 2)(q \Vdash_{\mathbb{P}} \dot{g} = \check{f} )\}$$ is dense open, hence given $$p \in \mathbb{P}$$ we can find $$p_0 \le p$$ and $$f:V_\delta \rightarrow 2$$ such that, for every $$x \in V_\delta$$, $$p_0 \Vdash \dot{g}(\check{x}) = \check{f}(\check{x})$$ or equivalently, $$p_0 \Vdash_{\mathbb{P}} "\tau_x = \check{f}(\check{x})" \iff$$ $$f(x) = 1$$ and $$\{q_0 \in \mathbb{P}: (\exists s \in \mathcal{B}_x)( q_0 \le s)\}$$ is dense below $$p_0$$, or $$f(x)=0$$ and $$\{q_0 \in \mathbb{P}: (\forall s \in \mathcal{B}_x)( q_0 \perp s)\}$$ is dense below $$p_0$$.
For the converse, assume (2), then fix $$p\in\mathbb{P}$$ and $$\dot{a}\in V^{\mathbb{P}}$$ such that $$p \Vdash_\mathbb{P} \dot{a} \in \dot{R}_{\delta+1}$$. Then, for $$x\in V_\delta$$ defining $$\mathcal{B}_x = \{ q \in \mathbb{P}: q\Vdash_{\mathbb{P}} \check{x} \in \dot{a} \}$$, we can find $$p_0 \le p$$ (by 2) such that for every $$x$$, either $$\{q_0 \in \mathbb{P}: (\exists s \in \mathcal{B}_x)( q_0 \le s)\}\text{, or }\{q_0 \in \mathbb{P}: (\forall s \in \mathcal{B}_x)( q_0 \perp s)\},$$ is dense below $$p_0$$. As such, letting $$A = \{ x \in V_\delta : \{q_0 \in \mathbb{P}: (\exists s \in \mathcal{B}_x)( q_0 \le s)\} \text{ is dense below } p_0 \}$$ we have $$p_0 \Vdash \dot{a} = \check{A} \in \check{V}_{\delta+1}$$. $$\square$$
Remark: If in (2), the requirement "either $$A^x_1=\{q_0 \in \mathbb{P}: (\exists s \in \mathcal{B}_x)( q_0 \le s)\}$$, or $$A^x_0=\{q_0 \in \mathbb{P}: (\forall s \in \mathcal{B}_x)( q_0 \perp s)\}$$ is dense below $$p_0$$", is changed to "$$p_0$$ is an element of $$A^x_0$$ or $$A^x_1$$", then the new statement is equivalent to $$\mathbb{P}$$ being $$\vert V_\delta \vert$$-baire.
The distributivity properties such as $$(\kappa,\lambda)$$-distributivity that characterizes when a forcing extension does not add any new functions from $$\kappa$$ to $$\lambda$$ are actually instances of the property “the intersection of at most $$\kappa$$ many open sets is open ” and to a lesser extent the $$\kappa$$-Baire property when we consider the complete Boolean algebra as a topological space whose points are the generic ultrafilters. Since complete Boolean algebras are frames and not topological spaces, and since the points in complete Boolean algebras only exist in forcing extensions, we shall formulate this answer in terms of point-free topology. For a background in point-free topology, please see the textbook Frames and Locales-Topology without points by Jorge Picado and Ales Pultr. And be sure to be aware of the result in Will Brian's answer before reading this one.
To see how distributivity relates to the intersection of open sets, let us look at some point-free topology.
For this post, we shall always assume that $$\kappa$$ is an uncountabe regular cardinal. A regular space is said to be a $$P$$-space if the intersection of countably many open sets is open. It is not too hard to show that every regular $$P$$-space is zero-dimensional. More generally, if $$\kappa$$ is a regular cardinal, then a regular space $$X$$ is a $$P_{<\kappa}$$-space if whenever $$U_{i}$$ is open for $$i\in I$$ and $$|I|<\kappa$$, the intersection $$\bigcap_{i\in I}U_{i}$$ is open. We say that a space $$X$$ is a $$<\lambda$$-Baire space if the intersection of less than $$\lambda$$ many dense open sets is still dense.
For example, if $$X$$ is any $$P_{<\kappa}$$-space and $$C_{n}$$ is a subspace of $$X$$ for $$n\in\omega$$ where each $$C_{n}$$ is nowhere dense in $$C_{n+1}$$, then the union $$\bigcup_{n\in\omega}C_{n}$$ is a $$P_{<\kappa}$$-space which is not a Baire space. Therefore, the notions of being a $$<\kappa$$-Baire space is independent of the notion of being a $$P_{<\kappa}$$-space. See this answer for more details about the relations between the $$<\kappa$$-Baire spaces and the $$P_{<\kappa}$$-spaces.
There are several different ways to generalize the notion of a $$P_{<\kappa}$$-space to point-free topology, and some of these generalizations are simply versions of distributivity.
A frame is a complete lattice that satisfies the infinite distributivity law $$x\wedge\bigvee_{i\in I}y_{i}= \bigvee_{i\in I}(x\wedge y_{i}).$$ Frames are the objects that people study in the field of point-free topology; if $$(X,\mathcal{T})$$ is a topological space, then $$\mathcal{T}$$ is a frame.
We say that a frame $$L$$ is $$(<\kappa,\lambda)$$-distributive if whenever $$J$$ is a set with $$|J|<\kappa$$ and $$C_{j}\subseteq L,|C_{j}|\leq\lambda$$ for each $$j\in J$$, then $$\bigwedge_{j\in J}\bigvee C_{j}=\bigvee\{\bigwedge_{j\in J}x_{j}\mid x_{j}\in C_{j}\,\text{for}\,j\in J\}.$$ We say $$L$$ is $$<\kappa$$-distributive if it is $$(<\kappa,\lambda)$$-distributive for all $$\lambda$$.
We say that a frame $$L$$ is weakly $$<\kappa$$-distributive if whenever $$|I|<\kappa$$, we have $$x\vee\bigwedge_{i\in I}y_{i}=\bigwedge_{i\in I}(x\vee y_{i}).$$
Every complete Boolean algebra is weakly $$<\kappa$$-distributive for each regular cardinal $$\kappa$$.
Theorem: Let $$(X,\mathcal{T})$$ be a regular space. Then the following are equivalent:
1. $$(X,\mathcal{T})$$ is a $$P_{<\kappa}$$-space.
2. $$\mathcal{T}$$ is $$<\kappa$$-distributive.
3. $$\mathcal{T}$$ is weakly $$<\kappa$$-distributive.
4. $$\mathcal{T}$$ is $$(<\kappa,2)$$-distributive.
5. $$\mathcal{T}$$ is $$(<\kappa,\lambda)$$-distributive.
Observation: Suppose that $$L$$ is a regular frame and $$\lambda_{2}\leq \lambda_{1}$$. Each of the following conditions implies the next one.
1. $$L$$ is $$<\kappa$$-distributive.
2. $$L$$ is $$(<\kappa,\lambda_{1})$$-distributive.
3. $$L$$ is $$(<\kappa,\lambda_{2})$$-distributive.
4. $$L$$ is $$(<\kappa,2)$$-distributive.
5. $$L$$ is weakly $$<\kappa$$-distributive.
Theorem: A regular frame $$L$$ is weakly $$<\kappa$$-distributive if and only if whenever $$|I|<\kappa$$ and $$C_{i}$$ is a closed sublocale of $$L$$ for $$i\in I$$, the least upper bound $$\bigvee_{i\in I}C_{i}$$ in the lattice of sublocales of $$L$$ is closed.
Weak $$<\kappa$$-distributivity is too weak of a generalization of the notion of a $$P_{<\kappa}$$-space to distinguish between complete Boolean algebras.
Theorem: A regular frame $$L$$ is a complete Boolean algebra if and only if it is weakly $$<\kappa$$-distributive for all cardinals $$\kappa.$$
By the above results and observations, we realize that the notions of $$(<\kappa,2),(<\kappa,\lambda),<\kappa$$-distributivity are actually a generalizations of the notion of a $$P_{<\kappa}$$-space to point-free topology.
The notion of a $$P_{<\kappa}$$-space is quite unusual; most well-known notions of general topology extend without a problem to point-free topology to only one point-free notion. However, there are many ways of generalizing the notion of a $$P_{<\kappa}$$-space to point-free topology, and several of these notions (such as that of a $$P$$-frame and a $$P_{0}$$-frame) do not even refer to distributivity at all. Furthermore, all of these non-equivalent generalizations of the notion of a $$P_{<\kappa}$$-space all look like satisfactory notions. However, with everything being said, these generalizations of the notion of a $$P_{<\kappa}$$-space have barely been studied in the context of point-free topology, so much research still needs to be done in this area.
Baire and dense open elements
Every frame $$L$$ has a smallest dense sublocale, namely $$B_{L}=\{x|x^{**}=x\}$$. The set $$B_{L}$$ is a complete Boolean algebra and $$B_{L}$$ is closed under arbitrary meets. If $$L$$ is a frame, then let $$\mathfrak{B}(L)$$ denote the Boolean subalgebra of complemented elements in $$L.$$
If $$\kappa$$ is a regular cardinal and $$L$$ is a $$<\kappa$$-distributive frame, then we say that $$L$$ is a $$<\kappa$$-Baire frame if whenever $$|I|<\kappa$$ and $$x_{i}$$ is dense in $$L$$ for $$i\in I$$, then $$\bigwedge_{i\in I}x_{i}$$ is dense in $$L$$.
Theorem: Suppose that $$L$$ is a regular frame and $$\kappa$$ is an uncountable regular cardinal.
1. If $$B_{L}$$ is $$<\kappa$$-distributive, then whenever $$|I|<\kappa$$ and $$x_{i}$$ is dense in $$L$$ for $$i\in I$$, then $$\bigwedge_{i\in I}x_{i}$$ is also dense in $$L$$.
2. Suppose $$L$$ is a $$<\kappa$$-distributive regular frame. Then $$L$$ is a $$<\kappa$$-Baire frame if and only if $$B_{L}$$ is $$<\kappa$$-distributive.
Proof: 1. Suppose that $$B_{L}$$ is $$<\kappa$$-distributive, $$|I|<\kappa$$, and $$x_{i}$$ is dense in $$L$$ for each $$i\in I$$. Then for each $$i\in I$$, there is some partition $$p_{i}$$ of $$B_{L}$$ where $$\bigvee^{L}p_{i}\leq x_{i}$$. Since $$B_{L}$$ is $$<\kappa$$, distributive, there is a partition $$p$$ of $$B_{L}$$ that refines each $$p_{i}$$. However, if $$a\in p$$ and $$i\in I$$, there there is some $$b\in p_{i}$$ where $$a\leq b\leq x_{i}$$. Therefore, $$a\leq\bigwedge_{i\in I}x_{i}$$. Thus, $$\bigvee^{L}p\leq\bigwedge_{i\in I}x_{i}$$. Since $$p$$ is a partition of $$B_{L}$$, $$\bigvee^{L}p$$ is dense in $$L$$, so $$\bigwedge_{i\in I}x_{i}$$ is also dense in $$L$$.
1. Suppose that $$L$$ is $$<\kappa$$-Baire and $$<\kappa$$-distributive. Let $$|I|<\kappa$$ and let $$p_{i}$$ be a partition of $$B_{L}$$ for each $$i\in I$$. Let $$x_{i}=\bigvee^{L}p_{i}$$ for each $$i\in I$$. Then $$\bigwedge_{i\in I}x_{i}$$ is dense. Now, $$\bigwedge_{i\in I}x_{i}=\bigwedge_{i\in I}\bigvee^{L}p_{i}=\bigvee^{L}\{\bigwedge_{i\in I}r_{i}|r_{i}\in p_{i}\,\text{for}\,i\in I\}$$. Since $$\bigvee^{L}\{\bigwedge_{i\in I}r_{i}|r_{i}\in p_{i}\,\text{for}\,i\in I\}$$ is dense in $$L$$, we conclude that $$\{\bigwedge_{i\in I}r_{i}|r_{i}\in p_{i}\,\text{for}\,i\in I\}\setminus\{0\}$$ is a partition of $$L$$ that refines each $$p_{i}$$. Therefore, $$B_{L}$$ is $$<\kappa$$-distributive. Q.E.D.
We say that $$x\in L$$ is regularly dense open if there are $$r,s\in B_{L}$$ where $$r\wedge s=0,r\vee^{B_{L}}s=1$$ and where $$r\vee^{L}s\leq x$$.
Theorem: Suppose that $$L$$ is a frame.
1. If $$B_{L}$$ is $$(<\kappa,2)$$-distributive, then whenever $$x_{i}$$ is regularly dense open for $$i\in I$$ and $$|I|<\kappa$$, then $$\bigwedge_{i\in I}x_{i}$$ is dense.
2. If $$L$$ is $$(<\kappa,2)$$-distributive, then $$B_{L}$$ is $$(<\kappa,2)$$-distributive if and only if whenever $$x_{i}$$ is regularly dense open for $$i\in I$$ and $$|I|<\kappa$$, then $$\bigwedge_{i\in I}x_{i}$$ is dense.
Proof: 1. Suppose that $$B_{L}$$ is $$(<\kappa,2)$$-distributive and assume that $$x_{i}$$ is strongly dense for $$i\in I$$ and $$|I|<\kappa$$. Then for $$i\in I$$, there are $$r_{i,0},r_{i,1}\in B_{L}$$ which are complements in $$B_{L}$$ but where $$r_{i,0}\vee^{L}r_{i,1}\leq x_{i}$$. By the $$(<\kappa,2)$$-distributivity of $$B_{L}$$, we have $$1= \bigvee^{B_{L}}\{\bigwedge_{i\in I}r_{i,a(i)}|a:I\rightarrow 2\}$$ and therefore $$\bigvee^{L}\{\bigwedge_{i\in I}r_{i,a(i)}|a:I\rightarrow 2\}$$ is dense in $$L$$. However, $$\bigvee^{L}\{\bigwedge_{i\in I}r_{i,a(i)}|a:I\rightarrow 2\}\leq \bigwedge_{i\in I}x_{i},$$ so $$\bigwedge_{i\in I}x_{i}$$ is dense in $$L$$.
1. We have already proven the direction $$\rightarrow$$, so let's embark on a proof of the direction $$\leftarrow$$. Suppose that whenever $$|I|<\kappa$$ and $$x_{i}$$ is regularly dense open for $$i\in I$$, then $$\bigwedge_{i\in I}x_{i}$$ is regularly dense open. Now, let $$r_{i,0},r_{i,1}$$ be complements in $$B_{L}$$ for $$i\in I$$. Then let $$x_{i}=r_{i,0}\vee^{L}r_{i,1}$$. Then $$\bigwedge_{i\in I}x_{i}$$ is dense. Therefore, by the $$(<\kappa,2)$$-distributivity of $$L$$, we have $$\bigwedge_{i\in I}x_{i}=\bigwedge_{i\in I}(r_{i,0}\vee^{L}r_{i,1})=\bigvee^{L}\{\bigwedge_{i\in I}R_{i,a(i)}|a:I\rightarrow 2\},$$ so since $$\bigvee^{L}\{\bigwedge_{i\in I}r_{i,a(i)}|a:I\rightarrow 2\}$$ is dense in $$L$$, we conclude that $$\bigvee^{B_{L}}\{\bigwedge_{i\in I}r_{i,a(i)}|a:I\rightarrow 2\}=1.$$ Therefore, $$B_{L}$$ is $$(<\kappa,2)$$-distributive. Q.E.D.
I will leave it an an exercise to generalize the above result to $$(<\kappa,\lambda)$$-distributivity.
Corollary: Suppose that $$L$$ is an extremally disconnected frame. If $$L$$ is $$(<\kappa,2)$$-distributive, then $$B_{L}$$ is also $$(<\kappa,2)$$-distributive.
Proof: Observe that each element in $$B_{L}$$ is complemented in $$L$$ and if $$r,s\in B_{L}$$ are complements, then $$r\vee^{L}s=1$$. Therefore, the only regularly dense open element in $$L$$ is $$1$$. Therefore, by the above result, if $$R$$ is a collection of regularly dense open elements in $$L$$, then $$\bigwedge R\in L$$ as well. Q.E.D.
Example: Suppose that $$\mu$$ is a measurable cardinal and $$\mathcal{U}$$ is a normal ultrafilter on $$\mu$$. If $$A\in[\mu]^{<\omega},R\in\mathcal{U}$$, then let $$C_{A,R}=\{B\in[R]^{<\omega}\mid A\subseteq B\}.$$ Then give $$[\mu]^{<\omega}$$ the topology where the basic open sets are the sets of the form $$C_{A,R}$$ where $$\max(A)<\min(R),A\in[\mu]^{<\omega},R\in\mathcal{U}.$$ The space $$[\mu]^{<\omega}$$ is a $$P_{<\mu}$$-space.
We say that a topological space is extremally disconnected if the closure of every open set is open. Equivalently, a frame $$L$$ is extremally disconnected if each element of $$B_{L}$$ is complemented in $$L$$ and in this case $$\mathfrak{B}(L)=B_{L}$$. The motivation behind extremal disconnectedness is that a frame $$L$$ is extremally disconnected precisely when the Boolean algebra $$\mathfrak{B}(L)$$ of complemented elements of $$L$$ is a complete Boolean algebra. As observed in this answer, the space $$[\mu]^{<\omega}$$ is extremally disconnected, so $$\mathrm{Ro}([\mu]^{<\omega})$$ is precisely the set of all clopen subsets of $$[\mu]^{<\omega}$$. The Boolean algebra $$\mathrm{Ro}([\mu]^{\omega})$$ is the complete Boolean algebra for Prikry forcing. Furthermore, the extremal disconnectedness of $$\mathrm{Ro}([\mu]^{\omega})$$ is a reformulation of the lemma known as the Prikry condition which states that if $$P_{\mathcal{U}}$$ is the poset for Prikry forcing (i.e. $$(x,R)\in P_{\mathcal{U}}$$ if and only if $$x\in[\mu]^{<\omega},R\in\mathcal{U},\max(x)<\min(R)$$) , then whenever $$(x,R)\in P$$ and $$\theta$$ is a statement in the forcing language, then there is some $$S\subseteq R,S\in\mathcal{U}$$ where $$(x,S)$$ decides $$\theta$$.
We observe that each $$[\mu]^{\leq n}$$ is closed and $$[\mu]^{\leq n}$$ is nowhere dense in $$[\mu]^{\leq n+1}$$. Therefore, $$[\mu]^{<\omega}$$ is the union of countably many nowhere dense sets and therefore $$[\mu]^{<\omega}$$ is not a Baire space. Therefore, since $$[\mu]^{<\omega}$$ is not a Baire space, $$\mathrm{Ro}([\mu]^{<\omega})$$ is not $$<\aleph_{1}$$-distributive, so forcing with $$\mathrm{Ro}([\mu]^{<\omega})$$ adds a new sequence $$(x_{n})_{n\in\omega}$$. On the other hand, since $$\mathrm{Ro}([\mu]^{<\omega})$$ is extremally disconnected, the complete Boolean algebra $$\mathrm{Ro}([\mu]^{<\omega})$$ is $$(<\mu,2)$$-distributive and therefore $$\mathrm{Ro}([\mu]^{<\omega})$$ adds no new subsets to $$P(\alpha)$$ for $$\alpha<\mu$$.
• I downvoted this answer; it's a confused and overly complicated exposition of the elementary result that $\kappa$-representable $\kappa$-complete Boolean Algebras are $(\mu, \lambda)$-distributive for every $\mu, \lambda < \kappa$ (a Boolean Algebra is $\kappa$-representable, iff, it can be realized as, $CO(X)$ for some regular $P_{<\kappa}$-space $X$, iff, it can be realized as a $\kappa$-complete field of sets.) – Not Mike Apr 9 '19 at 13:55
• (See Proposition 14.4, p. 214; The Handbook of Boolean Algebras, Vol. 1. ) – Not Mike Apr 9 '19 at 14:09
• @NotMike. If you have a simplification of these results, then please post another answer. – Joseph Van Name Apr 9 '19 at 14:14
• @NotMike. You are incorrect about your claim about representability. So $<\kappa$-representability is typically defined as being isomorphic as a Boolean algebra to a $<\kappa$-field of sets. This is not equivalent to being $\mathrm{CO}(X)$ for a $P_{<\kappa}$-space (here I am assuming that $\mathrm{CO}(X)$ denotes the algebra of clopen sets). – Joseph Van Name Apr 9 '19 at 23:59
• If $B$ is any $<\kappa$-complete Boolean algebra which is not both complete and atomic and where every $<\kappa$-complete ultrafilter on $B$ is principal, then $B$ is not isomorphic to the algebra of all clopen sets of any $P_{<\kappa}$-space. If $X$ is a $P_{<\kappa}$-space and $M$ is the $<\kappa$-complete algebra of clopen sets and every $<\kappa$-complete ultrafilter on $M$ is principal, then for each $x\in X$, the set $\{R\in M|x\in R\}$ is a $<\kappa$-complete ultrafilter on $M$ and hence principal. Therefore, $X$ is the discrete space, and hence $B\simeq M=P(X)$. – Joseph Van Name Apr 9 '19 at 23:59 | 8,023 | 22,937 | {"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": 391, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-25 | latest | en | 0.841301 |
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If you are looking for the answer of what is whole number, you’ve got the right page. We have approximately 10 FAQ regarding what is whole number. Read it below.
## List down five whole numbers that round off to 1000.
Ask: List down five whole numbers that round off to 1000.
What is the greatest whole number that rounds off to 1000?
What is the least whole number that rounds off to 1000?
hi crush kapo ni migo
Step-by-step explanation:
## The sum of several whole numbers is equal to 16.
Ask: The sum of several whole numbers is equal to 16. What is
the greatest product of these whole numbers?
9+5+2
dwfewfegfefwfwfwfwf
## 2. What kind of number is P120P Is it a
Ask: 2. What kind of number is P120P Is it a whole number or
decimal? (Whole number)
it is a whole number
Step-by-step explanation:
basta
## what is a whole number but not a counting numbers?
Ask: what is a whole number but not a counting numbers?
There is only a number that is a whole number but not a counting number. That number is 0. It’s important to note that 0 (Zero) isn’t a Natural Number.
## 1. List the whole numbers that round off to 80.
Ask: 1. List the whole numbers that round off to 80.
a. What is the greatest whole number that rounds off to 80?
b. What is the least whole number that rounds off to 80?
Lint down five whole numbers that round off to 500.
1. 75, 76, 77, 78, 79, 80, 81, 82, 83, 84
2. 84
3. 75
495, 496, 497, 498, 499
#CarryOnLearning
## what is mixed number and whole number
Ask: what is mixed number and whole number
A mixed number is a fraction which there is a whole and a fraction
ex. 2 9/10
there is still a part with a whole part
A whole number is the opposite of mixed number, only a whole number is shown.
ex. 1, 2, 8
there are no fractions
## Mixed Number:
This are combination of whole numbers and fractions
3 1/2
2 and 3/4
5 1/8
## WholeNumbers:
A number without fraction
An Integer
1
8
2
#CarryOnLearning
#BagongAralin
## 3. What is the difference between whole numbers andcounting numbers?a.
Ask: 3. What is the difference between whole numbers and
counting numbers?
a. Whole numbers are not fractions while counting numbers
can be fractions.
b. Whole numbers are integers with counting numbers and
zero while counting numbers are whole numbers without
zero.
c. Whole numbers and counting numbers are just the same.
d. Whole numbers are real numbers while counting numbers
are natural numbers.
TAI
D. Is the one
Step-by-step explanation:
## if n is a whole number, for what values of
Ask: if n is a whole number, for what values of n is 24/n also a whole number?
Multiple of 24 because 24 divided by 24 is 1 and if its not a multiple by 24, the result will be a fraction of an irrational number
## What is the difference between whole numbers and counting numbers?A.
Ask: What is the difference between whole numbers and counting numbers?
A. Whole numbers are not fraction while counting numbers can be fraction.
B. Whole numbers are integers while with counting numbers and zero while counting numbers are whole numbers without zero.
C. Whole numbers and counting numbers are just the same.
D. Whole numbers and real numbers will counting numbers are natural numbers.
Letter A po
Step-by-step explanation:
not sure tho
A or B
Step-by-step explanation:
Because its a number
## 1.what are the whole number of 3.43 2. what are
Ask: 1.what are the whole number of 3.43
2. what are the whole number of 2.91
3. what are the whole number of 5.35
4. what are the whole number of 4.56
5. what are the whole number of 8.13
1. 3
2. 3
3. 5
4. 5
5. 8
Step-by-step explanation:
Do you mean rounding up numbers? If you do, here are the answers.
Here’s the general rule for rounding: If the number you are rounding is followed by 5, 6, 7, 8, or 9, round the number up.
1) 3.43
• a) 4 – round down
• b) 3 – the whole number
2) 2.91
• a) 9 – round up
• b) 3 – the whole number
3) 5.35
• a) 3 – round down
• b) 5 – the whole number
4) 4.56
• a) 5 – round up
• b) 5 – the whole number
5) 8.13
• a) 1 – round down
• b) 8 – the whole number
I hope it helps! Thank you! ❤️
Not only you can get the answer of what is whole number, you could also find the answers of 3. What is, what is a, what is mixed, 1.what are the, and List down five. | 1,235 | 4,356 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2023-50 | latest | en | 0.880495 |
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