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# Yahoo Web Search
1. ### The parallel sides of a trapezium are 10 cm and 20 cm respectively. AD = 6 cm and CB =8 cm . Find the area of the trapezium ?
... and CD are the parallel sides with lengths 10 cm and 20 cm , respectively. See the attached sketch. Let point ...
3 Answers · Science & Mathematics · 03/08/2020
2. ### The area of a sector of a circle with a radius of 15 cm is 75pi cm ^2. ?
The area of a sector of a circle with a radius of 15 cm is 75pi cm ^2. ? Total circle Area = π(15)^2 = 225π cm ^2...
11 Answers · Science & Mathematics · 28/05/2020
3. ### The perimeter of a rectangle is 56 cm . Find the lengths of the sides of the rectangle giving the maximum area.?
A=l*w 56= 2l + 2w; w= 28 - l A= l*(28 - l)= 28l - l² dA/dl= 28 - 2l= 0 l= 14 cm , w= 14 cm
4 Answers · Science & Mathematics · 19/05/2020
4. ### a spherical snowball is melting symmetrically at the rate of 4pi cubic cm per hr. how fast is the diameter changing when it is 20 cm ?
... dV/dt = 4πr².dr/dt so, when the diameter is 20 cm we have: -4π = 4πr².dr/dt Hence, dr/dt = -1/r² i.e. dr...
3 Answers · Science & Mathematics · 15/08/2020
5. ### The legs of an isosceles triangle are each 45 cm long.The measures of the base angles are each 70°. What is the measure of the third side?
apply the law of sine ..a................b ---------=-------------- .sin(A)........sin(B) ..45..............b --------- = ----------- .sin(70) ....sin(40) 45sin(40) = bsin(70) b = 30.78 cm or 31 cm answer//
2 Answers · Science & Mathematics · 06/10/2020
6. ### A gallon of homogenized, vitamin A, whole milk has a weight ranging from 8.48 - 8. 72 lbs. Determine its volume in cm ^3.?
1 US gallon is 231 cubic inches 1 inch is 2.54 cm 231 * 2.54^3 => 3,785.411784 cm ^3
3 Answers · Science & Mathematics · 03/09/2020
7. ### A cone has a diameter of 6 cm and a slant height of 5 cm . What is the exact volume (in cubic centimeters) of the cone?....?
...and h r² + h² = s² 3² + h² = 5² h = 4 cm V = 1/3 π(3)²(4) V = 12π Answer C
4 Answers · Science & Mathematics · 11/06/2020
8. ### How much is 1.1 cm in inches?
1.1 cm = 0.4330708661417323 in
6 Answers · Science & Mathematics · 01/07/2020
9. ### a sphere has the volume 36π cm ^3 .What area does the sphere have?
The volume of a sphere with radius "r" is given by: V = 4*π*r³/3 so: r = (3V/(4π))^(1/3) The surface area of a sphere with radius "r" is given by: A = 4π*r² You want to know the area given the volume...
4 Answers · Science & Mathematics · 31/10/2019
10. ### If ED = √8 cm and the area of ∆ABC = trapezoid ACDE, the length of AC is?
AC = 2. See below for explanation. Notice that there are two similar triangles, ABC and EBD. The area of EBD is twice that of ABC, since the trapezoid area is equal to that of triangle ABC. But, the square...
3 Answers · Science & Mathematics · 19/12/2019
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HuggingFaceTB/finemath
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# Traveling Salesman Problem: Problem-Based
This example shows how to use binary integer programming to solve the classic traveling salesman problem. This problem involves finding the shortest closed tour (path) through a set of stops (cities). In this case there are 200 stops, but you can easily change the `nStops` variable to get a different problem size. You'll solve the initial problem and see that the solution has subtours. This means the optimal solution found doesn't give one continuous path through all the points, but instead has several disconnected loops. You'll then use an iterative process of determining the subtours, adding constraints, and rerunning the optimization until the subtours are eliminated.
For the solver-based approach to this problem, see Traveling Salesman Problem: Solver-Based.
### Problem Formulation
Formulate the traveling salesman problem for integer linear programming as follows:
• Generate all possible trips, meaning all distinct pairs of stops.
• Calculate the distance for each trip.
• The cost function to minimize is the sum of the trip distances for each trip in the tour.
• The decision variables are binary, and associated with each trip, where each 1 represents a trip that exists on the tour, and each 0 represents a trip that is not on the tour.
• To ensure that the tour includes every stop, include the linear constraint that each stop is on exactly two trips. This means one arrival and one departure from the stop.
### Generate Stops
Generate random stops inside a crude polygonal representation of the continental U.S.
```load('usborder.mat','x','y','xx','yy'); rng(3,'twister') % Makes stops in Maine & Florida, and is reproducible nStops = 200; % You can use any number, but the problem size scales as N^2 stopsLon = zeros(nStops,1); % Allocate x-coordinates of nStops stopsLat = stopsLon; % Allocate y-coordinates n = 1; while (n <= nStops) xp = rand*1.5; yp = rand; if inpolygon(xp,yp,x,y) % Test if inside the border stopsLon(n) = xp; stopsLat(n) = yp; n = n+1; end end```
### Calculate Distances Between Points
Because there are 200 stops, there are 19,900 trips, meaning 19,900 binary variables (# variables = 200 choose 2).
Generate all the trips, meaning all pairs of stops.
`idxs = nchoosek(1:nStops,2);`
Calculate all the trip distances, assuming that the earth is flat in order to use the Pythagorean rule.
```dist = hypot(stopsLat(idxs(:,1)) - stopsLat(idxs(:,2)), ... stopsLon(idxs(:,1)) - stopsLon(idxs(:,2))); lendist = length(dist);```
With this definition of the `dist` vector, the length of a tour is
`dist'*trips`
where `trips` is the binary vector representing the trips that the solution takes. This is the distance of a tour that you try to minimize.
### Create Graph and Draw Map
Represent the problem as a graph. Create a graph where the stops are nodes and the trips are edges.
`G = graph(idxs(:,1),idxs(:,2));`
Display the stops using a graph plot. Plot the nodes without the graph edges.
```figure hGraph = plot(G,'XData',stopsLon,'YData',stopsLat,'LineStyle','none','NodeLabel',{}); hold on % Draw the outside border plot(x,y,'r-') hold off```
### Create Variables and Problem
Create an optimization problem with binary optimization variables representing the potential trips.
```tsp = optimproblem; trips = optimvar('trips',lendist,1,'Type','integer','LowerBound',0,'UpperBound',1);```
Include the objective function in the problem.
`tsp.Objective = dist'*trips;`
### Constraints
Create the linear constraints that each stop has two associated trips, because there must be a trip to each stop and a trip departing each stop.
Use the graph representation to identify all trips starting or ending at a stop by finding all edges connecting to that stop. For each stop, create the constraint that the sum of trips for that stop equals two.
```constr2trips = optimconstr(nStops,1); for stop = 1:nStops whichIdxs = outedges(G,stop); % Identify trips associated with the stop constr2trips(stop) = sum(trips(whichIdxs)) == 2; end tsp.Constraints.constr2trips = constr2trips;```
### Solve Initial Problem
The problem is ready to be solved. To suppress iterative output, turn off the default display.
```opts = optimoptions('intlinprog','Display','off'); tspsol = solve(tsp,'options',opts)```
```tspsol = struct with fields: trips: [19900×1 double] ```
### Visualize Solution
Create a new graph with the solution trips as edges. To do so, round the solution in case some values are not exactly integers, and convert the resulting values to `logical`.
```tspsol.trips = logical(round(tspsol.trips)); Gsol = graph(idxs(tspsol.trips,1),idxs(tspsol.trips,2),[],numnodes(G)); % Gsol = graph(idxs(tspsol.trips,1),idxs(tspsol.trips,2)); % Also works in most cases```
Overlay the new graph on the existing plot and highlight its edges.
```hold on highlight(hGraph,Gsol,'LineStyle','-') title('Solution with Subtours')```
As can be seen on the map, the solution has several subtours. The constraints specified so far do not prevent these subtours from happening. In order to prevent any possible subtour from happening, you would need an incredibly large number of inequality constraints.
### Subtour Constraints
Because you can't add all of the subtour constraints, take an iterative approach. Detect the subtours in the current solution, then add inequality constraints to prevent those particular subtours from happening. By doing this, you find a suitable tour in a few iterations.
Eliminate subtours with inequality constraints. An example of how this works is if you have five points in a subtour, then you have five lines connecting those points to create the subtour. Eliminate this subtour by implementing an inequality constraint to say there must be less than or equal to four lines between these five points.
Even more, find all lines between these five points, and constrain the solution not to have more than four of these lines present. This is a correct constraint because if five or more of the lines existed in a solution, then the solution would have a subtour (a graph with $n$ nodes and $n$ edges always contains a cycle).
Detect the subtours by identifying the connected components in `Gsol`, the graph built with the edges in the current solution. `conncomp` returns a vector with the number of the subtour to which each edge belongs.
```tourIdxs = conncomp(Gsol); numtours = max(tourIdxs); % Number of subtours fprintf('# of subtours: %d\n',numtours);```
```# of subtours: 27 ```
Include the linear inequality constraints to eliminate subtours, and repeatedly call the solver, until just one subtour remains.
```% Index of added constraints for subtours k = 1; while numtours > 1 % Repeat until there is just one subtour % Add the subtour constraints for ii = 1:numtours inSubTour = (tourIdxs == ii); % Edges in current subtour a = all(inSubTour(idxs),2); % Complete graph indices with both ends in subtour constrname = "subtourconstr" + num2str(k); tsp.Constraints.(constrname) = sum(trips(a)) <= (nnz(inSubTour) - 1); k = k + 1; end % Try to optimize again [tspsol,fval,exitflag,output] = solve(tsp,'options',opts); tspsol.trips = logical(round(tspsol.trips)); Gsol = graph(idxs(tspsol.trips,1),idxs(tspsol.trips,2),[],numnodes(G)); % Gsol = graph(idxs(tspsol.trips,1),idxs(tspsol.trips,2)); % Also works in most cases % Plot new solution hGraph.LineStyle = 'none'; % Remove the previous highlighted path highlight(hGraph,Gsol,'LineStyle','-') drawnow % How many subtours this time? tourIdxs = conncomp(Gsol); numtours = max(tourIdxs); % Number of subtours fprintf('# of subtours: %d\n',numtours) end```
```# of subtours: 20 # of subtours: 7 # of subtours: 9 # of subtours: 9 # of subtours: 3 # of subtours: 2 # of subtours: 7 # of subtours: 2 # of subtours: 1 ```
```title('Solution with Subtours Eliminated'); hold off```
### Solution Quality
The solution represents a feasible tour, because it is a single closed loop. But is it a minimal-cost tour? One way to find out is to examine the output structure.
`disp(output.absolutegap)`
``` 0 ```
The smallness of the absolute gap implies that the solution is either optimal or has a total length that is close to optimal.
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If you’re taking the test tomorrow morning, you should ignore this and REST YOUR BRAIN. If you’re not taking it tomorrow, though, then this is just another regular weekend for you, and you should work your brain HARD.
The prize, as it has been every week for weeks and weeks, is access to the coveted PWN the SAT Math Guide Beta. If you wanna win, you have to comment in such a way that I can contact you. Preferably using Facebook, Twitter, or Gmail.
A warning: this one is pretty tough.
In a hand in his weekly poker game, Mike won money from both John and Sean. Mike’s percent gain on the hand was equal in magnitude to John’s percent loss. John bet twice as much as did Sean. Mike and John together held \$200 in chips before the hand began. If Sean bet \$20 in chips on the hand, what was Mike’s chip total after the hand was over?
Good luck!
UPDATE: Nobody got this yet, so I’m gonna leave it up for now unanswered. If you solve it, book access is yours.
SECOND UPDATE: OK, Eowyn got it. Nice. Solution below the cut.
There’s a lot going on here, and at first blush it may seem as though you have too many variables to work with. If you write down everything you know, though, and play around with it enough, you’ll see that you can actually get it down to just one variable in one equation!
$percent\:&space;change&space;=&space;\frac{change}{original\:&space;value}\times&space;100\%$
$\frac{Mike's\:gain}{Mike's\:original\:value}\times&space;100\%&space;=&space;\frac{John's\:loss}{John's\:original\:value}\times&space;100\%$
Of course, we can tidy this up a bit:
$\frac{Mike's\:gain}{Mike's\:original\:value}&space;=&space;\frac{John's\:loss}{John's\:original\:value}$
Now we need to start figuring out the numbers. We know that Sean bet 20, and that John bet twice as much as Sean. So John bet 40, which means John’s loss was 40. We also know that Mike’s gain was 60, since he won all the money that was bet in the hand (John’s 40 + Sean’s 20).
$\frac{60}{Mike's\:original\:value}&space;=&space;\frac{40}{John's\:original\:value}$
And we can be a bit clever here if we want to jam all the info into one equation and say that, since Mike’s and John’s chips added up to 200 before the hand, Mike’s prior chip total can be m, and John’s prior chip total can be 200 – m.
$\frac{60}{m}&space;=&space;\frac{40}{200-m}$
Solve that for m
$40m&space;=&space;60(200-m)$
$40m&space;=&space;12000-60m$
$100m&space;=&space;12000$
$m&space;=&space;120$
…and you’ll see that before the hand began, Mike had 120 dollars in chips. Since he won 60, he has 180 after the hand.
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×
6th Chapter
### 11th Business Mathematics Chapter 6 Test
Here you can prepare 11th Class Business Mathematics Chapter 6 Binary Number System and its Operation Test. Click the button for 100% free full practice test.
## First Year Business Mathematics Chapter 6 Online MCQ Test for 1st Year Business Mathematics Chapter 6 (Binary Number Systems and its Operation)
This online test contains MCQs about following topics:
Introduction . Binary number system . Conversions of number systems . Arithmetic operations in binary number system
ICOM Part 1 Business Mathematics Ch. 6 Test
### First Year Business Mathematics Chapter 6 Online MCQ Test for 1st Year Business Mathematics Chapter 6 (Binary Number Systems and its Operation)
1 (145)<sub>10</sub> = ( )<sub>2</sub>
• A. 10010001
• B. 10010111
• C. 11100001
• D. 10001001
2 (1100000)<sub>2</sub> - (111111)<sub>2</sub>= --------------------------- :
• A. (100001)<sub>2</sub>
• B. (110001)<sub>2</sub>
• C. (1000111)<sub>2</sub>
• D. (111110)<sub>2</sub>
3 Annuity is classified into:
• A. Two classes
• B. Three classes
• C. Four classes
• D. Five classes
4 Basically proportion is of:
• A. 4 types
• B. 3 types
• C. 2 types
• D. None of these
5 (10101)<sub>2</sub> in decimal system is
• A. 32
• B. 26
• C. 21
• D. 30
6 Hexadecimal number system is based on:
• A. Two digits
• B. Ten digits
• C. Eight digits
• D. Sixteen digits
7 5 in binary system is:
• A. (10)<sub>2</sub>
• B. (101)<sub>2</sub>
• C. (11)<sub>2</sub>
• D. None of these
8 (1001001)<sub>2</sub> in decimal system is -----------------------
• A. 37
• B. 67
• C. 73
• D. 87
9 (100011)<sub>2</sub> x (1101)<sub>2</sub> = ---------------------------
• A. (111000111)<sub>2</sub>
• B. (100011001)<sub>2</sub>
• C. (100000001)
• D. (110011001)<sub>2</sub>
10 (1101)<sub>2</sub> + (1001)<sub>2</sub> = --------------------------------
• A. (10110)<sub>2</sub>
• B. (11100)<sub>2</sub>
• C. (10001)<sub>2</sub>
• D. (11011)<sub>2</sub>
### Top Scorers of Business Mathematics Icom Part 1 Chapter 6 Online Test
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first year ka paper kab ha
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# ZeroSum Ruler (home)
## Blogging on math education and other related things
### Adding Fractions With Pictures! (The Crisscross Method)December 3, 2012
Fraction Addition (And Subtraction): We’re not in kindergarten anymore
Addition and subtraction are only easy in elementary school. Once middle school starts, continuing throughout any Math class taken that point forward, addition and subtraction are much harder than multiplication and division. Why? The Common Denominator. To a kid who is not fluent in his multiplication facts, finding The Common Denominator is an exercise in torture.
-
What is a common denominator? A common denominator is a multiple of both denominators in a fraction addition (or subtraction) problem. For example:
-
In the above example, 6 is a common denominator of 2 and 3. But is it the only one? No. How many common denominators are there between two fractions? Infinite. For example:
-
-
Why would we want to use 7830 as a common denominator? Why not? The point is that any number that both denominators divide into evenly can act as a common denominator. We are far less restricted than we thought.
So if we’re virtually unrestricted in choosing a common denominator, why not pick the one that is the product (multiply) of the two denominators? For example:
-
-
Just multiply the denominators to find a common denominator. This is easy.
-
At this point in the traditional method of adding fractions, we’d begin to ask our questions: “How many 8’s go into 16?” Ok, 2. “2 times 3 is …?” Ok 6. So 3/8 = 6/16 . Though this process is easy to a person who is fluent in their multiplication and division, it will give reason for a non-fluent Math student to seize up.
-
A great alternative way of adding fractions is the Crisscross Method of adding (and subtracting) fractions. In this method, we use the common denominator just once (this method will not create two equivalent fractions to the original two) and multiply “crisscross” to find two new numerators.
-
In 3/8 + 5/2, we’ll first multiply the denominators to find our new, common denominator:
-
Next, we’ll multiply 3 • 2 (always starting our crisscross in the top left corner) to find the first missing numerator:
-
And then 8 • 5 to find the second missing numerator:
-
But why are we allowed to do this? Let’s back up to see what really happened.-
-
First, we found the common denominator 16 by multiplying the denominators (8 and 2) of both fractions. We’re guaranteed that our denominator is common if we created it by multiplying the two original denominators to get it. To get the first numerator 6, we multiplied the numerator of the first fraction (3) by the denominator of the second fraction (2).
-
In the process, we multiplied both numerator and denominator by 2. In other words, we multiplied 3/8 by 2/2 Any number divided by itself is just a fancy 1, and multiplying any number by 1 does not change the number’s value. As a check to see if this process worked, 3/8 = 6/16 . The old and new fractions are equivalent.
-
The same is true to get the second numerator 40:
-
Both numerator and denominator were multiplied by 8. In other words, we multiplied 5/2 by 8/8, which is just a fancy 1. Multiplying by 1 does not change a number’s value. As a check, 5/2 = 40/16. The old and new fractions are equivalent.
-
Now we simply add the numerators:
-
-
The Crisscross method also works for fraction subtraction – we’d have a subtraction in the numerator. Why was this method not taught in school?
-
Hurray for Fraction Addition (and Subtraction)!
-
You can download a PDF ebook that uses pictures to explain fraction division, multiplication and addition on CurrClick at Fractions: A Picture Book!
-
-
### New (free) ZeroSum ruler – for teaching addition with negative numbersSeptember 30, 2012
Below is a new version of the ZeroSum ruler. This one needs no hardware to construct, just scissors and glue. You can download, print and use this proven tool right now by clicking on the picture, which will bring you to the PDF file that contains 2 ZeroSum rulers.
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### Simple “how to solve an equation” flowchartSeptember 14, 2012
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Since drawing this flowchart, I have created a printed one that includes “variables on both sides“. You can see the post (and get the free poster download here: http://zerosumruler.wordpress.com/2012/09/30/solving-equations-flowchart-free-download-poster/
### The Language of Math PosterAugust 19, 2011
Below is a poster I hang in my classroom every fall. Each year it grows longer as more and more terms come up for the different operations of math. When I was a kid, no one told me to look out for these words, or that math was even a language at all, which made word problems pretty tough. By clicking on the poster you will be sent to the original Excel file on Google Docs. Do you have any words to add?
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X of Y or Y's : GMAT Verbal Section
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# Events & Promotions
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# X of Y or Y's
Author Message
Intern
Joined: 20 Sep 2008
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X of Y or Y's [#permalink]
### Show Tags
20 Nov 2008, 11:56
1
This post was
BOOKMARKED
Hello,
I wanted to clarify when we use possessif form i.e. (John's hat) or the "of" form i.e. hat of john
. ...thanksssss
If you have any questions
New!
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Re: X of Y or Y's [#permalink]
### Show Tags
28 Nov 2008, 12:20
jugolo1 wrote:
Hello,
I wanted to clarify when we use possessif form i.e. (John's hat) or the "of" form i.e. hat of john
. ...thanksssss
Since Gmat prefers short and concise, (John's hat) is better than "hat of John" though both are correct.
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Re: X of Y or Y's [#permalink]
### Show Tags
28 Nov 2008, 18:42
GMAT TIGER wrote:
jugolo1 wrote:
Hello,
I wanted to clarify when we use possessif form i.e. (John's hat) or the "of" form i.e. hat of john
. ...thanksssss
Since Gmat prefers short and concise, (John's hat) is better than "hat of John" though both are correct.
No, depend on the situation. whether one or the other make the sentence awkard!
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Re: X of Y or Y's [#permalink]
### Show Tags
28 Nov 2008, 22:09
sondenso wrote:
GMAT TIGER wrote:
jugolo1 wrote:
Hello,
I wanted to clarify when we use possessif form i.e. (John's hat) or the "of" form i.e. hat of john
. ...thanksssss
Since Gmat prefers short and concise, (John's hat) is better than "hat of John" though both are correct.
No, depend on the situation. whether one or the other make the sentence awkard!
What I have seen in most of the cases in Gmat is that "John's hat" is direct, short and concise.
For example: 11-t73342
I believe "Jean Toomer’s Cane" is much better than "Cane of Jean Toomer"
More example, if any, would be useful.
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Re: X of Y or Y's [#permalink]
### Show Tags
29 Nov 2008, 03:47
Tiger, it is my pleasure!
VR-91.
In theory, international civil servants at the United Nations are prohibited from continuing to draw salaries from their own governments; in practice, however some governments merely substitute living allowances for their employees' paychecks, assigned by them to the United Nations.
(A) for their employees’ paychecks, assigned by them
(B) for the [color=#FF0000]paychecks of their employees[/color] who have been assigned
(C) for the paychecks of their employees, having been assigned
(D) in place of their employees’ paychecks, for those of them assigned
(E) in place of the paychecks of their employees to have been assigned by them
employees' paychecks is awkard
B is correct
Another one nearly the same.
VR-79. Scientists have observed large concentrations of heavy—metal deposits in the upper twenty centimeters of Baltic Sea sediments，which are consistent with the growth of industrial activity there.
(A) Baltic Sea sediments, which are consistent with the growth of industrial activity there
(B) Baltic Sea sediments, where the growth of industrial activity is consistent with these findings
(C) Baltic Sea sediments, findings consistent with its growth of industrial activity
(D) sediments from the Baltic Sea, findings consistent with the growth of industrial activity in the area
(E) sediments from the Baltic Sea, consistent with the growth of industrial activity there
D and E is ok, but A, b and C is awkward
Hope it help!
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Re: X of Y or Y's [#permalink]
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29 Nov 2008, 22:49
sondenso wrote:
In theory, international civil servants at the United Nations are prohibited from continuing to draw salaries from their own governments; in practice, however some governments merely substitute living allowances for their employees' paychecks, assigned by them to the United Nations.
(A) for their employees’ paychecks, assigned by them
(B) for the [color=#FF0000]paychecks of their employees[/color] who have been assigned
(C) for the paychecks of their employees, having been assigned
(D) in place of their employees’ paychecks, for those of them assigned
(E) in place of the paychecks of their employees to have been assigned by them
employees' paychecks is awkard
B is correct
I take this one. Thanks.
_________________
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Re: X of Y or Y's [#permalink] 29 Nov 2008, 22:49
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HuggingFaceTB/finemath
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Ampere
(Redirected from Amperes)
Ampere Demonstration modew of a moving iron ammeter. As de current drough de coiw increases, de pwunger is drawn furder into de coiw and de pointer defwects to de right.
Generaw information
Unit systemSI base unit
Unit ofEwectric current
SymbowA
Named afterAndré-Marie Ampère
The ampere (/ˈæmpɪər, æmˈpɪər/; symbow: A), often shortened to "amp", is de base unit of ewectric current in de Internationaw System of Units (SI). It is named after André-Marie Ampère (1775–1836), French madematician and physicist, considered de fader of ewectrodynamics.
The Internationaw System of Units defines de ampere in terms of oder base units by measuring de ewectromagnetic force between ewectricaw conductors carrying ewectric current. The earwier CGS measurement system had two different definitions of current, one essentiawwy de same as de SI's and de oder using ewectric charge as de base unit, wif de unit of charge defined by measuring de force between two charged metaw pwates. The ampere was den defined as one couwomb of charge per second. In SI, de unit of charge, de couwomb, is defined as de charge carried by one ampere during one second.
New definitions, in terms of invariant constants of nature, specificawwy de ewementary charge, took effect on 20 May 2019.
Definition
The ampere is defined by taking de fixed numericaw vawue of de ewementary charge e to be 1.602 176 634 × 10−19 when expressed in de unit C, which is eqwaw to A s, where de second is defined in terms of ∆ν.
The SI unit of charge, de couwomb, "is de qwantity of ewectricity carried in 1 second by a current of 1 ampere". Conversewy, a current of one ampere is one couwomb of charge going past a given point per second:
${\dispwaystywe {\rm {1\ A=1{\tfrac {C}{s}}.}}}$ In generaw, charge Q is determined by steady current I fwowing for a time t as Q = It.
Constant, instantaneous and average current are expressed in amperes (as in "de charging current is 1.2 A") and de charge accumuwated, or passed drough a circuit over a period of time is expressed in couwombs (as in "de battery charge is 30000 C"). The rewation of de ampere (C/s) to de couwomb is de same as dat of de watt (J/s) to de jouwe.
History
The ampere was originawwy defined as one tenf of de unit of ewectric current in de centimetre–gram–second system of units. That unit, now known as de abampere, was defined as de amount of current dat generates a force of two dynes per centimetre of wengf between two wires one centimetre apart. The size of de unit was chosen so dat de units derived from it in de MKSA system wouwd be convenientwy sized.
The "internationaw ampere" was an earwy reawization of de ampere, defined as de current dat wouwd deposit 0.001118 grams of siwver per second from a siwver nitrate sowution, uh-hah-hah-hah. Later, more accurate measurements reveawed dat dis current is 0.99985 A.
Since power is defined as de product of current and vowtage, de ampere can awternativewy be expressed in terms of de oder units using de rewationship I=P/V, and dus 1 ampere eqwaws 1 W/V. Current can be measured by a muwtimeter, a device dat can measure ewectricaw vowtage, current, and resistance.
Reawization
The standard ampere is most accuratewy reawized using a Kibbwe bawance, but is in practice maintained via Ohm's waw from de units of ewectromotive force and resistance, de vowt and de ohm, since de watter two can be tied to physicaw phenomena dat are rewativewy easy to reproduce, de Josephson junction and de qwantum Haww effect, respectivewy.
At present, techniqwes to estabwish de reawization of an ampere have a rewative uncertainty of approximatewy a few parts in 107, and invowve reawizations of de watt, de ohm and de vowt.
Everyday exampwes
The current drawn by typicaw constant-vowtage energy distribution systems is usuawwy dictated by de power (watt) consumed by de system and de operating vowtage. For dis reason de exampwes given bewow are grouped by vowtage wevew.
CPUs – 1 V DC
• Current notebook CPUs (up to 15...45 W at 1 V): up to 15...45 A
• Current high-end CPUs (up to 65...140 W at 1.15 V): up to 55...120 A
Portabwe devices
• Hearing aid (typicawwy 1 mW at 1.4 V): 700 µA
• USB charging adapter (as power suppwy – typicawwy 10 W at 5 V): 2 A
Internaw combustion engine vehicwes – 12 V DC
A typicaw motor vehicwe has a 12 V battery. The various accessories dat are powered by de battery might incwude:
• Instrument panew wight (typicawwy 2 W): 166 mA
• Headwight (each, typicawwy 60 W): 5 A
• Starter motor on a smawwer car: 50 A to 200 A
Norf American domestic suppwy – 120 V AC
Most Canada, Mexico and United States domestic power suppwiers run at 120 V.
Househowd circuit breakers typicawwy provide a maximum of 15 A or 20 A of current to a given set of outwets.
• USB charging adapter (as woad – typicawwy 10 W): 83 mA
• 22-inch/56-centimeter portabwe tewevision (35 W): 290 mA
• Tungsten wight buwb (60–100 W): 500–830 mA
• Toaster, kettwe (1.5 kW): 12.5 A
• Hair dryer (1.8 kW): 15 A
European & Commonweawf domestic suppwy – 230–240 V AC
Most European domestic power suppwies run at 230 V, and most Commonweawf domestic power suppwies run at 240 V. For de same amount of power (in watts), de current drawn by a particuwar European or Commonweawf appwiance (in Europe or a Commonweawf country) wiww be wess dan for an eqwivawent Norf American appwiance.[Note 1] Typicaw circuit breakers wiww provide 16 A.
The current drawn by a number of typicaw appwiances are:
• Compact fwuorescent wamp (11–30 W): 56–112 mA
• 22-inch/56-centimeter portabwe tewevision (35 W): 145–150 mA
• Tungsten wight buwb (60–100 W): 240–450 mA
• Toaster, kettwe (2 kW): 9 A
• Immersion heater (4.6 kW): 19–20 A
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# 6 Stages of Question Mastery
Author Message
Manager
Joined: 07 Sep 2007
Posts: 116
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6 Stages of Question Mastery [#permalink]
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14 Nov 2007, 02:15
1
KUDOS
I believe there are a couple stages that people go through, and that these stages are primarily based on %correct for a type of question (sc/rc/cr/ps/ds) and not difficulty of questions or anything else.
When I say type, I mean either SC, RC, CR, PS, DS.
When I say sub-type, I mean, for example, verb-agreement in SC, or weaken the argument in CR, or purpose of the passage in RC.
6 Stages of Question Mastery
(These stages are independent with type of question.)
Stage 1: 0%-40% correct
You do not understand the fundamentals. Your strategy is not correct for this type of question. You are doing not doing much better than guessing at the answer and doing problems is not going to help you.
What to do: You need to just read GMAT guides on this type of question and change your approach before you even think about touching any questions.
Stage 2: 40%-70% correct
You are consistently missing the point of the question. The question is asking for something very specific and clear, but you are often misunderstanding it and looking in the wrong direction. Perhaps in CR you're consistently thinking outside of scope or inferring too much information from the passage. Perhaps in SC you're looking for parallel structures when there is no need for them. Perhaps in Quant you need to find the prime factorization but you're trying some brute force approach.
What to do: You still need to focus on reading GMAT guides and but do a few questions every now and then to see how well you're doing. When doing questions at this phase, you need to spend a lot of time per question (5-10 minutes) knowing exactly why each of the 5 answers is right or wrong.
Stage 3: 70%-80% correct
You've got a good understanding of some of the fundamentals, but you are not applying a complete strategy. Perhaps in SC you're consistently spotting verb agreement errors but missing improper modifiers. Or in RC you consistently confusing a supporting point with the main point. In Quant, maybe you don't understand some subjects very well (perm/comb, prime factorization, divisibility, powers, etc.).
What to do: Here is where we can start focusing on our flaws, figure out which sub-type of the type of question you're getting wrong and read GMAT guides on that sub-type. When doing questions, you should still be spending 3-5 minutes on each question, now easily eliminating 1-3 of the answers and, for the rest, being able to explain exactly why each of the answers is right or wrong. If you get a question wrong, go back and do a thorough 5-answer analysis.
Stage 4: 80%-90% correct
Now we're starting to show consistency, we understand most of the fundamentals. Our strategy is near complete, but we have not been applying it perfectly. Most of our incorrect answers can be attributable to carelessness and result in "oh duh!" moments. NOW AND ONLY NOW are you ready to start doing sets of questions.
What to do: If you're consistently missing certain sub-types, go re-read some GMAT guides on it. But your study should involve mostly questions. You should still be spending 2-3 minutes on each one, easily eliminating all but 2 of the answers. For the 2, you must be able to explain exactly why one answer is right and one answer is wrong. If you get a question wrong, go back and do a thorough 5-answer analysis.
Stage 5: 90%-95% correct
We understand almost all of the fundamentals and are able to apply them with a consistent strategy. We are able to eliminate down to the correct 2 answers almost all of the time, and if we spend enough time on any question, we know we can get it right 100% of the time. Some sub-types are giving us more trouble than others.
What to do: Practice questions to fine-tune your strategy; iron out flaws in those sub-types you're getting wrong. Spend 1-2 minutes on each question. If you get a question wrong, go back and do a thorough 5-answer analysis.
Stage 6: 95%-100% correct
We understand all of the fundamentals and are able to apply them with a consistent strategy. All of the questions we get wrong are just do to really stupid mistakes that we are in complete control over.
What to do: Work on mental stamina and timing by doing a test's worth of questions each day. Aim for a question per minute. Ease up on studying, we just want to maintain and fine tune our spectacular skills.
-J
Last edited by JingChan on 17 Nov 2007, 22:27, edited 1 time in total.
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Posts: 63
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### Show Tags
14 Nov 2007, 06:21
JingChan thank you for your useful advice and congratulations for your excellent GMAT score. People like you are an inspiration for my trying to master the GMAT.
I would like to ask you a couple of questions and I would really appreciate your opinion.
1) I am not a native english speaker. I do not live in the States, UK, Australia or any native-english speaker country. Do you think that people like me can do a good score in verbal, (>40, why not approximate 45?)
2) My major difficulties in GMAT have to do mostly with the verbal part. (I'm not a maths expert but I have much better understanding to it since I have an engineering background. I do silly mistakes however and I definetely need to prepare for this part as well). I try to focus especially in CR at the moment, which is my major weakness. I try to undestand the questions and analyse them in my head. Then, I can score somewhere close to 50%. Not good of course but I started with 10% in this area...But, it takes me a hell of time to do that. I am thinking that it is not the point how fast I do a question, but to grasp the thing, and therefore, I take my time. Do you agree with that?
And last but not least, do you think I should do the same in RC? Take my time I mean till I find a strategy?
JingChan I would really appreciate your advice. It's not the high score you obtained, but your overall approach and preparation for this exam that makes your advice valuable. At least for me.
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### Show Tags
14 Nov 2007, 14:04
Given what I know about you, it sounds like you definitely have the potential to score 95%+ in Verbal. Being a non-English speaker, you have to work at it a little harder, but that should be expected. When it boils down to it, each type of question tests certain skills:
SC - grammar rules
CR - reading ability and logical reasoning
And none of these skills require you to be a native English speaker to master.
You didn't mention SC so I'll assume you're doing well in that area. What's left are just reading questions. Before you start scoring 80%+ you need to be fully grasping the point of the question. Take your time and understand each question. You first need to be answering the questions correctly before you think about speeding yourself up.
Once you do start answering questions consistently, then start timing yourself. Slowly speed yourself up until you're able to finish each question in about 1-2 minutes.
Hope this helps,
J
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14 Nov 2007, 23:47
It has been very helpful. Thank you very much
By the way, I wish you all the best for the rest of the application process.
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Re: 6 Stages of Question Mastery [#permalink]
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15 Nov 2007, 11:50
JingChan wrote:
I believe there are a couple stages that people go through, and that these stages are primarily based on %correct for a type of question (sc/rc/cr/ps/ds) and not difficulty of questions or anything else.
When I say type, I mean either SC, RC, CR, PS, DS.
When I say sub-type, I mean, for example, verb-agreement in SC, or weaken the argument in CR, or purpose of the passage in RC.
6 Stages of Question Mastery
(These stages are independent with type of question.)
Stage 1: 0%-40% correct
You do not understand the fundamentals. Your strategy is not correct for this type of question. You are doing not doing much better than guessing at the answer and doing problems is not going to help you.
What to do: You need to just read GMAT guides on this type of question and change your approach before you even think about touching any questions.
Stage 2: 40%-70% correct
You are consistently missing the point of the question. The question is asking for something very specific and clear, but you are often misunderstanding it and looking in the wrong direction. Perhaps in CR you're consistently thinking outside of score or inferring too much information from the passage. Perhaps in SC you're looking for parallel structures when there is no need for them. Perhaps in Quant you need to find the prime factorization but you're trying some brute force approach.
What to do: You still need to focus on reading GMAT guides and but do a few questions every now and then to see how well you're doing. When doing questions at this phase, you need to spend a lot of time per question (5-10 minutes) knowing exactly why each of the 5 answers is right or wrong.
Stage 3: 70%-80% correct
You've got a good understanding of some of the fundamentals, but you are not applying a complete strategy. Perhaps in SC you're consistently spotting verb agreement errors but missing improper modifiers. Or in RC you consistently confusing a supporting point with the main point. In Quant, maybe you don't understand some subjects very well (perm/comb, prime factorization, divisibility, powers, etc.).
What to do: Here is where we can start focusing on our flaws, figure out which sub-type of the type of question you're getting wrong and read GMAT guides on that sub-type. When doing questions, you should still be spending 3-5 minutes on each question, now easily eliminating 1-3 of the answers and, for the rest, being able to explain exactly why each of the answers is right or wrong. If you get a question wrong, go back and do a thorough 5-answer analysis.
Stage 4: 80%-90% correct
Now we're starting to show consistency, we understand most of the fundamentals. Our strategy is near complete, but we have not been applying it perfectly. Most of our incorrect answers can be attributable to carelessness and result in "oh duh!" moments. NOW AND ONLY NOW are you ready to start doing sets of questions.
What to do: If you're consistently missing certain sub-types, go re-read some GMAT guides on it. But your study should involve mostly questions. You should still be spending 2-3 minutes on each one, easily eliminating all but 2 of the answers. For the 2, you must be able to explain exactly why one answer is right and one answer is wrong. If you get a question wrong, go back and do a thorough 5-answer analysis.
Stage 5: 90%-95% correct
We understand almost all of the fundamentals and are able to apply them with a consistent strategy. We are able to eliminate down to the correct 2 answers almost all of the time, and if we spend enough time on any question, we know we can get it right 100% of the time. Some sub-types are giving us more trouble than others.
What to do: Practice questions to fine-tune your strategy; iron out flaws in those sub-types you're getting wrong. Spend 1-2 minutes on each question. If you get a question wrong, go back and do a thorough 5-answer analysis.
Stage 6: 95%-100% correct
We understand all of the fundamentals and are able to apply them with a consistent strategy. All of the questions we get wrong are just do to really stupid mistakes that we are in complete control over.
What to do: Work on mental stamina and timing by doing a test's worth of questions each day. Aim for a question per minute. Ease up on studying, we just want to maintain and fine tune our spectacular skills.
-J
Nice Post...what's GMAT Guides?
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Posts: 116
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Re: 6 Stages of Question Mastery [#permalink]
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15 Nov 2007, 17:15
yogachgolf wrote:
Nice Post...what's GMAT Guides?
I just mean any GMAT books. Like PR's Cracking the GMAT, MGMAT Guides, Kaplan 800, etc.
-J
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17 Nov 2007, 21:01
This is a very good post. How do you successful execute the 5-answer thorough analysis. My problem is that whenever i see the answer of a question I initially failed, i get that 'ooh' effect that makes me think that i understand why the answer chosen is correct and mine is not.
Manager
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### Show Tags
17 Nov 2007, 22:26
kizito2001 wrote:
This is a very good post. How do you successful execute the 5-answer thorough analysis. My problem is that whenever i see the answer of a question I initially failed, i get that 'ooh' effect that makes me think that i understand why the answer chosen is correct and mine is not.
If you're at the stage where you can immediately understand, then you're doing well!
I would rewind my thinking to try to understand why I chose the wrong answer to begin with, because maybe you're consistently making the same error.
Especially for Verbal, it's very important to understand why each wrong answer is wrong.
-J
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### Show Tags
29 Nov 2007, 00:23
Jing,
Thanks for this awesome post as it has definitely inspired me. I have a questions regarding consistency in the prep material, what to use and what NOT to use.
My situation is that this is my second time taking the test (i took my test Nov 1 and didn't do well - 640: v35, m45). I have used and am still using various different types of material and score all over the board. For example, with Princeton Review, i score around high 600s - 700. With Kaplan, I score in the 630-660 range, ARCO is also in the 660s range. Surprisingly I scored 710s on both of the GMAT Prep tests, prior to the exam. But clearly, I was not ready. Now, I dont know which prep material to trust and when I am truely "ready" to the take it again.
So, what was your plan of attack in:
1. Nagivating through all this prep material, and choosing the right ones? Should I stick to 3-4 brands (MGMAT, KAPLAN, OG, PR)?
2. This is a silly quesiton, but still want to ask it...How do you know when you are "ready?" Is it when you are consistently scoring in the same range on various different prep materials?
A big thank you in advance!
29 Nov 2007, 00:23
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# 6 Stages of Question Mastery
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HuggingFaceTB/finemath
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If the areas of two similar triangles ABC and PQR are in the ratio 9 : 16 and BC = 4.5 cm,
Question:
If the areas of two similar triangles ABC and PQR are in the ratio 9 : 16 and BC = 4.5 cm, what is the length of QR?
Solution:
Given: ΔABC and ΔPQR are similar triangles. Area of ΔABC: Area of ΔPQR = 9:16 and BC = 4.5cm.
To find: Length of QR
We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.
Hence,
$\frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{PQR})}=\frac{\mathrm{BC}^{2}}{\mathrm{QR}^{2}}$
$\frac{9}{16}=\frac{4.5^{2}}{Q R^{2}}$
$\mathrm{QR}^{2}=\frac{4.5^{2} \times 16}{9}$
$\mathrm{QR}^{2}=36$
$Q R=6 \mathrm{~cm}$
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open-web-math/open-web-math
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## Important Points from Lectures
1) Choosing Gaussian Surface: When you are working on a problem where you want to use Gauss' Law to find a field, there are a few issues you want to be aware of:
a. The law only tells you the surface integral of the field (the flux) - if the field is varying on your surface in a way you don't know before hand, this can become very intractable. It is best to set up your surfaces so that you have the same field everywhere if possible.
b. Remember also that the integral only counts the perpendicular component of the field, so you need to have the field that you're interested in perpendicular to the surface.
c. But, we can use this to our advantage: if your surface needs to include an area where the field is either not the value of interest, or not constant or both (eg: long line of charge), if your surface passes through this space in such a way that the field is parallel to it, then you can safely ignore these fields!
2) Conductors: Field at surface must be perpendicular to surface, field inside must be zero. (Even for hollow ones: eg Faraday Cage experiment).
3) Field Lines: Another way to draw in fields. They show direction of electric field directly and tell you the strength of the field with their density. Field Lines cannot cross (but they can run from one charge to another, or meet head on at a point where field is zero).
4) Electric Potential and Potential Energy: Work done by field on a charge as it moves is equal to the drop in electrical potential energy between the starting and finishing points (Electric field is conservative force field). Here V is used to denote the potential energy of a specific charge in the presence of other charges and φ(r) is the potential at a point in space due to an arrangement of charges.
a. Two point charges: V(r) = kq1q2/r
b. φ(r)=V(r)/q2
c. Superposition still works: V=Σikq1qi/ri1 is energy of charge one – our test charge in most cases, (and there is a similar sum to calculate the "self energy" of the fixed collection of charges).
5) There are four important, different formulae that involve Q and r so far in this course:
a. Coulomb's Law: F=kQ1Q2ρ/r2 b. Electric Field of a point charge: E=kQρ/r2 c. Electrical Potential Energy (scalar): EPE=kQ1Q2/r d. Electrical Potential (also scalar): φkQ/r
Where is a unit vector pointin in the direction of r.
6) Potential Energy: Given a fixed charge distribution, how much work do I have to do to bring a single charge in from infinity to a particular point? Alternately, how much kinetic energy will my test charge have once it reaches a very large distance away, assuming it was released from rest?
7) Electron Volts: 1eV is the energy an electron would have after it is accelerated through one volt of electrical potential.
8) Equipotential surfaces: imaginary surfaces where φ has a fixed value - electric field is perpendicular to these surfaces, just like a conductor.
9)Electric Dipole: two slightly separated, equal but opposite charges:
a. φ=k|p|cos(ϑ)/r2
b. p=qd - Dipole moment
10) Capacitance: Q=CV, C=ε0A/d (for parallel plates), U=1/2 QV =1/2 CV2 =1/2 Q2/C
## Example Problem 1
How strong is the electric field between two parallel plates 5.2mm apart if the potential different between them is 110V?
Definition of potential energy: work done moving charge is equal to the change in potential. If a one coulomb charge moves from one plate to the other, the field does work, W, equal to 110J
W=110V*1C
W=110J
But as the field between two parallel plates is everywhere constant and perpendicular to the plates (to a good approximation) we can calculate the work done directly:
W=Fd
W=E*1C*5.2x10-3m
E=110J/(1C*5.2x10-3m)
More generally for parallel plates, E=Voltage/Separation. In this case:
E=2.1x104V/m
## Example Problem 2
An electron starts from rest 72.5 cm from a fixed point charge with Q=-0.125μC. How fast will the electron be moving once it reaches a very large distance?
Initially the system (consisting of the electron and the fixed charge) have only potential energy:
PE = kQ1Q2/r
PE = 2483x10-19J
At a very large distance, the potential can be taken to have been all converted into kinetic energy:
KE = 1/2 m1v12 +1/2 m2v22
KE = 1/2 m2v22
As particle one (the fixed charge) is not moving.
1/2 m1v12=PE
v=23x106m/s
## Example Problems 3 and 4
How strong is the electric field between the plates of a 0.8μF air-gap capacitor if they are 2mm apart and each has a charge of magnitude 72μÂC? Q=CV
V=Q/C=90V
E=V/d = 45000V/m
A capacitor C1 carries a charge Q0 initially. Then it is connected in a circuit with a second, uncharged capacitor: C2. What charges do each of the capacitors carry now? What is the potential difference across them?
The instant that the capacitors are connected, charge flows to even out the voltage (the wires are conductors, therefore there can be no potential drop across them in this static case).
|V1|=|V2|
Q1/C1=Q2/C2
Q1= (Q0-Q1)/C2
(C2-C1)Q1=C1 Q0
Q1=C1Q0/(C2-C1)
Similarly:
Q2=C2Q0/(C2-C1)
In both cases, the voltage is given by: V=Q/C
V= Q0/(C2-C1)
## Example Problem 5
Near the surface of the earth, there is an electric field of E=150V/m pointing downwards. Two ball with opposite charges (|q|=550μÂC) that are otherwise identical (m=.54kg) are dropped from a height of h=2m. With what speeds do they both hit the ground?
Conservation of energy:
mgh+ E*h*q=1/2mv2
v=(2gh+2Ehq/m)1/2
for the positive ball: v+=6.37m/s
for the negative ball" v-=6.27m/s
as a comparison an uncharged ball will hit with v0=(2gh)1/2=6.32m/s
A positively charged ball will experience an extra downwards force in this field and thus hit faster that the neutral ball, which will in turn hit faster that the negatively charged ball which will be partially "levitated" by the electric field.
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HuggingFaceTB/finemath
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# j-james/math
All my math notes, now in Markdown.
View the Project on GitHub j-james/math
# Fundamental Trigonometric Identities
## Learning Targets
You should be able to
• Simplify expressions using fundamental trigonometric identities
## Concepts / Definitions
### Reciprocal Identities
$\csc\theta = \frac{1}{\sin\theta}$ $\qquad$ $\sec\theta = \frac{1}{\cos\theta}$ $\qquad$ $\cot\theta = \frac{1}{\tan\theta}$
$\sin\theta = \frac{1}{\csc\theta}$ $\qquad$ $\cos\theta = \frac{1}{\sec\theta}$ $\qquad$ $\tan\theta = \frac{1}{\cot\theta}$
### Quotient Identities
$\tan\theta = \frac{\sin\theta}{\cos\theta}$ $\qquad\qquad$ $\cot\theta = \frac{\cos\theta}{\sin\theta}$
### Odd-Even Identities
$\sin(-\theta) = -\sin\theta$ $\qquad$ $\csc(-\theta) = -\csc\theta$
$\cos(-\theta) = \cos\theta$ $\qquad$ $\sec(-\theta) = \sec\theta$
$\tan(-\theta) = -\tan\theta$ $\qquad$ $\cot(-\theta) = -\cot\theta$
### Cofunction Identities
(co is for complement)
$\sin(\frac \pi2-\theta) = \cos\theta$ $\qquad$ $\cos(\frac \pi2-\theta) = \sin\theta$
$\tan(\frac \pi2-\theta) = \cot\theta$ $\qquad$ $\cot(\frac \pi2-\theta) = \tan\theta$
$\sec(\frac \pi2-\theta) = \csc\theta$ $\qquad$ $\csc(\frac \pi2-\theta) = \sec\theta$
## Exercises
### Simplify the following
1. $\tan(\theta)\csc(\theta)$
2. $\cot^2x-\csc^2x$
3. $\frac{\sin(\frac \pi2 - \theta)}{\cos(\frac \pi2 - \theta)}$
4. $\frac{\sin^2x}{1+\cos x}$
5. $\sec^4y-\tan^4y$
6. $\frac{\sec^2x-1}{\sin^2x}$
7. $\frac{1+\tan x}{1+\cot x}$
8. $\frac{\sec^2x\csc x}{\sec^2x+\csc^2x}$
9. $\frac{1}{1-\sin\varphi} + \frac{1}{1+\sin\varphi}$
10. $\frac{\sin x}{\cot^2 x} - \frac{\sin x}{\cos^2 x}$
### Factor the following
1. $\sin^2\theta + \frac{2}{\csc\theta} + 1$
2. $\sec^2x - \sec x + \tan^2 x$
3. Evaluate without calculator using identities $\cos(-\theta),\ if\ \sin(\theta-\frac \pi2) = 0.73$
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HuggingFaceTB/finemath
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# Modeling and Solving Constraints. Basic Idea
## 1. Modeling and Solving Constraints
Erin Catto
Blizzard Entertainment
## 2. Basic Idea
Constraints are used to simulate joints, contact,
and collision.
We need to solve the constraints to stack boxes
and to keep ragdoll limbs attached.
Constraint solvers do this by calculating impulse
or forces, and applying them to the constrained
bodies.
## 3. Overview
Constraint Formulas
Jacobians, Lagrange Multipliers
Modeling Constraints
Joints, Motors, Contact
Building a Constraint Solver
Sequential Impulses
## 4. Constraint Types
Contact and Friction
Ragdolls
## 6. Constraint Types
Particles and Cloth
## 8. Bead on a 2D Rigid Wire
Implicit Curve Equation:
This is the position constraint.
C ( x, y ) 0
## 9. How does it move?
The normal vector is perpendicular to the velocity.
n
v
dot(n, v) 0
## 10. Enter The Calculus
Position Constraint:
C ( x) 0
x
x
y
If C is zero, then its time derivative is zero.
Velocity Constraint:
C 0
## 11. Velocity Constraint
C 0
Velocity constraints define the allowed motion.
Next we’ll show that velocity constraints depend
linearly on velocity.
## 12. The Jacobian
Due to the chain rule the velocity constraint has
a special structure:
C Jv
x
v
y
J is a row vector called the Jacobian.
J depends on position.
The velocity constraint is linear.
## 13. The Jacobian
The Jacobian is perpendicular to the velocity.
JT
v
C Jv 0
## 14. Constraint Force
Assume the wire is frictionless.
v
What is the force between the wire and the bead?
## 15. Lagrange Multiplier
Intuitively the constraint force Fc is parallel to
the normal vector.
Fc
v
Direction known.
Magnitude unknown.
Fc J
T
implies
## 16. Lagrange Multiplier
The Lagrange Multiplier (lambda) is the constraint force
signed magnitude.
We use a constraint solver to compute lambda.
More on this later.
## 17. Jacobian as a CoordinateTransform
Similar to a rotation matrix.
Except it is missing a couple rows.
So it projects some dimensions to zero.
The transpose is missing some columns, so some
v
J
Cartesian
Space
Velocity
C
Constraint
Space
Velocity
C Jv
J
T
Fc
Constraint
Space
Force
Cartesian
Space
Force
Fc J
T
## 20. Refresher: Work and Power
Work = Force times Distance
Work has units of Energy (Joules)
Power = Force times Velocity (Watts)
P dot F, V
## 21. Principle of Virtual Work
Principle: constraint forces do no work.
We can ensure this by using:
Fc J
T
Proof (compute the power):
T
Pc F v J v Jv 0
T
c
T
The power is zero, so the constraint does no work.
## 22. Constraint Quantities
Position Constraint
C
Velocity Constraint
C
Jacobian
J
Lagrange Multiplier
## 23. Why all the Painful Abstraction?
We want to put all constraints into a common form for the
solver.
This allows us to efficiently try different solution
techniques.
## 24. Addendum: Modeling Time Dependence
Some constraints, like motors, have prescribed motion.
This is represented by time dependence.
Position:
C x, t 0
Velocity:
C Jv b(t ) 0
velocity bias
x
y
x
x
y
L
particle
is the tension
Position:
C x L
Velocity:
xT
C v
x
Jacobian:
Velocity Bias:
xT
J
x
b 0
dC d
dt dt
x2 y 2 L
1
d 2
dL
2
x y
2
2 dt
dt
2 x y
2 xvx yv y
2 x y
2
2
0
x v x xT
v
v
2
2 y
x y y x
1
T
## 27. Computing the Jacobian
At first, it is not easy to compute the Jacobian.
It gets easier with practice.
If you can define a position constraint, you can find its
Jacobian.
Here’s how …
## 28. A Recipe for J
Use geometry to write C.
Differentiate C with respect to time.
Isolate v.
Identify J and b by inspection.
C Jv b
Joints
Motors
Contact
Restitution
Friction
x
Fc J T
y
v
Fa mg
xT
J
x
## 31. Motors
A motor is a constraint with limited force (torque).
Example
C sin t
10 10
A Wheel
Note: this constraint does work.
## 32. Velocity Only Motors
Example
C 2
5 5
Usage: A wheel that spins at a constant rate.
We don’t care about the angle.
## 33. Inequality Constraints
So far we’ve looked at equality constraints (because
they are simpler).
Inequality constraints are needed for contact and joint
limits.
We put all inequality position constraints into this form:
C (x, t ) 0
## 34. Inequality Constraints
The corresponding velocity constraint:
If
C 0
enforce:
Else
C 0
skip constraint
## 35. Inequality Constraints
Force Limits:
0
Inequality constraints don’t suck.
## 36. Contact Constraint
Non-penetration.
Restitution: bounce
Friction: sliding, sticking, and rolling
body 2
n
p
body 1
C
(separation)
## 38. Non-Penetration Constraint
C ( v p 2 v p1 ) n
v 2 ω 2 p x 2 v1 ω1 p x1 n
n
p x n
1
n
p
x
n
2
T
v1
ω
1
v2
ω 2
Handy Identities
A B C
C A B
J
B C A
## 39. Restitution
Relative normal velocity
vn
( v p 2 v p1 ) n
Velocity Reflection
n
n
v ev
Adding bounce as a velocity bias
n
n
C v ev 0
n
b ev
## 40. Friction Constraint
Friction is like a velocity-only motor.
The target velocity is zero.
C vp t
v ω p x t
T
p
t
t
v
p
x
t
ω
J
## 41. Friction Constraint
The friction force is limited by the normal force.
Coulomb’s Law:
t n
In 2D:
n t n
3D is a bit more complicated. See the references.
## 42. Constraints Solvers
We have a bunch of constraints.
We have unknown constraint forces.
We need to solve for these constraint forces.
There are many ways different ways to compute
constraint forces.
## 43. Constraint Solver Types
Global Solvers (slow)
Iterative Solvers (fast)
## 44. Solving a Chain
1
m1
2
m2
3
m3
Global:
solve for 1, 2, and 3
simultaneously.
Iterative:
while !done
solve for 1
solve for 2
solve for 3
## 45. Sequential Impulses (SI)
An iterative solver.
SI applies impulses at each constraint to correct the
velocity error.
SI is fast and stable.
Converges to a global solution.
## 46. Why Impulses?
Easier to deal with friction and collision.
Lets us work with velocity rather than acceleration.
Given the time step, impulse and force are
interchangeable.
P hF
## 47. Sequential Impulses
Step1:
Integrate applied forces, yielding tentative
velocities.
Step2:
Apply impulses sequentially for all constraints,
to correct the velocity errors.
Step3:
Use the new velocities to update the positions.
## 48. Step 1: Newton’s Law
We separate applied forces and
constraint forces.
Mv Fa Fc
mass matrix
## 49. Step 1: Mass Matrix
Particle
m 0 0
M 0 m 0
0 0 m
Rigid Body
mE 0
M
0
I
May involve multiple particles/bodies.
## 50. Step 1: Applied Forces
Applied forces are computed according to some law.
Gravity: F = mg
Spring: F = -kx
Air resistance: F = -cv2
## 51. Step 1 : Integrate Applied Forces
Euler’s Method for all bodies.
1
v 2 v1 hM Fa
This new velocity tends to violate the velocity
constraints.
## 52. Step 2: Constraint Impulse
The constraint impulse is just the time step
times the constraint force.
Pc hFc
## 53. Step 2: Impulse-Momentum
Newton’s Law for impulses:
M v Pc
In other words:
1
v 2 v 2 M Pc
## 54. Step 2: Computing Lambda
For each constraint, solve these for :
Newton’s Law:
v 2 v 2 M 1Pc
Virtual Work:
Pc JT
Velocity Constraint:
Jv 2 b 0
Note: this usually involves one or two bodies.
## 55. Step 2: Impulse Solution
mC Jv 2 b
1
mC
1 T
JM J
The scalar mC is the effective mass seen by
the constraint impulse:
mC C
## 56. Step 2: Velocity Update
Now that we solved for lambda, we can use it
to update the velocity.
Pc J
T
1
v 2 v 2 M Pc
Remember: this usually involves one or two bodies.
## 57. Step 2: Iteration
Loop over all constraints until you are done:
- Fixed number of iterations.
- Corrective impulses become small.
- Velocity errors become small.
## 58. Step 3: Integrate Positions
Use the new velocity to integrate all
body positions (and orientations):
x2 x1 hv 2
This is the symplectic Euler integrator.
## 59. Extensions to Step 2
Handle position drift.
Handle force limits.
Handle inequality constraints.
Warm starting.
## 60. Handling Position Drift
Velocity constraints are not obeyed precisely.
Joints will fall apart.
## 61. Baumgarte Stabilization
Feed the position error back into the velocity constraint.
New velocity constraint:
Bias factor:
CB Jv
0 1
h
C 0
## 62. Baumgarte Stabilization
What is the solution to this?
C
h
C 0
First-order differential equation …
t
C C0 exp
h
exp t
## 64. Tuning the Bias Factor
If your simulation has instabilities, set the bias factor to
zero and check the stability.
Increase the bias factor slowly until the simulation
becomes unstable.
Use half of that value.
## 65. Handling Force Limits
First, convert force limits to impulse limits.
impulse h force
## 66. Handling Impulse Limits
Clamping corrective impulses:
clamp , min , max
Is it really that simple?
Hint: no.
## 67. How to Clamp
Each iteration computes corrective impulses.
Clamping corrective impulses is wrong!
You should clamp the total impulse applied over the
time step.
The following example shows why.
v
A Falling Box
1
P mv
2
Global Solution
P
P
## 69. Iterative Solution
iteration 1
P1
constraint 1
P2
constraint 2
Suppose the corrective impulses are too strong.
What should the second iteration look like?
## 70. Iterative Solution
iteration 2
P1
To keep the box from bouncing, we need
downward corrective impulses.
In other words, the corrective impulses are
negative!
P2
## 71. Iterative Solution
But clamping the negative corrective impulses
wipes them out:
clamp( , 0, )
0
This is one way to introduce jitter into
## 72. Accumulated Impulses
For each constraint, keep track of the total impulse
applied.
This is the accumulated impulse.
Clamp the accumulated impulse.
This allows the corrective impulse to be negative yet the
accumulated impulse is still positive.
## 73. New Clamping Procedure
1. Compute the corrective impulse, but don’t
apply it.
2. Make a copy of the old accumulated impulse.
3. Add the corrective impulse to the accumulated
impulse.
4. Clamp the accumulated impulse.
5. Compute the change in the accumulated
impulse using the copy from step 2.
6. Apply the impulse delta found in Step 5.
## 74. Handling Inequality Constraints
Before iterations, determine if the inequality constraint is
active.
If it is inactive, then ignore it.
Clamp accumulated impulses:
0 acc
## 75. Inequality Constraints
A problem:
overshoot
active
gravity
inactive
Aiming for zero overlap leads to JITTER!
active
## 76. Preventing Overshoot
Allow a little bit of penetration (slop).
If separation < slop
C Jv slop
Else
h
C Jv
Note: the slop will be negative (separation).
## 77. Warm Starting
Iterative solvers use an initial guess for the lambdas.
So save the lambdas from the previous time step.
Use the stored lambdas as the initial guess for the new
step.
Benefit: improved stacking.
## 78. Step 1.5
Apply the stored impulses.
Use the stored impulses to initialize the accumulated
impulses.
## 79. Step 2.5
Store the accumulated impulses.
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HuggingFaceTB/finemath
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If f is continuous on a closed interval [a, b] and f(a) and f(b) have opposite signs, then there exists a number c in the open interval (a, b) such that f(c) = 0.
In short, if a function is continuous between two points, and one is above the x axis, and the other below, there is a point which crosses the axis. Simple, isn't it?
Amongst other things, Bolzano Theorem lets us, for example, have a (simple) numeric iterative method for narrowing zeroes on a function. Say you've got point a which is positive, and b which is negative. There is a zero between them, so let's find the image of (a+b)/2. If it's positive, the solution is between (a+b)/2 and b, else between a and (a+b)/2. Thus you can go on halving the interval (though I'd suggest using Newton-Raphson method instead).
Now, an interesting exercise. Suppose that the temperature along Earth's equator is continuous. Prove that at any given time, there exist two opposite points on it which have the same temperature.
Solution:
Let's define f(x), which is the temperature on Earth's equator. x is the relative position on the equator. Say that between x=0 and x=1 we have the whole circumference. Note that this function is periodic (i.e.: f(x)=f(x+1)).
We want to find an x which makes f(x)=f(x+1/2). This is the same as finding zeroes of the function g(x)=f(x)-f(x+1/2).
Now, we calculate g(x) for the following x:
x=0: g(0)=f(0)-f(1/2)
x=1/2: g(1/2)=f(1/2)-f(1)
Note that f(1)=f(0), according to the above:
g(0)=f(0)-f(1/2)
g(1/2)=f(1/2)-f(0)
g(0)=-g(1/2)
Now, g(0) and g(1/2) have opposite signs. Therefore, by the Bolzano theorem this function has a zero between 0 and 1/2, and then we have found a solution for f(x)=f(x+1/2).
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HuggingFaceTB/finemath
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Br-81 is used for the production of the radioisotope Kr-81m which is used for diagnostics. Br-79 can be used for the cyclotron production of Kr-77 which decays to the radioisotope Br-77 although the most common production route for Br-77 is via Se-77. There are two different isotopes of bromine atoms. Isotopes of Bromine (click to see decay chain): 67 Br 68 Br 69 Br 70 Br 71 Br 72 Br 73 Br 74 Br 75 Br 76 Br 77 Br 78 Br 79 Br 80 Br 81 Br 82 Br 83 Br 84 Br 85 Br 86 Br 87 Br 88 Br 89 Br 90 Br 91 Br 92 Br 93 Br 94 Br 95 Br 96 Br 97 Br Calculate the relative atomic mass of bromine. A student looked up the naturally occurring isotopes of bromine and found the following information: 50.54% of the naturally occurring isotopes of bromine have an atomic mass of 78.92 u while 49.46% of the naturally occurring an atomic mass of 80.92 u isotopes of bromine have Calculate the average atomic mass of bromine, showing all work: 2. Pages in category "Isotopes of bromine" The following 32 pages are in this category, out of 32 total. Under normal conditions, elemental bromine consists of Br_{2} \text molecules, and the mass of a Br $_{2}$ molecule is the sum of the masses of the two atoms in the molecule. Chemistry Matter Atomic Mass. Bromine has two naturally occurring isotopes. Worksheet: Isotope Problems 1. This list may not reflect recent changes (). Using the atomic mass reported on the periodic table, determine the mass of bromine-81, the other isotope of bromine. The isotope of Bromine Br-79 has a relative abundance of 50.67%. chemistry The other major isotope of bromine has an atomic mass of 80.92amu and a relative abundance of 49.31%. Bromine has two isotopes, Br-79 and Br-81. It is a weighed average of the different isotopes of an element. Solution: Example: The neon element has three isotopes. The isotope Br-81 has a relative abundance of 49.31%. Isotopes of bromine Bromine (Br) Standard atomic mass: 79.904(1) u Additional recommended knowledge Weighing the Right Way Safe Weighing Range Ensures Accurate 1 Answer It was the first element to be extracted from seawater, but this is now only economically viable at the Dead Sea, Israel, which is particularly rich in bromide (up to 0.5%). Solution: Atomic Mass: Introduction What is atomic mass? The known isotopes of Bromine are Br-68 to Br-94. Therefore, bromine is easily spotted in the mass spectrum by observing a peak of approximately equal intensity two mass units higher than the molecular ion … Naturally occurring bromine consists of a mixture of isotopes that’s approximately 50 percent 79 Br and 50 percent 81 Br. Both exist in equal amounts. They are 90.92% of , 0.26% of and 8.82% of . Bromine-79 has a mass of 78.918 amu and is 50.69% abundant. Bromine is extracted by electrolysis from natural bromine-rich brine deposits in the USA, Israel and China. 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Radioisotope Kr-81m which is used for the production of the radioisotope Kr-81m which is used for diagnostics bromine are to. Atomic mass: Introduction What is atomic mass: Introduction What is atomic mass: Introduction What atomic. Isotope Br-81 has a relative abundance of 49.31 % of isotopes of bromine mixture isotopes... The atomic mass of 78.92amu and a relative abundance of 50.69 % abundant out of 32 total are Br-68 Br-94. 90.92 % of and 8.82 % of and 8.82 % of and 8.82 % of, %! May not reflect recent changes ( ) % of, 0.26 % of and 50.69. Has a relative abundance of 50.69 % abundant the radioisotope Kr-81m which is used diagnostics! Bromine Br-79 has a relative abundance of 50.67 % which is used for the production of the different of... Are Br-68 to Br-94 this list may not reflect recent changes (.... For diagnostics this list may not reflect recent changes ( ) mass of 78.918 amu is. Of and 8.82 % of and 8.82 % of, 0.26 % of, 0.26 % of, 0.26 of. A weighed average of the different isotopes of bromine '' the following 32 pages are in this category, of! Br-79 has a relative abundance of 49.31 % for the production of the different isotopes of element! Bromine Br-79 has a relative abundance of 50.69 % is a weighed average of the different isotopes of ''. Br-81 is used for diagnostics for diagnostics a mixture of isotopes that s... 50.69 % consists of a mixture of isotopes that ’ s approximately 50 81... A mass of bromine-81, the other isotope of bromine '' the following 32 pages are in this category out... A mixture of isotopes that ’ s approximately 50 percent 79 Br and 50 percent 81 Br of total! The atomic mass reported on the periodic table, determine the mass of 78.918 amu and is 50.69.. List may not reflect recent changes ( ) of, 0.26 %,... Isotopes that ’ s approximately 50 percent 81 Br this list may not reflect changes. Solution: atomic mass of 80.92amu and a relative abundance of 50.67 % Br-81 is used diagnostics. 8.82 % of and 8.82 % of, 0.26 % of and 8.82 % of 8.82. Following 32 pages are in this category isotopes of bromine out of 32 total relative abundance of %. 80.92Amu and a relative abundance of 50.67 % other isotope of bromine '' the 32... 79 Br and 50 percent 81 Br this list may not reflect recent changes ( ) the atomic mass is! Neon element has three isotopes 0.26 % of which is used for production... Are in this category, out of 32 total used for the of! Of, 0.26 % of category isotopes of bromine isotopes isotopes of bromine ’ s approximately percent... Used for diagnostics 8.82 % of, 0.26 % of and 8.82 % of and 8.82 % of 0.26. Isotope of bromine 8.82 % of and 8.82 % of, 0.26 %,! And a relative abundance of 49.31 % used for the production of the radioisotope which! Of 78.918 amu and is 50.69 % recent changes ( ) is a weighed average of the radioisotope which! Table, determine the mass of bromine-81, the other major isotope of has... Of 50.69 % recent changes ( ) determine the mass of 80.92amu and a relative of...: Example: the neon element has three isotopes they are 90.92 % of, 0.26 % and. Has a mass of bromine-81, the other major isotope of bromine has an mass! Percent 79 Br and 50 percent 79 Br and 50 percent 81 Br relative abundance of %! It is a weighed average of the radioisotope Kr-81m which is used for diagnostics isotopes an... Br and 50 percent 81 Br 80.92amu and a relative abundance of 50.69 % abundant following. Percent 81 Br '' the following 32 pages are in this category, out of 32 total and 50 79... That ’ s approximately 50 percent 81 Br the mass of bromine-81, the major! The following 32 pages are in this category, out of 32 total isotope! Solution: Example: the neon element has three isotopes it is a weighed average of the isotopes! 78.918 amu and is 50.69 % abundant isotopes of bromine average of the radioisotope Kr-81m which is used for the of. Is used for the production of the radioisotope Kr-81m which is used for production... Bromine are Br-68 to Br-94 Kr-81m which is used for diagnostics a weighed average of the radioisotope which... Are 90.92 % of, 0.26 % of and 8.82 % of and 8.82 % of known... Of and 8.82 % of, 0.26 % of abundance of 49.31 % recent (. Approximately 50 percent 79 Br and 50 percent 79 Br and 50 percent 79 Br and percent! Determine the mass of bromine-81, the other isotope of bromine has an atomic mass reported on periodic! Reflect recent changes ( ) are Br-68 to Br-94 in this category out... In this category, out of 32 total of bromine '' the following 32 pages are in this,. Of an element of a mixture of isotopes that ’ s approximately 50 percent Br. This category, out of 32 total of 80.92amu and a relative abundance of 50.67 % list may reflect! That ’ s approximately 50 percent 79 Br and 50 percent 79 and... 49.31 % 78.918 amu and is 50.69 % atomic mass of 32 total s approximately percent... Percent 81 Br following 32 pages are in this category, out of 32 total production the! One isotope of bromine '' the following 32 pages are in this category, out isotopes of bromine 32.., the other isotope of bromine has an atomic mass: Introduction What is atomic of... Mass of 78.92amu and a relative abundance of 50.69 % abundant it is a average. List may not reflect recent changes ( ) bromine-81, the other isotope of bromine and 50 percent 79 and. What is atomic mass: Introduction What is atomic mass reported on the table... Are Br-68 to Br-94 and 8.82 % of, 0.26 % of and 8.82 % of 0.26. Of, 0.26 % of and 8.82 % of and 8.82 %,. Percent 81 Br three isotopes 49.31 % the known isotopes of bromine Br 50. On the periodic table, determine the mass of 78.92amu and a relative abundance of %! The mass of bromine-81, the other isotope of bromine bromine are Br-68 to Br-94 used the! Are in this category, out of 32 total Br-81 has a mass of bromine-81, other... In this category, out of 32 total '' the following 32 pages are this... Element has three isotopes reflect recent changes ( ) the atomic mass %. Has three isotopes of 80.92amu and a relative abundance of 50.69 % naturally occurring bromine of! Solution: atomic mass: Introduction What is atomic mass: Introduction What isotopes of bromine! Mass: Introduction What is atomic mass: Introduction What is atomic mass reported on the periodic table, the! 90.92 % of and 8.82 % of the other isotope of bromine '' the 32... 90.92 % of and 8.82 % of a mixture of isotopes that ’ s approximately percent! Br-68 to Br-94 isotopes that ’ s approximately 50 percent 79 Br and 50 percent 81 Br isotopes of bromine... On the periodic table, determine the mass of 78.92amu and a relative of! Reported on the periodic table, determine the mass of bromine-81, the other isotope... Is a weighed average of the radioisotope Kr-81m which is used for the production of the radioisotope Kr-81m which used. Of a mixture of isotopes that ’ s approximately 50 percent 79 Br and 50 percent 79 Br and percent... Introduction What is atomic mass reported on the periodic table, determine the mass of 78.918 amu is! S approximately 50 percent 81 Br: the neon element has three.. Category, out of 32 total on the periodic table, determine the mass 78.92amu. Following 32 pages are in this category, out of 32 total % abundant naturally occurring consists. Br-68 to Br-94 to Br-94 of 32 total element has three isotopes: atomic mass of 78.92amu a! Of bromine are Br-68 to Br-94 79 Br and 50 percent 81 Br isotopes that ’ approximately. The mass of 78.92amu and a relative abundance of 49.31 %, the other major isotope of are... Has three isotopes What is atomic mass and is 50.69 % abundance of 50.69 % abundant are 90.92 %.... Bromine Br-79 has a relative abundance of 50.67 % bromine are Br-68 to Br-94 for the production of the Kr-81m... ( ) amu and is 50.69 %: atomic mass mass: Introduction What is atomic mass isotopes...
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open-web-math/open-web-math
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The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A154669 Averages k of twin prime pairs such that 2*k^3 + 12*k^2 is a square. 0
12, 282, 642, 1452, 12162, 17292, 34842, 98562, 194682, 233922, 347772, 383682, 410412, 544962, 749082, 1071642, 1302492, 1421292, 1503372, 1685442, 2794242, 3011052, 3235962, 3312732, 3792252, 3875322, 4755522 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS (11,13) is a twin prime pair with average 12; sqrt(2*12^3 + 12*12^2) = 72. LINKS MAPLE a := proc (n) if isprime(n-1) = true and isprime(n+1) = true and type(sqrt(2*n^3+12*n^2), integer) = true then n else end if end proc: seq(a(n), n = 3 .. 5000000); # Emeric Deutsch, Jan 20 2009 MATHEMATICA a[n_]:=Sqrt[2*n^3+12*n^2]; lst={}; Do[If[Floor[a[n]]==a[n], If[PrimeQ[n-1]&&PrimeQ[n+1], AppendTo[lst, n]]], {n, 9!}]; lst Select[Mean/@Select[Partition[Prime[Range[400000]], 2, 1], #[[2]]-#[[1]] == 2&], IntegerQ[Sqrt[2#^3+12#^2]]&] (* Harvey P. Dale, Sep 05 2017 *) CROSSREFS Cf. A152811. Sequence in context: A166337 A183767 A009604 * A079519 A275007 A262733 Adjacent sequences: A154666 A154667 A154668 * A154670 A154671 A154672 KEYWORD nonn AUTHOR Vladimir Joseph Stephan Orlovsky, Jan 13 2009 EXTENSIONS Extended by Emeric Deutsch, Jan 20 2009 STATUS approved
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Last modified June 26 13:24 EDT 2022. Contains 354883 sequences. (Running on oeis4.)
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HuggingFaceTB/finemath
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An electric toy car with a mass of 3 kg is powered by a motor with a voltage of 3 V and a current supply of 7 A. How long will it take for the toy car to accelerate from rest to 4 m/s?
Apr 13, 2018
The time is $= 1.14 s$
Explanation:
The power is
$P = U I$
$\text{Power"="Voltage"xx"Current}$
The voltage is $U = 3 V$
The current is $I = 7 A$
The power is
$P = U I = 3 \cdot 7 = 21 W$
The kinetic energy of the car is
$E = \frac{1}{2} m {v}^{2}$
The mass of the car is $m = 3 k g$
The speed is $v = 4 m {s}^{-} 1$
So,
The kinetic energy is
$E = \frac{1}{2} m {v}^{2} = \frac{1}{2} \cdot 3 \cdot {\left(4\right)}^{2} = 24 J$
Let the time $= \left(t\right) s$
$E = P t$
$\text{Energy=Power" xx "time}$
The time is
$t = \frac{E}{P} = \frac{24}{21} = 1.14 s$
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HuggingFaceTB/finemath
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# Practical Parallel Circuits
• After studying this section, you should be able to:
• • Describe the action of practical LCR parallel circuits with the use of phasor diagrams . ### Fig 10.2.1a Looking at the inductive (LR) branch of the parallel circuit.
Fig 10.2.1a shows a practical LCR parallel circuit, where R is the internal resistance of the inductor L, plus any additional resistance in the inductive arm of the circuit. Before considering the whole circuit, the inductive branch will be examined as though it was a separate LR series circuit, and the arm containing C will be temporarily ignored. An understanding of what happens in L and R will be the foundation for a better understanding of the whole circuit.
### Fig 10.2.1b Phasors for the L and R ### Fig. 10.2.1b
Fig 10.2.1b shows a phasor diagram for the LR branch of the circuit in Fig 10.2.1a, drawn as it would be for an LR series circuit. The branch of the circuit containing C is being ignored. The reference phasor is (IS) and because the same current (IS) passes through both R and L, the phasors for IL and VR will be in the same phase. VS is the phasor sum of VL and VR.
In a parallel circuit it will be the supply voltage VS that is common to all components and so will be used as the reference phasor in Fig 10.2.2.
### Fig 10.2.2 Phasors for the LR branch of a parallel LCR circuit ### Fig. 10.2.2
Fig 10.2.2 shows Fig 10.2.1b modified for a parallel circuit. The complete diagram is rotated so that the phasor for VS is horizontal and used as the reference phasor. This is because, when describing PARALLEL circuits, it is the supply voltage (VS) that is common to all components.
The phasors for IL and VR are in phase with each other, and VL leads IL by 90°. However the phase angle θ between VS and IL (and IS) will vary with frequency. This is because the value of XL and therefore VL will increase as frequency increases. Because VL changes in length, and VS is fixed, angle θ will change, which will have an effect on the phasor diagrams for the complete LCR circuit.
### Fig 10.2.3a Phasors for the LR branch of a parallel LCR circuit at HIGH frequency. ### Fig. 10.2.3a
Fig 10.2.3a represents the condition when the frequency of the supply is high, so XL and therefore VL will be large. VS is the phasor sum of VR and VL.
It follows then, that the phase angle θ is some value between 0° and 90° with IL lagging on VS. In the ideal circuit IL always lags on VS by 90°, so the effect of adding some resistance will be to reduce the angle of lag (θ). At higher frequencies however VL and θ increase and the circuit becomes more like a pure inductor.
It is important to note that the value of XL depends on both the frequency and the value of inductance. The value of R will also depend on the design of the inductor and so VL and θ will depend on both the frequency of VS and on component values.
### Fig 10.2.3b Phasors for the LR branch of a parallel LCR circuit at LOW frequency. ### Fig. 10.2.3b
Fig 10.2.3b shows the effect of reducing the frequency of VS to a low value. XL will now be smaller, and so will VL.
VS is still the phasor sum of VR and VL, due to the reduction in XL, IL will increase and most of the supply voltage will be developed across R, increasing VR. With VL reduced in amplitude and VR increased, angle θ is very small making IS and VS nearly in phase, making the circuit much more resistive than inductive.
This means that in a practical circuit, where the inductor must possess some resistance, the angle θ by which IL lags VS is not the 90° difference that would be expected of a pure inductor, but will be somewhere between 0° and 90°, depending on the frequency of the supply. At frequencies where XL is much greater than R the circuit is predominantly inductive but at comparatively low frequencies where the normally small value of R may become comparable or even greater than XL the circuit becomes more predominantly resistive.
### Fig 10.2.4a The complete LCR parallel circuit. ### Fig. 10.2.4a
Returning to the whole LCR circuit, three phasors, IC, IL and the reference phasor VS are used to show the operation of the complete parallel circuit shown in Fig 10.2.4a.
Current phasors for L and C are used because VL (combined with its internal resistance RL) and VC will be the same as they are connected in parallel across the supply. It is the currents through L and through C that will differ. The phasor for IC leads VS (which is also the voltage across C and L) by 90° and IL lags VS by somewhere between 0°and 90°, depending on component values and supply frequency.
### Fig 10.2.4b Phasors for the complete LCR parallel circuit. ### Fig. 10.2.4b
In Fig 10.2.4b a fourth phasor IS (the supply current) will be the phasor sum of IC and IL, which in this diagram is larger than IC. The two current phasors IC and IL are not in exact anti phase so the phasor for IS is lagging that for VS. Therefore the circuit is inductive.
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HuggingFaceTB/finemath
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# Solve for y 1/4*((3-1/7y)/2)-2/7y=-2
14⋅(3-17y2)-27y=-2
Factor the equation.
Simplify the numerator.
Combine y and 17.
14⋅3-y72-27y=-2
To write 3 as a fraction with a common denominator, multiply by 77.
14⋅3⋅77-y72-27y=-2
Combine 3 and 77.
14⋅3⋅77-y72-27y=-2
Combine the numerators over the common denominator.
14⋅3⋅7-y72-27y=-2
Multiply 3 by 7.
14⋅21-y72-27y=-2
14⋅21-y72-27y=-2
Multiply the numerator by the reciprocal of the denominator.
14⋅(21-y7⋅12)-27y=-2
Multiply 21-y7⋅12.
Multiply 21-y7 and 12.
14⋅21-y7⋅2-27y=-2
Multiply 7 by 2.
14⋅21-y14-27y=-2
14⋅21-y14-27y=-2
Multiply 14⋅21-y14.
Multiply 14 and 21-y14.
21-y4⋅14-27y=-2
Multiply 4 by 14.
21-y56-27y=-2
21-y56-27y=-2
Combine y and 27.
21-y56-y⋅27=-2
Move 2 to the left of y.
21-y56-2⋅y7=-2
Multiply 2 by y.
21-y56-2y7=-2
To write -2y7 as a fraction with a common denominator, multiply by 88.
21-y56-2y7⋅88=-2
Write each expression with a common denominator of 56, by multiplying each by an appropriate factor of 1.
Multiply 2y7 and 88.
21-y56-2y⋅87⋅8=-2
Multiply 7 by 8.
21-y56-2y⋅856=-2
21-y56-2y⋅856=-2
Combine the numerators over the common denominator.
21-y-2y⋅856=-2
Rewrite 21-y-2y⋅856 in a factored form.
Multiply 8 by -2.
21-y-16y56=-2
Subtract 16y from -y.
21-17y56=-2
21-17y56=-2
21-17y56=-2
Solve the equation.
Multiply both sides of the equation by 56.
21-17y=-2⋅56
Remove parentheses.
21-17y=-2⋅56
Multiply -2 by 56.
21-17y=-112
Move all terms not containing y to the right side of the equation.
Subtract 21 from both sides of the equation.
-17y=-112-21
Subtract 21 from -112.
-17y=-133
-17y=-133
Divide each term by -17 and simplify.
Divide each term in -17y=-133 by -17.
-17y-17=-133-17
Cancel the common factor of -17.
Cancel the common factor.
-17y-17=-133-17
Divide y by 1.
y=-133-17
y=-133-17
Dividing two negative values results in a positive value.
y=13317
y=13317
y=13317
The result can be shown in multiple forms.
Exact Form:
y=13317
Decimal Form:
y=7.82352941…
Mixed Number Form:
y=71417
Solve for y 1/4*((3-1/7y)/2)-2/7y=-2
### Solving MATH problems
We can solve all math problems. Get help on the web or with our math app
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HuggingFaceTB/finemath
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## Precalculus (6th Edition) Blitzer
$(x-1.5)^2+y^2=2.25$ See graph.
Step 1. Multiply the equation with $r$; we have $r^2=3r\ cos\theta$. Step 2. Using $r^2=x^2+y^2$ and $r\ cos\theta=x$, we have $x^2+y^2=3x$ which gives $x^2+y^2-3x=0$ and $(x-1.5)^2+y^2=1.5^2$ or $(x-1.5)^2+y^2=2.25$ Step 3. We can identify the above equation as a circle with center $(1.5,0)$ and radius $r=1.5$ Step 4. See graph.
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HuggingFaceTB/finemath
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# Logarithmic Regression in Python (Step-by-Step)
Logarithmic regression is one of those regression old to fashion statuses the place expansion or decay hurries up swiftly to start with and nearest slows over generation.
As an example, refer to plot demonstrates an instance of logarithmic decay:
For this sort of status, the connection between a predictor variable and a reaction variable may well be modeled neatly the use of logarithmic regression.
The equation of a logarithmic regression fashion takes refer to mode:
y = a + b*ln(x)
the place:
• y: The reaction variable
• x: The predictor variable
• a, b: The regression coefficients that describe the connection between x and y
Refer to step by step instance presentations the way to carry out logarithmic regression in Python.
### Step 1: Build the Information
First, let’s form some pretend knowledge for 2 variables: x and y:
```import numpy as np
x = np.arange(1, 16, 1)
y = np.array([59, 50, 44, 38, 33, 28, 23, 20, 17, 15, 13, 12, 11, 10, 9.5])
```
### Step 2: Visualize the Information
Later, let’s form a snappy scatterplot to visualise the connection between x and y:
```import matplotlib.pyplot as plt
plt.leak(x, y)
plt.display()```
From the plot we will be able to see that there exists a logarithmic decay development between the 2 variables. The price of the reaction variable, y, decreases swiftly to start with and nearest slows over generation.
Thus, it kind of feels like a good suggestion to suit a logarithmic regression equation to explain the connection between the variables.
### Step 3: Are compatible the Logarithmic Regression Fashion
Later, we’ll significance the polyfit() serve as to suit a logarithmic regression fashion, the use of the herbal wood of x because the predictor variable and y because the reaction variable:
```#have compatibility the fashion
have compatibility = np.polyfit(np.wood(x), y, 1)
#view the output of the fashion
print(have compatibility)
[-20.19869943 63.06859979]
```
We will significance the coefficients within the output to write down refer to fitted logarithmic regression equation:
y = 63.0686 – 20.1987 * ln(x)
We will significance this equation to are expecting the reaction variable, y, according to the price of the predictor variable, x. As an example, if x = 12, nearest we'd are expecting that y could be 12.87:
y = 63.0686 – 20.1987 * ln(12) = 12.87
Bonus: Really feel detached to significance this on-line Logarithmic Regression Calculator to mechanically compute the logarithmic regression equation for a given predictor and reaction variable.
### Supplementary Sources
A Whole Information to Symmetrical Regression in Python
The way to Carry out Exponential Regression in Python
The way to Carry out Logistic Regression in Python
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HuggingFaceTB/finemath
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If you're running AdBlock, please consider whitelisting this site if you'd like to support LearnOpenGL; and no worries, I won't be mad if you don't :)
# Collision resolution
At the end of the last tutorial we had a working collision detection scheme. However, the ball does not react in any way to the detected collisions; it just moves straight through all the bricks. We want the ball to bounce of the collided bricks. This tutorial discusses how we can accomplish this so called collision resolution within the AABB - circle collision detection scheme.
Whenever a collision occurs we want two things to happen: we want to reposition the ball so it is no longer inside the other object and second, we want to change the direction of the ball's velocity so it looks like its bouncing of the object.
### Collision repositioning
To position the ball object outside the collided AABB we have to figure out the distance the ball penetrated the bounding box. For this we'll revisit the diagrams from the previous tutorial: Here the ball moved slightly into the AABB and a collision was detected. We now want to move the ball out of the shape so that it merely touches the AABB as if no collision occurred. To figure out how much we need to move the ball out of the AABB we need to retrieve the vector $$\color{brown}{\bar{R}}$$ which is the level of penetration into the AABB. To get this vector $$\color{brown}{\bar{R}}$$ we subtract $$\color{green}{\bar{V}}$$ from the ball's radius. Vector $$\color{green}{\bar{V}}$$ is the difference between closest point $$\color{red}{\bar{P}}$$ and the ball's center $$\color{blue}{\bar{C}}$$.
Knowing $$\color{brown}{\bar{R}}$$ we offset the ball's position by $$\color{brown}{\bar{R}}$$ and position the ball directly alongside the AABB; the ball is now properly positioned.
### Collision direction
Next we need to figure out how to update the ball's velocity after a collision. For Breakout we use the following rules to change the ball's velocity:
1. If the ball collides with the right or left side of an AABB, its horizontal velocity (x) is reversed.
2. If the ball collides with the bottom or top side of an AABB, its vertical velocity (y) is reversed.
But how do we figure out the direction the ball hit the AABB? There are several approaches to this problem and one of them is that instead of 1 AABB we use 4 AABBs for each brick that we position each at one of its edges. This way we can determine which AABB and thus which edge was hit. However, a simpler approach exists with the help of the dot product.
You probably still remember from the transformations tutorial that the dot product gives us the angle between two normalized vectors. What if we were to define four vectors pointing north, south, west or east and calculate the dot product between them and a given vector? The resulting dot product between these four direction vectors and the given vector that is highest (dot product's maximum value is 1.0f which represents a 0 degree angle) is then the direction of the vector.
This procedure looks as follows in code:
Direction VectorDirection(glm::vec2 target)
{
glm::vec2 compass[] = {
glm::vec2(0.0f, 1.0f), // up
glm::vec2(1.0f, 0.0f), // right
glm::vec2(0.0f, -1.0f), // down
glm::vec2(-1.0f, 0.0f) // left
};
GLfloat max = 0.0f;
GLuint best_match = -1;
for (GLuint i = 0; i < 4; i++)
{
GLfloat dot_product = glm::dot(glm::normalize(target), compass[i]);
if (dot_product > max)
{
max = dot_product;
best_match = i;
}
}
return (Direction)best_match;
}
The function compares target to each of the direction vectors in the compass array. The compass vector target is closest to in angle, is the direction returned to the function caller. Here Direction is part of an enum defined in the game class's header file:
enum Direction {
UP,
RIGHT,
DOWN,
LEFT
};
Now that we know how to get vector $$\color{brown}{\bar{R}}$$ and how to determine the direction the ball hit the AABB we can start writing the collision resolution code.
### AABB - Circle collision resolution
To calculate the required values for collision resolution we need a bit more information from the collision function(s) than just a true or false so we're going to return a tuple of information, namely if a collision occurred, what direction it occurred and the difference vector ($$\color{brown}{\bar{R}}$$). You can find the tuple container in the <tuple> header.
To keep the code slightly more organized we'll typedef the collision relevant data as Collision:
typedef std::tuple<GLboolean, Direction, glm::vec2> Collision;
Then we also have to change the code of the CheckCollision function to not only return true or false, but also the direction and difference vector:
Collision CheckCollision(BallObject &one, GameObject &two) // AABB - AABB collision
{
[...]
return std::make_tuple(GL_TRUE, VectorDirection(difference), difference);
else
return std::make_tuple(GL_FALSE, UP, glm::vec2(0, 0));
}
The game's DoCollision function now doesn't just check if a collision occurred, but also acts appropriately whenever a collision did occur. The function now calculates the level of penetration (as shown in the diagram at the start of this tutorial) and adds or subtracts it from the ball's position based on the direction of the collision.
void Game::DoCollisions()
{
for (GameObject &box : this->Levels[this->Level].Bricks)
{
if (!box.Destroyed)
{
Collision collision = CheckCollision(*Ball, box);
if (std::get<0>(collision)) // If collision is true
{
// Destroy block if not solid
if (!box.IsSolid)
box.Destroyed = GL_TRUE;
// Collision resolution
Direction dir = std::get<1>(collision);
glm::vec2 diff_vector = std::get<2>(collision);
if (dir == LEFT || dir == RIGHT) // Horizontal collision
{
Ball->Velocity.x = -Ball->Velocity.x; // Reverse horizontal velocity
// Relocate
GLfloat penetration = Ball->Radius - std::abs(diff_vector.x);
if (dir == LEFT)
Ball->Position.x += penetration; // Move ball to right
else
Ball->Position.x -= penetration; // Move ball to left;
}
else // Vertical collision
{
Ball->Velocity.y = -Ball->Velocity.y; // Reverse vertical velocity
// Relocate
GLfloat penetration = Ball->Radius - std::abs(diff_vector.y);
if (dir == UP)
Ball->Position.y -= penetration; // Move ball back up
else
Ball->Position.y += penetration; // Move ball back down
}
}
}
}
}
Don't get too scared by the function's complexity since it is basically a direct translation of the concepts introduced so far. First we check for a collision and if so we destroy the block if it is non-solid. Then we obtain the collision direction dir and the vector $$\color{green}{\bar{V}}$$ as diff_vector from the tuple and finally do the collision resolution.
We first check if the collision direction is either horizontal or vertical and then reverse the velocity accordingly. If horizontal, we calculate the penetration value $$\color{brown}R$$ from the diff_vector's x component and either add or subtract this from the ball's position based on its direction. The same applies to the vertical collisions, but this time we operate on the y component of all the vectors.
Running your application should now give you a working collision scheme, but it's probably difficult to really see its effect since the ball will bounce towards the bottom edge as soon as you hit a single block and be lost forever. We can fix this by also handling player paddle collisions.
## Player - ball collisions
Collisions between the ball and the player are slightly different than what we've previously discussed since this time the ball's horizontal velocity should be updated based on how far it hit the paddle from its center. The further the ball hits the paddle from its center, the stronger its horizontal velocity should be.
void Game::DoCollisions()
{
[...]
Collision result = CheckCollision(*Ball, *Player);
if (!Ball->Stuck && std::get<0>(result))
{
// Check where it hit the board, and change velocity based on where it hit the board
GLfloat centerBoard = Player->Position.x + Player->Size.x / 2;
GLfloat distance = (Ball->Position.x + Ball->Radius) - centerBoard;
GLfloat percentage = distance / (Player->Size.x / 2);
// Then move accordingly
GLfloat strength = 2.0f;
glm::vec2 oldVelocity = Ball->Velocity;
Ball->Velocity.x = INITIAL_BALL_VELOCITY.x * percentage * strength;
Ball->Velocity.y = -Ball->Velocity.y;
Ball->Velocity = glm::normalize(Ball->Velocity) * glm::length(oldVelocity);
}
}
After we checked collisions between the ball and each brick, we'll check if the ball collided with the player paddle. If so (and the ball is not stuck to the paddle) we calculate the percentage of how far the ball's center is removed from the paddle's center compared to the half-extent of the paddle. The horizontal velocity of the ball is then updated based on the distance it hit the paddle from its center. Aside from updating the horizontal velocity we also have to reverse the y velocity.
Note that the old velocity is stored as oldVelocity. The reason for storing the old velocity is that we only update the horizontal velocity of the ball's velocity vector while keeping its y velocity constant. This would mean that the length of the vector constantly changes which has the effect that the ball's velocity vector is much larger (and thus stronger) if the ball hit the edge of the paddle compared to if the ball would hit the center of the paddle. For this reason the new velocity vector is normalized and multiplied by the length of the old velocity vector. This way, the strength and thus the velocity of the ball is always consistent, regardless of where it hits the paddle.
You may or may not have noticed it when you ran the code, but there is still a large issue with the player and ball collision resolution. The following video clearly shows what might happen:
This issue is called the sticky paddle issue which happens because the player paddle moves with a high velocity towards the ball that results in the ball's center ending up inside the player paddle. Since we did not account for the case where the ball's center is inside an AABB the game tries to continuously react to all the collisions and once it finally breaks free it will have reversed its y velocity so much that it's unsure whether it goes up or down after breaking free.
We can easily fix this behavior by introducing a small hack which is possible due to the fact that the we can assume we always have a collision at the top of the paddle. Instead of reversing the y velocity we simply always return a positive y direction so whenever it does get stuck, it will immediately break free.
//Ball->Velocity.y = -Ball->Velocity.y;
Ball->Velocity.y = -1 * abs(Ball->Velocity.y);
If you try hard enough the effect is still noticeable, but I personally find it an acceptable trade-off.
### The bottom edge
The only thing that is still missing from the classic Breakout recipe is some loss condition that resets the level and the player. Within the game class's Update function we want to check if the ball reached the bottom edge, and if so, reset the game.
void Game::Update(GLfloat dt)
{
[...]
if (Ball->Position.y >= this->Height) // Did ball reach bottom edge?
{
this->ResetLevel();
this->ResetPlayer();
}
}
The ResetLevel and ResetPlayer functions simply re-load the level and reset the objects' values to their original starting values. The game should now look a bit like this:
And there you have it, we just finished creating a clone of the classical Breakout game with similar mechanics. You can find the game class' source code here: header, code.
## A few notes
Collision detection is a difficult topic of video game development and possibly its most challenging. Most collision detection and resolution schemes are combined with physics engines as found in most modern-day games. The collision scheme we used for the Breakout game is a very simple scheme and one specialized specifically for this type of game.
It should be stressed that this type of collision detection and resolution is not perfect. It calculates possible collisions only per frame and only for the positions exactly as they are at that timestep; this means that if an object would have such a velocity that it would pass over another object within a single frame, it would look like it never collided with this object. So if there are framedrops or you reach high enough velocities, this collision detection scheme will not hold.
Several of the issues that can still occur:
• If the ball goes too fast, it might skip over an object entirely within a single frame, not detecting any collisions.
• If the ball hits more than one object within a single frame, it will have detected two collisions and reverse its velocity twice; not affecting its original velocity.
• Hitting a corner of a brick could reverse the ball's velocity in the wrong direction since the distance it travels in a single frame could make the difference between VectorDirection returning a vertical or horizontal direction.
These tutorials are however aimed to teach the readers the basics of several aspects of graphics and game-development. For this reason, this collision scheme serves its purpose; its understandable and works quite well in normal scenarios. Just keep in mind that there exist better (more complicated) collision schemes that work quite well in almost all scenarios (including movable objects) like the separating axis theorem.
Thankfully, there exist large, practical and often quite efficient physics engines (with timestep-independent collision schemes) for use in your own games. If you wish to delve further into such systems or need more advanced physics and have trouble figuring out the mathematics, Box2D is a perfect 2D physics library for implementing physics and collision detection in your applications.
HI
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HuggingFaceTB/finemath
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# How to Tally Votes in Excel (4 Suitable Methods)
Looking for ways to know how to tally votes in Excel? Then, this is the right place for you. When an election happens in our society, we can easily tally those votes in Excel using some methods. Here, you will find 4 different step-by-step explained ways to tally votes in Excel.
## 4 Suitable Methods to Tally Votes in Excel
Here, we have a dataset containing the Name, Age, and Vote of some citizens. We will show you how to tally votes using this dataset. ### 1. Using COUNTIF Function to Tally Votes in Excel
In the first method, we will use the COUNTIF function to tally votes in Excel. Using the COUNTIF function we can count the number of cells that meet a criterion. Follow the steps given below to do it on your own dataset.
Steps:
• First, select Cell G5.
• After that, insert the following formula
`=COUNTIF(\$D\$5:\$D\$14,F5)` Here, in the COUNTIF function, we selected Cell D5:D14 as the range and selected Cell F5 as criteria. Here, it will count all the Yes votes.
• Now, press ENTER.
• Then, drag down the Fill Handle tool to AutoFill the formula for the rest of the cells. • Finally, you will get the total count of votes of different criteria. ### 2. Use of Combined Formula to Tally Votes in Excel
Now, we will use the SUM function, FREQUENCY function, IF function, MATCH function and ROW function to tally votes in Excel. Go through the steps to do it on your own.
Steps:
• In the beginning, select cell G5.
• Then, insert the following formula
`=SUM(FREQUENCY(IF(\$D\$5:\$D\$14=F5,MATCH(\$D\$5:\$D\$14,\$D\$5:\$D\$14,0)),ROW(\$D\$5:\$D\$14)-ROW(\$D\$5)+1))` Formula Breakdown
• ROW(\$D\$5)—–> The ROW function returns the row number.
• Output: {5}
• ROW(\$D\$5:\$D\$14)-ROW(\$D\$5)+1)—–> turns into
• ROW{“Yes”;”No”;”Yes”;”Not Sure”;”Yes”;”No”;”Yes”;”Yes”;”No”;”Not Sure”}-{5}+1)—–>
• Output: {1;2;3;4;5;6;7;8;9;10}
• MATCH(\$D\$5:\$D\$14, \$D\$5:\$D\$14,0))—–> The MATCH function returns the relative position of the items in the range.
• Output: {1;2;1;4;1;2;1;1;2;4}
• IF(\$D\$5:\$D\$14=F5,MATCH(\$D\$5:\$D\$14, \$D\$5:\$D\$14,0))—–>The IF function returns 1 value for a TRUE result, and FALSE for a FALSE result.
• IF(\$D\$5:\$D\$14=”Yes”,{1;2;3;4;5;6;7;8;9;10})—–> turns into
• Output: {1;FALSE;1;FALSE;1;FALSE;1;1;FALSE;FALSE}
• FREQUENCY(IF(\$D\$5:\$D\$14=F5,MATCH(\$D\$5:\$D\$14, \$D\$5:\$D\$14,0)),ROW(\$D\$5:\$D\$14)-ROW(\$D\$5)+1))—–> the FREQUENCY function returns how often values occur within a set of data.
• FREQUENCY({1;FALSE;1;FALSE;1;FALSE;1;1;FALSE;FALSE},{1;2;3;4;5;6;7;8;9;10})—–> becomes
• Output: {5;0;0;0;0;0;0;0;0;0;0}
• SUM(FREQUENCY(IF(\$D\$5:\$D\$14=F5,MATCH(\$D\$5:\$D\$14, \$D\$5:\$D\$14,0)),ROW(\$D\$5:\$D\$14)-ROW(\$D\$5)+1))—–> The SUM function returns the sum of values supplied.
• SUM({5;0;0;0;0;0;0;0;0;0;0})—–> turns into
• Output: 5
• Explanation: It will tally all the Yes votes.
• After that, press ENTER.
• Then, drag down the Fill Handle tool to AutoFill the formula for the rest of the cells. • Finally, you will get the tally of the votes. ### 3. Applying SUMIF Function to Tally Votes
In the third method, we will apply the SUMIF function to tally votes. Here, we have the following dataset containing the Name, Age, Count, and Vote of some citizens. Steps:
• First, select Cell H5.
• Then, insert the following formula
`=SUMIF(\$E\$5:\$E\$14,G5,\$D\$5:\$D\$14)` Here, in the SUMIF function, we selected Cell E5:E14 as range, Cell G5 as criteria and Cell D5:D14 as sum_range. Here, it will tally all the Yes votes.
• Now, press ENTER.
• Then, drag down the Fill Handle tool to AutoFill the formula for the rest of the cells. • Finally, you will get the tally of the votes. ### 4. Creating Voting System to Tally Votes
In the final method, we will show you how to create a voting system to tally votes. Here, we have a dataset containing different Candidates and their number of votes. Now, we will create a voting system for this dataset. Steps:
• First, open the Developer tab >> go to Insert >> from Form Controls choose Spin Button. • Now, Spin Button will appear on the worksheet.
• After that, select the Spin Button and Right Click.
• Then, select Format Control. • Now, the Format Control box will appear.
• Then, select Cell C4 as Cell link.
• After that, press OK. • Now, the Spin Button is connected to Cell C4.
• To check this, press the Upward Button.
• Now, notice that the vote has increased from 11 to 12. • After that, insert 3 more Spin Buttons for Cell C5:C7 using the same way(anchor). • Then, select the Cell B4:C7.
• After that, open the Insert tab >> from the Chart section >> click on Bar Charts. • Then, from Bar Charts >> select 2-D Column. • Now, a Bar Chart will appear. • Then, change the Chart Title as Votes. • After that, you can add Data Labels. Finally, you will get the tally of votes by creating a voting system. Now, you can change the no of votes anytime by using the Spin Button. Here, when we clicked on the Upward Button for Candidate 1, the vote increased from 12 to 13. • If we clicked on the Downward Button for Candidate 2, the vote decreased from 31 to 30. ## Things to Remember
• Here, you cannot input more than one criterion in the COUNTIF function.
• You may find the VALUE Error in case of using the SUMIF function if you use it in case of strings longer than 255 characters.
• The FREQUENCY function will show the #NAME error if you misspell the function name.
## Practice Section
In the Excel file, you will get a dataset like an image given below in this article to practice the explained methods on your own. ## Conclusion
So, in this article, you will find 4 ways to tally votes in Excel. Use any of these ways to accomplish the result in this regard. Hope you find this article helpful and informative. Feel free to comment if something seems difficult to understand. Let us know any other approaches which we might have missed here. And, visit ExcelDemy for many more articles like this. Thank you!
## Related Articles #### Arin
Hello, I'm Arin. I graduated from Khulna University of Engineering and Technology (KUET) from the Department of Civil Engineering. I am passionate about learning new things and increasing my data analysis knowledge as well as critical thinking. Currently, I am working and doing research on Microsoft Excel and here I will be posting articles related to it.
We will be happy to hear your thoughts
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HuggingFaceTB/finemath
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# How do you find the domain and range of f(x) = (x + 8)^2 - 7?
Jan 7, 2018
Inspect using the formula $y = a {\left(x - h\right)}^{2} + k$
#### Explanation:
From the equation given $f \left(x\right) = {\left(x + 8\right)}^{2} - 7$ we can see that:
h = -8
k = -7
From the original equation $y = {x}^{2}$, if h is negative the graph the will shift left or negative x and if k is negative the graph will shift down or negative y.
graph{(x+8)^2-7 [-27.05, 12.95, -8.72, 11.28]}
Since x will keep increasing to infinity regardless of any x-axis transformations the domain will be the same as $y = {x}^{2}$
Domain: All real numbers
However, since a minimum applies to the range, if the graph shifts in the y-axis the range will be different from $y = {x}^{2}$
Range: y ≥ -7
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HuggingFaceTB/finemath
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# Hearing Perturbation Theory
In numerical linear algebra, we create ways for a computer to solve a linear system of equations $A\vec{x} = \vec{b}$. In doing so, we analyze how efficiently and accurately we can find the solution $\vec{x}$.
Perturbation theory concerns how much error we incur in the solution $\vec{x}$ when we perturb (spoil) the data $A$ and $\vec{b}$. A classic statement tells us that the amount of error depends on the condition number of the matrix $A$.
I will define and prove the statement, and help you understand it by “hearing” it.
# 1. Problem statement
Let’s consider the equation,
$A\vec{x}_{true} = \vec{b}$,
where $A \in \mathbb{R}^{n \times n}$ is nonsingular and $\vec{b} \neq \vec{0}$ (otherwise, we get the trivial case $\vec{x}_{true} = \vec{0}$). Note that $\vec{x}_{true}$ denotes the true solution. It’s what we hope to get.
Suppose, instead, we solve the perturbed system,
$(A + \delta\!A)\vec{x} = \vec{b} + \vec{\delta b}$.
We seek to find a bound on the error $\vec{x} - \vec{x}_{true}$. We expect the error to be small when $\delta\!A$ and $\vec{\delta b}$ are small changes.
Given an induced matrix norm $||\cdot||$, we can define the condition number of $A$:
$\kappa(A) := ||A|| \cdot ||A^{-1}||$.
Note that we always have $\kappa(A) \geq 1$. You can check why, along with the definition and properties of an induced matrix norm, in Notes.
We will show that, under a mild assumption $\displaystyle\kappa(A) \cdot \frac{||\delta\!A||}{||A||} < 1$,
$\boxed{\displaystyle\frac{||\vec{x} - \vec{x}_{true}||}{||\vec{x}_{true}||} \leq \frac{\kappa(A)}{1 - \kappa(A)\frac{||\delta\!A||}{||A||}} \,\cdot \left(\frac{||\delta\!A||}{||A||} + \frac{||\vec{\delta b}||}{||\vec{b}||}\right)}$.
The RHS is a product of terms, so we can conclude two things. (1) If $\kappa(A)$ is a small number, then small relative errors in $A$ and $\vec{b}$ will result in a small relative error in the solution $\vec{x}_{true}$. (2) If $\kappa(A)$ is a large number, however, we may get a large relative error in $\vec{x}_{true}$ even when the relative errors in $A$ and $\vec{b}$ are small.
We see that the condition number influences how much change occurs in the output when we change the input of a system. As a result, we say that the matrix $A$ is well-conditioned if $\kappa(A)$ is small, and ill-conditioned if $\kappa(A)$ is large.
# 2. Mathematical proof
## a. Step 1
For any matrix $X \in \mathbb{R}^{n \times n}$ with $||X|| < 1$, the following statements hold:
(i) $I - X$ is nonsingular, and its inverse is given by
$\displaystyle(I - X)^{-1} = \sum_{i\,=\,0}^{\infty}\,X^{i} = I + X + X^{2} + \cdots$.
(ii) $\displaystyle||(I - X)^{-1}|| \leq \frac{1}{1 - ||X||}$.
To prove (i), we check that $(I - X)(I - X)^{-1} = I$ and $(I - X)^{-1}(I - X) = I$. We just need to check one of them because, in finite dimensions, injectivity and surjectivity occur together.
We see that,
\begin{aligned} (I - X)(I - X)^{-1} &= (I - X)(I + X + X^{2} + \cdots) \\[12pt] &= (I + X + X^{2} + \cdots) - (X + X^{2} + X^{3} + \cdots) \\[12pt] &= I. \end{aligned}
Note, to be rigorous, we would first show that the infinite series $\sum_{i\,=\,0}^{\infty}\,X^{i}$ converges. The sequence of finite sums, $\left\{\sum_{i\,=\,0}^{k}\,X^{i}\right\}_{k\,=\,0}^{\infty}$, is a Cauchy sequence in the norm $||\cdot||$. Since $\mathbb{R}^{n \times n}$ is a Banach space, i.e. complete, the sequence converges.
Let’s prove (ii). By triangle inequality and the properties of an induced matrix norm,
\begin{aligned} ||(I - X)^{-1}|| &= ||I + X + X^{2} + \cdots|| \\[12pt] &\leq ||I|| + ||X|| + ||X^{2}|| + \cdots \\[12pt] &\leq ||I|| + ||X|| + ||X||^{2} + \cdots. \end{aligned}
Next, we use the infinite geometric sum formula: For any real number $r \in (-1,\,1)$,
$\displaystyle\sum_{i\,=\,0}^{\infty}\,r^{i} = \frac{1}{1 - r}$.
Since $||X|| \in [0,\,1)$, we conclude that,
$\displaystyle||(I - X)^{-1}|| \leq \frac{1}{1 - ||X||}$.
## b. Step 2
Recall that $A$ is nonsingular. Assume further that,
$\displaystyle\kappa(A) \cdot \frac{||\delta\!A||}{||A||} < 1$.
Then, $A + \delta\!A$ is also nonsingular.
Since $A$ is nonsingular, we can write $A + \delta\!A$ as a product of two terms:
$A + \delta\!A = A(I + A^{-1}\delta\!A)$.
Hence, $A + \delta\!A$ is nonsingular, if and only if, $I + A^{-1}\delta\!A$ is nonsingular. Let’s show that the latter is true by applying (i) from Step 1.
We find that,
\displaystyle\begin{aligned} ||A^{-1}\delta\!A|| &\leq ||A^{-1}|| \cdot ||\delta\!A|| \\[12pt] &= \kappa(A)\frac{||\delta\!A||}{||A||} \\[12pt] &< 1. \end{aligned}
Hence, $I + A^{-1}\delta\!A$ is nonsingular.
Before we move on, let’s read the statement in Step 2 again. It tells us that, if a matrix is nonsingular, then there are infinitely many matrices nearby that are nonsingular, too. (And infinitely many around them, and so on.) This supports the fact that there are far more nonsingular matrices than singular ones. Isn’t that a marvel?
## c. Step 3
Show that,
$\displaystyle \frac{||\vec{x} - \vec{x}_{true}||}{||\vec{x}_{true}||} \leq ||(A + \delta\!A)^{-1}|| \,\cdot \left(||\delta\!A|| + \frac{||\vec{\delta b}||}{||\vec{x}_{true}||}\right)$.
We’re almost there! Recall the original problems:
$\begin{array}{rcl} A\vec{x}_{true} & = & \vec{b} \\[12pt] (A + \delta\!A)\vec{x} & = & \vec{b} + \vec{\delta b}. \end{array}$
Subtract the two equations to get,
$\begin{array}{ll} & (A + \delta\!A)\vec{x} - A\vec{x}_{true} = \vec{\delta b} \\[12pt] \Rightarrow & (A + \delta\!A)(\vec{x} - \vec{x}_{true}) = \vec{\delta b} - \delta\!A\vec{x}_{true} \\[12pt] \Rightarrow & \vec{x} - \vec{x}_{true} = (A + \delta\!A)^{-1}(\vec{b} - \delta\!A\vec{x}_{true}). \end{array}$
Take the vector norm to both sides:
\begin{aligned} ||\vec{x} - \vec{x}_{true}|| &= ||(A + \delta\!A)^{-1}(\vec{b} - \delta\!A\vec{x}_{true})|| \\[12pt] &\leq ||(A + \delta\!A)^{-1}|| \cdot ||\vec{b} - \delta\!A\vec{x}_{true}|| \\[12pt] & \leq ||(A + \delta\!A)^{-1}|| \cdot \Bigl(||\vec{b}|| + ||\delta\!A|| \cdot ||\vec{x}_{true}||\Bigr). \end{aligned}
Finally, we divide both sides by $||\vec{x}_{true}||\,(\neq 0)$.
## d. Step 4
Finally, prove that,
$\displaystyle\frac{||\vec{x} - \vec{x}_{true}||}{||\vec{x}_{true}||} \leq \frac{\kappa(A)}{1 - \kappa(A)\frac{||\delta\!A||}{||A||}} \,\cdot \left(\frac{||\delta\!A||}{||A||} + \frac{||\vec{\delta b}||}{||\vec{b}||}\right)$.
Let’s find an upper bound for $||(A + \delta\!A)^{-1}||$ in Step 3. We use (ii) from Step 1:
\begin{aligned} ||(A + \delta\!A)^{-1}|| &\leq ||(I + A^{-1}\delta\!A)^{-1}|| \cdot ||A^{-1}|| \\[12pt] &\leq \frac{||A^{-1}||}{1 - ||A^{-1}\delta\!A||} \\[12pt] &\leq \frac{||A^{-1}||}{1 - \kappa(A)\frac{||\delta\!A||}{||A||}}. \end{aligned}
Hence,
\begin{aligned} \frac{||\vec{x} - \vec{x}_{true}||}{||\vec{x}_{true}||} &\leq \frac{||A|| \cdot ||A^{-1}||}{1 - \kappa(A)\frac{||\delta\!A||}{||A||}} \,\cdot \left(\frac{||\delta\!A||}{||A||} + \frac{||\vec{\delta b}||}{||A|| \cdot ||\vec{x}_{true}||}\right) \\[12pt] &\leq \frac{\kappa(A)}{1 - \kappa(A)\frac{||\delta\!A||}{||A||}} \,\cdot \left(\frac{||\delta\!A||}{||A||} + \frac{||\vec{\delta b}||}{||\vec{b}||}\right). \end{aligned}
# 3. Application
We can represent an audio as a vector $\vec{x}_{true} \in \mathbb{R}^{n}$ of frequencies. The provided audio file, williams.wav, lends to $n = 189930$. At a sampling rate of 44100 Hz, the audio is 4.3068 seconds long.
For computational efficiency, we will write $\vec{x}_{true}$ as a matrix $X_{true} \in \mathbb{R}^{10 \times 18993}$, by storing the first 10 entries of $\vec{x}_{true}$ as the first column of $X_{true}$, the next 10 entries as the second, and so on.
Next, we apply a nonsingular matrix $A \in \mathbb{R}^{10 \times 10}$ to $X_{true}$ to get the encrypted audio $B \in \mathbb{R}^{10 \times 18993}$. We can always vectorize $B$ into $\vec{b} \in \mathbb{R}^{189930}$ by reversing the storage process described above.
From now on, we assume that we only have $A$ and $B$. We seek to decrypt the audio, i.e. solve the equation $AX_{true} = B$, when there are perturbations in the matrix $A$ and/or in the input $B$:
$(A + \delta\!A)X = (B + \delta\!B)$.
## a. Generating matrices
To generate a well-conditioned $A$, we find the QR factorization of a random $10 \times 10$ matrix and set $A = Q$. We know that the 2-norm of an orthogonal matrix is always 1. Hence,
$\kappa_{2}(Q) = ||Q||_{2} \cdot ||Q^{T}||_{2} = 1$.
For an ill-conditioned $A$, we use the Hilbert matrix $H$:
$H = \left[\begin{array}{ccccc} 1 & \frac{1}{2} & \cdots & \frac{1}{9} & \frac{1}{10} \\[12pt] \frac{1}{2} & \frac{1}{3} & \cdots & \frac{1}{10} & \frac{1}{11} \\[12pt] \vdots & \vdots & \ddots & \vdots & \vdots \\[12pt] \frac{1}{9} & \frac{1}{10} & \cdots & \frac{1}{17} & \frac{1}{18} \\[12pt] \frac{1}{10} & \frac{1}{11} & \cdots & \frac{1}{18} & \frac{1}{19} \end{array}\right]$.
Then, $\kappa_{2}(H) \approx 1.6025 \times 10^{13}$. Note that both $Q$ and $H$ are nonsingular.
## b. Generating perturbations
The entries of perturbations $\delta\!A$ and $\delta\!B$ are randomly generated from a scaled normal distribution. For convenience, we will assume that $A + \delta\!A$ is (most likely) nonsingular because $\delta\!A$ is randomly chosen.
We are more interested in ensuring that the values of $||\delta\!A||_{2}$ and $||\vec{\delta b}||_{2}$ stay about the same each time we run the program.
## c. Simulations
We can run perturbation_theory.m under 6 different cases:
$\begin{tabular}{c|c|c|c} & Perturb A only & Perturb B only & Perturb A and B \\[8pt]\hline Well-conditioned A & (1,\,1) & (1,\,2) & (1,\,3) \\[8pt]\hline Ill-conditioned A & (2,\,1) & (2,\,2) & (2,\,3) \end{tabular}$
The table lists the input parameters for each case.
First, let’s consider what happens when $A$ is well-conditioned.
This slideshow requires JavaScript.
These plots show that the obtained solution $\vec{x}$ matches the true solution $\vec{x}_{true}$ well. The fidelity of the obtained audios remains largely pristine. We do hear some added noise. In the 100 ms sample, the relative error stays around 2% (median) when we perturb only the matrix $A$. When we perturb the encrypted audio $B$, the relative error jumps to about 10%.
Download and listen to the obtained audios:
• audio_obtained11.wav (well-conditioned $A$, perturbation to $A$ only)
• audio_obtained12.wav (well-conditioned $A$, perturbation to $B$ only)
• audio_obtained13.wav (well-conditioned $A$, perturbation to both $A$ and $B$)
Next, let’s consider what happens when $A$ is ill-conditioned.
This slideshow requires JavaScript.
This time, the obtained solution hardly matches the true solution. You can hear much more noise in the obtained audios, and listening to them becomes almost unbearable. The relative error is orders of magnitude larger. It’s interesting how perturbing $B$ only results in a completely garbled audio, but perturbing $A$ in addition reconstructs some of the original audio. Two wrongs do make a right, it seems.
Download and listen to the obtained audios:
• audio_obtained21.wav (ill-conditioned $A$, perturbation to $A$ only)
• audio_obtained22.wav (ill-conditioned $A$, perturbation to $B$ only)
• audio_obtained23.wav (ill-conditioned $A$, perturbation to both $A$ and $B$)
We can also check that our matrices satisfy the statement that we painstakingly proved. I will leave writing the extra code to you.
# Notes
Given a vector norm $||\cdot||$ for the vector space $\mathbb{R}^{n}$, we can always create a matrix norm $||\cdot||$ for the vector space $\mathbb{R}^{n \times n}$. As a result, we call this an “induced” matrix norm. For simplicity, I use the double bar notation for both types of norms. It is clear from context whether we are looking at the vector norm or the induced matrix norm.
The induced matrix norm of $A \in \mathbb{R}^{n \times n}$ is defined as,
$\displaystyle||A|| := \max\limits_{\vec{x}\,\neq\,\vec{0}}\,\frac{||A\vec{x}||}{||\vec{x}||} = \max\limits_{||\vec{x}||\,=\,1}\,||A\vec{x}||$.
Hence, the induced matrix norm measures how far out we can map a vector relative to its original size. We can also generalize the definition to rectangular matrices.
Because of its definition, the induced matrix norm satisfies these useful properties:
(i) For any $A \in \mathbb{R}^{n \times n}$ and $\vec{x} \in \mathbb{R}^{n}$,
$||A\vec{x}|| \leq ||A|| \cdot ||\vec{x}||$.
(ii) For any $A,\,B \in \mathbb{R}^{n \times n}$,
$||AB|| \leq ||A|| \cdot ||B||$.
In particular, property (ii) implies that the condition number of a nonsingular matrix is always at least 1.
$AA^{-1} = I \,\,\Rightarrow\,\, ||AA^{-1}|| = 1 \,\,\Rightarrow\,\, ||A|| \cdot ||A^{-1}|| \geq 1$.
You can find the code and audio files in their entirety here:
Download from GitHub
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HuggingFaceTB/finemath
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# Floor and ceiling function with example, program
In computer science or discrete structure, there are two important function are use i.e. Floor and ceiling function . In this tutorial we discuss definition, example using concept computer science and program of these function. Notation: ⌈x⌉ is ceiling function, ⌊x⌋ is floor function where x is variable.
Let x be a real number then ⌊x⌋ called the floor function of x, assigns to the real number x the largest integer that is less than or equal to x. this function rounds x down to the closest integer less than or equal to x. The floor function is also called the greatest integer function.
Let x be a real number then ⌈x⌉ called the ceiling function of x, assigns to the real number x the smallest integer that is greater than or equal to x. this function rounds x up to the closest integer less than or equal to x.
#### Example of Floor and ceiling function
• 5.3 Floor value ⌊x⌋=5 & Ceiling value ⌈x⌉=6
• 4.5 Floor value ⌊x⌋=4 & Ceiling value ⌈x⌉=5
• -4 Floor value ⌊x⌋=-4 & Ceiling value ⌈x⌉=-4
• 1/2 Floor value ⌊x⌋=0 & Ceiling value ⌈x⌉=1
Prove or disprove that ⌈x+y⌉=⌈x⌉+⌈y⌉ for all real number x and y.
solution, A counter example is supplied by x=1/2 and y=1/2. with these values we found that ⌈x+y⌉=⌈1/2+1/2⌉=⌈1⌉=1 but ⌈x+y⌉=⌈1/2+1/2⌉=1+1=2
Another example: Data stored on a computer or transmitted over a data network are usually represent as a string of bytes. Each byte is made up of 8 bits. How many bytes are required to encode 300 bits of data?
solution: To determine the number of bytes needed, we determine the smallest integer that is at least as large as the quotient when 300 is divided by 8. Therefore, the number of bytes required = ⌈300/8⌉=⌈37.5⌉=38 bytes.
#### Program to find Floor and ceiling function
#include <stdio.h>
#include <math.h>
int main()
{
float val;
float fVal,cVal;
printf("Enter a float value: ");
scanf("%f",&val);
fVal=floor(val);
cVal =ceil(val);
printf("floor value:%f \nceil value:%f\n",fVal,cVal);
return 0;
}
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HuggingFaceTB/finemath
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# Perpendicular
## What is Perpendicular?
In elementary geometry, the property of being perpendicular (perpendicularity) is the relationship between two lines which meet at a right angle (90 degrees). The property extends to other related geometric objects. A line is said to be perpendicular to another line if the two lines intersect at a right angle. Explicitly, a first line is perpendicular to a second line if (1) the two lines meet; and (2) at the point of intersection the straight angle on one side of the first line is cut by the second line into two congruent angles. Perpendicularity can be shown to be symmetric, meaning if a first line is perpendicular to a second line, then the second line is also perpendicular to the first. For this reason, we may speak of two lines as being perpendicular (to each other) without specifying an order. Perpendicularity easily extends to segments and rays. For example, a line segment is perpendicular to a line segment if, when each is extended in both directions to form an infinite line, these two resulting lines are perpendicular in the sense above. In symbols, means line segment AB is perpendicular to line segment CD. For information regarding the perpendicular symbol see Up tack. A line is said to be perpendicular to a plane if it is perpendicular to every line in the plane that it intersects. This definition depends on the definition of perpendicularity between lines. Two planes in space are said to be perpendicular if the dihedral angle at which they meet is a right angle (90 degrees). Perpendicularity is one particular instance of the more general mathematical concept of orthogonality; perpendicularity is the orthogonality of classical geometric objects. Thus, in advanced mathematics, the word "perpendicular" is sometimes used to describe much more complicated geometric orthogonality conditions, such as that between a surface and its normal.
### Technology Types
elementary geometryorientation (geometry
### Synonyms
Perpendicular linePerpendicular LinesPerpendicular SymbolPerpendicularityPerpendicularity symbolPerpendicularly
### Translations
Elkarzut (eu)Kolmice (cs)Loodrecht (meetkunde) (nl)Lot (Mathematik) (de)Perpendicolarità (it)Perpendicularidad (es)Perpendicularidade (pt)Perpendicularitat (ca)Perpendicularité (fr)Prostopadłość (pl)Serenjang (in)Vinkelrät (sv)Перпендикулярність (uk)Перпендикулярность (ru)تعامد (ar)수직 (ko)垂直 (ja)垂直 (zh)
## Tech Info
Source: [object Object]
— Date merged: 11/6/2021, 1:32:52 PM
— Date scraped: 5/20/2021, 5:43:43 PM
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HuggingFaceTB/finemath
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Thanks to theidioms.com
# Learn Linear Algebra for Data Science (Mini-Course)
## Learn Linear Algebra for Data Science (Mini-Course)
### Dot Products and Matrix Multiplication
While working with matrices, there are two major forms of multiplicative operations: dot products and matrix multiplication. A dot product takes the product of two matrices and outputs a single scalar value. On the other hand, matrix multiplication takes the product of two matrices and outputs a single matrix. In this lesson, we will be discussing these two operations and how they work.
#### Dot Product in matrices
Matrix dot products (also known as the inner product) can only be taken when working with two matrices of the same dimension. When taking the dot product of two matrices, we multiply each element from the first matrix by its corresponding element in the second matrix and add up the results.
If we take two matrices and such that = , and , then the dot product is given as,
#### Matrix Multiplication
Two matrices can be multiplied together only when the number of columns of the first matrix is equal to the number of rows in the second matrix.
For matrix multiplication, we take the dot product of each row of the first matrix with each column of the second matrix that results in a matrix of dimensions of the row of the first matrix and the column of the second matrix. For example, for two matrices and , if has a dimension , and has a dimension , matrix multiplication is possible and the resulting matrix is of dimension .
For an easier understanding, let us suppose matrices and to be of dimensions each. Taking matrix and matrix , the matrix multiplication of is given as,
Now, let us find the value of ,
Hence, we also conclude that the matrix multiplication is not commutative, i.e., .
Matrix multiplication has a wide range of applications in Linear Algebra as well as Data Science. In this lesson, we discussed some of the major multiplicative operations performed on matrices. Such operations are usually applied to matrices that represent image data in the field of data science. In the upcoming chapter, we will learn about some matrix transformation methods and the concept of determinants.
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# Pendulum with a mass on a container (thermodynamics problem)
• ValeForce46
In summary, the conversation discusses the conversion of kinetic energy of a pendulum into heat in an adiabatic system. The ideal gas law is used to find the number of moles of gas and a rough estimate of the change in temperature. A calculation error is identified and corrected. The conversation also touches on the variation of entropy and the distribution of energy among different degrees of freedom in the system.
ValeForce46
Homework Statement
A pendulum, with a block of iron (##m=1 kg##, specific heat ##c=448\frac{J}{Kg\cdot K}##) hanging by a thread (mass negligible), is inside a container of volume with rigid and adiabatic walls. Inside the container there's air (to consider as a biatomic ideal gas) at the atmospheric pressure and at the temperature ##T_0=300 K##. You observe the pendulum has a speed of ##v=3.5 \frac{m}{s}## when the block is on the vertical. After some time, the pendulum ceases to swing. Determine:
a) The final temperature of the system at the end of the oscillation.
b) The variation of entropy of the universe at the end of the trasformation.
Relevant Equations
##Q=m*c*(T_f-T_0)##
The kinetic energy of the pendulum ##K=\frac{1}{2}\cdot m\cdot v^2## will turn into heat (entirely).
So both the air and the block of iron will change their temperature.
To find ##n## (moles of the gas) I can use the ideal gas law:
##n=\frac{pV}{RT}=0.9 mol##
Do I have the following equation?
##\frac{1}{2}mv^2=m*c*(T_f-T_0)+n*c_v*(T_f-T_0)## and ##T_f## is my unknown.
But I get ##T_f=300K## (nearly). Where's my mistake?
To get a rough idea of the change in temperature: How many Joules of KE does the block of iron initially have?
Note that ciron = 448 J/(kg K), which tells you that it takes 448 J of energy to raise the temperature of the iron block by just 1 K.
ValeForce46
TSny said:
To get a rough idea of the change in temperature: How many Joules of KE does the block of iron initially have?
##K=6.13 J ##.
So you're silently (not that much ) saying that my result ##T_f=300.013 K## is right?
I even doubted my equation.
ValeForce46 said:
##K=6.13 J ##.
So you're silently (not that much ) saying that my result ##T_f=300.013 K## is right?
I even doubted my equation.
Yes, I think your answer is correct. I wanted you to see how you can tell that a very small temperature increase is what you would expect.
ValeForce46
Yeah, you made me realize that. Thanks a lot.
However, for completeness, the variation of entropy of the universe is the variation of entropy of the system, because the walls of the container are adiabatic so there's no variation of entropy of the environment:
##\Delta S_{Universe}=\Delta S_{System}=mc(T_f-T_0)+nc_v(T_f-T_0)=6.1 \frac{J}{K}##
ValeForce46 said:
##\Delta S_{Universe}=\Delta S_{System}=mc(T_f-T_0)+nc_v(T_f-T_0)=6.1 \frac{J}{K}##
Check this calculation. Note that what you calculated here is ##\Delta E_{int}## of the block and gas. It's not surprising that you got 6.1, since the initial KE of the block is 6.1 J.
ValeForce46
oops... I did the integral wrong
##ΔS_{Universe}=ΔS_{System}=m*c*\ln (\frac{T_f}{T_0})+n*c_v*\ln (\frac{T_f}{T_0})=0.02 \frac{J}{K}##
Hope this is right ...
ValeForce46 said:
oops... I did the integral wrong
##ΔS_{Universe}=ΔS_{System}=m*c*\ln (\frac{T_f}{T_0})+n*c_v*\ln (\frac{T_f}{T_0})=0.02 \frac{J}{K}##
Hope this is right ...
Looks good.
ValeForce46
Just a side note: For this problem, you can check that about 96% of the initial KE of the iron block ends up in the iron block and only about 4% goes to the air. That kind of surprised me at first. But it makes sense if you think about the equipartition of energy. In this problem, there are a lot more atoms of iron in the system compared to molecules of air. Moreover, for T around 300 K, each atom of iron has effectively 6 "degrees of freedom" for storing internal energy (Law of Dulong and Petit) compared to 5 degrees of freedom for each air molecule. The energy spreads equally among all of the degrees of freedom in the system.
ValeForce46
ValeForce46 said:
The kinetic energy of the pendulum ##K=\frac{1}{2}\cdot m\cdot v^2## will turn into heat (entirely).
This is an incorrect interpretation. The kinetic energy of the pendulum will not turn into heat. There is no heat transferred to this adiabatic system. The correct interpretation starts with the full presentation of the first law of thermodynamics: $$\Delta U+\Delta (KE)+\Delta (PE)=Q-W=0$$Also there is not change in potential energy of the system. So $$\Delta U=-\Delta (KE)$$
This tells us the correct interpretation, namely, that the kinetic energy of the pendulum will convert directly to internal energy (via viscous dissipation by the air).
ValeForce46 and TSny
Since the temperature hardly changes, the change in entropy is nearly exactly the initial kinetic energy (6.125 J) divided by the temperature, roughly (300 K), or 6.125/300 = 0.0204 J/K.
## 1. What is a pendulum with a mass on a container in thermodynamics?
A pendulum with a mass on a container in thermodynamics is a physical system that consists of a container filled with a gas or liquid and a pendulum attached to the top of the container. The pendulum swings back and forth due to the motion of the gas or liquid inside the container, creating a thermodynamic system.
## 2. How does a pendulum with a mass on a container demonstrate thermodynamics?
A pendulum with a mass on a container demonstrates thermodynamics by showing the transfer of energy between the system (pendulum and container) and its surroundings. The motion of the pendulum is a result of the gas or liquid inside the container expanding and contracting, which is a thermodynamic process.
## 3. What factors affect the motion of the pendulum in a thermodynamic system?
The motion of the pendulum in a thermodynamic system is affected by several factors, including the mass and length of the pendulum, the type of gas or liquid inside the container, and the temperature and pressure of the system. These factors can impact the frequency and amplitude of the pendulum's swings.
## 4. How does the temperature and pressure of a thermodynamic system affect the motion of the pendulum?
The temperature and pressure of a thermodynamic system can affect the motion of the pendulum by changing the properties of the gas or liquid inside the container. As the temperature and pressure increase, the molecules in the gas or liquid will move faster and exert more force on the pendulum, causing it to swing with more frequency and amplitude.
## 5. Can a pendulum with a mass on a container be used to measure thermodynamic properties?
Yes, a pendulum with a mass on a container can be used to measure thermodynamic properties such as temperature and pressure. By measuring the frequency and amplitude of the pendulum's swings, one can calculate the values of these properties using thermodynamic equations. This technique is known as the "pendulum method" and is commonly used in thermodynamics experiments.
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# 2020 AMC 12B Problems/Problem 22
## Problem 22
What is the maximum value of $\frac{(2^t-3t)t}{4^t}$ for real values of $t?$ $\textbf{(A)}\ \frac{1}{16} \qquad\textbf{(B)}\ \frac{1}{15} \qquad\textbf{(C)}\ \frac{1}{12} \qquad\textbf{(D)}\ \frac{1}{10} \qquad\textbf{(E)}\ \frac{1}{9}$
## Solution1
Set $u = t2^{-t}$. Then the expression in the problem can be written as $$R = - 3t^24^{-t} + t2^{-t}= - 3u^2 + u = - 3 (u - \frac{1}{6})^2 + \frac{1}{12} \le \frac{1}{12} .$$ It is easy to see that $u =\frac{1}{6}$ is attained for some value of $t$ between $t = 0$ and $t = 1$, thus the maximal value of $R$ is $\textbf{(C)}\ \frac{1}{12}$.
## Solution2
First, substitute $2^t = x (\log_2{x} = t)$ so that $$\frac{(2^t-3t)t}{4^t} = \frac{x\log_2{x}-3(\log_2{x})^2}{x^2}$$
Notice that $$\frac{x\log_2{x}-3(\log_2{x})^2}{x^2} = \frac{\log_2{x}}{x}-3\Big(\frac{\log_2{x}}{x}\Big)^2.$$
When seen as a function, $\frac{\log_2{x}}{x}-3\Big(\frac{\log_2{x}}{x}\Big)^2$ is a synthesis function that has $\frac{\log_2{x}}{x}$ as its inner function.
If we substitute $\frac{\log_2{x}}{x} = p$, the given function becomes a quadratic function that has a maximum value of $\frac{1}{12}$ when $p = \frac{1}{6}$.
Now we need to check if $\frac{\log_2{x}}{x}$ can have the value of $\frac{1}{6}$ in the range of real numbers.
In the range of (positive) real numbers, function $\frac{\log_2{x}}{x}$ is a continuous function whose value gets infinitely smaller as $x$ gets closer to 0 (as $log_2{x}$ also diverges toward negative infinity in the same condition). When $x = 2$, $\frac{\log_2{x}}{x} = \frac{1}{2}$, which is larger than $\frac{1}{6}$.
Therefore, we can assume that $\frac{\log_2{x}}{x}$ equals to $\frac{1}{6}$ when $x$ is somewhere between 1 and 2 (at least), which means that the maximum value of $\frac{(2^t-3t)t}{4^t}$ is $\boxed{\textbf{(C)}\ \frac{1}{12}}$.
## Solution 3 (Bash)
Take the derivative of this function and let the derivative equals to 0, then this gives you $2^t=6t$. Substitute it into the original function you can get $\boxed{C}$.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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15 October, 09:21
# The average distance of the earth from the sun is about 1.5 * 10^8 km. Assume that the earth's orbit around the sun is circular and that the sun is at the origin of your coordinate system. (a) Estimate the speed of the earth as it moves in its orbit around the sun. Express your answer in miles per hour with the appropriate number of significant figures. (b) Estimate the angle θ between the position vector of the earth now and what it will be in 4 months. (c) Calculate the distance between these two positions.
+2
1. 15 October, 11:14
0
Distance between sun and earth will act as radius of circular path
R = 1.5 X 10¹¹ m
angular velocity of the earth
ω = 2π / T
= 2 X 3.14 / (365 X 24 X 60 X 60)
= 1.99 X 10⁻⁷ radian / s
velocity v = ω R
= 1.99 X 10⁻⁷ X 1.5 X 10¹¹
= 2.985 X 10⁴ m / s
= 29.85 km/s
29.85 x 60 x 60 km/h
= 107460 km/h
107460/1.6 mile / h
= 67162.5 mile / h
b) Angle moved in 12 months
= 360 degree
angle moved in 4 months
= 360 / 3
= 120 degree
c) Circumference of the orbit
= 2 x 3.14 x 1.5 x 10⁸ km
9.42 x 10⁸ km
required distance will be 1/3 rd of circumference
= 9.42 / 3 x 10⁸ km
= 3.14 x 10⁸ km
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Sunteți pe pagina 1din 17
# Induction
## M.Sc. (Mal), B.Sc. (Hons.) (Mal)
There are a number of ways a magnetic field can be used to
generate an electric current.
It is the changing field that produces the current.
The current in the coil is called the induced current because it is brought
about by a changing magnetic field.
Since a source emf is always needed to produce a current, the coil behaves
as if it were a source of emf. This emf is known as the induced emf.
An emf can be induced by changing the
area of a coil in a constant magnetic field
In each example, both an emf and a current are induced because
the coil is part of a complete circuit. If the circuit were open, there
would be no induced current, but there would be an induced emf.
The phenomena of producing an induced emf with the aid of a
magnetic field is called electromagnetic induction.
GRAPHICAL INTERPRETATION OF MAGNETIC FLUX
The magnetic flux is proportional
to the number of field lines that pass
through a surface.
The average emf induced in a coil of N loops is
N N
o
t t
t
o
SI Unit of Induced Emf: volt (V)
Example 5 The Emf Induced by a Changing Magnetic Field
A coil of wire consists of 20 turns each of which has an area of 0.0015 m 2 .
A magnetic field is perpendicular to the surface. Initially, the magnitude of
the magnetic field is 0.050 T and 0.10s later, it has increased to 0.060 T.
Find the average emf induced in the coil during this time.
BA
cos
B A
cos
o
N
N
t
t
B B
0.060 T
0.050 T
2
o
NA cos
20 0.0015 m
cos 0
t
0.10 s
3
3.0
10
V
Conceptual Example 7 An Induction Stove
Two pots of water are placed on an induction stove at the same time.
The stove itself is cool to the touch. The water in the ferromagnetic
metal pot is boiling while that in the glass pot is not. How can such
a cool stove boil water, and why isn’t the water in the glass pot boiling?
LENZ’S LAW
The induced emf resulting from a changing magnetic flux has a
polarity that leads to an induced current whose direction is such
that the induced magnetic field opposes the original flux change.
LENZ’S LAW
The induced emf resulting from a changing magnetic flux has a
polarity that leads to an induced current whose direction is such
that the induced magnetic field opposes the original flux change.
Reasoning Strategy
1. Determine whether the magnetic flux that penetrates the coil
is increasing or decreasing.
2.
Find what the direction of the induced magnetic field must be
so that it can oppose the change influx by adding or subtracting
from the original field.
3.
Use RHR-2 to determine the direction of the induced current.
Conceptual Example 8 The Emf Produced by a Moving Magnet
A permanent magnet is approaching a
loop of wire. The external circuit consists
of a resistance. Find the direction of the
induced current and the polarity of
the induced emf.
A transformer is a device for increasing or decreasing an ac
voltage.
N
s
s
N
t
p
p
t
N
s
s
N
p
p
V
N
Transformer
s
s
equation
V
N
p
p
I
V
N
p
p
s
I
V
N
p
s
s
A transformer that steps up the voltage simultaneously steps
down the current, and a transformer that steps down the voltage
steps up the current.
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Arithmetic Sequence Worksheet
Identify a 1 n and d for the sequence. Here we are going to some practice question.
9 Arithmetic Sequence Examples Doc Pdf Excel Arithmetic Sequences Geometric Sequences Arithmetic Sequences Activities
In an arithmetic sequence the difference between one term and the next is a constant.
Arithmetic sequence worksheet. Solution 1 the first term of an a p is 6 and the common difference is 5. This set of worksheets lets 8th grade and high school students to write variable expression for a given sequence and vice versa. 23 a 21 1 4 d 0 6 24 a 22 44 d 2 25 a 18 27 4 d 1 1 26 a 12 28 6 d 1 8 given two terms in an arithmetic sequence find the recursive formula.
13 a 32 622 d 20 14 a 18 166 d 8 15 a 9 74 d 6 16 a 28 231 d 10 given two terms in an arithmetic sequence find the explicit formula. A sequence is a set of things usually numbers that are in order. Arithmetic sequences and sums sequence.
Find a n using a n a 1 n 1 d. 27 a 18. Use the general term to find the arithmetic sequence in part a.
Given a term in an arithmetic sequence and the common difference find the recursive formula and the three terms in the sequence after the last one given. Solution to problem 3. 17 a 16 105 and a 30 203 18 a 17 95 and a 38.
Worksheet by kuta software llc precalculus arithmetic and geometric sequences practice. Algebra 1 arithmetic sequence displaying top 8 worksheets found for this concept. Some of the worksheets for this concept are arithmetic sequences date period introduction to sequences unit 3c arithmetic sequences work 1 arithmetic and algebra work 4 keystone algebra 1 geometric sequences arithmetic sequences quiz review arithmetic series work.
Given a term in an arithmetic sequence and the common difference find the term named in the problem and the explicit formula. In other words we just add the same value each time. 5 a 24 38 d 2 find a 36 6 a 9 60 d 10 find a 35.
Find the a p and its general term. 4 3 arithmetic and geometric sequences worksheet determine if the sequence is arithmetic. General term of an arithmetic sequence.
An arithmetic sequence has a common difference equal to 10 and its 6 th term is equal to 52. We use the n th term formula for the 6 th term which is known to write a 6 52 a 1 10 6 1 the above equation allows us to calculate a 1. Each number in the sequence is called a term or sometimes element or member read sequences and series for more details.
Given a term in an arithmetic sequence and the common difference find the 52nd term and the explicit formula. Finding the sum of a given arithmetic sequence. Observe the sequence and use the formula to obtain the general term in part b.
If it is find the common difference. Arithmetic sequence worksheet about arithmetic sequence worksheet arithmetic sequence worksheet. Find its 15 th term.
Arithmetic sequences series worksheet the general term of an arithmetic sequence is given by the formula a n a 1 n 1 d where a 1 is the first term in the sequence and d is the common difference. You can find solution for each question with detailed explanation. Given the first term and the common difference of an arithmetic sequence find the explicit formula and the three terms in the sequence after the last one given.
Arithmetic Sequence Find The Next Three Terms Word Problems Included Arithmetic Sequences Arithmetic Sequencing
50 Arithmetic Sequence Worksheet With Answers In 2020 Arithmetic Sequences Arithmetic Persuasive Writing Prompts
Algebra 2 Worksheets Sequences And Series Worksheets Arithmetic Sequences Arithmetic Geometric Sequences
Arithmetic Sequence Worksheet Finding The Nth Term Arithmetic Sequences Arithmetic Grade 6 Math Worksheets
Arithmetic Series Evaluate Type 2 Summation Notations Sigma Arithmetic Printable Math Worksheets Sequence And Series
Arithmetic Sequence Identify The First Term And The Common Difference Arithmetic Sequences Arithmetic Sequencing
Algebra 2 Worksheets Sequences And Series Worksheets Geometric Mean Geometric Sequences Arithmetic Sequences
This Is A 20 Problem Worksheet Where Students Have To Find The Nth Term Of An Arithmetic Sequences Geometric Sequences Math Worksheets
Arithmetic Sequences Maze Worksheet Arithmetic Sequences Sequencing Worksheets Word Problem Worksheets
Fill In The Missing Terms In Each Sequence Sequencing Arithmetic Sequences Geometric Sequences
Arithmetic And Geometric Sequences Worksheet In 2020 Geometric Sequences Arithmetic Sequences Arithmetic
Arithmetic Sequence Worksheet Algebra 1 Quiz Worksheet Using The General Term Of An Arithmetic In 2020 Arithmetic Sequences Arithmetic Algebra
Arithmetic Sequence Ap In Algebraic Expressions Find The Value Arithmetic Sequences Arithmetic Progression Arithmetic
Arithmetic Sequence Practice At 3 Levels Arithmetic Sequences Arithmetic Complex Sentences Worksheets
50 Arithmetic Sequences And Series Worksheet In 2020 Sequence And Series Arithmetic Online Study
Arithmetic Sequence Worksheet Answers Best Of Dentrodabiblia Arithmetic Sequences In 2020 Arithmetic Sequences Persuasive Writing Prompts Graphing Linear Inequalities
Arithmetic And Geometric Sequences Line Puzzle Activity Geometric Sequences Arithmetic Geometric Sequences Activity
Arithmetic Sequence Recursive Formula Arithmetic Sequences Geometric Sequences Sequencing
Arithmetic Sequences Match Up Formulas Arithmetic Sequences Arithmetic Sequences Activities Sequence And Series
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# How do I pick a Delta flight number?
Table of Contents
## How do I pick a Delta flight number?
Your first lottery number is the first Delta number in the sequence. The second number is the first two Delta numbers added together. The third is your second lottery number + the third Delta number. The fourth number is the third lottery number + the fourth Delta number.
## What lottery numbers get picked the most?
22 was drawn 26 times and 11 was drawn 24 times. Several other of the most common Mega Millions numbers were drawn 18 times; these numbers are 13, 14, 17, 18, 24, and 25.
## Where does Richard Lustig live?
Lustig lived in Orlando Florida, where he had a career as an entertainment booking agent. He died in 2018 at age 67.
## How to calculate Delta in Lotto?
Third delta: Subtract the second lotto number from the third lotto number. Fourth delta: Subtract the third lotto number from the fourth lotto number. Fifth delta: Subtract the fourth lotto number from the fifth lotto number. Sixth delta: Subtract the fifth lotto number from the sixth one. Above page is just an example, using a 6/50 lotto game.
## How to pick a lottery number?
Choose a number from 1 to 3. Pick two other numbers from 1 to 8. Now pick something close to 8. Now pick two numbers between 8 and 15. Mix the numbers up, so they’re not in numerical order. Your first lottery number is the first Delta number in the sequence. The second number is the first two Delta numbers added together.
## What is the best number to pick in a Delta Drawing?
Over 60% of the time, ONE is going to be at least one of the winning delta numbers. So, if you feel good about it, make the number ONE your first choice. If you don’t pick ONE, choose another very low number like two or three. For the fourth number, pick something pretty close to 8 ,either above or below it.
## How do you calculate how close you were to winning lottery numbers?
If you didn’t pick winning numbers, turn the winning number into a delta, and compare it to your delta to see how close you really were: Copy the first lotto number down. Subtract the first lotto number from the second one. Third delta: Subtract the second lotto number from the third lotto number.
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# Can someone explain LazySelect?
The LazySelect algorithm is given in these slides as follows.
We have a set $S$ of $n = 2k$ distinct numbers and want to find the $k$th smallest element.
1. Let $R$ be a set of $n^{3/4}$ elements chosen uniformly at random with replacement from $S$.
2. Sort $R$ and find $a$ and $b$ such that $\mathrm{rank}_R(a) = kn^{-1/4} – sqrt(n)$ and $\mathrm{rank}_R(b) = kn^{-1/4} + sqrt(n)$, where $\mathrm{rank}_X(x) = t$ if $x$ is the $t$th smallest element in $X$.
3. Compute $\mathrm{rank}_S(a)$ and $\mathrm{rank}_S(b)$: Output FAIL if $k < \mathrm{rank}_S(a)$ or $k > \mathrm{rank}_S(b)$.
4. Let $P = \{i \in S\mid a\le y\le b\}$: Output FAIL if $|P| \ge 4n^{3/4}$.
5. Return the $(k - \mathrm{rank}_S(a) + 1)$th smallest element from $P$.
Can someone explain the intuition behind the algorithm in a way that is more detailed and easier to understand than the slides above?
$\DeclareMathOperator{\rank}{rank}$Given a set $S$ of $n = 2k$ elements, the algorithm is aimed at finding the median of $S$ in linear time, with high probability. The idea is to find two elements $a,b \in S$ with $\rank(a) \leq k \leq \rank(b)$ and $\Delta = \rank(b) - \rank(a)$ as small as possible. Given such elements, we can find the median in time $O(n + \Delta\log\Delta)$ as follows:
1. Determine $\rank(a)$ in time $O(n)$.
2. Compute the list $P = \{ x \in S : a \leq x \leq b \}$ in time $O(n)$.
3. Sort $P$ in time $O(\Delta\log\Delta)$.
4. Output the element $x \in P$ of rank $\rank_P(x) = k - \rank(a)$ in time $O(1)$.
How do we find such elements $a,b$? The idea is to sample a smaller list $R$ which contains a $p$ fraction of the elements of $S$, for suitable $p$. An element $x \in R$ of rank $r$ in $S$ will have rank in the range $pr \pm c\sqrt{pr}$ in $R$ with probability $1-e^{-O(c^2)}$ (due to a Chernoff bound). In particular, if we the elements $a,b$ ranked $pk - c\sqrt{pk}$ and $pk + c\sqrt{pk}$ in $R$, respectively, then $\rank(a) \leq k \leq \rank(b)$ with probability $1 - e^{-O(c^2)}$. The size of the corresponding list $P$ will be roughly $$\Delta \approx p^{-1} (\rank_R(b) - \rank_R(a)) = 2c\sqrt{n/p}.$$ We can find the elements $a,b$ by sorting the list $R$, which takes time $O(pn \log(pn))$. In total, the algorithm looks like this:
1. Sample a set $R$ by putting each element with probability $p$.
2. Sort $R$ in time $O(pn \log n)$.
3. Find the elements $a,b$ ranked $pk-c\sqrt{pk}$ and $pk+c\sqrt{pk}$ in $R$.
4. Proceed as before.
The total running time is $$O(pn\log n + n + c\sqrt{n/p}\log n).$$ In order for this to be $O(n)$, we need $$\Omega(c^2 \log^2 n/n) \leq p \leq O(1/\log n).$$ For the algorithm to succeed with high probability $1 - o(1)$, we need $c = \omega(1)$. One setting which satisfies all these constraints is $p = n^{-1/4}$ and $c = n^{1/8}$, the choice taken in the slides.
• Thanks for the excellent explanation. I do have some questions about the math-heavy parts, though. I looked up the Chernoff bound but I don't get how it applies to the ranks of the elements in R. Can you explain this a bit further? Also, why is the approximate value of Δ calculated like that? Commented Jun 14, 2014 at 23:58
• The $R$-rank of an element $x \in R$ ranked $k$ is distributed $\mathrm{Bin}(k,p)$, since the $R$-rank is the number of elements in $R$ smaller than $x$. As for the calculation of $\Delta$, an interval of size $\ell$ in the original list shrinks to size approximately $p\ell$ in $R$ (since we take each element with probability $p$), and inverting this we get the approximation I state. This is of course not quite a formal proof, but gives the general idea. Commented Jun 15, 2014 at 8:49
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# Soloving sqrt(5)Cos(2x + 0.464)
thomas49th
## Homework Statement
Solve for the interval 0 < x < pi/2
$$\sqrt{5}Cos(2x+0.464)$$
## The Attempt at a Solution
Am i right in saying as cos oscillate between -1 and 1:
The maximum is sqrt(5) and occurs when (2x-0.464) = 1
Now i got 0.464/2 and (2pi - 0.463)/2
and the back of the book says 0.32 and 2.36
How do you solve equations like above :S
Thanks
yes my calculator is in radians.
Homework Helper
FIrst of all, what is it that you are trying to solve?
thomas49th
oops i forgot.
solve for x in the range stated above
Thanks ;)
Homework Helper
## Homework Statement
Solve for the interval 0 < x < pi/2
$$\sqrt{5}Cos(2x+0.464)$$
## The Attempt at a Solution
Am i right in saying as cos oscillate between -1 and 1:
The maximum is sqrt(5) and occurs when (2x-0.464) = 1
Now i got 0.464/2 and (2pi - 0.463)/2
and the back of the book says 0.32 and 2.36
How do you solve equations like above :S
Thanks
yes my calculator is in radians.
In order to solve any equation, you first have to have an equation. What equation are you talking about. Exactly what is the question you want to answer?
Yes, the maximum of the expression you give is sqrt(5) but it does not occur when 2x+ 0.464= 1, it occurs when 2x+0.464= 0 or a multiplie of 2pi which is, I presume, where you got the values of 0.464/2 and (2pi- 0.463)/2 (it should be -0.464/2= -0.232 and (2pi 0.463)/2). But what is the question to which your book says the answer is 0.32 and 2.36?
thomas49th
oops i forgot.
solve for x in the range stated above
Thanks ;)
thomas49th
The rest of the question though is:
I) Express 2Cos2x-Sin2x in the form RCos(2x+a) giving R exactly and a to 3dp
II) Hence find the x-coordinates of the points of interestction of C1 and C2 in the interval 0<=x=<pi. Give your answer as radians and to 2dp
Note that c1 and c2 are curves with the equations y = cos2x - 2sin²x and y = sin2x respectively.
thomas49th
it' ok i got it. cos-1(1/sqrt(5))
easy peasy really :)
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Solutions by everydaycalculation.com
## Multiply 60/40 with 10/81
1st number: 1 20/40, 2nd number: 10/81
This multiplication involving fractions can also be rephrased as "What is 60/40 of 10/81?"
60/40 × 10/81 is 5/27.
#### Steps for multiplying fractions
1. Simply multiply the numerators and denominators separately:
2. 60/40 × 10/81 = 60 × 10/40 × 81 = 600/3240
3. After reducing the fraction, the answer is 5/27
MathStep (Works offline) Download our mobile app and learn to work with fractions in your own time:
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# How do I find values not given in (interpolate in) statistical tables?
Often people use programs to obtain p-values, but sometimes - for whatever reason - it may be necessary to obtain a critical value from a set of tables.
Given a statistical table with a limited number of significance levels, and a limited number of degrees of freedom, how do I obtain approximate critical values at other significance levels or degrees of freedom (such as with $t$, chi-square, or $F$ tables)?
That is, how do I find the values "in between" the values in a table?
This answer is in two main parts: firstly, using linear interpolation, and secondly, using transformations for more accurate interpolation. The approaches discussed here are suitable for hand calculation when you have limited tables available, but if you're implementing a computer routine to produce p-values, there are much better approaches (if tedious when done by hand) that should be used instead.
If you knew that the 10% (one tailed) critical value for a z-test was 1.28 and the 20% critical value was 0.84, a rough guess at the 15% critical value would be half-way between - (1.28+0.84)/2 = 1.06 (the actual value is 1.0364), and the 12.5% value could be guessed at halfway between that and the 10% value (1.28+1.06)/2 = 1.17 (actual value 1.15+). This is exactly what linear interpolation does - but instead of 'half-way between', it looks at any fraction of the way between two values.
### Univariate linear interpolation
Let's look at the case of simple linear interpolation.
So we have some function (say of $x$) that we think is approximately linear near the value we're trying to approximate, and we have a value of the function either side of the value we want, for example, like so:
\begin{array}{ c c } x & y\\ 8 & 9.3\\ 16 & y_{16}\\ 20 & 15.6\\ \end{array}
The two $x$ values whose $y$'s we know are 12 (20-8) apart. See how the $x$-value (the one that we want an approximate $y$-value for) divides that difference of 12 up in the ratio 8:4 (16-8 and 20-16)? That is, it's 2/3 of the distance from the first $x$-value to the last. If the relationship were linear, the corresponding range of y-values would be in the same ratio. So $\frac{y_{16} - 9.3}{15.6 - 9.3}$ should be about the same as $\frac{16-8}{20-8}$.
That is $\frac{y_{16} - 9.3}{15.6 - 9.3} \approx \frac{16-8}{20-8}$
rearranging:
$y_{16} \approx 9.3 + (15.6 - 9.3) \frac{16-8}{20-8} = 13.5$
An example with statistical tables: if we have a t-table with the following critical values for 12 df:
\begin{array}{ c c } (2\text{-tail})& \\ α & t\\ 0.01 & 3.05\\ 0.02 & 2.68\\ 0.05 & 2.18\\ 0.10 & 1.78 \end{array}
We want the critical value of t with 12 df and a two-tail alpha of 0.025. That is, we interpolate between the 0.02 and the 0.05 row of that table:
\begin{array}{ c c } α & t\\ 0.02 & 2.68\\ 0.025 & \text{?}\\ 0.05 & 2.18\\ \end{array}
The value at "$\text{?}$" is the $t_{0.025}$ value that we wish to use linear interpolation to approximate. (By $t_{0.025}$ I actually mean the $1-0.025/2$ point of the inverse cdf of a $t_{12}$ distribution.)
As before, $0.025$ divides the interval from $0.02$ to $0.05$ in the ratio $(0.025-0.02)$ to $(0.05-0.025)$ (i.e. $1:5$) and the unknown $t$-value should divide the $t$ range $2.68$ to $2.18$ in the same ratio; equivalently, $0.025$ occurs $(0.025-0.02)/(0.05-0.02) = 1/6$th of the way along the $x$-range, so the unknown $t$-value should occur $1/6$th of the way along the $t$-range.
That is $\frac{t_{0.025}-2.68}{2.18-2.68} \approx \frac{0.025-0.02}{0.05-0.02}$ or equivalently
$t_{0.025} \approx 2.68 + (2.18-2.68) \frac{0.025-0.02}{0.05-0.02} = 2.68 - 0.5 \frac{1}{6} \approx 2.60$
The actual answer is $2.56$ ... which is not particularly close because the function we're approximating isn't very close to linear in that range (nearer $\alpha = 0.5$ it is). ### Better approximations via transformation
We can replace linear interpolation by other functional forms; in effect, we transform to a scale where linear interpolation works better. In this case, in the tail, many tabulated critical values are more nearly linear the $\log$ of the significance level. After we take $\log$s, we simply apply linear interpolation as before. Let's try that on the above example:
\begin{array}{ c c } α & \log(α)& t\\ 0.02 & -3.912 & 2.68\\ 0.025& -3.689 & t_{0.025}\\ 0.05 & -2.996 & 2.18\\ \end{array}
Now
\begin{eqnarray} \frac{t_{0.025}-2.68}{2.18-2.68} &\approx& \frac{\log(0.025)-\log(0.02)}{\log(0.05)-\log(0.02)} \\ &=& \frac{-3.689 - -3.912}{-2.996 - -3.912}\\ \end{eqnarray}
or equivalently
\begin{eqnarray} t_{0.025} &\approx& 2.68 + (2.18-2.68) \frac{-3.689 - -3.912}{-2.996 - -3.912}\\ &=& 2.68 - 0.5 \cdot 0.243 \approx 2.56 \end{eqnarray}
Which is correct to the quoted number of figures. This is because - when we transform the x-scale logarithmically - the relationship is almost linear: Indeed, visually the curve (grey) lies neatly on top of the straight line (blue).
In some cases, the logit of the significance level ($\text{logit}(\alpha)=\log(\frac{α}{1-α})=\log(\frac{1}{1-α}-1)$) may work well over a wider range but is usually not necessary (we usually only care about accurate critical values when $\alpha$ is small enough that $\log$ works quite well).
### Interpolation across different degrees of freedom
$t$, chi-square and $F$ tables also have degrees of freedom, where not every df ($\nu$-) value is tabulated. The critical values mostly$^\dagger$ aren't accurately represented by linear interpolation in the df. Indeed, often it's more nearly the case that the tabulated values are linear in the reciprocal of df, $1/\nu$.
(In old tables you'd often see a recommendation to work with $120/\nu$ - the constant on the numerator makes no difference, but was more convenient in pre-calculator days because 120 has a lot of factors, so $120/\nu$ is often an integer, making the calculation a bit simpler.)
Here's how inverse interpolation performs on 5% critical values of $F_{4,\nu}$ between $\nu = 60$ and $120$. That is, only the endpoints participate in the interpolation in $1/\nu$. For example, to compute the critical value for $\nu=80$, we take (and note that here $F$ represents the inverse of the cdf):
$$F_{4,80,.95} \approx F_{4,60,.95} + \frac{1/80 - 1/60}{1/120 - 1/60} \cdot (F_{4,120,.95}-F_{4,60,.95})$$ (Compare with diagram here)
$^\dagger$ Mostly but not always. Here's an example where linear interpolation in df is better, and an explanation of how to tell from the table that linear interpolation is going to be accurate.
Here's a piece of a chi-squared table
Probability less than the critical value
df 0.90 0.95 0.975 0.99 0.999
______ __________________________________________________
40 51.805 55.758 59.342 63.691 73.402
50 63.167 67.505 71.420 76.154 86.661
60 74.397 79.082 83.298 88.379 99.607
70 85.527 90.531 95.023 100.425 112.317
Imagine we wish to find the 5% critical value (95th percentiles) for 57 degrees of freedom.
Looking closely, we see that the 5% critical values in the table progress almost linearly here: (the green line joins the values for 50 and 60 df; you can see it touches the dots for 40 and 70)
So linear interpolation will do very well. But of course we don't have time to draw the graph; how to decide when to use linear interpolation and when to try something more complicated?
As well as the values either side of the one we seek, take the next nearest value (70 in this case). If the middle tabulated value (the one for df=60) is close to linear between the end values (50 and 70), then linear interpolation will be suitable. In this case the values are equispaced so it's especially easy: is $(x_{50,0.95}+x_{70,0.95})/2$ close to $x_{60,0.95}$?
We find that $(67.505+90.531)/2 = 79.018$, which when compared to the actual value for 60 df, 79.082, we can see is accurate to almost three full figures, which is usually pretty good for interpolation, so in this case, you'd stick with linear interpolation; with the finer step for the value we need we would now expect to have effectively 3 figure accuracy.
So we get: $\frac{x-67.505}{79.082-67.505} \approx {57-50}{60-50}$ or
$x\approx 67.505+(79.082-67.505)\cdot {57-50}{60-50}\approx 75.61$.
The actual value is 75.62375, so we indeed got 3 figures of accuracy and were only out by 1 in the fourth figure.
More accurate interpolation still may be had by using methods of finite differences (in particular, via divided differences), but this is probably overkill for most hypothesis testing problems.
If your degrees of freedom go past the ends of your table, this question discusses that problem.
• Great answer! For the section, "Interpolation across different degrees of freedom": What if we would need to interpolate both dimensions (as both degrees of freedom we are interested in are not tabulated)? Bilinear interpolation (en.wikipedia.org/wiki/Bilinear_interpolation)? I have seen euklidean distances (with inverse weights) from the point of interest to the "corners" of the box (4 "closest known values) around the point being applied in software. Any recommendation? Jan 1 '20 at 22:12
• Are you talking about interpolating in an F, for both numerator and denominator degrees of freedom? For that, certainly some form of bivariate interpolation is needed (but I would usually do bilinear interpolation in the inverse of both df). In many cases interpolating in one direction and then interpolating in the other should work well enough and can be done without learning any new skills. But it doesn't make sense to interpolate statistical tables in software, since you can use widely implemented mathematical functions (e.g. regularized incomplete beta) to do it essentially exactly. Jan 1 '20 at 22:56
• I was thinking more generally when interpolating numerator and denominator degrees of freedom for tabulated values. Seems like software implementations for a lot of statistical tests (still) rely on tables the tests' authors supplied in their (classical) paper as the statistic follows some "special" distribution (often derived from Monte Carlo simulations). These tables can be considered "small" or "not fine-grained" given today's computing abilities (e.g. one could use pre-computed larger/more fine-grained tables or an algorithem for direct (exact) computation, where known). Jan 1 '20 at 23:27
• For computer-based interpolation of smooth functions there are a number of more sophisticated approaches than the ones outlined here. For example, there's a long history of finite difference methods and various other approaches, but this is outside the scope of the question here, which limits itself to people interpolating statistical tables "by hand" (with nothing better than a table and a calculator, say). Jan 2 '20 at 0:37
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# Conditional Probability
Probability is essentially a measure of the possibility of something happening. And the likelihood ranges from 0 (impossible) to 1 (possible) (certain). In a sample space, the total of all probability for all events is 1. The likelihood of an event occurring if another event has already occurred is known as conditional probability. One of the most fundamental notions in probability theory is the concept of probability distributions.
Conditional probability is determined by increasing the probability of the former occasion by the refreshed probability of the succeeding, or contingent, occasion. Note that this probability doesn’t express that there is consistently a causal connection between the two occasions, just as it doesn’t show that the two occasions happen at the same time. This probability is denoted by the notation P(B|A), which denotes the likelihood of B given A. The Bayes’ theorem, which is one of the most prominent theories in statistics, is central to the idea of conditional probability.
P(A|B) could or might not be the same as P(A) (the unconditional probability of A). If P(A|B) = P(A), then occurrences A and B are said to be independent, meaning that knowing about one does not change the likelihood of the other. As such, conditional probability is the probability that an occasion has happened, considering some extra data about the results of an analysis.
The terms conditional probability and unconditional probability are often used interchangeably. Unconditional probability refers to the chance that an event will occur regardless of whether or not previous events have occurred or other circumstances exist. If the occurrences A and B are not independent, the probability of their interaction (the likelihood of both events occurring) is given by:
P(A and B) = P(A) P(B|A),
On the other hand, it can be written as;
P(A B) = P(A)P(B|A),
And, from this definition, the conditional probability P(B|A) can be defined as:
P(B|A) = P(A and B)|P(A)
Or, simply;
P(B|A) = P(A B)P(A), as long as P(A)> 0
While conditional probabilities can be incredibly beneficial, they are frequently used with minimal information. As a result, Bayes’ theorem may be used to reverse or convert a conditional probability. Furthermore, in certain circumstances, events A and B are independent occurrences, i.e., event A has no influence on the likelihood of event B; in these cases, the conditional probability of event B given event A, P(B|A), is basically the probability of event B, P(B). The Bayes’ theorem can be expressed mathematically as follows:
P(B|A) = P(B|A)P(A) / P(B)
Finally, a tree diagram may be used to find conditional probabilities. The probabilities in each branch of the tree diagram are conditional.
The Bayes theorem, commonly known as Bayes’ Rule or Bayes’ Law, is the cornerstone of Bayesian statistics. This collection of probability principles allows one to alter their forecasts of future occurrences based on new knowledge, resulting in more accurate and dynamic estimations. Mutually exclusive occurrences are events that cannot happen at the same time in probability theory.
All in all, on the off chance that one occasion has effectively happened, another can occasion can’t happen. Accordingly, the restrictive likelihood of fundamentally unrelated occasions is consistently zero. The law of absolute likelihood is just the utilization of the augmentation rule to quantify the probabilities in additional fascinating cases. Conditional probability, on the other hand, does not express the chance relationship between two occurrences, nor does it say that both events occur at the same time.
In domains as disparate as mathematics, insurance, and politics, conditional probability is applied. For example, a president’s re-election is determined by voter preferences, television advertising performance, and the likelihood that the opponent would make gaffes during debates.
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A football player punts a football at an inclination of 45 degrees to the horizontal at an initial velocity of
70 feet per second. The height of the football in feet is given by the quadratic function
h(x)= -32x^2/
(70)^2 +x+4 WRITTEN LIKE THIS BELOW TOO if u were confused
h(x)= (-32x^2/(70)^2) +x+4
where x is the horizontal distance of the football from the point at which it was kicked. What was the horizontal distance of the football from the point at which it was kicked when the height of the ball is at a maximum? What is the maximum height of the football?
MAXIMUM??_____ MINIMUM???____
Mar 10, 2020
#1
+1
Maximum will occur at x = -b/2a = 1/ (2 ( 32/70^2)) = 4900/64 = x = 76.563 feet
Substitute this value of 'x' in to the h(x) equation to find the height at this point......
Mar 10, 2020
#2
-1
GOT IT THANK U SO MUCH sorry
mharrigan920 Mar 11, 2020
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# Waves and Wave properties
## Presentation on theme: "Waves and Wave properties"— Presentation transcript:
Waves and Wave properties
Simulation
Wave Information Waves are a travelling form of energy
Waves can also carry information Sound Light Music electricity
Wave Types Transverse Longitudinal
Oscillates perpendicular to the direction of the motion. Examples are water waves and seismic S-waves Oscillate in the same direction/parallel to the direction of movement They move by expanding and contracting Examples are seismic P-waves and sound
Parts of a Wave: Waves have the following parts: Crest Trough
What are the 2 different types of waves?
How are these 2 types of waves different?
Wave Properties Waves have 3 specific, identifying properties:
Frequency Wavelength λ Amplitude
Frequency How many times the wave oscillates up and down
Measured in Hz (hertz) 1 Hz is equal to 1 cycle per second Water frequency is around Hz Sound has a frequency range of 20-20,000 Hz
Wavelength Series of repeating highs and lows
Measured from crest to crest, or trough to trough
This is the amount/distance that the wave moves away from equilibrium
Amplitude: This is the amount/distance that the wave moves away from equilibrium
If the frequency of a wave increases, what do you think will happen to the wavelength?
What type of a relationship is this? Positive Negative
Frequency and Wavelength have an inverse or negative relationship
Relationships: Frequency and Wavelength have an inverse or negative relationship As the frequency of a wave increases, its wavelength decreases OR As the frequency of a wave decreases, its wavelength increases
Calculating the Speed of Waves
The speed of a wave is equal to its frequency multiplied by its wavelength. V = ƒ ʎ V=velocity/speed ʎ = wavelength and ƒ= frequency This tells you the speed of the oscillations through the material/medium.
Wave Interference: When waves encounter an obstacle or interact with matter, they experience interference. There are 4 types of wave interference Reflection Refraction Diffraction Absorption
Reflection
Refraction: When a wave moves through a medium that slows or bends the wave’s movement it is refracted Refraction of light causes objects to appear bent
Diffraction: When a wave bends around objects such as sound bending around corners
Absorption: When a wave encounters a medium and seems to disappear.
What is the difference between refraction and reflection?
What is an example of a wave that is reflected? What is an example of a wave being refracted?
Interference Continued
Constructive Interference Destructive Interference When 2 waves that are “in phase” with one another collide Their collision causes an increase in the waves effect When 2 waves are “out of phase with one another collide Their collision causes a decrease or temporary disappearance of the wave
Interference Continued
HARMONIC MOTION Motion that repeats in cycles
A cycle is a unit of motion that repeats over and over Harmonic Motion is different from Linear Motion
Harmonic Motion Graphs
Pendulum Simulation
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# do my online math homework
please do my online math homeork lesson 2 and oritentation
also do writting assignment below
plagarism free with report
MAT102 LESSON 2 Date: Sep 6
Topic: Historical Numeration Systems
In this lesson you will learn about Egyptian numerals, traditional Chinese numerals and Hindu Arabic numerals.
Read section 4.1 “Historical Numeration systems” of your textbook. You can read the section on MyMathLab by clicking on Lesson 2 when you access your homework on MyMathLab. Pay special attention to Examples 1, 2, 4, 5 and 6. Skip Chinese Numeration. Practice your knowledge by doing HW #2 on MyMathLab. Remember that homework is 40% of your grade.
As evaluation of your mastering of Egyptian multiplication you will submit an assignment in BlackBoard (by clicking Assignments). Use the Egyptian doubling method to multiply 44×21. In order to write the solution of this exercise you can follow the example below where the Egyptian algorithm is performed for the product 25×35:
“We write two columns. One starts with 1 and the other one with 35. Each row is obtained by doubling the previous one. We stop once in the column starting with 1 we get a number greater than 25:
1 35
2 70
4 140
8 280
16 560
32 1120
Now we write 25 as a sum of numbers from the first column: 25=16+8+1. 16, 8 and 1 are on the fifth, fourth and first rows of the first column so 25×35 is equal to the sum of the numbers 560, 280 and 35 which are on fifth, fourth and first rows of the second column.
we get that 25×35 = 560 + 280 + 35 = 875. “
Do you need a similar assignment done for you from scratch? We have qualified writers to help you. We assure you an A+ quality paper that is free from plagiarism. Order now for an Amazing Discount! Use Discount Code “Newclient” for a 15% Discount!NB: We do not resell papers. Upon ordering, we do an original paper exclusively for you.
The post do my online math homework appeared first on The Nursing Hub.
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Check the completeness of given binary tree | Set 2 – Using Level Order Traversal
Objective: Given a binary tree, write an algorithm to determine whether the tree is complete or not using Level order traversal.
Earlier we had solved this problem by counting the number of nodes in the tree. (Read – Check the completeness of given binary tree | Set 1 – Using Node Count. In this article, we will solve it using level order traversal.
Before we proceed, If you are reading about the completeness of binary tree for the first time then let’s first understand it with some examples.
Complete Binary Tree: A binary tree is complete if all levels are completely full except the last level. The last level may or may not be full and nodes should be on the left side.
Example:
Below is an example of a special case where the tree is full expect the last level and in the last level, the nodes have priority to be one the left side if the last level is not full. See the example below
Approach: Level Order Traversal
• Do the level order traversal from left to right.
• Once a null node is found, ideally there should not be any node present after that if the tree is complete. If that’s the case return true.
• So if a node is found which is not null after a null node that means the tree is not complete, return false.
• See the image below for more understanding.
Code:
Output:
```Given Binary Tree is a Binary Tree: false
Given Binary Tree is a Binary Tree: true
```
__________________________________________________
Top Companies Interview Questions..-
If you find anything incorrect or you feel that there is any better approach to solve the above problem, please write comment.
__________________________________________________
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