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Anonymous
Anonymous asked in Science & MathematicsMathematics · 5 years ago
# Area of a sector of a circle?
The area of a sector of a circle with a central angle of 60° is 14 yd2. Find the radius of the circle. Do not round any intermediate computations. Round your answer to the nearest tenth.
Thank you for your help! :)
Relevance
• Anonymous
5 years ago
A complete circle has three hundred sixty degrees, so a sixty degree angle sector has area that is one-sixth the area of a full circle.
Area of circle: A = Pi*r^2
Area of sector: S = A/6 = (Pi/6)r^2
solve the last equation for r:
r = sqrt(6S/Pi)
put the value of 14 in for S and do the calculation on your calculator app. answer will be in units of yards.
• Jay Be
Lv 5
5 years ago
You simply multiply the area of the circle inversely times the portion of the circle it is.
360/60 = 6
6 * 14 = 84
• Thomas
Lv 7
5 years ago
60/360=14/pr^2=1/6
pr^2=84
r^2=84/p
r=(84/p)^(1/2)
r~5.2yd
• alex
Lv 7
5 years ago
(pi/6)r^2=14 --->r = ...
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# The "One Recipe" Theory - Question
### Help Support Soapmaking Forum:
#### linne1gi
Supporting Member
I will give it a try. Read so much of Deanna's stuff on Classic Bells about the chelators so would have to decide whether to go with citric acid and deal with adjusting the lye or go the easier route and just use sodium citrate. Seems cheap enough to just buy sodium citrate and have one less calculation to deal with.
The math is simple (even for me) Say you have 32 ounces of oils (907 grams, I use grams for ease). Multiply that 907 by .02 (2%) The recommendation is 1-2%, so I go with the 2%. So 907X.02=18.14. We know that 10 grams of citric neutralizes 6 grams of NaOH (It's different for KOH). So we have 18.14 grams of citric, now multiply that by 6 and divide by 10. So 18.14X6=104.84 and divide by 10=10.884 (I round that up to 11). So 18 grams of citric and 11 grams of NaOH. Now that's not much NaOH, so you could just allow your superfat to increase by a little. But the basic formula is always OilsX.02X6/10. It's not too difficult to understand.
#### SPowers
##### Well-Known Member
I use ciitric acid but have never adjusted the lye amount. Are we talking about a big amount? I'm guessing that since I haven't done an adjustment it basically increased the superfat - is this correct? And if so but how many percentage point would the sf increase. Thanks.
#### linne1gi
Supporting Member
I use ciitric acid but have never adjusted the lye amount. Are we talking about a big amount? I'm guessing that since I haven't done an adjustment it basically increased the superfat - is this correct? And if so but how many percentage point would the sf increase. Thanks.
I don’t know percentage points, but I add about 8 grams extra NaOH, that’s not much.
Thanks.
#### KiwiMoose
Supporting Member
The math is simple (even for me) Say you have 32 ounces of oils (907 grams, I use grams for ease). Multiply that 907 by .02 (2%) The recommendation is 1-2%, so I go with the 2%. So 907X.02=18.14. We know that 10 grams of citric neutralizes 6 grams of NaOH (It's different for KOH). So we have 18.14 grams of citric, now multiply that by 6 and divide by 10. So 18.14X6=104.84 and divide by 10=10.884 (I round that up to 11). So 18 grams of citric and 11 grams of NaOH. Now that's not much NaOH, so you could just allow your superfat to increase by a little. But the basic formula is always OilsX.02X6/10. It's not too difficult to understand.
Over My Head GIF - OverMyHead No Nope - Discover & Share GIFs
#### linne1gi
Supporting Member
#### linne1gi
Supporting Member
I know, but when I see it in writing like that I just glaze over
Silly girl.
#### Zany_in_CO
##### Saponifier
Moi aussi! Me too! My mind glazed over at the second sentence. LOL
This reminds me of when I was driving carpool. Brian wanted to quit French because he just didn't get it. My son. Peter, said, "Don't quit! It's easy!" I had to take him aside later and explain. Like anything else, some have a head for languages, some don't. Some have a head for math, some don't.
#### linne1gi
Supporting Member
Moi aussi! Me too! My mind glazed over at the second sentence. LOL
This reminds me of when I was driving carpool. Brian wanted to quit French because he just didn't get it. My son. Peter, said, "Don't quit! It's easy!" I had to take him aside later and explain. Like anything else, some have a head for languages, some don't. Some have a head for math, some don't.
I’m no math whiz, I barely got through the courses I had to take in college and for my masters degree, but honestly this isn’t that difficult. You have to want to learn it though.
#### cmzaha
Supporting Member
It actually took me several years to come up with my actual vegan and non-vegan base recipes only changing my liquid oils. While I have sold for over 10 yrs and everyone was happy the last 5 yrs the comments were that my soaps were even better than before. I soap with vinegar, Sodium Gluconate, Edta, and Sorbitol with a 2% superfat on average with dual lye. When I started using chelators I started with Citric Acid but did not like the crystal coating that would form after a few months on the bars.
#### MaryinOK
##### Member
I have been making soap for many years. I keep to the same 2 recipes for the soaps I sell. A milk soap with OO, Rice Bran, CO and a couple of fancy things and a very slightly tweaked Castille (thanks, Zany!!) That doesn't mean I don't experiment, but when I do it is tiny batches, in tiny increments, changing one thing at a time. If I prefer the "new and improved", then that is phased in as I sell out of the old. I have tried a few recipes that are very different to my usual ones, but none have made me decide to change what I am already doing. Of course my current recipes look totally different from those I started with, that is why soapmaking is as much an art as a science.
#### linne1gi
Supporting Member
It actually took me several years to come up with my actual vegan and non-vegan base recipes only changing my liquid oils. While I have sold for over 10 yrs and everyone was happy the last 5 yrs the comments were that my soaps were even better than before. I soap with vinegar, Sodium Gluconate, Edta, and Sorbitol with a 2% superfat on average with dual lye. When I started using chelators I started with Citric Acid but did not like the crystal coating that would form after a few months on the bars.
Hm, I’ve been using citric acid for just over 2 years and I’ve never noticed a crystal coating from the citric. Are you perhaps talking about soda ash?
#### Arimara
Supporting Member
Moi aussi! Me too! My mind glazed over at the second sentence. LOL
This reminds me of when I was driving carpool. Brian wanted to quit French because he just didn't get it. My son. Peter, said, "Don't quit! It's easy!" I had to take him aside later and explain. Like anything else, some have a head for languages, some don't. Some have a head for math, some don't.
I love when people say French is easy. I never had a head for French but I can follow some Spanishes (I have my reasons for saying that), Italian (minimally) and I can recognize a bit of Korean (I have to if I want to get certain good products).
Hm, I’ve been using citric acid for just over 2 years and I’ve never noticed a crystal coating from the citric. Are you perhaps talking about soda ash?
There could be something about her environment that produces that. She is in West Coast, after all.
#### Steve85569
Supporting Member
I use 2% citric to the weight of oils. .02 x weight of oils = weight of citric acid to be added.
To compensate to neutral the multiplier that I use is 0.624.
Weight of citric acid x 0.624 is the weight of sodium hydroxide to be added.
Example:
Batch is 794 grams of oil.
794 x .02 = 15.9 grams of citric.
15.9 x 0.624 = 9.9 grams of sodium hydroxide.
The multiplier for potassium hydroxide is 0.842 because the mol weight is different. DeeAnna can explain that. She has to me ( thank you!).
And yes I have a base recipe that I use for most of my soap. I will change it a bit from time. See the post on soy wax users.
#### Zany_in_CO
##### Saponifier
a very slightly tweaked Castille (thanks, Zany!!)
You're so very welcome!
#### Zany_in_CO
##### Saponifier
I’m no math whiz, I barely got through the courses I had to take in college and for my masters degree,
I don't have a masters degree but I do have a work-around for adding citric acid -- I soap at 0% SF. Think about it.
#### cmzaha
Supporting Member
Hm, I’ve been using citric acid for just over 2 years and I’ve never noticed a crystal coating from the citric. Are you perhaps talking about soda ash?
I do know the difference with soda ash, it was a coating from the citric. With my large selection due to selling many times, my soaps would stay in stock for 6-12 months or more. Citric acid would form a layer of crystals on the soap. If you refer to DeeAnna Soapy Stuff she also refers to it, in fact, she and I had a couple of discussions about it.
ETA Hm, 2% also did not cut the scum to my liking...
#### cmzaha
Supporting Member
I use ciitric acid but have never adjusted the lye amount. Are we talking about a big amount? I'm guessing that since I haven't done an adjustment it basically increased the superfat - is this correct? And if so but how many percentage point would the sf increase. Thanks.
You multiply the amount of citric acid you are using by 0.624 to get your extra lye usage. When I did use citric acid I used 12g citric acid and 7g extra NaOH 12 x 0.624 = 7.488 I rounded to 7 g
#### Gaisy59
Supporting Member
Just for my two cents worth...i hot process and i have used patchouli, tea tree, lavender eo’s and i add them after the cook before the mold and the scents stay and even last on my skin for a bit.
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# how to multiply every single element in a matrix to entire of another matrix?
3 views (last 30 days)
majid on 7 Apr 2024
Edited: majid on 7 Apr 2024
Hello.
I want to multiply every single element in a matrix to entire of another matrix.
x = [100,500,900,1300,1700;
120,600,1080,1560,2040;
140,700,1260,1820,2380;
160,800,1440,2080,2720;
180,900,1620,2340,3060];
z = rand (4,20);
I want to multiply 100 to entire z and then 500 and so on... but when each multipication occured saved it in new indexed parameter.
how can I do this?
Paul on 7 Apr 2024
Here's one option to store each result in a cell array that's the same size as x
x = [100,500,900,1300,1700;
120,600,1080,1560,2040;
140,700,1260,1820,2380;
160,800,1440,2080,2720;
180,900,1620,2340,3060];
%z = rand (4,20);
z = [1 2;3 4]; % small z for example
S = cellfun(@(x) x*z,mat2cell(x,ones(1,size(x,1)),ones(1,size(x,2))),'Uni',false)
S = 5x5 cell array
{2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double}
% check
S{1,1} - x(1,1)*z
ans = 2x2
0 0 0 0
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
S{3,3} - x(3,3)*z
ans = 2x2
0 0 0 0
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
Bruno Luong on 7 Apr 2024
Edited: Bruno Luong on 7 Apr 2024
Shorter
x = [100,500,900,1300,1700;
120,600,1080,1560,2040;
140,700,1260,1820,2380;
160,800,1440,2080,2720;
180,900,1620,2340,3060];
%z = rand (4,20);
z = [1 2;3 4]; % small z for example
S = arrayfun(@(xij) xij*z,x,'Uni',false)
S = 5x5 cell array
{2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double}
majid on 7 Apr 2024
thank you for guiding me.
Steven Lord on 7 Apr 2024
If you're asking can you dynamically create variables with numbered names like x1, x2, x3, etc.? Yes.
Should you do this? The general consensus is no. That Discussions post explains why this is generally discouraged and offers several alternative approaches.
One such approach is to use pages of a 3-dimensional array.
x = [100,500,900,1300,1700;
120,600,1080,1560,2040;
140,700,1260,1820,2380;
160,800,1440,2080,2720;
180,900,1620,2340,3060];
z = rand (4,20);
zv = reshape(z, 1, 1, []);
result = x.*zv;
To check:
isequal(result(:, :, 42), x.*z(42))
ans = logical
1
majid on 7 Apr 2024
Edited: majid on 7 Apr 2024
Bruno Luong on 7 Apr 2024
Edited: Bruno Luong on 7 Apr 2024
x = [100,500,900,1300,1700;
120,600,1080,1560,2040;
140,700,1260,1820,2380;
160,800,1440,2080,2720;
180,900,1620,2340,3060];
%z = rand (4,20);
z = [1 2;3 4]; % small z for example
[mx,nx] = size(x);
[mz,nz] = size(z);
C = mat2cell(kron(x,z),repelem(mz,mx),repelem(nz,nx))
C = 5x5 cell array
{2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double} {2x2 double}
majid on 7 Apr 2024
Edited: majid on 7 Apr 2024
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# WARM UP! J AN. 25, 2013 Q UESTION 1 A card is drawn from a deck of 52 cards. Find the probability of drawing a picture card, (includes jacks, queens,
## Presentation on theme: "WARM UP! J AN. 25, 2013 Q UESTION 1 A card is drawn from a deck of 52 cards. Find the probability of drawing a picture card, (includes jacks, queens,"— Presentation transcript:
WARM UP! J AN. 25, 2013
Q UESTION 1 A card is drawn from a deck of 52 cards. Find the probability of drawing a picture card, (includes jacks, queens, and kings.
Q UESTION 2 Rotation Rules: State the rule for each, a. 180 degree rotation b. 90 degree clockwise rotation. c. 90 degree counter clockwise rotation.
Q UESTION 3 A right triangle has legs with length of 5 inches and 12 inches. What is the area of a square drawn on the hypotenuse?
Q UESTION 4 Dilations: In Math the word dilate means to _____________ or _________________ a figure. If a scale factor is less than 1 then your figure gets ____________. If a scale factor is greater than 1 then your figure gets ______________.
ANSWERS 1. 3/13 2. a. change signs of both the x & y coordinate. b. Switch the coordinates of each point and then change the new 2 nd coordinate to the opposite sign. c. Switch the coordinates of each point and then change the new 1 st coordinate to the opposite sign. 3. 169 sq inches 4. Enlarged or reduced Less than 1 your figure gets reduced. Greater then 1 your figure gets enlarged
C RITICAL T HINKING 1 Subtracting UP! Can you figure out a way to take out 1 from 14 so that you end up with 15?
ANSWER Use Roman Numerals: If you take away I from XIV you are left with XV or 15.
C RITICAL T HINKING 2 A Clown-Undrum: Baron Chillarado the Clown earned 3 gold bricks after working an 8 th grade b-day party. While walking home he came to an old foot bridge. The sign said, “Weight Capacity 150 pounds.” Chillarado weighs 148 pounds & each brick weighs 1 pound. How did he get himself and all three bricks across safely?
ANSWER He juggled the gold bricks as he walked. That way at least one was in the air at all times so only two bricks (2pds) and the Baron (148pds) had weight on the bridge at any given time.
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## Tuesday, September 20, 2011
### Vector Analysis and Calculus - General math
This post highlights vector calculus and vector analysis. Such topics and discussions will cover ideas and concepts encountered when dealing with vector analysis. Vectors are a very important concept in science and math and appear very early in calculus and physics classes in college or in advanced high school classes. Vectors are used to describe a direction and magnitude of important concepts in our world such as gravity and other forces. Vectors can also group together many sets of equations into a compact form such as the three-dimensional equations of motion expressed as a single line of a vector equation instead of expanding for each of the three components. Vectors appear very often in fluid dynamics and are a fundamental basis for the subject. Many fluid dynamic books introduce the reader into and with vector analysis [1-5] or contain an appendix at the end of the book [5] while the manuscript by Aris [6] is based upon vectors and fluid dynamics altogether.
As mentioned previously, vectors make up the fundamental basis of mathematics and physics in the world. Many mathematical books which generalize math topics which scientist, engineers, physicists, etc, should be familiar with often place vectorial analysis in the beginning or at least include it in the book type encyclopedia (a collection of topics and methods) of math for applied uses [7].
Vectors not only make up fundamental basis for fluid dynamics, but also for math and physics. In fact it is an underlying foundation in math and physics sciences. Einstein used vector notation to develop and describe his relativity theories. Many in the early stages of the study of math and physics fields generally see vectors in essential courses such as calculus and general physics. In calculus, vector analysis might appear in a II or III semester of the course as the concepts of divergence, curl, gradient, etc. are brought up and discussed. Vectors appear very early at the very beginning in general college physics courses because almost everything is based upon the vector theory. Forces, acceleration, velocity, position, momentum, electrostatics, fluid mechanics and dynamics, etc. all utilize vector analysis.
Specifically, since my background is in fluid dynamics from a graduate aerospace engineering education, I will use my experiences to discuss vectors and other topics. Vectors in fluid dynamics/mechanics can be very useful to describe many fluid concepts and phenomena. Additionally, vectors are very useful as they are coordinate independent or in other words the user does not have to worry about a coordinates system.
A general representation of a vector can be
$\vec{A} = A_1 \vec{e}_1 + A_2\vec{e}_2 + A_3 \vec{e}_3 + \ldots + A_n \vec{e}_n = \sum_{i = 1}^n$
where the coordinate system can be anything and the number of dimensions $$n$$. This simple relation is why vectors are so useful, powerful, and why they make up the foundations of our math, physics, and related sciences.
The vector may also be written without the arrow above where boldface font is available.
$\mathbf{A} = A_1 \mathbf{e}_1 + A_2 \mathbf{e}_2 + A_3 \mathbf{e}_3 + \ldots + A_n \mathbf{e}_n$
Scalars according to Serre [ ],
"...are elements of some field $$k$$ (or $$K$$), or sometimes of a ring $$R$$."
In progress...
References:
[1] K. Karamcheti. Principles of Ideal-Fluid Aerodynamics. John Wiley & Sons, Inc., New York, NY. 1966
[2] R. L. Panton. Incompressible Flow. (3rd ed.). John Wiley & Sons, Inc. Hoboken, NJ. 2005
[3] M. E. O’Neill & F. Chorlton. Viscous and Compressible Fluid Dynamics. Ellis Horwood Limited. Chichester, UK. 1989
[4] T. C. Papanastasiou, G. C. Georgiou, & A. N. Alexandrou. Viscous Fluid Flow. CRC Press. Boca Raton, FL. 2000
[5] M. T. Schobeiri. Fluid Mechanics for Engineers - A Graduate Textbook. Springer.
Berlin, Germany. 2010
[6] R. Aris. Vectors, Tensors and the Basic Equations of Fluid Mechanics. Dover Publications.
New York, NY. 1990
[7] G. B. Arfken & H.-J. Weber, Mathematical Methods for Physicists (6th ed.). Elsevier Academic Press. Burlington, MA. 2005
[ ] D. Serre. Matrices: Theory and Applications. Springer-Verlag New York, Inc., New York, NY. 2002
[ ] D. Serre. Matrices: Theory and Applications. 2nd ed. Springer Science+Business Media, LLC., New York, NY. 2010
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# An algorithm for incrementally building separation graphs
You know that annoying thing when you’re reading a paper and it’s fairly clear they’ve not actually tried that algorithm in practice? Don’t you hate that?
Anyway, here’s an algorithm I’ve not tried in practice. I haven’t even written code for it yet. This post is as much to get the ideas clear in my head as anything else. This algorithm may turn out to be worse than doing it the stupid way in practice.
This is the problem we’re trying to solve:
x
We have n points, $$X = \{x_1, \ldots, x_n\}$$, and a metric function d. We are trying to build the unordered graph with edges $$E = \{ (i, j) : d(x_i, x_j) > t \}$$.
This is motivated by trying to find small diameter partitions of X, which we will be doing by trying to either prove (X, E) is not bipartite or finding a bipartition of it (see this previous post)
The question is how to do this efficiently? The obvious brute force approach is $$O(n^2)$$, and indeed any solution must have $$O(n^2)$$ worst-case. We’d like to do well in practice by applying the following two key ideas:
1. By using the triangle inequality, we may often be able to escape having to do distance calculations because we can infer the distance will be too large / too small
2. We are likely not to need the entire graph – a proof that a graph is not bipartite may involve only a very small subset of the nodes, so if we build the graph incrementally we may be able to stop early
The way we do this will be as follows:
For each node x, we keep track of two sets, Definite(x) and Candidates(x). We will also index distances as we find them by tracking Distances(x), which will be track all points we’ve already calculated the distance from x to.
### Roles and implementation requirements
Definite(x) is a set containing all nodes we know are > t away from x and starts empty. We require efficient addition of new elements to it without introducing duplicates and the creation of an iterator which is guaranteed to iterate over all elements in the set even if new ones are added during iteration, even after it has previously claimed there are no elements left. One way to do this would be to have it consist of both an array and a hash set.
Candidates(x) contains all points which we don’t yet know for sure whether or not they are > t from x and starts equal to X. As a result we want a set implementation which is cheap to allocate an instance containing the set and cheap to delete from. This is where the previous post about integer sets comes in (we represent a node by its index).
Distances(x) is some sort of ordered map from distances to lists of points. It needs to support efficient range queries (i.e. give me everything nearer than s or further than s).
### The Algorithm
Our two basic operations for manipulating the data we’re tracking are Close(x, y) and Far(x, y). Close(x, y) removes x from Candidates(y) and y from Candidates(x). Far does the same thing but also adds x to Definite(y) and y to Definite(x).
Our iterator protocol is that an iterator has an operation Next which returns either None or an Element. We will construct Neighbours(x), which creates an iterator that incrementally returns all points > t away from the x and uses information discovered whilst doing so to flesh out the information we know about other nodes too.
Here is the algorithm, written in some bizarre pseudocode language that looks like nothing I actually write (no, I don’t know why I’m using it either):
Neighbours(X) = NI(x, Iterator(Definite(x))) Next(NI(x, found)) # If we while ((result = Next(found)) == None && not IsEmpty(Candidates)) do candidate = Pop(Candidates(x)) s = d(x, candidate) if s > t then Far(x, candidate) else Near(x, candidate) # Because d(y, candidate) <= d(y, x) + d(candidate, x) <= t - s + s = t for(y in Distances(x) where d(y, x) <= t - s) Near(candidate, y) # Because if d(y, candidate) <= t then d(y, x) <= d(candidate, x) + t <= s + t for(y in Distances(x) where d(y, x) > t + s) Far(candidate, y) # Because if d(candidate, y) <= t then d(candidate, x) <= d(candidate, x) + d(y, x) <= t - s + s = t if s > t then for(y in Distances(x) where d(y, x) <= t - s) Far(candidate, y) # If we have no more candidates left then Distances(x) will never be used again and is wasted space if IsEmpty(Candidates(x)) then delete Distances(x) else Put(Distances(x), s, y)) done return result
The idea being that as we're searching for edges for a given node we're also using this to fill out the edges for the other nodes. This should work particularly well for doing depth first search, because in particular it often means that the nodes we're going to transition to will have some of their neighbours filled in already, and we may in fact be able to spend most of our time just chasing through Definite sets rather than having to do new distance calculations. Even where we haven't been able to do that, hopefully we've managed to prune the lists of elements to consider significantly before then.
### Notes
• It should be fairly evident that this is not only $$O(n^2)$$ worst case, it's probably $$O(n^2)$$ expected case for exploring the whole graph, even in reasonable metric spaces - it's likely that one or the other of the two sets of points we're looking at during each iteration step will be O(n) - so the only possible win over the brute force algorithm is if it does significantly fewer expensive distance calculations
• I've not yet written any real code for this, so I'm sure some of the details are wrong or awkward
• Even if it's not wrong or awkward it may well turn out to have sufficiently bad constant factors, or save few enough distance invocations, that it's worthless in practice
This entry was posted in Code, programming on by .
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# Manhattan GMAT Challenge Problem of the Week – 15 January 2013
by on January 15th, 2013
Here is a new Challenge Problem! If you want to win prizes, try entering our Challenge Problem Showdown. The more people that enter our challenge, the better the prizes!
## Question
If = 25, = 64, and = 216, and xy > 0, which of the following is true?
A. x > y > z
B. y > x > z
C. y > z > x
D. z > y > x
E. z > x > y
First, rewrite the right sides of the given equations as powers of small integers:
= 25 =
= 64 =
= 216 = (you might not have recognized this one, but you could get there with prime factorization)
Take roots as necessary to isolate each variable. You know that x and z must be positive, because their odd powers are positive; moreover, since xy > 0, y must also be positive.
Third root:
x =
y = =
z =
Now compare the results in pairs, putting “??” between to indicate the unknown inequality. First, compare x and y:
??
Raise both sides to the 6th power, to remove the fractional powers:
??
??
625 ?? 512
Since 625 is bigger than 512, x is bigger than y. You can now eliminate B, C, and D.
The remaining two answers differ as to whether z is largest or smallest. Since y seems simpler than x (both in the base and in the exponent), compare y and z:
??
Raise both sides to the 10th power, to remove the fractional powers:
??
??
?? ×
??
512 ?? = = 81 × 9 = 729
So z is larger than y, eliminating A. The answer must therefore be E.
Incidentally, it’s very difficult to prove by hand that z is bigger than x. To do so, you need to show that is bigger than , very much a non-trivial task! Fortunately, you don’t need to determine whether z or x is larger in order to get the right answer.
The correct answer is E.
Special Announcement: If you want to win prizes for answering our Challenge Problems, try entering our Challenge Problem Showdown. Each week, we draw a winner from all the correct answers. The winner receives a number of our our Strategy Guides. The more people enter, the better the prize. Provided the winner gives consent, we will post his or her name on our Facebook page.
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https://www.brilliantmaths.com/lessons/lesson-8-rounding-decimals-3/
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# Lesson 8- Rounding Decimals
Objective
At the end of this lesson, students should be able round decimal numbers to the nearest whole number, tenth, hundredth or thousandth.
Rounding a decimal means representing it by a number to which the decimal is nearest to.
Rounding a decimal numbers allows us to write the number in a way to suit our purposes. Decimals can be rounded to the nearest whole number, tenth, hundredth, thousandth, etc.
If we are to round to the nearest whole number:
a) 3.54 = 4 b) 3.49 = 3 c) 3.504 = 4 d) 3.497 = 3
A similar idea is what we use to round figures. Below are some examples:
Example 1
Round these decimal numbers to the nearest whole number.
a) 2.63 b) 13.429 c) 238.532
Example 2
Round to the nearest tenth (1 decimal place):
a) 4.829 b) 13.499 c) 6.952
Example 3
Round to the nearest hundredth (2 decimal places)
a) 4.826 b) 267.199 c) 10.841
Example 4
Round to the nearest thousandth (3d.p)
a) 2.8469 b) 18.2195 c) 9.1111
Solution
a) 2.8469 = 2.847
b) 18.2195 = 18.220
c) 9.1111 = 9.111
Lesson Content
error: Content is protected !!
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https://www.kofastudy.com/courses/ss1-maths-3rd-term/lessons/trigonometry-i-week-5-6/topic/pythagoras-theorem-3/
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Lesson 4, Topic 2
In Progress
# Pythagoras Theorem
Lesson Progress
0% Complete
From the geometrical point of view, Pythagoras theorem gives the relation between areas of squares on the sides of a right-angled triangle. Also, from an algebraic point of view, the formula below can be used to solve right-angled triangles.
⇒ $$\scriptsize a^2 = b^2 \: + \: c^2$$
which is $$\scriptsize Hyp^2 = Opp^2 \: + \: Adj^2$$
### Pythagorean Triple:
In ΔABC above, the sides are 3cm, 4cm and 5cm. This is an example of a Pythagorean triple, i.e. a set of 3 whole numbers which can be taken as the lengths of the sides
$$\scriptsize 5^2 = 4^2 \: + \: 3^2$$
$$\scriptsize 25 = 16 \: + \: 9$$
$$\scriptsize 25 = 25$$
When a triangle’s sides are a Pythagorean Triple it is a right-angled triangle.
Another example is 5, 12, 13
$$\scriptsize 13^2 = 5^2 \: + \: 12^2$$
$$\scriptsize 169 = 25 \: + \: 144$$
$$\scriptsize 169 = 169$$
Here is a list of pythagorean triples
### Example 1:
If the diagonal of a square is 8cm, what is the area of the square? (WAEC)
Solution:
Recall, that the diagonals of a square bisect at right angles.
Thus, $$\scriptsize \overline{EB} = \scriptsize \overline{EA} = 4cm$$
We can then find x using ΔAEB (Figure 3)
Using Pythagoras theorem
$$\scriptsize x^2 = \overline{EB}^2 \: + \: \overline{EA}^2$$
⇒ $$\scriptsize x^2 = 4^2 \: + \: 4^2$$
$$\scriptsize x^2 = 16 \: + \: 16$$
$$\scriptsize x^2 = 32$$
Take the square root of both sides
$$\scriptsize \sqrt{x^2} = \sqrt{32}$$
$$\scriptsize x = \sqrt{32} \: cm$$
Using Figure 1
Area of square = $$\scriptsize x \: \times \: x$$
Area of square = $$\scriptsize \sqrt{32} \: cm \: \times \: \sqrt{32} \: cm$$
Area of square = $$\scriptsize 32 cm^2$$
OR (2nd Method)
Using Figure 1, Consider ΔABC
Using Pythagoras Theorem
⇒ $$\scriptsize AC^2 = BC^2 \: + \: AB^2$$
$$\scriptsize 8^2 = x^2 \: + \: x^2$$
$$\scriptsize 64 = 2x^2$$
$$\scriptsize 2x^2 = 64$$
$$\scriptsize x^2 = \frac{64}{2}$$
$$\scriptsize x^2 = 32 cm^2$$
Same as above
### Example 2
In the diagram below, O is the centre of circle HKL, |HK| = 16cm, |HL| = 10cm, and the perpendicular from O to |HK| is 4cm. What is the length of the perpendicular from O to |HL|? (WAEC)
Solution
Redraw the diagram using the information provided
r is the radius and is the distance from the centre O to H
the perpendicular from O to |HK| = |OP| = 4cm
If we find r we can find |OM| which is the length of the perpendicular from O to |HL|
Consider ΔHPO
Using Pythagoras theorem
$$\scriptsize r^2 = opp^2 \: + \: adj^2$$
$$\scriptsize r^2 = 8^2 \: + \: 4^2$$
$$\scriptsize r = \sqrt{64 \: + \: 16}$$
$$\scriptsize r = \sqrt{64 \: + \: 16}$$
$$\scriptsize r = \sqrt{80}$$
$$\scriptsize r = 4\sqrt{5}\: cm$$
Consider ΔHMO
Using Pythagoras theorem
$$\scriptsize r^2 = HM^2 \: + \: OM^2$$
$$\scriptsize r^2 = 5^2 \: + \: OM^2$$
$$\scriptsize OM^2 = r^2 \: – \: 5^2$$
$$\scriptsize r = \sqrt{80}$$
Therefore $$\scriptsize OM^2 = \sqrt{80}^2 \: – \: 5^2$$
$$\scriptsize OM^2 = 8.9443^2 \: – \: 5^2$$
$$\scriptsize OM^2 = 80 \: – \: 25$$
Take the square root of both sides
$$\scriptsize \sqrt{OM^2} = \sqrt{80 \: – \: 25}$$
$$\scriptsize OM = \sqrt{80 \: – \: 25}$$
$$\scriptsize OM = \sqrt{55}$$
$$\scriptsize OM = 7.416$$
$$\scriptsize OM \approx 7.4 \: cm$$
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https://causeweb.org/cause/statistical-topic/conditional
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# Conditional
• ### Conditional Logistic Regression
This presentation is a part of a series of lessons on the Analysis of Categorical Data. This lecture covers the following: unconditional likelihood, elimination of nuisance parameters, and Mantel-Haenzsel estimate.
• ### Dataset Example: Student's T
This resource explains the t-distribution and hypothesis testing (informally) using an example on laptop quality.
• ### Concepts of Confidence Intervals
This PowerPoint presentation dicusses general concepts of confidence intervals and interprets confidence intervals for a mean, difference in two means, and the relative risk. The original presenation is available for download.
• ### Confidence Interval for a Mean
This PowerPoint lecture presenation explains confidence intervals for a mean when using a small sample. It discusses the t-distribution, compares the t-statistic to the z-statistic, and provides an example of a small sample confidence interval. The original presentation is available for download.
• ### Star Library: An Unusual Episode
This article describes an activity that illustrates contingency table (two-way table) analysis. Students use contingency tables to analyze the "unusual episode" (the sinking of the ocean liner Titanic)data (from Dawson 1995) and attempt to use their analysis to deduce the origin of the data. The activity is appropriate for use in an introductory college statistics course or in a high school AP statistics course. Key words: contingency table (two-way table), conditional distribution
• ### Star Library: Random Rendezvous
This activity leads students to appreciate the usefulness of simulations for approximating probabilities. It also provides them with experience calculating probabilities based on geometric arguments and using the bivariate normal distribution. We have used it in courses in probability and mathematical statistics, as well as in an introductory statistics course at the post-calculus level. Students are expected to approximate the solution through simulation before solving it exactly. They are also expected to employ graphical as well as algebraic problem-solving strategies, in addition to their simulation analyses. Finally, students are asked to explain intuitively why it makes sense for the probabilities to change as they do. Key words: simulation, probability, geometry, independence, bivariate normal distribution
• ### Video: Against All Odds: 26. Small Sample Inference for One Mean
In this free online video, students discover an improved technique for statistical problems that involves a population mean: the t statistic for use when sigma is not known. Emphasis is on paired samples and the t confidence test and interval. The program covers the precautions associated with these robust t procedures, along with their distribution characteristics and broad applications."
• ### Video: Against All Odds: 14. The Question of Causation
In this free online video program, students will learn that "causation is only one of many possible explanations for an observed association. This program defines the concepts of common response and confounding, explains the use of two-way tables of percents to calculate marginal distribution, uses a segmented bar to show how to visually compare sets of conditional distributions, and presents a case of Simpson's Paradox. The relationship between smoking and lung cancer provides a clear example."
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https://www.12000.org/my_notes/CAS_integration_tests/reports/rubi_4_16_1_graded/test_cases/0_Independent_test_suites/Stewart_Problems/rese176.htm
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### 3.176 $$\int \frac{x^2}{1+x} \, dx$$
Optimal. Leaf size=15 $\frac{x^2}{2}-x+\log (x+1)$
[Out]
-x + x^2/2 + Log[1 + x]
________________________________________________________________________________________
Rubi [A] time = 0.00575, antiderivative size = 15, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 9, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.111, Rules used = {43} $\frac{x^2}{2}-x+\log (x+1)$
Antiderivative was successfully verified.
[In]
Int[x^2/(1 + x),x]
[Out]
-x + x^2/2 + Log[1 + x]
Rule 43
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Rubi steps
\begin{align*} \int \frac{x^2}{1+x} \, dx &=\int \left (-1+x+\frac{1}{1+x}\right ) \, dx\\ &=-x+\frac{x^2}{2}+\log (1+x)\\ \end{align*}
Mathematica [A] time = 0.0030243, size = 19, normalized size = 1.27 $\frac{1}{2} (x+1)^2-2 (x+1)+\log (x+1)$
Antiderivative was successfully verified.
[In]
Integrate[x^2/(1 + x),x]
[Out]
-2*(1 + x) + (1 + x)^2/2 + Log[1 + x]
________________________________________________________________________________________
Maple [A] time = 0.001, size = 14, normalized size = 0.9 \begin{align*} -x+{\frac{{x}^{2}}{2}}+\ln \left ( 1+x \right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^2/(1+x),x)
[Out]
-x+1/2*x^2+ln(1+x)
________________________________________________________________________________________
Maxima [A] time = 0.950728, size = 18, normalized size = 1.2 \begin{align*} \frac{1}{2} \, x^{2} - x + \log \left (x + 1\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2/(1+x),x, algorithm="maxima")
[Out]
1/2*x^2 - x + log(x + 1)
________________________________________________________________________________________
Fricas [A] time = 1.80622, size = 35, normalized size = 2.33 \begin{align*} \frac{1}{2} \, x^{2} - x + \log \left (x + 1\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2/(1+x),x, algorithm="fricas")
[Out]
1/2*x^2 - x + log(x + 1)
________________________________________________________________________________________
Sympy [A] time = 0.067622, size = 10, normalized size = 0.67 \begin{align*} \frac{x^{2}}{2} - x + \log{\left (x + 1 \right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**2/(1+x),x)
[Out]
x**2/2 - x + log(x + 1)
________________________________________________________________________________________
Giac [A] time = 1.04549, size = 19, normalized size = 1.27 \begin{align*} \frac{1}{2} \, x^{2} - x + \log \left ({\left | x + 1 \right |}\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2/(1+x),x, algorithm="giac")
[Out]
1/2*x^2 - x + log(abs(x + 1))
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## ◂Math Worksheets and Study Guides Fourth Grade. Fractions
### The resources above correspond to the standards listed below:
#### Idaho Content Standards
4.MD. Number and Operations – Fractions – 4. NF
Extend understanding of fraction equivalence and ordering.
4.MD.1. Explain why a fraction a/b is equivalent to a fraction (n × a)/(n × b) by using visual fraction models, with attention to how the number and size of the parts differ even though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions.
4.MD.2. Compare two fractions with different numerators and different denominators, e.g., by creating common denominators or numerators, or by comparing to a benchmark fraction such as 1/2. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model.
Build fractions from unit fractions by applying and extending previous understandings of operations on whole numbers.
4.MD.4. Apply and extend previous understandings of multiplication to multiply a fraction by a whole number.
4.MD.4.a. Understand a fraction a/b as a multiple of 1/b. For example, use a visual fraction model to represent 5/4 as the product 5 × (1/4), recording the conclusion by the equation 5/4 = 5 × (1/4).
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13,822 views
sir can u solve this?
you can see the answer by Umang below. Just do the changes in it and you will get it.
what should be the relation between T(1),T(2),T(3) so that the algorithm time complexity becomes constant?
n should be the correct answer.
T(n) = T(n - 1) + T(n - 2) - T(n - 3)
T(n - 1) = T(n - 2) + T(n - 3) - T(n - 4)
T(n - 2) = T (n - 3) + T(n - 4) - T(n - 5)
T(n - 3) = T (n - 4) + T(n - 5) - T(n - 6)
T(n - 4) = T (n - 5) + T(n - 6) - T(n - 7)
.....................................................
.....................................................
T(6) = T(5) + T(4) - T(3)
T(5) = T(4) + T(3) - T(2)
T(4) = T(3) + T(2) - T(1)
On Adding all the above recurrences, all the red terms of a recurrence will be cancelled out with the red terms of consecutive recurrences, similarly all the blue terms of a recurrence will be cancelled out with the blue terms of consecutive recurrences.
Thus we get
T(n) = T(n - 2) + T(3) - T(1),
but since T(3) =3 & T(1) = 1
so our final recurrence will be T(n) = T(n - 2) + 2
Solution of T(n) = T(n - 2) + 2 will be n.
T(n) = n if n ≤ 3 means T(1) = 1 T(2) = 2 T(3) = 3
after subtracting you will be left T(6)=T(4)-2 . right!
Yes umang, I made a little mistake here. T(n) = T(n - 2) + 2. & The answer should be n.
Thanks for pointing that out. Edited it.
we are talking aboue time complexity of the program but not the returned value of the program.
we can not cancel one time complexity with other time complexity.
Can you prove that this program will be executed only n times. and how do you know the complexity of T(n-2) will be (n-2) or O(n)?
$$T(n) = \begin{cases} T(n-1) + T(n-2) - T(n-3) & \mbox{ if } n > 3\\n & \mbox{ if } n \leq 3\end{cases}$$
From the definition, we can see that:
$T(1) = 1$
$T(2) = 2$
$T(3) = 3$
$T(4) = T(3) + T(2) - T(1) = 3 + 2 - 1 = 4$
$T(5) = T(4) + T(3) - T(2) = 4 + 3 - 2 = 5$
$T(6) = T(5) + T(4) - T(3) = 5 + 4 - 3 = 6$
We can observe the pattern. It looks like $T(n) = n$, but we need to prove it.
Proof by strong induction:
Hypothesis: $T(n) = n$
Base Case: $T(1)$ to $T(6)$ have been shown above.
Assume, for the sake of induction $T(x), \forall x \in [1, n-1]$
Then,
\begin{align}T(n) &= T(n - 1) + T(n - 2) - T(n - 3)\\&= (n - 1) + (n - 2) - (n - 3)\\&= n\end{align}
QED
So, the answer is option A) n
It's a homogeneous recurrence relation, and can be solved easily:
\begin{align*} T(n) &= T(n-1) + T(n-2) - T(n-3) \\ a_{n} &= a_{n-1} + a_{n-2} - a_{n-3} \qquad \text{; rewriting recurrence relation} \end{align*}
The characteristic equation will be of the form:
\begin{align*} r^3 &= r^2 + r -1\\ (r+1) (r-1)^3 &= 0 \end{align*}
You can now solve it on your own and get $T(n) = \mathcal{O}(n)$
i think ur going for dynamic programing..
Does the length of the longest path in the recursion tree is equal to Time complexity??
updated.
evry time 3 call ..and maximum we have n level.. because T(n-1) is given.
No.
The time complexity of the following function is $O \left (3^n \right )$ and $\Omega \left ( 3^{n/3} \right )$
def A(n):
if n <= 3:
return n
return A(n-1) + A(n-2) - A(n-3)
This function is a very inefficient function which calculates the value of the time complexity described by the recurrence in the question!
The recurrence for the time complexity of the function $A$ is:
$$T_A(n) = T_A(n-1) + T_A(n-2) \;{\boldsymbol{\Large\color{red}+}}\; T_A(n-3)\; \color{red}{+ O(1)}\\$$
Note: The question already gives us the recurrence. We just have to find the solution.
The recurrence in the question solves to $T(n) = n$ as answered by Amar.
What you are thinking is about an algorithm that computes the value described by the recurrence.
An those two things are very different! :)
ok got it. (y)
1
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How Are Percentages Calculated?
# How Are Percentages Calculated?
Percentages are calculated by finding what portion a fractional amount takes up out of 100. By comparing a fractional amount to x/100, a percentage can be found.
"Percent" comes from the latin word for "out of (per) 100". If there were 100 of some item, and a person removed 10 items, that person would have taken 10% of the items - they would have taken 10/100. For items which do not equal exactly 100, they can be compared to it. Take for instance a group of four people, in which three are female and one is male. 3/4 people are female, and 1/4 people are male. By multiplying 4 (the denominator) by 25, we get 100. Afterwards, 3 and 1 can be multiplied by 25 to give 75 and 25; these are the percentages of men and women in the example.
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# How to write reciprocal squared function shifted right by $3$ and down by $4$
Given a reciprocal squared function that is shifted right by $$3$$ and down by $$4$$, write this as a rational function.
I am uncertain how to denote this. My attempt: $$\frac{1}{x^2-3}-4$$
But I need to show this as a rational function. Right now the $$-4$$ is disconnected from the fraction part.
Is it just this?: $$\frac{1}{x^2-3-4}$$ $$\frac{1}{x^2-7}$$
How can I write the reciprocal squared function as a rational function where it has been shifted right by $$3$$ and down by $$4$$?
edit: Is it?: $$\frac{1}{(x-3)^2}+4$$
• $\frac1{x-3}-4\ne\frac1{x-3-4}$ Aug 29, 2020 at 15:12
• What's a reciprocal square function? I think the confusion here stems from the fact that the wording is vague. I really can't guess what is intended. None of your functions reflect the "squared" so I assume they are all wrong, but who knows?
– lulu
Aug 29, 2020 at 15:14
• Is the reciprocal squared function referring to $\frac1{x^2}$? If so, then all your expressions are wrong. Aug 29, 2020 at 15:15
• Also if you want to shift a function $f(x)$ by $b$ units to the right, do $f(x+b)$. If you want to shift a function $g(x)$ by $b$ units down, then do $g(x)-b$. This should be enough information to determine the answer, no matter what your function is. Aug 29, 2020 at 15:19
• I suspect what they mean is the function $f(x) = \frac{1}{(x - 3)^2} - 4$. However, the way the question is phrased makes the sequence of transformations unclear. Note that replacing $x$ by $x - 3$ shifts the graph to the right three units and subtracting $4$ from the expression shifts it down by $4$ units. Aug 29, 2020 at 17:11
$$f$$ is a reciprocal squared function: $$f(x) = \frac{1}{x^2}$$
$$g$$ is $$f$$ shifted by $$a$$ units to the right: $$g(x)=f(x-a)\\g(x)=\frac{1}{(x-a)^2}$$ $$h$$ is $$g$$ shifted by $$b$$ units down $$h(x) = g(x)-b\\h(x)=\frac{1}{(x-a)^2}-b$$ So if you shift $$f$$ by 3 units to the right and 4 units down you would get the following function $$h$$: $$h(x)=\frac{1}{(x-3)^2}-4$$ Now to simplify the expression of $$h$$ or to make it a "rational function" you just have to find the common denominator of the 2 summands which is in this case $$(x-3)^2$$: $$h(x)=\frac{1}{(x-3)^2}-\frac{4(x-3)^2}{(x-3)^2}=\frac{1-4(x^2-6x+9)}{(x-3)^2}\\h(x)=\frac{-4x^2+24x-35}{(x-3)^2}$$
For simplicity call $$u=(x-3)^2$$ so that $$h(x)=1/u + 4 = 1/u + 4u/u=(1+4u)/u$$ and now substituting back in we have $$h(x)=(1+4(x-3)^2)/(x-3)^2$$ which is the quotient of two polynomials as desired.
$$f(x\pm k)$$ shifts a function to the left/right by $$k$$. $$f(x) \pm m$$ shifts a function up/down by $$m$$.
So $$f(x-3) + 4$$ will shift a function to the right by $$3$$ and up by $$4$$.
So if $$f([\color{blue}x]) = \frac 1{[\color{blue}x]^2}$$
then $$f([\color{red}{x-3}])+ 4 = \frac 1{[\color{red}{x-3}]^2} + 4$$
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# How to Solve[] for rational constants?
I am trying to solve a system equations for rational, constant coefficients:
ord = 3;
vars = {a, b, c, d, e};
Do[tt[nn] = Total[vars^nn], {nn, 1, ord}];
mm = (Expand[tt[1]^ord] /. mm_^nn_ -> 0)/ord! // Expand; (*Sum of products of all unique sets of 3*)
expr = cc[1]*tt[1]^3 + cc[2]*tt[2]*tt[1] + cc[3]*tt[3] - mm;
aa = Solve[{expr == 0}, {cc[1], cc[2], cc[3]}]
(*{{cc[3] -> -((-a b c - a b d - a c d - b c d - a b e - a c e - b c e - a d e - b d e - c d e)/(a^3 + b^3 + c^3 + d^3 + e^3)) - ((a + b + c + d + e)^3 cc[1])/(a^3 + b^3 + c^3 + d^3 + e^3) - ((a + b + c + d + e) (a^2 + b^2 + c^2 + d^2 + e^2) cc[2])/(a^3 + b^3 + c^3 + d^3 + e^3)}} *)
This is the correct result for an under-determined system like this one, with 1 eqn and 3 unknowns. However, if we require that the cc[]'s be rational constants, i.e. independent of {a, b, c, d, e}, then the only solution (which I calculated by hand) is:
bb = {cc[1] -> 1/6, cc[2] -> -1/2, cc[3] -> 1/3};
expr /. bb // Expand
(* 0 *)
How can I get such solns from MMa? How does one impose the condition that the soln coefficients must be rational constants?
Ultimately, I need to do this with arbitrarily large vars lists and larger values of ord where ord < Length[vars].
Take a look at SolveAlways (documentation page). In this case:
SolveAlways[expr == 0, vars]
(* Out: {{cc[1] -> 1/6, cc[2] -> -(1/2), cc[3] -> 1/3}} *)
It may be, however, that in a more complicated expression you might obtain multiple sets of values for the parameters, and you may have to refine the results to pick those in which all cc[] parameters are rational.
SolveAlways is actually just a "shortcut" for the following Solve expression (see the "Details and Options" section of its documentation):
SolveAlways[expr == 0, vars]
(* is equivalent to *)
Solve[!Eliminate[!expr == 0, vars], Array[cc, 3]]
so you could use the version written with Solve and explicitly request that solutions be rational by specifying a domain:
Solve[!Eliminate[!expr == 0, vars], Array[cc, 3], Rationals]
• Why can't all Answers be this simple? Thanks! – Jerry Guern Jun 15 '15 at 3:26
• @JerryGuern glad it helped, and thank you for the accept! – MarcoB Jun 15 '15 at 3:31
• I have never seen exclamation points used the way you've used them in the code you added. Is "!expr==0" the same a "expr!=0"? What do the two !'s in that code do? Very intriguing. – Jerry Guern Jun 15 '15 at 6:46
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Solutions
Modern Digital and Analog Communication Systems (Oxford Series in Electrical and Computer Engineering)
Modern Digital and Analog Communication Systems (Oxford Series in Electrical and Computer Engineering) (3rd Edition) View more editions Solutions for Chapter 5.1
• 358 step-by-step solutions
• Solved by publishers, professors & experts
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SAMPLE SOLUTION
Chapter: Problem:
70% (23 ratings)
• Step 1 of 9
• Step 2 of 9
• Step 3 of 9
• Anonymous
8000 is correct, m(t) is 2 /(1/4 * 10^-3)t
• Anonymous
Rise: The graph goes from -1 to 1. Thus, Rise = 1 - (-1) = 2 Run = (1/4 * 10^-3) or (10^-3 / 4) 1/Run = 4 * 10^3 = 4000 Rise/Run = Rise * (1/Run) = (2)(4000) = 8000
• l g
You have to use ^ function, not EE function
• Andrew Deluna
I see that there is (10^-3)/4. Where does the 1/4 come from, specifically the 4?
• Anonymous
It should be 800 and -800 slope = rise/run => 2/((10^-3)/4) = 800. The rise is 2 because it goes from -2 to 2 and the run 10^-3/4 because it is just one a out of the total 4.
• Step 4 of 9
• Anonymous
^ This guys is right. Have to divide through by 2*pi to get f_i
• Anonymous
You have to divide the answer by 2*pi to get Hz. The answer given is 99.9 X 10^6 rads/sec not Mhz
• Step 5 of 9
• Step 6 of 9
• Anonymous
wrong
• Step 7 of 9
• Anonymous
Correct!
• InvincibleNinja7202
why do you not divide by 2pi to get from rad/s to Hz?
• Step 8 of 9
• Step 9 of 9
• Dylan James
the drawing doesnt seem to match the correct labels on it... anyone else agree?
• Krystle Villagran
Agree
Corresponding Textbook
Modern Digital and Analog Communication Systems (Oxford Series in Electrical and Computer Engineering) | 3rd Edition
9780195110098ISBN-13: 0195110099ISBN: B.P.LathiAuthors:
Alternate ISBN: 9781618127365
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login home contents what's new discussion bug reports help links subscribe changes refresh edit
# Edit detail for SandBoxDifferentialEquations revision 1 of 1
1 Editor: Bill Page Time: 2009/10/16 19:44:19 GMT-7 Note: merge
changed:
-
Differential Equations
\begin{axiom}
y := operator y
deq := D(y(x),x)=(1+y(x)^2)/(1+2*x*y(x))
solve(deq,y,x)
deq := D(y(x),x,3)+D(y(x),x,2)+D(y(x),x,1)+y(x)=x*sin(x)
solve(deq,y,x)
\end{axiom}
test solution
\begin{axiom}
deq := D(y(x),x)=(a+y(x)^2)/(b+2*x^2*y(2*x))
solve(deq,y,x)
deq := D(y(x),x,3)+D(y(x),x,2)+D(y(x),x,1)+y(x)=x*sin(x)
solve(deq,y,x)
\end{axiom}
Linear non homog equation
\begin{axiom}
deq := D(y(x),x) = y(x)^2 + 2 * y(x) + 1
solve(deq,y,x)
z := operator z
deq2 := D(z(x),x) = exp(y(x))/(1 - d * y(x))
solve(deq2,z,x)
\end{axiom}
Differential Equations
fricas
y := operator y
(1)
Type: BasicOperator?
fricas
deq := D(y(x),x)=(1+y(x)^2)/(1+2*x*y(x))
(2)
Type: Equation(Expression(Integer))
fricas
solve(deq,y,x)
(3)
Type: Union(Expression(Integer),...)
fricas
deq := D(y(x),x,3)+D(y(x),x,2)+D(y(x),x,1)+y(x)=x*sin(x)
(4)
Type: Equation(Expression(Integer))
fricas
solve(deq,y,x)
(5)
Type: Union(Record(particular: Expression(Integer),basis: List(Expression(Integer))),...)
test solution
fricas
deq := D(y(x),x)=(a+y(x)^2)/(b+2*x^2*y(2*x))
(6)
Type: Equation(Expression(Integer))
fricas
solve(deq,y,x)
>> Error detected within library code:
getfreelincoeff: not a linear ordinary differential equation
Linear non homog equation
fricas
deq := D(y(x),x) = y(x)^2 + 2 * y(x) + 1
(7)
Type: Equation(Expression(Integer))
fricas
solve(deq,y,x)
(8)
Type: Union(Expression(Integer),...)
fricas
z := operator z
(9)
Type: BasicOperator?
fricas
deq2 := D(z(x),x) = exp(y(x))/(1 - d * y(x))
(10)
Type: Equation(Expression(Integer))
fricas
solve(deq2,z,x)
(11)
Type: Union(Record(particular: Expression(Integer),basis: List(Expression(Integer))),...)
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### The Next Big Thing in boolean numbers
This is the second article in a row I have written a bit about the use of boolean numbers for a real estate agent. I believe that the number of units in a home is a major determinant in how much home you can afford. In a home where there are fewer units than what you are planning on buying, the price and value are likely to be much more dependent on the number of units you have available.
So if you’re buying a home, you have to make sure the number of units you’re buying is adequate to ensure you can afford the home. So we’re going to talk about the number of children, and how to calculate this number.
Number of children is the number of children in a home. A number of children is the number of children in a home. When a home has six children, it means it has six kids to play with, and one less baby than what you could want. These two numbers together mean how many people will sleep in the home. In a home where there are fewer children, you have to be a bit more careful with how you plan on spending your time.
While there is no one number that can be said to be the answer to a number of kids, we can use the word “number” to mean the number of children in a home. You can use the word “number” to mean the number of children in a family as well. In other words, we can say that a home has six children because there are six kids to play with. In a home where there are fewer children you can plan your time better.
One way to plan your time is to put it into a budget. When you’re budgeting you look at the number of children in your home and the number of children in your family. The reason for this is the children should all be able to spend equal amounts of time with each other. With six children in a family with four adults, one adult should spend most of his or her time with one child.
This can be done in a number of ways. With just one child, you just put that child in your household. With two children, you use the first one as a babysitter, the second one as a nanny, and the third one as a nanny. With three children, you take turns being the nanny. With four children, you take turns being the nanny. With five children, you take turns being the nanny.
This can be a really helpful method when you’re trying to figure out who to do what to who. I’m a big fan of this idea because it is very easy to remember and it is a very effective way to allocate your time and energy.
I am going to be making a lot of this up. I think because I like the idea of “boolean numbers.” So you take a number and you put it in a box to figure out how many of you need to do something to get that number done. For example, I put 10 as the number of nannies in this house. If you have two nannies, you need to put two nannies in this house.
I think that makes sense if you think about it. So, for instance, you have 10 nannies and you need to feed them, you have 10 nannies and you need to clean up for them, so you put 10 into a box and you put those in a box and you put that box in a box and you put that box in a box and that’s it.
The general rule is that you can’t put things in boxes unless you can figure out how many of them there are. And you can’t put things into boxes unless you know the total of how many things there are. I’ve also seen it stated that there are 10 boxes in the house, but there are actually 50 boxes and 50 boxes, so I’ve just ignored that.
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Free Algebra Tutorials!
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#### MAT-05.GM.G.02
5th Grade (MAT) Targeted Standard (GM) Geometry and Measurement (G) Geometry Learners will compose and classify figures and shapes based on attributes and properties; represent and solve problems using a coordinate plane.
## Progressions
Coordinate Plane
• MAT-03.GM.G.01 In two-dimensional shapes, identify lines, angles (right, acute, obtuse), and perpendicular and parallel lines.
• MAT-05.GM.G.02 Identify the x-coordinate and y-coordinate to graph and name points in the first quadrant of the coordinate plane.
• MAT-05.GM.G.03 Form ordered pairs and graph points in the first quadrant of the coordinate plane to solve authentic word problems.
• MAT-06.GM.GF.01 Identify and position ordered pairs of rational numbers in all four quadrants of a coordinate plane.
• MAT-06.GM.GF.02 Draw polygons in the coordinate plane given coordinates for vertices. Determine the length of a side joining points with the same first or second coordinate, including authentic problems.
• MAT-10.GM.27 Develop and verify the slope criteria for parallel and perpendicular lines. Apply the slope criteria for parallel and perpendicular lines to solve geometric problems using algebra.
• MAT-10.GM.28 Verify simple geometric theorems algebraically using coordinates. Verify algebraically, using coordinates, that a given set of points produces a particular type of triangle or quadrilateral.
• MAT-10.GM.29 Determine the midpoint or endpoint of a line segment using coordinates. (+) Find the point on a directed line segment between two given points that partitions the segments in a given ratio.
• MAT-12.NO.10 Represent complex numbers on the complex plane in rectangular, trigonometric, and polar forms. Find the modulus (absolute value) of a complex number. Explain why the rectangular, trigonometric, and polar forms of a given complex number represent the same number.
• MAT-12.NO.11 Represent addition, subtraction, multiplication, conjugation, powers, and roots of complex numbers geometrically on the complex and/or polar plane; use properties of this representation for computation.
• MAT-12.NO.14 Recognize vector quantities as having both magnitude and direction, writing them in polar form.
• MAT-12.NO.15 Find the components of a vector by subtracting the coordinates of an initial point from the coordinates of a terminal point.
• MAT-12.NO.16 Solve problems involving magnitude and direction that can be represented by vectors.
• MAT-12.NO.17 Add and subtract vectors.
• MAT-12.NO.18 Multiply a vector by a scalar.
• MAT-12.AR.F.18 Explain how the unit circle in the coordinate plane enables the extension of trigonometric functions to all real numbers, interpreted as radian measures of angles traversed counterclockwise around the unit circle.
• MAT-12.AR.F.19 Use the unit circle to express the values of sine, cosine, and tangent for π - x, π + x, and 2π - x in terms of their values for x, where x is any real number.
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# A train 50m long moving on a straight track passes a pole in 5 seconds. Find the speed of the train time it will take to cross a bridge 500m long.
pohnpei397 | Certified Educator
If the train passes the pole in 5 seconds, and the train is 50 meters long, its speed is 10 meters per second. You can find that by dividing its length by the time that it takes to pass a given point.
From the time that the front of the train reaches the bridge, it will take 50 seconds (500 meters divided by 10 meters per second) for the front to cross the bridge. However, the rest of the train will still be on the bridge.
It will be 5 more seconds (50 meter train divided by 10 meters per second) for the rest of the train to get off the bridge.
Therefore, from the time that the front of the train reaches the bridge, it will be 55 seconds until the end of the train gets off the bridge on the other side.
thewriter | Student
The train is 50m in length and crosses a pole in 5 seconds. Therefore its speed which is given by distance/ time is 50/5 or 10 m/s.
Now if the train has to cross the bridge which is 500m long make sure of not assuming that the total distance to be travelled is 500m. As the length of the train is 50m, only when the front end of the train moves 50m ahead after crossing the bridge will the rear end also cross the bridge. Therefore the total distance that has to be travelled is 550m.
As time is given by distance/speed it is 550/10=55s.
neela | Student
The length of the train = 50m.
To pass the pole the entire length of 50 meter the train has to pass the pole for which time taken = 5 secs.
Therefore train has moved 50 m in 5 seconds. So the speed of the train = 50m/5 sec = 10meter per sec = 50m*60*60 per hour = 180000m/s = 180 km/hour.
To cross the bridge of 500 m the train has to move over from one end to the other end of the bridge and the train has to clear its length. So the train has to move by bridge length+it's(train's) length = 500m+50m = 550m
So the time taken by train to pass the bridge = 550m/speed of train = 550m/10m/s = 55seconds.
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GMAT Math : Calculating the length of an edge of a tetrahedron
Example Questions
Example Question #1 : Rectangular Solids & Cylinders
Suppose the volume of the tetrahedron is . What is the edge of the tetrahedron?
Explanation:
Write the formula to solve for the edge of a tetrahedron.
Substitute the volume and solve.
Example Question #2 : Rectangular Solids & Cylinders
The height of one of the equilateral triangle faces on a tetrahedron is . What is the side length of the tetrahedron?
Explanation:
Because a tetrahedron has four congruent equilateral triangles as its faces, we know the three equal angles of each face, as with all equilateral triangles, are each . We are given the height of a face, which is the length of a line that bisects one of the angles and forms a right triangle with the side length as its hypotenuse. This means the angle between the height and the side length is , whose cosine is equal to the adjacent side, the height, over the hypotenuse, the side length. This gives us:
Example Question #1 : Calculating The Length Of An Edge Of A Tetrahedron
The volume of a tetrahedron is . What is the edge of the tetrahedron?
Explanation:
Write the formula to solve for the edge of a tetrahedron.
Substitute the given volume to the equation and solve.
Example Question #1 : Tetrahedrons
Find the exact edge length of a tetrahedron if the volume is .
Explanation:
Write the tetrahedron formula to solve for the edge.
Substitute the volume and simplify.
Example Question #5 : Rectangular Solids & Cylinders
A certain tetrahedron has a surface area of . What is the length of an edge of the tetrahedron?
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# Essay On Jumping Jacks
714 Words3 Pages
What is the effect of that time has on pushups compared to jumping jacks on the heart rate? If participants do pushups and jumping jacks then the heart rate will be at a higher rate doing jumping jacks then the participant doing pushups, because jumping jacks activate more muscles when performing the exercise compared to pushups so the heart rate will be more.To insure success in a experiment the scientist must make a null hypothesis. For example, the heartrate will not increase when doing jumping jacks rather than push ups. A good experiment always has a null hypothesis, this experiment is no exception.
Step One:Everyone who is particating in the trails must find a flat surface to do the exercises on.One paricipanmt must record the time using a stopwatch.
Step Two:The
Although the data shows that both Pushups and Jumping Jacks increase BPM over time. Jumping Jacks increased BPM at a growth rate of 52% between 20 seconds and 60 seconds, while Pushups had a slightly lower growth rate at 43% between 20 seconds and 60 seconds.
The trendline shows that at zero timed exercise the heart rate for jumping jacks was below the pushup zero timed exercise. Jumping jacks BPM overlaps Pushups BPM at 20 seconds of exercise and is know increasing at a faster rate than pushups BPM. At 40 and 60 seconds the BPM for Jumping jacks is increasing upward, while pushups BPM trendline stayed the same throughout.
If the trend of the data continues then I could predict that at 100 seconds Jumping Jacks BPM will be at 171.425 BPM. I 'm very confident in my prediction because the r2 value is 0.874 which means the line is very accurate to data interpretation. If the trend of the data continues for push ups as well. I can predict at 100 seconds pushups BPM will be at 160 BPM. Im also confident in my prediction because the r2 value is at 0.8333 which is moderately close to data
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## 261 Reputation
16 years, 113 days
## Illustration...
I will try to illustrate the problem some more. With laplace and in the time domain I get as solution: y(t) = 15 (-exp(-2 t) + 2 exp(-t) - 1) exp(-t) This is illustrated enough above. When I construct the transferfunction for the differential equation I get: H(s) = s / (s^2+3s+2) The transform of x(t) is given by: X(s) = 10/(s+3) De output is given by: Y(s) = X(s) H(s) y(t) = -15exp(-3t)+20exp(-2t)-5exp(-t) (Zero state response) This solution is different from the solution found with previous methods. I think that the fact that x(0)=10 is lost somewhere in the calculation but I can't find a way to add it. Normally when there are initial conditions voor y(t) and D(y)(t) that are different form zero I would add a zero input response to the zero state response calculated with the transferfunction. But in this case this is no solution because in the zero input response I can only plug conditions for y(t) because the input is zero. What is the problem here?
## Illustration...
I will try to illustrate the problem some more. With laplace and in the time domain I get as solution: y(t) = 15 (-exp(-2 t) + 2 exp(-t) - 1) exp(-t) This is illustrated enough above. When I construct the transferfunction for the differential equation I get: H(s) = s / (s^2+3s+2) The transform of x(t) is given by: X(s) = 10/(s+3) De output is given by: Y(s) = X(s) H(s) y(t) = -15exp(-3t)+20exp(-2t)-5exp(-t) (Zero state response) This solution is different from the solution found with previous methods. I think that the fact that x(0)=10 is lost somewhere in the calculation but I can't find a way to add it. Normally when there are initial conditions voor y(t) and D(y)(t) that are different form zero I would add a zero input response to the zero state response calculated with the transferfunction. But in this case this is no solution because in the zero input response I can only plug conditions for y(t) because the input is zero. What is the problem here?
## Strange...
What you say is true but when you construct the transfertfunction for example, all initial condition are null. Then you get something like: H(s) = s / (s^2+3s+2) The transform of x(t) is given by: X(s) = 10/(s+3) De output is given by Y(s) = X(s) H(s) y(t) = -15exp(-3t)+20exp(-2t)-5exp(-t) When I now solve this in the time domain manual or with Maple I have as solution: restart: DV:=diff(y(t),t\$2)+3*diff(y(t),t)+2*y(t)=diff(10*exp(-3*t),t); > Cond:=y(0)=0,D(y)(0)=0: > Opl:=dsolve({DV,Cond},y(t)): > Opl:=simplify(Opl); Opl := y(t) = 15 (-exp(-2 t) + 2 exp(-t) - 1) exp(-t) This is totally different from what I get with Laplace. I do not find what I'm doing wrong. I followed these steps quite some times for other differential equations. Equations where x(t) was not derived. But for this equation I can't get two times the same solution.
## Strange...
What you say is true but when you construct the transfertfunction for example, all initial condition are null. Then you get something like: H(s) = s / (s^2+3s+2) The transform of x(t) is given by: X(s) = 10/(s+3) De output is given by Y(s) = X(s) H(s) y(t) = -15exp(-3t)+20exp(-2t)-5exp(-t) When I now solve this in the time domain manual or with Maple I have as solution: restart: DV:=diff(y(t),t\$2)+3*diff(y(t),t)+2*y(t)=diff(10*exp(-3*t),t); > Cond:=y(0)=0,D(y)(0)=0: > Opl:=dsolve({DV,Cond},y(t)): > Opl:=simplify(Opl); Opl := y(t) = 15 (-exp(-2 t) + 2 exp(-t) - 1) exp(-t) This is totally different from what I get with Laplace. I do not find what I'm doing wrong. I followed these steps quite some times for other differential equations. Equations where x(t) was not derived. But for this equation I can't get two times the same solution.
## The problem is clear now....
The problem is clear now. I've found the right solutions thanks to method from JacquesC.
## The problem is clear now....
The problem is clear now. I've found the right solutions thanks to method from JacquesC.
## I see you were able to find...
I see you were able to find a general solution. For my problem it is sufficient enough to have an exact solution. Have some difficulties getting an exact solution out of this. By the way I've typed out a manually example from my handbook to show how It should about solved. It's kinda hard to translate it to maple. Download 2714_Example.mws
View file details
## I see you were able to find...
I see you were able to find a general solution. For my problem it is sufficient enough to have an exact solution. Have some difficulties getting an exact solution out of this. By the way I've typed out a manually example from my handbook to show how It should about solved. It's kinda hard to translate it to maple. Download 2714_Example.mws
View file details
## It's indeed also working for...
It's indeed also working for opl now. There must have been some confusion here when using restart and preassigning a value to total which made it look like opl was the trouble maker.
## It's indeed also working for...
It's indeed also working for opl now. There must have been some confusion here when using restart and preassigning a value to total which made it look like opl was the trouble maker.
## Sorry...
My excuses. Previous time I've posted only the worksheet and few reacted to the topic because most would not open it. Download 2714_Curl.mws
View file details Edit: I just rememberd the command dsolve. Hmzz looks like it doesn't work for a vectorfunction.
## Sorry...
My excuses. Previous time I've posted only the worksheet and few reacted to the topic because most would not open it. Download 2714_Curl.mws
View file details Edit: I just rememberd the command dsolve. Hmzz looks like it doesn't work for a vectorfunction.
## It was indeed the name that...
It was indeed the name that caused the trouble. My problem is solved now. Thank you,
## It was indeed the name that...
It was indeed the name that caused the trouble. My problem is solved now. Thank you,
## I've been able to make...
I've been able to make following sum: "for i from 1 to 4 do total:=total+i end do;" The result is logic: 1,3,6,10. Now I want to make the sum of some terms I've already calculated and named by opl[i]. With I a posint. Now I write the command: "for i from 1 to 4 do total:=total+opl[i] end do;" This gives me the error of to many levels of recursion. I am not able to see the problem in what I've programmed. Thank you in advance,
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# Sayavong
Sayavong is making cookies for the class. He has a recipe that calls for 3 and 1/2 cups of flour. He has 7/8 of a cup of wheat flour, and 2 and 1/2 cups of white flour. Does Mr. Sayavong have enough flour to make the cookies?
Result
x=
### Step-by-step explanation:
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REAL NUMBERS
(as opposed to fake numbers?)
Objective • TSW identify the parts of the Real Number System • TSW define rational and irrational numbers • TSW classify numbers as rational or irrational
Real Numbers • Real Numbers are every number. • Therefore, any number that you can find on the number line. • Real Numbers have two categories.
What does it Mean? • The number line goes on forever. • Every point on the line is a REAL number. • There are no gaps on the number line. • Between the whole numbers and the fractions there are numbers that are decimals but they don’t terminate and are not recurring decimals. They go on forever.
Real Numbers REAL NUMBERS
154,769,852,35 4
1.333 -8
5,632.1010101256849765 … 61
π 549.2378 9
49 %
Two Kinds of Real Numbers • Rational Numbers • Irrational Numbers
Rational Numbers • A rational number is a real number that can be written as a fraction. • A rational number written in decimal form is terminating or repeating.
Examples of Rational Numbers
•16 •1/2 •3.56
•-8 •1.3333 … •- 3/4
Integers One of the subsets of rational numbers
What are integers? • Integers are the whole numbers and their opposites. • Examples of integers are 6 -12 0 186 -934
• Integers are rational numbers because they can be written as fraction with 1 as the denominator.
Types of Integers • Natural Numbers(N): Natural Numbers are counting numbers from 1,2,3,4,5,................ N = {1,2,3,4,5,................} • Whole Numbers (W): Whole numbers are natural numbers including zero. They are 0,1,2,3,4,5,............... W = {0,1,2,3,4,5,..............} W=0+N
REAL NUMBERS NATURAL Numbers IRRATIONAL Numbers
WHOLE Number s
INTEGERS RATIONAL Numbers
Irrational Numbers • An irrational number is a number that cannot be written as a fraction of two integers. • Irrational numbers written as decimals are nonterminating and nonrepeating.
Irrational numbers can be written only as decimals that do not terminate or repeat. They cannot be written as the quotient of two integers. If a whole number is not a perfect square, then its square root is an irrational number. Caution! A repeating decimal may not appear to repeat on a calculator, because calculators show a finite number of digits.
Examples of Irrational Numbers • Pi
Try this! a) b) c) d) e)
2
• a) Irrational
12
• b) Irrational
25 5
• c) Rational • d) Rational
11
66
• e) Irrational
Additional Example 1: Classifying Real Numbers Write all classifications that apply to each number.
A.
5 is a whole number that is not a perfect square. irrational, real 5
B. –12.75 –12.75 is a terminating decimal. rational, real 16 16 4 = =2 C. 2 2 2 whole, integer, rational, real
A fraction with a denominator of 0 is undefined because you cannot divide by zero. So it is not a number at all.
Additional Example 2: Determining the Classification of All Numbers
State if each number is rational, irrational, or not a real number. A.
21 irrational
B.
0 3
0 =0 3
rational
Additional Example 2: Determining the Classification o All Numbers
State if each number is rational, irrational, or not a real number. C.
4 0 not a real number
Objective • TSW compare rational and irrational numbers • TSW order rational and irrational numbers on a number line
Comparing Rational and Irrational Numbers • When comparing different forms of rational and irrational numbers, convert the numbers to the same form. 3 7
Compare -3
and -3.571
(convert -337 to -3.428571…
-3.428571… > -3.571
Practice
Ordering Rational and Irrational Numbers • To order rational and irrational numbers, convert all of the numbers to the same form. • You can also find the approximate locations of rational and irrational numbers on a number line.
Example • Order these numbers from least to greatest. ¹/₄, 75%, .04, 10%, ⁹/₇ ¹/₄ becomes 0.25 75% becomes 0.75 0.04 stays 0.04 10% becomes 0.10
⁹/₇ becomes 1.2857142… Answer:
0.04, 10%, ¹/₄, 75%, ⁹/₇
Practice Order these from least to greatest:
Objectives • TSW identify the rules associated computing with integers. • TSW compute with integers
Examples: Use the number line if necessary. -5
0
1) (-4) + 8 = 4 2) (-1) + (-3) = -4 3) 5 + (-7) = -2
5
1) When the signs are the same, ADD and keep the sign. (-2) + (-4) = -6 2) When the signs are different, SUBTRACT and use the sign of the larger number. (-2) + 4 = 2 2 + (-4) = -2
Karaoke Time!
Addition Rule: Sung to the tune of “Row, row, row, your boat� Same signs add and keep, different signs subtract, keep the sign of the higher number, then it will be exact! Can your class do different
-1 + 3 = ? 1. -4 2. -2 3. 2 4. 4 Answer Now
-6 + (-3) = ? 1. -9 2. -3 3. 3 4. 9 Answer Now
The additive inverses (or opposites) of two numbers add to equal zero. Example: The additive inverse of 3 is -3 Proof: 3 + (-3) = 0 We will use the additive inverses for subtraction problems.
What’s the difference between 7 - 3 and 7 + (-3) ? 7 - 3 = 4 and 7 + (-3) = 4 The only difference is that 7 - 3 is a subtraction problem and 7 + (-3) is an addition problem. “SUBTRACTING IS THE SAME AS ADDING THE OPPOSITE.” (Keep-change-change)
When subtracting, change the subtraction to adding the opposite (keep-change-change) and then follow your addition rule. Example #1:
- 4 - (-7) - 4 + (+7) Diff. Signs --> Subtract and use larger sign. 3 Example #2: -3-7 - 3 + (-7) Same Signs --> Add and keep the sign. -10
Which is equivalent to -12 – (-3)? 1. 12 + 3 2. -12 + 3 3. -12 - 3 4. 12 - 3 Answer Now
7 – (-2) = ? 1. -9 2. -5 3. 5 4. 9 Answer Now
Review 1) If the problem is addition, follow your addition rule. 2) If the problem is subtraction, change subtraction to adding the opposite (keep-change-change) and then follow the addition rule.
State the rule for multiplying and dividing integers‌. If the signs are the same, the answer will be positive.
If the signs are different,
+
the answer will be negative.
1. -8 * 3 -24
Different What’s Signs The Negative Rule? Answer
2. -2 * -61 122
Tw Jus o tt at ak a e tim e
3. (-3)(6)(1) (-18)(1) -18
4. 6 ÷ (-3) -2 Start inside ( ) first
5. - (20/-5) - (-4) 4 Same Signs Positive Answer
6.
−408 −6 68
7. At midnight the temperature is 8°C. If the temperature rises 4°C per hour, what is the temperature at 6 am? How long Is it from Midnight to 6 am?
6 hours
How much does the temperature rise each hour?
+4 degrees
(6 hours)(4 degrees per hour) = 24 degrees
Add this to the original temp.
8° + 24° = 32°C
8. A deep-sea diver must move up or down in the water in short steps in order to avoid getting a physical condition called the bends. Suppose a diver moves up to the surface in five steps of 11 feet. Represent her total movements as a product of integers, and find the product.
Wha doe t s This mea n?
Multiply
(5 steps) (11 feet) (55 feet) 5 * 11 = 55
numeros reales
numeros reales
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# Stuck on a probability question
1. Mar 13, 2012
### silkdigital
Question:
Dreamboat cars are produced at three different factories A, B, and C. Factory A produces 20% of the total output of Dreamboats, B 50%, and C 30%. However, 5% of the cars at A are lemons, 2% at B are lemons, 10% at C are lemons. If you buy a Dreamboat and it turns out to be a lemon, what is the probability that it was produced at factory A?
My workings were:
0.05 / (0.05 + 0.02 + 0.1) = 29.41%
I ignored the output percentages because the base condition is that the car is a lemon already. Is my logic on the right track? Not sure if I'm correct!
Another tough question is this:
Suppose that 10 cards, of which 7 are red and 3 are green, are put at random into 10 envelopes, of which seven are red and three are green, so that each envelope contains one card. Determine the probability that exactly k envelopes will contain a card with a matching color.
I've managed to obtain k=4
[7C4 * 3C3] / 10C7
and k=10
[7C7 * 3C0] / 10C7
Not sure what the next steps are to express in terms of any k. Hope that makes sense.
Thanks for any help in advance, having some difficulty!
2. Mar 13, 2012
### awkward
Silkdigital,
For the dreamboats, you need
$$P(\text{produced at A} | \text{is a lemon}) = \frac{P(\text{produced at A and is a lemon)}}{P(\text{is a lemon})}$$
for which I think you need the output percentages.
For the cards, it may be easier to answer a closely related question: How many green cards are in green envelopes? This is exactly like taking a random sample of three cards without replacement and asking how many of them are green.
3. Mar 14, 2012
### SW VandeCarr
Note, you failed to weight your error percentages by the production percentages. The weights can be assigned in various ways provided the proportions are preserved. So for factory A: (0.2)(0.05)/[(0.2)(0.05)+(0.5)(0.02)+(0.3)(0.10)] = 0.01/(0.01+0.01+0.03)=0.2
So if you have a defect, the probabilities are 0.2 from A, 0.2 from B and 0.6 from C. Note they sum to 1.
4. Mar 14, 2012
### silkdigital
Yes I realised it now. I guess I misinterpreted the question. I did the same thing and ended with 0.2, for the envelope question I realised matches only exist for even k when k>=4 (ie 4,6,8,10) and all marginal probabilities add to 1.
Thanks for the help guys!
5. Mar 14, 2012
### SW VandeCarr
You're welcome.
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# math
posted by .
the trip took 2 hours longer to cover the 160 miles than it took to come back. if the total time of the round trip was 6 hours, what was the rate going and what was the rate coming back?
• math -
160/2 = 80 mph
160/4 = 40 mph
## Respond to this Question
First Name School Subject Your Answer
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Wave Nature of Light
# Wave Nature of Light
## Wave Nature of Light
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. Wave Nature of Light AP Physics B Chapter 24 Notes
2. Wave Theory Huygens supported the wave theory of light, and developed a principle that says every point on a wave can be thought of as a mini-wave that all spread out at the same speed and are enveloped by a new wave front
3. Wave Theory This principle is consistent with the phenomenon of diffraction—the bending of waves into the “shadow” region behind an obstacle
4. Wave Theory Refraction can also be explained by Huygens’ Principle. When light changes media, v andλchange but f does not. So
5. Diffraction—Superposition Review Recall what can happen when two waves run into each other—constructive or destructive interference. In phase Out of phase
6. Diffraction—Interference Patterns When light OR sound is produced by TWO sources a pattern results as a result of interference.
7. Diffraction—Double Slit If a wave is of monochromatic light (single wavelength) it should produce an interference pattern. Young conducted just such an experiment…and in fact did see such a pattern
8. Double Slit Diffraction Pattern Figure 2 Figure 1 Here is the pattern you will see. Notice in figure 2 that there are several bright spots and dark areas in between. The spot in the middle is the BRIGHTEST and thus the CENTRAL MAXIMUM. We call these spots FRINGES. The intensity of the light decreases as you move away from the central bright spot—these are called ORDERS away from the center.
9. Diffraction – Orders, m We use the letter, m, to represent the ORDER of the fringe from the bright central. Second Order Bright Fringe m = 2 Second Order Bright Fringe m = 2 First Order Bright Fringe m = 1 First Order Bright Fringe m = 1 Central Maximum It is important to understand that we see these bright fringes as a result of CONSTRUCTIVE INTERFERENCE.
10. Diffraction—Pattern Explained If two waves enter each slit, they will meet at the screen having travelled the same distance. They will be in phase and create a central maxima (bright spot).
11. Diffraction—Pattern Explained Now consider two diffracted waves moving at an angle θ. The blue wave will travel an additional λ compared to the red wave, so they will constructively interfere at the screen. Another bright spot
12. Diffraction –Path Difference Explained Notice that these 2 waves are IN PHASE. When they hit they screen they both hit at the same relative position, at the bottom of a crest. How much farther did the red wave have to travel? Exactly – ONE WAVELENGTH l The call this extra distance the PATH DIFFERENCE. The path difference and the ORDER of a fringe help to form a pattern.
13. Diffraction—Pattern Explained Now consider two diffracted waves moving at another angle θ. The blue wave will travel an additional 1/2λ compared to the red wave, so they will destructively interfere at the screen. A dark spot
14. Diffraction –Path Difference Explained Notice that these 2 waves are OUT OF PHASE. When they hit they screen they both hit at the different relative positions, one at the bottom of a crest and the other coming out of a trough, causing destructive interference. How much farther did the red wave have to travel? Exactly – ONE HALF OF A WAVELENGTH 0.5 l This PATH DIFFERENCE and the ORDER of a fringe help to form another pattern.
15. Path Difference Redux Simple geometry tells us that the path difference is equal to: d sinθ
16. Diffraction Calculations We already determined the path difference was λ for constructive interference and 1/2λ for destructive interference, so for all integers m: d sinθ=mλd sinθ=(m+ ½)λ
17. Diffraction Calculations Suppose we have 2 slits separated by a distance, d and a distance L from a screen. Let point P be a bright fringe. We can find the angle q using tanq= y/L This can be used in our calculations or if q is small enough sinq≈tanq≈ q But q must be in radians! P y q d B.C. L
18. Double Slit Example A viewing screen is separated from a double slit source by 1.2 m. The distance between the two slits is 0.030 mm. The second -order bright fringe ( m=2) is 4.5 cm from the central maximum. Determine the wavelength of light.
19. Diffraction Summary The diffraction pattern gives a nice smooth interference pattern between maxima and minima
20. Visible Spectrum and Dispersion Intensity of light is related to the energy carried (proportional to A2). Color of light is related to f and λ (recall c=fλ) Visible light ranges from 400 nm (below is UV) to 750 nm above is (IR) Blue/UV have higher fthan red/IR.
21. Visible Spectrum and Dispersion A prism separates light into a rainbow because the index of refraction for a material depends on λ.
22. Visible Spectrum and Dispersion Another nice benefit is the rainbow, which occurs because of dispersion inside a water drop (so how special are double rainbows!?)
23. Diffraction by a Single Slit Double slit interference seems reasonable, but diffraction patterns caused by single slits are less intuitive…but its effects were predicted by Poisson
24. Diffraction by a Single Slit The diffraction of waves will cause bright(ish) bands as the angle of diffraction increases so PD is 3/2, 5/2, … longer for the top rays. The central maxima is brightest by far.
25. Diffraction by a Single Slit The dark bands (minima) are found by: Dsinθ=mλ Where D is the slit width and θ is one half the angular width of the central maxima
26. Single Slit Example P. 22, pg, 693 How wide is the central diffraction peak on a screen 2.30 m behind a 0.0348 mm-wide slit illuminated by 589 nm light?
27. Diffraction Gratings Diffraction gratings have many slits—light will constructively interfere at an angle θ that makes a PD of Δl=mλ. This creates an interference pattern maxima given by: dsinθ=mλ Maxima are sharper and narrower than with double slit.
28. Thin Film Interference Another way interference can be seen is in thin films (must be order of few λ thick)—light reflected off top and bottom layers will interfere
29. Thin Film Interference Reflected light wave exhibit the same behavior as mechanical waves—they are inverted when travelling from faster to slower media (n1< n2), upright when n2< n1 This is key to understanding the interference
30. Constructive Thin Film Interference If the thickness of the film is 1/4λ then PD is 1/2λ which constructively interferes with the reflected ray
31. Thin Film Interference Generally, for constructive interference to occur film thickness, t, needs to be: 2t=(m + ½)λn or2nt=(m + ½)λ Destructive interference occurs at thickness: 2t=(m)λn or 2nt=(m)λ Thickness= t
32. Thin Film Interference Problem solving strategy: • Identify the film causing interference and the indices of refraction for the two media • Select appropriate equation to use for constructive interference • Remember to use λn=λ/n • Calculate thickness or wavelength as needed
33. Thin Film Interference--Example P 41 pg. 693 What is the smallest thickness of soap film (n=1.42) that would appear black if illuminated with 480 nm light?
34. Newton’s Rings A similar effect can be seen for a curved piece of glass placed on a flat one—these are called Newton’s Rings
35. Thin Wedge Interference Likewise, two plates of glass can be wedged open by a small diameter object and the wedge of air will cause an interference pattern
36. Nonreflective Coating Thin films can be used to reduce reflection by creating destructive interference—useful for optical instruments Often designed to eliminate middle of visible spectrum; 550nm Makes lenses look “purplish”
37. Polarization The electric field of unpolarized light vibrates in all directions perpendicular to the direction of motion—polarizing filters eliminated all but one direction of vibration, thus reducing intensity of the light
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Case Based Questions (MCQ)
Class 10
Chapter 10 Class 10 - Light - Reflection and Refraction
## A student focuses the image of a candle flame, placed at about 2 m from a convex lens of focal length 10 cm, on a screen. After that he gradually moves the flame towards the lens and each time focuses its image on the screen.
### (iv) away from flame
In a convex lens
As the object and lens come closer (‘u’ decreases) , the image distance (‘v’) increases.
Let us check
• Moving lens away from screen
• Choosing a point much closer to the lens, let’s say 30 cm (beyond 2F), we see the image move closer to the screen, but still not properly focused.
• Now, we move the lens further away from the screen (close to the objects), and away from the screen (let’s say at 15 cm, b/w C and F), we can see the proper image forming on the screen (beyond 2F)
• To move the object closer to the lens, the lens must be moved away from the screen so the image can be focused.
• Moving towards the screen, the image to form closer to the lens, resulting in blurring
• If we do not move , we will see no change in the image.
• Moving the lens away from the flame takes us closer to the screen (same as option (ii))
Thus, the correct answer is (i) away from screen
### (iv) size will become too small
Ans.
In a convex lens
• As the object moves towards the lens, the size of the image increases
• until the object crosses the focus.
• Between F and P, the image becomes virtual.
Thus, the correct answer is (ii) size of image will increase
### (iv) the flame disappears
Ans.
Intensity is the amount of energy per unit area per unit time .
Thus
• Intensity is inversely proportional to area
• As we move closer, the size of the image increases
• As size increases, area increases
• Thus, intensity decreases
Thus, the correct answer is (iii) intensity of flame reduces
### (iv)no image
Ans.
Now,
• Given focal length of convex lens is 10 cm
• Thus, 5 cm means that object is between F and P.
• Convex lens produces a virtual image when object is between F and P .
• Hence, this image cannot be seen on screen.
Thus the correct answer is (iv) no image
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MAE102_QUIZ_SOL
# MAE102_QUIZ_SOL - Name MAE 102 WINTER 2008 Quiz 1 A plane P...
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Unformatted text preview: Name MAE 102 WINTER 2008 Quiz 1. A plane, P, flys in a Vertical circle with a radius of 1500m. The plane’s forward velocity is 500 km/hr. At the instant shown, r = 1 km and the plane is at an altitude of 500 m. Express the plane’s velocity using both Cartesian (x-y) and Polar (r-9) coordinates. If the plane maintains a constant speed, find F and 65. - m/ SideView V: gown)“ ’ INN s all P ~ ~r1d 5:700 ) 0-51 4 (7:30” \/ a W fl vx "' r" ' W7: ow"; (on/+380”; V: V/osfiff VS;.,,{5;‘ :’,qbq‘%kaY—;:—”f‘-————‘j 7 “MN 4 : max”): 1‘ + arm'wsj m Y 3 VHS/wag, ~ Vs.n/V‘#)fzr : H“‘a'31”:z:7;7:271”W 2 "(3 o.é35);”“EIMT'EV. m V: 5» Vr ~ ’" ‘ V/Us/wrs) => Zr : Maw/MA, = triflzr/s I V0 T Y‘V 3 ‘49.}; “"/g :3 2 ' 0- 0‘” "// ( I L‘ W“- /\/I L )3 ram) ) r ~~~~~~~~ NW 0 3 \_ K- ‘- I M 1 V’ 33 , Wow. 3 6’ 6 /5 \Mmm _______ VW~M'«’f*1_”WTTT‘-AWT "‘y", 2. What must 9 be so that the velocity of A relative to B (VA , B) is purely in the x direction? What is V A , B ? If B accelerates in the direction of its velocity by 5 "g; , how much must A accelerate in the direction of its velocity in order to have A A , B be purely in the x direction? What is A A ,8 45‘ YA =1 5103!? T + 6S’sinflfjwynr Q A \) ‘24, r Q‘A/fl =3 Vfc‘Hr :q/uafnn m m +, 33%;. 3. The Sliders A and B are connected by a massless rigid bar of length l m and move with negligible friction in their slots, which lie in a vertical plane. At the instant shown, the velocity ofB is VB = 0.5% and XB = 0.8m. The mass ofA is 2 kg and the mass of B is 5 kg. Find the Tension in the rod and the accelerations of A and B. im 0 V XD-gm ‘ A X3 : "VG 1‘04”): in111‘ X82 :) Xfl‘o-W" {5: @0‘0 3 SS.’3° /0W\$‘lroini 91,: 12 : (A? + xv: )\$++;M(A41H/QLV‘L 1) 0 : ; X‘A x19 + yxfi X's mil; V} _ d -'z ~~ . x A "Robin't —) 0-xfl+xflxfl+*;+*8x8 £6 = “(gnu i31+xa3<~A) Re @3; 4 M Q) \ Ne, Z FxA: "Trusf5 'Mchtusz’ : MA XA A‘ 71' v {2 o"\\\ 37m (“Ag "‘ H W (D, @I 8) (’q’uuifoms XA’ £6 T UWKnUWnS :3 ‘Hu bur )S in («mph/55in” 25w "mam n N AHMMHW MHW‘} Fur Fthirs VP’I(}+y W4] A([{14u1+‘l\n Rehhomshps USIn9 cham Vflo‘h’y /A((( If”; tfim 'PvrAu/n 3 51v, 1 Y“ * \l's/A [M 3 GA (us(3 3, fan. SIMB£¢ gr “a‘lusfi: D‘Arosfi +037} ‘las’s _ as 3 GA ...
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# Simplify 3sqrt(125/36)?
$\frac{5 \sqrt{5}}{2}$
#### Explanation:
I'm reading this as $3 \sqrt{\frac{125}{36}}$
First thing to do is see that we can rewrite the terms under the square root sign into terms that have perfect squares:
$3 \sqrt{\frac{5 \times 25}{36}}$
$3 \left(\frac{\sqrt{5} \times \sqrt{25}}{\sqrt{36}}\right)$
Let's take square roots of the perfect squares:
$3 \times \frac{\sqrt{5} \times 5}{6}$
$\frac{3 \times \sqrt{5} \times 5}{6}$
Let's cancel the 6 in the denominator against the 3 in the numerator and reorder the terms:
$\frac{5 \sqrt{5}}{2}$
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# Octal to binary
Source: Internet
Author: User
Bit, also known as bit
Byte 1B = 8B minimum unit of memory storage
Length: The number of binary digits that can be processed by a computer within the same time period.
The word length determines the computing precision of the computer. The longer the word length, the higher the computing precision of the computer. Therefore, high-performance computers have long characters,
For computers with poor performance, the word length is relatively shorter.
Secondly, the word length determines the direct addressing capability of commands. Generally, the font length of a machine is 1, 2, 4, or 8 times that of a byte. The word length of a microcomputer is 8-bit and 16-bit.
, 32-bit, 64-bit, for example, 286 machines are 16-bit machines, 386 and 486 are 32-bit machines, and the latest piII is a 64-bit high-end machine.
The word length also affects the operation speed of the machine. The longer the word length, the faster the operation speed.
Word: is the basic unit for processing data or information in a computer. A word consists of several bytes. It is usually a word that is grouped into a single digit.
Long.
An octal digit can be expressed by three binary numbers. A hexadecimal number can be expressed by four binary numbers,
The hexadecimal conversion is very simple (001/010/011/100/101/110/111) (0001/0010/0011/0100/0101/0110/0111/1000/
/1001/1010 A/1011b/1100c/1101d/1110e/1111f)
For example, convert (1010111.01101) 2 to the octal number.
1010111.01101 = 001 010 111. 011 010
Too many other users
1 2 7 3 2
So (1010111.011.1) 2 = (127.32) 8
Convert (327.5) 8 to binary
011 010 111. 101
So (327.5) 8 = (11010111.101) 2
Convert (110111101.011101) 2 to hexadecimal number
(110111101.011101) 2 = 0001 1011 1101. 0111 0100
Too many other users
1 B D 7 4
So (110111101.011101) 2 = (1bd. 74) 16
Convert (27.fc) 16 to binary
2. f c
0010 0111 1111 1100
So (27.fc) 16 = (100111.111111) 2
Binary representation
Original code: Each digit represents a symbol.
Negative code: positive number is the same as the original code. Negative number is the opposite of other digits except the symbol.
Complement: the positive number is the same as the original code. The negative number is obtained by the reverse code + 1 except the symbol.
The address bus width determines the physical address space that the CPU can access. Simply put, it means that the CPU can use a large amount of memory.
Eight-bit address bus: an eight-bit binary number can represent up to eight power data of 2, from 00000000 to 11111111, in decimal format: 0-255.
In this example, the maximum address that can be distinguished by an 8-bit address bus is from 0 to 255. We say his addressing capability is 256 bytes, that is, 256 bytes.
20 bits: 1 m
32-bit: 4G
The above is the physical memory that can be accessed by different address bus. Note: During computing, for example, the addressing capability of the 16-bit address bus is not a binary consisting of 16 cores.
The result of the number, but add 1, because there is a 00000000000000000
That is, the 16 power of 2, and the binary number of 16 1 is reduced by 1 to the 16 power of 2
Others:
Convert decimal to binary:
The result is 1 after 2 rounds.
Write the remainder and the last 1 in descending order as the result.
For example, 302
302/2 = 151 + 0
151/2 = 75 + 1
75/2 = 37 + 1
37/2 = 18 + 1
18/2 = 9 + 0
9/2 = 4 + 1
4/2 = 2 + 0
2/2 = 1 + 0
The binary value is 100101110.
Convert binary to decimal
Start from the last digit and column it as 0th, 1, 2...
The number (0 or 1) of the nth digit multiplied by the Npower of 2
Example: 01101011. Convert to decimal:
0th bits: 1 multiplied by 2 to the power of 0 = 1
1 to the power of 2 = 2
0 multiplied by the 2's power = 0
1 multiplied by the power of 2 = 8
0 multiplied by the power of 2 = 0
1 multiplied by the power of 2 = 32
1 multiplied by the power of 2 = 64
0 multiplied by the power of 2 = 0
Then: 1 + 2 + 0
+ 8 + 0 + 32 + 64 + 0 = 107.
Binary 01101011 = decimal 107.
1. Convert binary to decimal
The basic practice of converting a binary number to a decimal number is to write the binary number as the expansion type of the weighting coefficient, and then sum the values according to the decimal addition Rules.
. This method is called "adding weights.
Ii. Convert decimal number to binary number
When converting a decimal number to a binary number, because the integer and decimal number conversion methods are different, the integer and decimal part of the decimal number are first divided
After conversion, merge them.
1. convert a decimal integer to a binary integer
To convert a decimal integer to a binary integer, use the "Division 2 remainder, reverse order" method. The specific method is: remove the decimal integer with 2 to get
The quotient and the remainder. Then, remove the quotient with 2, and then get a quotient and the remainder until the quotient is zero, and then take the first remainder
The lower-level valid bits of the binary number, and the remainder obtained as the upper-level valid bits of the binary number are arranged in sequence.
2. Convert decimal to binary decimal
To convert decimal places into binary decimal places, use the "take 2 as an integer and arrange them in sequence" method. The specific method is to use 2 decimal places to obtain the product
The integer part of the product is taken out, and then 2 is used to multiply the remaining fractional part, and a product is obtained. Then, the integer part of the product is taken out until the product is in progress.
The decimal part of is zero or reaches the required precision.
Then, sort the retrieved integers in order. The first integer is used as the high valid bit of the binary decimal, And the last integer is used as the low valid bit.
Bit.
1. Binary and decimal conversion
(1) binary to decimal <br> method: "sum by right expansion"
Example:
(1011.01) 2 = (1x23 + 0x22 + 1x21 + 1x20 + 0x2-1 + 1X2-2) 10
= (8 + 0 + 2 + 1 + 0 + 0.25) 10
= (11.25) 10
(2) convert decimal to binary
· Convert a decimal integer to a binary number: "divide by 2 to return the remainder, and output in reverse order"
Example: (89) 10 = (1011001) 2
2 89
2 44 ...... 1
2 22 ...... 0
2 11 ...... 0
2 5 ...... 1
2 2 ...... 1
2 1 ...... 0
0 ...... 1
· Decimal to binary: "multiply by 2 to get an integer and output it in sequence"
Example:
(0.625) 10 = (0.101) 2
0.625
X 2
1.25
X 2
0.5
X 2
1.0
2. Conversion between octal and binary
For example, convert 37.416 of octal to binary:
37. 4 1 6
011 111. 100 001 110
That is, (37.416) 8 = (11111.10000111) 2
For example, convert binary 10110.0011 to octal:
0 1 0 1 1 0. 0 0 1 0 0
That is, (10110.011) 2 = (26.14) 8
3. hexadecimal and binary conversion <br> for example, convert hexadecimal number 5df. 9 to binary:
5 d F. 9
0101 1101 1111.1001
That is: (5df. 9) 16 = (10111011111.1001) 2
For example, convert binary 1100001.111 to hexadecimal:
0110 0001. 1110
6. e
That is, (1100001.111) 2 = (61.e) 16
Related Keywords:
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## All-in-one learning app
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In everyday life, one comes across various types of motions of objects, such as the trajectory of a basketball or a volleyball, etc. Such trajectories are in the shape of a ‘parabola’, a curve that can be modeled using a Quadratic Equation of single variable, and such equations of single variables are Quadratic Functions.
Quadratic equations are applied to a variety of practical problems, such as monitoring the path of a projectile, predicting a financial stock model, statistical mechanics, and so on. The part of the equation ax2 is known as the quadratic part, bx as the linear part, and c as the constant part of the function.
## Graphing Quadratic Functions - the Parabola
The graph of every Quadratic function is called a parabola.
A parabola is a set of points equidistant from a point and a line.
where the point is called the Focus of a parabola and the line is known as the directrix. Another important point on the parabola is called the Vertex of the parabola. It is the point where the axis of symmetry of a parabola meets the parabola.
Here the axis of symmetry is an imaginary line and the function replicates itself on either side of the line. The graph of a parabola is like a mirror image of a curve, below is a diagram to illustrate this,
The graph of a quadratic equation - StudySmarter Originals
Here is what the graph of a quadratic function looks like, which is the quadratic function. It can be seen that the curve on the right side of the blue line and the other side of that line, are exactly the same. In mathematical terms, we say that the graph is symmetrical along that blue line. That is why that line is called the axis of symmetry. It is important to note that the axis of symmetry is an imaginary axis, it is not a part of the graph plotted.
Graph of a Parabola - StudySmarter Originals
It can be seen that the axis of symmetry is parallel to the y-axis and so we say that the parabola is symmetrical to the y-axis. And the point where the parabola meets the axis of symmetry is known as the Vertex of the parabola. It is also the minima of the function. In other words, a vertex is a point where the value of the quadratic function is minimum, hence the name, minima. In the above diagram, point A is the vertex of the parabola.
And for the parabola , the axis of symmetry turns out to be which is symmetrical to y-axis.
There is another crucial point on the parabola, which is the y-intercept of the parabola. It is the point where the parabola meets the y-axis, i.e. where it intercepts the y-axis. Hence, the word, y-intercept. In the above diagram, point C is the y-intercept of the parabola. To find out the coordinates of C, all we need to do is calculate y at x=0. We get,
which gives y=c. Hence, the coordinates of C are (0,c).
We can write quadratic function equations in 3 different forms. Let's look at them in more detail
There are three commonly used forms of quadratic functions.
• Standard or General Form:
• Factored or Intercept Form:
• Vertex Form:
Each of these forms can be used to determine different information about the path of a projectile. Understanding the benefits of each form of a quadratic function will be useful for analyzing different situations that come your way.
As the name suggests, the general form is what most quadratic functions are in. The intercept form is useful to easily read off the x and y intercepts of the given curve. The vertex form is especially used when the vertex of the curve has to be read off and determine the related properties.
### Standard Form of a Quadratic Function
Quadratic equations in one variable are equations that can be expressed in the form
This is the shape of a parabola, as seen in the image below.
Graph of a standard parabola - StudySmarter originals
Essentially, these are the equations that have a degree more than Linear equations. Linear equations have a degree of one and quadratic equations have a degree of 2. Here a, b, and c are constants where a≠0. If a=0, then we would only have , which is a linear equation.
So the condition to form a quadratic equation should be that the coefficient of x2 should be non-zero. The other constants b and c can be zero as they won’t affect the degree of the equations.
### Vertex Form of a Quadratic Function
The general form of a quadratic may not be the most convenient form to work with, and so we have the Vertex form of a Quadratic Equation. As the name suggests, it is a form based upon the vertex of the parabola formed by the quadratic equation. The vertex is the most important point of a parabola, using which, we can construct the parabola.
The Vertex Form of a Quadratic Equation is given as follows:
where the vertex of the parabola lies at the point (h,k). This form is especially useful when we are given the coordinates of the vertex and are asked to find the equation of the parabola.
### Factored Form of a Quadratic Equation
The Factored Form of a Quadratic Equation is a form where the quadratic is factored into its linear factors. Just as we had the vertex form to identify the vertex of a parabola formed by the quadratic equation, the factored form is used to identify the intercepts of the parabola formed.
The Factored or Intercept Form of a Quadratic Equation is given as follows:
where the two x-intercepts are given by . This can be easily verified by setting y=0 and finding the roots of the quadratic equation. Alternatively, one can use the given x-intercepts and a point on the parabola to figure out the quadratic equation.
Which of the following are quadratic functions?
(i) (ii) (iii)
Solution:
Recognize the highest degree of each of the functions, if the highest degree is 2 then only it is a quadratic function.
(i)
It can be seen that the highest degree of this function is and it is trivial that and so it is NOT a quadratic function.
(ii)
It is clear that the highest degree of this function is 2 and hence it is a Quadratic function.
(iii)
One can see that the second term has a degree 2 but only the highest degree should be taken into consideration which is 3, and so it is NOT a quadratic function.
Quadratic functions are a generalized form of quadratic equations. When for the quadratic function defined earlier, for some real constant d then the equation formed is known as a Quadratic equation. In general form, a quadratic equation has the form,
where and where represents the set of real numbers. The solution of a quadratic equation is the value of x for which the equation is satisfied. In other words, the solution of a quadratic function is the value of x for which f(x)=0.
We already know that a linear equation has a unique solution, in the case of quadratic equations, there are always two solutions. The solutions need not be unique, they can be the same and solutions may even be complex. However, we will be looking at real solutions and not complex one.
The solutions are also called the zeros of a function. They should not be confused as they are the same thing. To find the zeros, we can simply solve the quadratic using the quadratic formula for zeros, and we get
For practice on how to solve quadratic equations, see our article on Solving quadratic equations and Graph and solve quadratic equations.
## The Inverse of a Quadratic Function
Given that a function is Bijective (Injective and Surjective), the inverse exists. For a quadratic function, which is bijective, the inverse of it can be easily calculated. Every inverse is related to the function as follows,
To find the inverse of , we first equate the RHS to y,
The aim is to solve the above quadratic equation in terms of x, i.e., solve for x and express x in terms of y. The above equation can be rearranged to get,
which is quadratic in x, and we can find its roots using the quadratic formula, which gives us,
which is the inverse of y,
Now replacing the variable y with x, we get the inverse in x
where b2+4ax > 4ac for real values of the function.
## Quadratic functions - Key takeaways
• A quadratic function is a function whose highest power is 2. That is the highest degree of the equation is 2.
• The graph of a quadratic function is called a parabola, with a parent equation of .
• The solutions (zeros or roots) of a quadratic equation can be calculated using the quadratic formula or factoring the equation into its linear factors.
• Each quadratic equation has two zeros (they need not be unique). They can be real or imaginary.
• The graph of a quadratic function is a parabola that can have its axis of symmetry on the y-axis or x-axis.
• A parabola is defined as the set of points equidistant from a point and a line.
• One can find the axis of symmetry and the coordinates of the vertex by setting the y=0 and x=0 respectively.
A quadratic function is a function whose highest power is 2. That is the highest degree of the equation is 2.
Calculate the x-coordinate of the vertex for the quadratic equation y=ax2+bx+c using the formula -b/2a, then substitute this value of x in the original quadratic equation to get the value of y coordinate of the vertex.
The zeros of the quadratic equation y=ax2+bx+c can be found by plugging y=0 in the equation. That is ax2+bx+c=0
Linear and quadratic functions can be solved by plotting graphs. The solution for them would be the point of intersection of both graphs.
The factored form of the quadratic equation is y=a(bx+c)(dx+e)
Question
True or False: The equation -3x²-5x-6=0 is a quadratic equation.
True.
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Question
True/False: x²-3x=-5 is a quadratic equation in standard form.
False.
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Question
Given the equation x²-x-6=0, find the y-intercept
-6
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Question
The solution set to the equation x²-4x-5=0 is _____
{-1, 5}
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The vertex of the parabola from the function f(x)=x²-4x+5 is _____
(2, 1)
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A function that can be written in the form f(x)=ax²+bx+c for real numbers a,b and c, with a≠0 is a _____
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If the discriminant of an equation is 49, what type of solution will it have?
Two rational solutions
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Question
What is the discriminant for the equation 4x²+x+1=0?
-15
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Question
(x-1)(x+1)=0 is known as _____
The zero-factor property
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Question
Which equation has the roots (3,9)?
x²-12x+27=0
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a general formula for solving any quadratic equation.
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If a quadratic equation has only one rational solution what effect does this have on its discriminant?
The discriminant is 0.
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The solution to a quadratic equation are synonymous to the ____
roots, zeros, or the x-intercepts.
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What is the factor of the perfect square binomial x²-2x+1?
(x-1)²
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Question
If the highest degree of an equation is 3, is it a quadratic function?
No.
Show question
Question
No.
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What kind of curve is obtained by plotting a quadratic function?
Parabola
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Question
What is a Parabola?
A parabola is a set of points equidistant from a line and point.
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Question
How many zeros does a quadratic function has?
Two
Show question
Question
What are the conditions for a quadratic function (or any function in general) to have an inverse?
It should be Injective and Surjective.
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Question
What is the purpose of completing the square of a quadratic equation?
To simplify a quadratic equation and determine the maximum or minimum values
Show question
Question
What are the five steps to solving a quadratic equation by completing the squares?
1. Divide the expression by the coefficient of x2
2. Move the third term to the right-hand side
3. Complete the square and balance the equation
4. Take the square roots of both sides
5. Solve for x
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Question
If the coefficient of xis positive in a given quadratic equation, what kind of turning point do we obtain?
Minimum value
Show question
Question
If the coefficient of xis negative in a given quadratic equation, what kind of turning point do we obtain?
Maximum value
Show question
Question
What is the maximum value? What are other terms that describe this?
The maximum value is the highest point of the curve in a graph. This is also known as the maximum turning point or local maxima.
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Question
What is the minimum value? What are other terms that describe this?
The minimum value is the lowest point of the curve in a graph. This is also known as the minimum turning point or local minima.
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What is a vertex? Is there another word that can be used to describe this?
The vertex is a point at which the curve on a graph turns. This is also known as a turning point.
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Question
There are three forms for quadratic equations. What form is: y=ax² + bx +c?
Standard form
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Question
The U shape of the graph of a quadratic function is called a...
Parabola
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Question
For a quadratic equation with a positive term in x2, what do the coordinates (-ad,e) tell us about the graph?
The curve has a minimum value of e at x=-ad
Show question
Question
For a quadratic equation with a negative term in x2, what do the coordinates (ad,e) tell us about the graph?
The curve has a maximum value of e at x=ad
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Question
What is the lowest or highest point of a parabola called?
Vertex
Show question
Question
The highest point of a parabola is......
The maximum
Show question
Question
The lowest point of a parabola is......
Minimum
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Question
Can any quadratic equation be solved by completing the square?
Yes, all quadratic equation be solved by completing the square
Show question
Question
The line that splits the parabola in two is called.......
Axis of symmetry
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Question
The x-intercepts of a parabola have three other names. What are they?
Zeros, roots and solutions
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Question
In the quadratic equation y=ax²-bx+c, when "a" is positive means?
Upward parabola
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Question
In the quadratic equation y=ax2+bx-c when "a" is negative means?
Downward parabola
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Question
All real numbers are a domain of the quadratic function. True or false
True
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Question
The equation y=-(x-h)²+k is in .............. form.
Vertex
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Question
Find the vertex of the equation y=(x+5)²-24
(-5,-24)
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Question
Convert the equation y=x²+6x+2 into vertex form.
(x+3)²=7
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Find the x-intercepts of the factored form of the equation y=(x+1)(x-5)
(-1,0), (5,0)
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Question
From the factored quadratic equation form y=-(x-3)(x+5), find the values that represent a, b and c in standard form.
a=-1, b=-2 and c=15
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Question
What is the Quadratic Formula used for?
To find solutions of a given quadratic equation
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Question
How many solutions does the Quadratic Formula produce? What is the sign in the Quadratic Formula that gives that particular number of solutions?
Two solutions. The '±' sign indicates that there are two solutions when we apply the Quadratic Formula.
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How can the Quadratic Formula help us plot the graph of a given quadratic equation?
Since the Quadratic Formula determines the roots of a quadratic equation, we can locate them and plot the graph more accurately
Show question
Question
When can we use the Quadratic Formula to solve a given quadratic equation?
We can use this for quadratic equations that cannot be factored (however, we can indeed use it to solve any quadratic equation)
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What is another term used to describe a graph with one real root?
Repeated real root
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` there are many in progression who to know these pattern.can u give me all the pattern that i will come across.`
5 years ago
SHAIK AASIF AHAMED
74 Points
``` Hello student,There are basically 4 types of progressionsThey are arithmetic progression,If the first term of the Arithmetic Progression which is generally called as A.P. is ‘a’ and the common difference is ‘d’ then the nth term is given by tn = a+(n-1)d.A sequence of numbers <an> is said to be in arithmetic progression if the difference between the terms is constant i.e. tn – tn-1 is a constant for all n ∈ N. the constant difference between the terms is called the common difference and is denoted by‘d’.Similarly, the sum of first ‘n’ terms of the A.P. is given bySn= n/2 [2a + (n-1) d] = n/2 [a + l], where l is the last term of the A.P.Geometric progression,If the first term of the geometricprogression is ‘a’ and the common difference is ‘r’ then the nth term is given by tn = arn-1If every succesiveterm is a fixed multiple of the preceding term then the series is said to be a geometric series. A progression of the form a, ar, ar2, ….. is a geometric progression where the common multiple r is called the fixed ratio. it follows from the definition itself that no term of the G.P can be zero because in such a case the series will get reducedto a zero series.The sum Snof the first ‘n’ terms of the GP is given bySn= a(rn-1)/ (r-1) , if r ≠1 = na, if r =1.Harmonic progressionThe nth term of H.P. is given by the sequence a1, a2, …. an, where ai ≠ 0 for every i is said to be in harmonicprogression (H.P.) if the sequence 1/a1, 1/a2, …. ,1/an is in A.P.an=1 / [a+(n-1)d], where a = 1/a1and d = 1/a2-1/a1.Arithmetico geometric progression.Suppose a1, a2, a3, …. is an A.P. and b1, b2, b3, …… is a G.P. Then the sequence a1b1, a2b2, …, anbnis said to be an arithmetic-geometric progression. An arithmetic-geometric progression is of the form ab, (a+d)br, (a + 2d)br2, (a + 3d)br3, ……Its sum Snto n terms is given bySn= ab + (a+d)br + (a+2d)br2+……+ (a+(n–2)d)brn–2+ (a+(n–1)d)brn–1.Multiply both sides by r, so thatrSn= abr+(a+d)br2+…+(a+(n–3)d)brn–2+(a+(n–2)d)brn–1+(a+(n–1)d)brn.Subtracting we get(1 – r)Sn= ab + dbr + dbr2+…+ dbrn–2+ dbrn–1– (a+(n–1)d)brn. = ab + dbr(1–rn–1)/(1–r) (a+(n–1)d)brn ⇒ Sn= ab/1–r + dbr(1–rn–1)/(1–r)2 – (a+(n–1)d)brn/1–r.If –1 < r < 1, the sum of the infinite number of terms of the progression is limn→∞Sn= ab/1–r + dbr/(1–r)2.Thanks and RegardsShaik AasifaskIITians faculty
```
3 years ago
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how to add 3 matrices with one dimension same for all and other dimension different?
4 views (last 30 days)
summyia qamar on 24 Jan 2017
Commented: Niels on 24 Jan 2017
If I have
A=[1;
2;
3]
B=[2 3 4;
5 6 7;
8 9 0]
C=[3 4 5 6;
9 8 7 6;
5 4 2 1]
all 3 matrices have 3 rows but columns different. I want to add elements of rows of all matrices in this way for example I want
D=[(1+2+3) (1+2+4) (1+2+5) (1+2+6) (1+3+3) (1+3+4) (1+3+5) (1+3+6) (1+4+3) (1+4+4) (1+4+5) (1+4+6);
(2+5+9) (2+5+8) (2+5+7) (2+5+6) (2+6+9) (2+6+8) (2+6+7) (2+6+6) (2+7+9) (2+7+8) (2+7+7) (2+7+6);
(3+8+5) (3+8+4) (3+8+2) (3+8+1) (3+9+5) (3+9+4) (3+9+2) (3+9+1) (3+0+5) (3+0+4) (3+0+2) (3+0+1)]
2 CommentsShowHide 1 older comment
summyia qamar on 24 Jan 2017
to any kind..although my specific problem is
A=64x1
B=64x1806
C=64x4602
so I need a generalized version
Sign in to comment.
Accepted Answer
Niels on 24 Jan 2017
Edited: Niels on 24 Jan 2017
[rows,col_B]=size(B);
[~,col_C]=size(C);
result=zeros(rows,col_B*col_C);
for i=1:col_B
for j=1:col_C
result(:,(i-1)*col_C+j)=A+B(:,i)+C(:,j);
end
end
11 CommentsShowHide 10 older comments
Niels on 24 Jan 2017
the result will be that large, so even if the calculations are minimzed, the problem with the size of the result is still the same.
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More Answers (1)
John Chilleri on 24 Jan 2017
Edited: John Chilleri on 24 Jan 2017
Hello,
I only tested this with your example and it worked, but I think it's generally correct:
%
% Strangely adding matrices
%
function D = strangelyAddingMatrices(A,B,C)
n = size(C,2);
nB = size(B,2);
for i = 1:n
C(:,i) = C(:,i)+A;
end
D = [];
for i = 1:nB
C2 = C;
for j = 1:n
C2(:,j) = C2(:,j)+B(:,i);
end
D = [D C2];
end
end
Hope this helps!
9 CommentsShowHide 8 older comments
John Chilleri on 24 Jan 2017
What do you mean by minimum sum? Like which columns summed produce the minimum? Or something else?
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# Sampling
An airline was concerned about passengers arriving too late at the airport to allow for the additional security measures. The airline collected survey data from 1,000 passengers on their time from arrival at the airport to reaching the boarding gate. The sample mean was 1 hour and 20 minutes, with a sample standard deviation of 30 minutes. Based on this sample, how long prior to a flight should a passenger arrive at the airport to have a 95% probability of making it to the gate on time? A) Two hours, ten minutes. B) Two hours, thirty minutes. C) One hour, fifty minutes.
is it C
is it A ?? I used the z dist = (x- sample mean)/s.d the question says 95% prob … if its option A …then (130-80)/30 …(converted into mins) = 1.66 whats the right answer ?
Answer A/B… Explaination: ->> Sample mean = 1Hr 20 min = 80 min 80 +/- 1.96(30) = 138 min…
A. Here, we do not divide by the standard error, because we are interested in a point estimate of making our flight. The answer is one hour, twenty minutes + 1.65(30 minutes) = 2 hours, 10 minutes. (We use a one-tailed test because we are not concerned with passengers arriving too early, only arriving too late.)
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# WATCH: What Is The Largest Number?
Infinity plus one doesn't count.
BEC CREW
28 FEB 2016
Alright guys, we're about to crunch some serious numbers, so I hope you're ready. Sharkee on YouTube is going to help us figure out what the biggest number is, but he's got some conditions so things don't get too crazy. First off, the number must be useful in some way, and secondly, it can't just be an arbitrary number - you have to show how you got it. Thirdly, no, you can't just say, "A number plus one," because that's cheating. Oh yeah, and infinity doesn't count - the number must be finite.
Okay, now that we've got the rules out of the way, here we go. First up, we need to break one of those rules and pick an arbitrary number - if only to see if our non-arbitrary numbers are larger or smaller than it. We're starting off with the very impressive googol, which is 10100 (or if you're writing the actual number out, it's 1, followed by 100 zeros).
To illustrate how enormous a googol is, the video above explains that it's actually larger than the number of atoms in your body. Your mind isn't boggled yet? Okay then, a googol is larger than all the atoms ON EARTH. Still not cool enough? How about the fact that a googol is more than the all the atoms that make up the observable Universe (which is about 1080). Yep, a googol is really freaking big.
So what number is bigger than a googol, but is still useful in mathematics? To figure that out, Sharkee gets the help of a hypothetical Rubik's Cube with 1.95 x 10160 sides, which is definitely bigger than a googol.
But can we find a number that's bigger than a googolplex, which is 10 to the power of a googol, so 10 to the power of 10100? As the video explains, the number of different possible universes is smaller than a googolplex, but if we discount the limitation that we have to be about to observe these universes, then we can beat the googolplex.
If you thought a googolplex was big, how about a googolplexian - 10 to the power of a googolplex. Yep, Sharkee goes there, and while these numbers are near-impossible to wrap your head around, it's weirdly addictive stuff.
Watch the video above to find out what the "biggest, largest, meanest, and most terrifying number ever to be conceived" actually is, and then send out good thoughts to all those mathematians and philosophers out there whose job it is to know what that actually means. They're the real heroes.
H/T: Scientific American
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SPSS TUTORIALS BASICS DATA ANALYSIS REGRESSION ANOVA T-TESTS
# SPSS Wilcoxon Signed-Ranks Test – Simple Example
For comparing two metric variables measured on one group of cases, our first choice is the paired-samples t-test. This requires the difference scores to be normally distributed in our population. If this assumption isn't met, we can use Wilcoxon S-R test instead. It can also be used on ordinal variables -although ties may be a real issue for Likert items.
Don't abbreviate “Wilcoxon S-R test” to simply “Wilcoxon test” like SPSS does: there's a second “Wilcoxon test” which is also known as the Mann-Whitney test for two independent samples.
## Wilcoxon Signed-Ranks Test - How It Basically Works
1. For each case calculate the difference between score_1 and score_2. Ties (cases whose two values are equal) are excluded from this test altogether.
2. Calculate the absolute difference for each case.
3. Rank the absolute differences over cases. Use mean ranks for ties (different cases with equal absolute difference scores).
4. Create signed ranks by applying the signs (plus or minus) of the differences to the ranks.
5. Compute the test statistic Wilcoxon W+, which is the sum over positive signed ranks. If score_1 and score_2 really have similar population distributions, then W+ should be neither very small nor very large.
6. Calculate the p-value for W+ from its exact sampling distribution or approximate it by a standard normal distribution.
So much for the theory. We'll now run Wilcoxon S-R test in SPSS on some real world data.
## Adratings Data - Brief Description.
A car manufacturer had 18 respondents rate 3 different commercials for one of their cars. They first want to know which commercial is rated best by all respondents. These data -part of which are shown below- are in adratings.sav.
## Quick Data Check
Our current focus is limited to the 3 rating variables, ad1 through ad3. Let's first make sure we've an idea what they basically look like before carrying on. We'll inspect their histograms by running the syntax below.
## Basic Histograms Syntax
*Check if data look plausible with histograms.
/format notable
/histogram.
## Histograms - Results
First and foremost, our 3 histograms don't show any weird values or patterns so our data look credible and there's no need for specifying any user missing values.
Let's also take a look at the descriptive statistics in our histograms. Each variable has n = 18 respondents so there aren't any missing values at all. Note that ad2 (the “Youngster car commercial”) has a very low average rating of only 55. It's decided to drop this commercial from the analysis and test if ad1 and ad3 have equal mean ratings.
## Difference Scores
Let's now compute and inspect the difference scores between ad1 and ad3 with the syntax below.
*Compute difference scores.
*Inspect histogram difference scores for normality.
frequencies diff
/format notable
/histogram normal.
## Result
Our first choice for comparing these variables would be a paired samples t-test. This requires the difference scores to be normally distributed in our population but our sample suggests otherwise. This isn't a problem for larger samples sizes (say, n > 25) but we've only 18 respondents in our data.For larger sample sizes, the central limit theorem ensures that the sampling distribution of the mean will be normal, regardless of the population distribution of a variable. Fortunately, Wilcoxon S-R test was developed for precisely this scenario: not meeting the assumptions of a paired-samples t-test. Only now can we really formulate our null hypothesis: the population distributions for ad1 and ad3 are identical. If this is true, then these distributions will be slightly different in a small sample like our data at hand. However, if our sample shows very different distributions, then our hypothesis of equal population distributions will no longer be tenable.
## Wilcoxon S-R test in SPSS - Menu
Now that we've a basic idea what our data look like, let's run our test. The screenshots below guide you through.
2 Related Samples refers to comparing 2 variables measured on the same respondents. This is similar to “paired samples” or “within-subjects” effects in repeated measures ANOVA.
Optionally, reverse the variable order so you have the highest scores (ad1 in our data) under Variable2.
“Wilcoxon” refers to Wilcoxon S-R test here. This is a different test than Wilcoxon independent samples test (also know as Mann-Whitney test).
may or may not be present, depending on your SPSS license. If you do have it, we propose you fill it out as below.
## Wilcoxon S-R test in SPSS - Syntax
Following these steps results in the syntax below (you'll have one extra line if you requested exact statistics).
*Wilcoxon Signed Ranks Test Syntax.
NPAR TESTS
/MISSING ANALYSIS.
## Wilcoxon S-R Test - Ranks Table Ouput
Let's first stare at this table and its footnotes for a minute and decipher what it really says. Right. Now, if ad1 and ad3 have similar population distributions, then the signs (plus and minus) should be distributed roughly evenly over ranks. If you find this hard to grasp -like most people- take another look at this diagram.
This implies that the sum of positive ranks should be close to the sum of negative ranks. This number (159 in our example) is our test statistic and known as Wilcoxon W+.
Our table shows a very different pattern: the sum of positive ranks (indicating that the “Family car” was rated better) is way larger than the sum of negative ranks. Can we still believe our 2 commerials are rated similarly?
## Wilcoxon S-R Test - Test Statistics Ouput
Oddly, our ”Test Statistics“ table includes everything except for our actual test statistic, the aforementioned W+.
We prefer reporting Exact Sig. (2-tailed). Its value of 0.001 means that the probability is roughly 1 in 1,000 of finding the large sample difference we did if our variables really have similar population distributions.
If our output doesn't include the exact p-value, we'll report Asymp. Sig. (2-tailed) instead, which is also 0.001. This approximate p-value is based on the standard normal distribution (hence the “Z” right on top of it).“Asymp” is short for asymptotic. It means that as the sample size approaches infinity, the sampling distribution of W+ becomes identical to a normal distribution. Or more practically: this normal approximation is more accurate for larger sample sizes.
It's comforting to see that both p-values are 0.001. Apparently, the normal approximation is accurate. However, if we increase the decimal places, we see that it's almost three times larger than the exact p-value.A nice tool for doing so is downloadable from Set Decimals for Output Tables Tool.
The reason for having two p-values is that the exact p-value can be computationally heavy, especially for larger sample sizes.
## How to Report Wilcoxon Signed-Ranks Test?
The official way for reporting these results is as follows: “A Wilcoxon Signed-Ranks test indicated that the “Family car” commercial (mean rank = 10.6) was rated more favorably than the “Youngster car” commercial (mean rank = 4.0), Z = -3.2, p = 0.001.” We think this guideline is poor for smaller sample sizes. In this case, the Z-approximation may be unnecessary and inaccurate and the exact p-value is to be preferred.
I hope this tutorial has been helpful in understanding and using Wilcoxon Signed-Ranks test in SPSS. Please let me know by leaving a comment below. Thanks!
# Tell us what you think!
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Anonymous
Anonymous asked in Science & MathematicsMathematics · 10 months ago
# Two hundred tickets are being sold for a school draw. What is your chance of winning with one ticket? Express your answer as a percent.?
Relevance
• 10 months ago
TotalTickets sold is 200
winning percentage is 1/200/100 = 0.00005%
• 10 months ago
1/200 x 100% = 0.5% chance to win 1 ticket..//
• Jim
Lv 7
10 months ago
I agree with Zirp: how many prizes are there???
Assuming only 1:
1/200 *(100%) = 5%
Notes: Notice I multiplied by 100%? 100% =1 so this is the correct way to convert to percentage!
• Zirp
Lv 7
10 months ago
how many prizes are there?
• David
Lv 7
10 months ago
Chance of winning: 1/200 times 100 = 0.5%
• RR
Lv 7
10 months ago
1 in 200
or
1/200
to convert to percentage
1/200 x 100
=100/200
=1/2
=0.5%
• 10 months ago
The chance of winning with one ticket is 0.5%.
• Anonymous
10 months ago
1 / 200
You do know how to convert a fraction to a percent I assume.
• Anonymous
10 months ago
What part of this is confusing you?
• 10 months ago
1/200 = 0.005 = 0.5%
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# The Lorenz Attractor ¶
We start with exploring Lorenz differential equations using Python through examples and look at how we create machine learning models.
Every book on chaos or non-linear dynamics contain the Lorenz differential equations, prints the unmistakable image of the attractor, or mentions the "butterfly effect". It has been intensely studied by scientist from many disciplines since its publication by Edward Lorenz in 1963. In it, he described a simplified mathematical model for atmospheric convection [1] . Nowadays, these equations are simply known as Lorenz attractor. However, for many years scientist have argued if Lorenz attractor was truly chaos or an artifact of exponential and explosive amplifications of numerical truncation errors. The existence of Lorenz attractor was finally settled by Tucker in 2002 [2]
Lorenz was a meteorologist and a mathematician in search of a model that was capable of generating long stretches of data statistically similar to atmospheric conditions. He set out to build a nonlinear model with an aperiodic solution that was simple enough for the computing power available at the time. After a long search, Lorenz came up with the now-famous set of equations [3] :
\begin{aligned} \dot{x} & = \sigma(y-x) \\ \dot{y} & = \rho x - y - xz \\ \dot{z} & = -\beta z + xy \end{aligned}
where $\sigma$, $\beta$, $\rho$ are real numbers.
Like many nonlinear equations, there isn't an analytical solution for Lorenz attractor. So, with practicality in mind, I will use numerical methods (more specifically discrete numerical methods with finite time-steps) to solve these equations and share some of the surprising features using the generated time-series. Besides, time-series are often the only observable data available to quantify the complicated dynamics in real life situations or experiments. So, let's jump right into it!
In [1]:
%matplotlib inline
import utils
from IPython.core.pylabtools import figsize
t, x_t = utils.solve_lorenz(10.0, 8./3, 28.0)
I will not get into what variables ($x$, $y$, $z$) and parameters ($sigma$, $rho$, $beta$) represent. However, for wide range of values of parameters numerical solutions of the equations above generate extremely complicated trajectories. The figure above, plotted in 3 dimensions, shows 30 random trajectories computed for $sigma = 10$, $rho = 2.\overline{6}$, and $beta = 28$. Notice that each trajectory exhibits aperiodic behavior, not cross itself. This is more apparent if we plot each variable separately.
In [2]:
import matplotlib.pyplot as plt
plt.style.use('bmh')
plt.figure(figsize=(12, 18))
x, y, z = x_t[20,:,:].T
ax1 = plt.subplot(311)
ax1.plot(z)
ax1.set_ylabel('$z$')
ax2 = plt.subplot(312, sharex=ax1)
ax2.plot(y)
ax2.set_ylabel('$y$')
ax3 = plt.subplot(313, sharex=ax1)
ax3.plot(x)
ax3.set_ylabel('$x$')
ax3.set_xlabel('$n$')
plt.show()
Creating histograms of the average positions (across different trajectories) show that, on average, the trajectories swirl about the attractors.
In [3]:
xyz_avg = x_t.mean(axis=1)
# xyz_avg.shape
plt.hist(xyz_avg[:,0], 20)
plt.title('Average $x(t)$');
In [4]:
plt.hist(xyz_avg[:,1], 20)
plt.title('Average $y(t)$');
Many physical, chemical, and biological processes in nature are described by a set of coupled first-order autonomous differential equations, or autonomous flows. A widely used technique in the study of these systems is the Poincaré surface-of-section technique. On a Poincaré surface of section, the dynamics can be described by a discrete map whose phase-space dimension is one less than that of the original continuous flow. This sectioning technique thus provides a natural link between continuous flows and discrete maps. With a tremendous facilitation in analysis, numerical computation, and visualization, maps also capture many fundamental dynamical proper- ties of flows. These advantages have made the Pioncaré surface-of-section technique one of the most popular analysis tools in nonlinear dynamics and chaos.
A remarkable tool that extracts the essence by throwing away information is a technique called the surface of section or Poincaré section. A surface of section is generated by looking at successive intersections of a trajectory or a set of trajectories with a plane in the phase space. Typically, the plane is spanned by a coordinate axis and the canonically conjugate momentum axis. We will see that surfaces of section made in this way have nice properties.
In [5]:
utils.plot_poincare_surface()
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# Birthday Paradox 2 pairs
I have followed the various birthday paradox posts. Can someone please assist with the logic for finding the probability of two pairs of people with the same birthday in a group of $23$. If $n=23$ gives us more than $0.5$ probability of one pair, what is the outcome for two pairs in the same population (assuming the second pair do not share the same birth date as the first pair)?
## 1 Answer
For the standard birthday problem, the number of possible sequences of $n$ birthdays that don't have two the same is $365\times364\times...\times(365-n+1)$ and so the probability that we don't have two the same after $n$ attempts is $\frac{365\times364\times...\times(365-n+1)}{365^n}=p_n$, which goes below $1/2$ at $n=23$.
The probability of either having all birthdays different or exactly one pair the same, all others different, or exactly three people the same, all others different is $$\frac{365\times...\times(365-n+1)+\binom n2365\times...\times(365-n+2)+\binom n3365\times...\times(365-n+3)}{365^n}\\=p_n\bigg(1+\binom n2\frac1{365-n+1}+\binom n3\frac1{(365-n+1)(365-n+2)}\bigg).$$ Thus we need to check when this is less than $1/2$. This first happens at $n=36$.
This assumes that you count any two pairs, including where both pairs have the same birthday. It would be more complicated to exclude this, and it shouldn't make very much difference, but as it happens the probability is actually very close to $1/2$ at $n=36$.
(edit: Actually, doing some more careful bounds, the probability of not having exactly one set of $2$, $3$ or $4$ the same, all others different at $n=36$ is $0.49944$. A simple bound on the probability of any configuration with $5$ or more people having the same birthday is the expected number of sets of $5$ people with the same birthday, and this is only $0.0000212$, not enough to tip the probability over $1/2$. So it is $n=36$ with either interpretation.)
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# Finding the generator matrix for x^3+x+1
27 visualizaciones (últimos 30 días)
Jonathan George el 8 de Dic. de 2022
Respondida: Sai el 30 de Dic. de 2022
Hello, I'm inexperienced with MATHLAB and just trying to handle the ropes.
I have managed to use
genpoly=[1 1 0 1];
[parmat,genmat]=cyclgen(7,genpoly,'nonsys')
[parmatsys,genmatsys]=cyclgen(7,genpoly)
to construct a generator matrix for the g( x ) =x^3+ x^2+1 polynominal with (7,4) cyclic code but I'm having difficulty using the same method to find the generator matrix for x^3+x+1. Could someone please help me out with an astute explanation so I may understand it better?
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### Respuestas (1)
Sai el 30 de Dic. de 2022
Hi Jonathan,
Let g(x) be generator polynomial.
If g(x) = x^3 + x + 1 = 1 + x + x^3 is a generator polynomial, then the corresponding generator polynomial vector(genpoly) would be [1 1 0 1] but not [1 0 1 1]. In that case, for g(x) = x^3 + x^2 + 1, then genpoly would be [1 0 1 1] but not [1 1 0 1]. ‘genpoly’ is obtained from g(x) considering g(x) is written in increasing powers of x.
The following code snippet helps you finding generator matrix for g(x) = x^3 + x^2 + 1 = 1 + x^2 + x^3
genpoly=[1 0 1 1];
[parmat,genmat]=cyclgen(7,genpoly,'nonsys')
[parmatsys,genmatsys]=cyclgen(7,genpoly,'system')
The following code snippet helps you finding generator matrix for g(x) = x^3 + x + 1 = 1 + x + x^3
genpoly=[1 1 0 1];
[parmat,genmat]=cyclgen(7,genpoly,'nonsys')
[parmatsys,genmatsys]=cyclgen(7,genpoly)
Hope the query is resolved
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# 2.5: Los números de 0—100
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
## 0-10
0…………………………………………………………………………………..…………...cero
1……………………………………………………………………………………...………...uno
2………………………………………………………………………………………………...dos
3………………………………………………………………………………………………...tres
4…………………………………………………………………………………………......cuatro
5……………………………………………………………………………………………...cinco
6……………………………………………………………………………………………......seis
7……………………………………………………………………………………………….siete
8…………………………………………………………………………………………….....ocho
9……………………………………………………………………………………………...nueve
10……………………………………………………………………………………………....diez
One through Nine are absolutely critical—master them!
## 11-15
11…………………………………………………………………………………..…………..once
12……………………………………………………………………………………..………..doce
13………………………………………………………………………………………..……..trece
14……………………………………………………………………………………….……catorce
15………………………………………………………………………………………….....quince
Eleven through Fifteen are tricky—spend extra time on them.
## 16-19
16…………………………………………………………………………………..…….....dieciséis*
17…………………………………………………………………………………………..diecisiete
18…………………………………………………………………………………………..dieciocho
19……………………………………………………………………………………….…diecinueve
*When a single-digit one-through-nine number (2, 3, 6) is attached to another, it takes an accent.
## 20-29
20…………………………………….…veinte 25………………………………...…veinticinco
21……………………………..……veintiuno** 26………………………………….…veintiséis
22…………………………………….veintidós 27………………………….……...…veintisiete
23…………………………………….veintitrés 28…………………….………......…veintiocho
24………………………………….veinticuatro 29……………………………......…veintinueve
**When a masculine noun immediate follows “uno,” the –o drops and the u take an accent.
- Tengo veintiún años.
The –e in veinte changes to an “i” when connecting a 1-9 number. This mimics the “and” sound: “y.”
## 30-39
30…………………………………….…treinta 35………………………………...treinta y cinco
31……………………………...treinta y uno** 36………………………………….treinta y seis
32………………………………...treinta y dos 37………………………………….treinta y siete
33………………………………...treinta y tres 38………………………………...treinta y ocho
34………………………………treinta y cuatro 39……………………………….treinta y nueve
Once we get to “treinta,” things get much easier: Thirty and one, thirty and two…. Thirty and six... etc.
## 40-100
40…………………….……………….cuarenta 70…………………….…………....……setenta
50…………………….………………cincuenta 80…………………….………..………ochenta
60…………………….……………...…sesenta 90…………………….…………..……noventa
100………………………………………………………………………………………..………cien
That pattern continues for the rest of the “tens”
##### Note
“Cien” works for 100 “on the dot” only. We will see larger numbers later, but it is worth noting that after 100 (101, etc), “hundred” is “ciento.”
Los números de teléfono. Phone numbers. The ways people will say their phone numbers vary throughout the Spanish speaking world, but a common way is to state the first number as a single digit, and then state three double-digit numbers for the rest.
Por ejemplo: 5-33-89-79. Cinco, treinta y tres, ochenta y nueve, setenta y nueve.
Mi teléfono es ________
Next, ask for the phone numbers of two potential study buddies using the question:
¿Cuál es tu teléfono?
________
________
Las sumas y restas: la frase secreta. Addition and subtraction: the secret phrase. Solve the addition and subtraction problems below, then use the key to solve the secret phrase!
1. Setenta menos sesenta son ________
2. Tres más catorce son ________
3. Once menos uno son ________
4. Trece más cuatro son ________
5. Ochenta más diez son ________
6. Treinta menos dos son ________
7. Noventa menos diecisiete son ________
8. Veinte más veintitrés son ________
9. Cincuenta menos treinta y uno son ________
10. Doce más tres son ________
Clave: key
Setenta y tres = A
Diez = L
Veintiocho = R
Noventa = G
Diecisiete = O
Quince = E
Diecinueve = T
Cuarenta y tres = S
La frase secreta es: ¡________ ________ ________ ________ ________ ________ ________ ________ ________ ________ !
1 2 3 4 5 6 7 8 9 10
De compras. Shopping. You just won a \$100 dollar gift card! Use the information below to determine the price of each item.
Tengo
I have
Compro
El precio
The price
Preguntas personales. Answer the following questions as completely as possible—be sure to write out the numbers. If you do not have something, put “no” in front of the verb. Ej. No tengo mascotas. Audio
1. ¿Cuántas clases tienes?
Tengo ________
2. ¿Cuántas horas trabajas?
Trabajo ________
3. ¿Cuántos créditos tienes?
Tengo ________
4. ¿Cuántos primos* tienes? *cousins
Tengo ________
5. ¿Cuántos hermanos tienes?
Tengo ________
6. ¿Cuántas mascotas* tienes? *pets
Tengo ________
Yo tengo tres mascotas: una gata (Miel) y dos perros (un perro, Kermit, y una perra, Harriet).
This page titled 2.5: Los números de 0—100 is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Erin Huebener (Independent) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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# Significant Figures Worksheet
```Significant Figures Worksheet
How many significant figures are in each of the following numbers?
1)
5.40 ____
6)
1.2 x 103 ____
2)
210 ____
7)
0.00120 ____
3)
801.5 ____
8)
0.0102 ____
4)
1,000 ____
9)
9.010 x 10-6 ____
5)
101.0100 ____
10)
2,370.0 ____
11)
Why are significant figures important when taking data in the laboratory?
12)
Why are significant figures NOT important when solving problems in your
math class?
13)
Using two different instruments, I measured the length of my foot to be 27
centimeters and 27.00 centimeters. Explain the difference between these
two measurements.
14)
I can lift a 20 kilogram weight over my head ten times before I get tired.
Write this measurement to the correct number of significant figures.
For chemistry help, visit www.chemfiesta.com
How many significant figures are in each of the following numbers?
1)
5.40
3
6)
1.2 x 103
2
2)
210
2
7)
0.00120
3
3)
801.5
4
8)
0.0102
3
4)
1,000
1
9)
9.010 x 10-6 4
5)
101.0100
7
10)
2,370.0
11)
Why are significant figures important when taking data in the laboratory?
Significant figures indicate the precision of the measured value to
anybody who looks at the data. For example, if a weight is measured
as being “1100 grams”, this means that the mass has been rounded
to the nearest hundred grams. If a weight is measured as being
“1100.0 grams”, this means that the mass has been rounded to the
nearest tenth of a gram. Though the numbers plug into the
calculator in exactly the same way, they mean very different things.
12)
Why are significant figures NOT important when solving problems in your
math class?
Math classes don’t deal with measured values. As a result, all of the
numbers are considered to be infinitely precise.
13)
Using two different instruments, I measured the length of my foot to be 27
centimeters and 27.00 centimeters. Explain the difference between these
two measurements.
As in problem 11, the first measurement implies that my foot is
somewhere between 26.5 and 27.4 cm long. The second
measurement implies that my foot is between 26.995 and 27.004 cm
long. Again, though the numbers plug into the calculator in the
same way, they imply different precisions.
14)
I can lift a 20 kilogram weight over my head ten times before I get tired.
Write this measurement to the correct number of significant figures.
The answer is written as 20.0, with a line drawn above the zero in the
tenths place. This is one of the few cases where you can measure
data with infinite significant figures. After all, I can either lift it or I
can’t – there’s no “half-lift” that would result in a decimal.
For chemistry help, visit www.chemfiesta.com
5
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kmmiles.com
# 1538.9 miles in km
## Result
1538.9 miles equals 2476.0901 km
You can also convert 1538.9 km to miles.
## Conversion formula
Multiply the amount of miles by the conversion factor to get the result in km:
1538.9 mi × 1.609 = 2476.0901 km
## How to convert 1538.9 miles to km?
The conversion factor from miles to km is 1.609, which means that 1 miles is equal to 1.609 km:
1 mi = 1.609 km
To convert 1538.9 miles into km we have to multiply 1538.9 by the conversion factor in order to get the amount from miles to km. We can also form a proportion to calculate the result:
1 mi → 1.609 km
1538.9 mi → L(km)
Solve the above proportion to obtain the length L in km:
L(km) = 1538.9 mi × 1.609 km
L(km) = 2476.0901 km
The final result is:
1538.9 mi → 2476.0901 km
We conclude that 1538.9 miles is equivalent to 2476.0901 km:
1538.9 miles = 2476.0901 km
## Result approximation
For practical purposes we can round our final result to an approximate numerical value. In this case one thousand five hundred thirty-eight point nine miles is approximately two thousand four hundred seventy-six point zero nine km:
1538.9 miles ≅ 2476.09 km
## Conversion table
For quick reference purposes, below is the miles to kilometers conversion table:
miles (mi) kilometers (km)
1539.9 miles 2477.6991 km
1540.9 miles 2479.3081 km
1541.9 miles 2480.9171 km
1542.9 miles 2482.5261 km
1543.9 miles 2484.1351 km
1544.9 miles 2485.7441 km
1545.9 miles 2487.3531 km
1546.9 miles 2488.9621 km
1547.9 miles 2490.5711 km
1548.9 miles 2492.1801 km
## Units definitions
The units involved in this conversion are miles and kilometers. This is how they are defined:
### Miles
A mile is a most popular measurement unit of length, equal to most commonly 5,280 feet (1,760 yards, or about 1,609 meters). The mile of 5,280 feet is called land mile or the statute mile to distinguish it from the nautical mile (1,852 meters, about 6,076.1 feet). Use of the mile as a unit of measurement is now largely confined to the United Kingdom, the United States, and Canada.
### Kilometers
The kilometer (symbol: km) is a unit of length in the metric system, equal to 1000m (also written as 1E+3m). It is commonly used officially for expressing distances between geographical places on land in most of the world.
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Associated Topics || Dr. Math Home || Search Dr. Math
### Number/Color Cube
```
Date: 09/13/2001 at 07:50:45
From: bill welden
Subject: Number/color cube
You have a cube that you want to make into a number cube by putting
the numbers 1,2,3,4,5,6 on the faces. The only restriction is that
you put 1 and 5, 3 and 6, 2 and 4 on opposite faces. Each face is a
different color (red, orange, yellow, green, blue, purple). How many
ways can you make the cube?
Answer I'm trying to solve: 6 sides to a cube, 6 different colors,
3 different number combinations, 2 different numbers in each pair.
```
```
Date: 09/13/2001 at 11:22:35
From: Doctor Mitteldorf
Subject: Re: Number/color cube
Dear Bill,
There may be powerful, general methods for attacking this problem, but
I'm going to think about it from the ground up, in a very concrete
way.
First, about the colors: how many ways are there to put colors on the
faces?
Answer: red can be paired opposite 5 different colors. Once you've
decided which color is opposite red, there are 4 colors left, which
can be paired on opposite faces in 3 different ways (e.g., {yellow-
green, and blue-purple}, {yellow-blue, and green-purple}, {yellow-
purple, and blue-green}). So 5*3 makes 15 distinct ways to assign
pairs of colors to the pairs of faces, and there's one more freedom
that we have: the mirror-image of each of these paintings is a
distinct way in itself.
For example, suppose the pairing is {red-orange, yellow-green, blue-
purple}. Then there's one corner that has red, yellow, and blue faces
coming together. Looking down on that corner, do those colors appear
(red,yellow,blue) clockwise or (red, yellow, blue) counter-clockwise?
These 2 mirror-image possibilities, multiplied by the 15 pairings,
indicates a total of 30 ways to paint the cube.
Second, once we've chosen a paint scheme, how many ways are there to
assign the numbers, consistent with the pairings (1-5, 3-6, 2-4)?
Again, we'll subdivide this question into two parts: first, how to
match the number-pairs with the color-pairs; second, how to choose
which number goes with which color within those pairings.
So, the first part of the second part: Each of these 3 number pairs
can be matched with any of the 3 color pairs. The number of ways of
doing this is the same as the number of ways of ordering 3 objects,
which is 3!=6. This is because you can regard the color-pairings as a
threesome in fixed order, and then each ordering of the number-pairs
corresponds to a way to match them against these.
And for the final part of the second part: Once you've decided which
of the 30 colorings to use, and you've decided which of the 6 color-
pair, number-pair matchings to use, you still have a choice within
each pair. For each color-number pair (e.g. red-orange labeled 1-2),
you can choose one of two labelings: (e.g. either 1 is red and 2 is
orange, or 1 is orange and 2 is red). You have 3 such choices to make,
and they are independent, so there are 2^3=8 ways of doing the
assignment.
Putting all this together, I get 5*3*2 * 3! * 2^3 = 1440 different
dice.
- Doctor Mitteldorf, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Discrete Math
High School Discrete Mathematics
High School Permutations and Combinations
Search the Dr. Math Library:
Find items containing (put spaces between keywords): Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words
Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/
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Education.com
Try
Brainzy
Try
Plus
# Waves: Of Special Interest to Physics C Students
(not rated)
By McGraw-Hill Professional
Updated on Feb 12, 2011
Practice problems for these concepts can be found at:
Waves Practice Problems for AP Physics B & C
### Maxwell's Equations
Okay, we'll get this out of the way right now: You will not have to solve Maxwell's equations on the AP Physics exam. These four equations include integrals the likes of which you will not be able to solve until well into college physics, if then. However, you can understand the basic point of each equation, and, most importantly, understand the equations' greatest consequence.
Maxwell obtained this wave speed as a mathematical result from the equations. He noticed that, when the experimentally determined constants were plugged in, the speed of his "electromagnetic waves" was identical to the speed of light.3 Maxwell's conclusion was that light must be an electromagnetic wave.
What are Maxwell's equations? We're not even going to write them out, for fear that you might throw down your book in trepidation. If you're really interested in the integral or differential form of the equations, you will find them in your physics book (or on a rather popular T-shirt). While we won't write the equations, we'll gladly summarize what they are and what they mean.
• Maxwell equation 1 is simply Gauss's law: The net electric flux through a closed surface is proportional to the charge enclosed by that surface.
• Maxwell equation 2 is sometimes called Gauss's law for magnetism: The net magnetic flux through a closed surface must always be zero. The consequence of this equation is that magnetic poles come in north/south pairs—you cannot have an isolated north magnetic pole.
• Maxwell equation 3 is simply Faraday's law: A changing magnetic flux through a loop of wire induces an EMF.
• Maxwell equation 4 is partly Ampére's law, but with an addition called "displacement current" that allows the equation to be valid in all situations. The principal consequence is that just as a changing magnetic field can produce an electric field, a changing electric field can likewise produce a magnetic field.
Practice problems for these concepts can be found at:
Waves Practice Problems for AP Physics B & C
### Ask a Question
150 Characters allowed
### Related Questions
#### Q:
See More Questions
### Today on Education.com
Top Worksheet Slideshows
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| 626,638,306
| 6,118
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# Search by Topic
#### Resources tagged with Calculating with fractions similar to Golden Fractions:
Filter by: Content type:
Age range:
Challenge level:
### There are 18 results
Broad Topics > Fractions, Decimals, Percentages, Ratio and Proportion > Calculating with fractions
### Not Continued Fractions
##### Age 14 to 18 Challenge Level:
Which rational numbers cannot be written in the form x + 1/(y + 1/z) where x, y and z are integers?
### Comparing Continued Fractions
##### Age 16 to 18 Challenge Level:
Which of these continued fractions is bigger and why?
### The Harmonic Triangle and Pascal's Triangle
##### Age 16 to 18
The harmonic triangle is built from fractions with unit numerators using a rule very similar to Pascal's triangle.
### There's a Limit
##### Age 14 to 18 Challenge Level:
Explore the continued fraction: 2+3/(2+3/(2+3/2+...)) What do you notice when successive terms are taken? What happens to the terms if the fraction goes on indefinitely?
### More Twisting and Turning
##### Age 11 to 16 Challenge Level:
It would be nice to have a strategy for disentangling any tangled ropes...
### All Tangled Up
##### Age 14 to 18 Challenge Level:
Can you tangle yourself up and reach any fraction?
### A Swiss Sum
##### Age 16 to 18 Challenge Level:
Can you use the given image to say something about the sum of an infinite series?
### Fair Shares?
##### Age 14 to 16 Challenge Level:
A mother wants to share a sum of money by giving each of her children in turn a lump sum plus a fraction of the remainder. How can she do this in order to share the money out equally?
### Fracmax
##### Age 14 to 16 Challenge Level:
Find the maximum value of 1/p + 1/q + 1/r where this sum is less than 1 and p, q, and r are positive integers.
### Lower Bound
##### Age 14 to 16 Challenge Level:
What would you get if you continued this sequence of fraction sums? 1/2 + 2/1 = 2/3 + 3/2 = 3/4 + 4/3 =
### Harmonic Triangle
##### Age 14 to 16 Challenge Level:
Can you see how to build a harmonic triangle? Can you work out the next two rows?
### Countdown Fractions
##### Age 11 to 16 Challenge Level:
Here is a chance to play a fractions version of the classic Countdown Game.
### Archimedes and Numerical Roots
##### Age 14 to 16 Challenge Level:
The problem is how did Archimedes calculate the lengths of the sides of the polygons which needed him to be able to calculate square roots?
##### Age 11 to 16 Challenge Level:
The items in the shopping basket add and multiply to give the same amount. What could their prices be?
### And So on - and on -and On
##### Age 16 to 18 Challenge Level:
Can you find the value of this function involving algebraic fractions for x=2000?
### Investigating the Dilution Series
##### Age 14 to 16 Challenge Level:
Which dilutions can you make using only 10ml pipettes?
### Reductant Ratios
##### Age 16 to 18 Challenge Level:
What does the empirical formula of this mixture of iron oxides tell you about its consituents?
### Ratios and Dilutions
##### Age 14 to 16 Challenge Level:
Scientists often require solutions which are diluted to a particular concentration. In this problem, you can explore the mathematics of simple dilutions
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# 132910 (number)
132,910 (one hundred thirty-two thousand nine hundred ten) is an even six-digits composite number following 132909 and preceding 132911. In scientific notation, it is written as 1.3291 × 105. The sum of its digits is 16. It has a total of 3 prime factors and 8 positive divisors. There are 53,160 positive integers (up to 132910) that are relatively prime to 132910.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 6
• Sum of Digits 16
• Digital Root 7
## Name
Short name 132 thousand 910 one hundred thirty-two thousand nine hundred ten
## Notation
Scientific notation 1.3291 × 105 132.91 × 103
## Prime Factorization of 132910
Prime Factorization 2 × 5 × 13291
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 132910 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 132,910 is 2 × 5 × 13291. Since it has a total of 3 prime factors, 132,910 is a composite number.
## Divisors of 132910
8 divisors
Even divisors 4 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 239256 Sum of all the positive divisors of n s(n) 106346 Sum of the proper positive divisors of n A(n) 29907 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 364.568 Returns the nth root of the product of n divisors H(n) 4.44411 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 132,910 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 132,910) is 239,256, the average is 29,907.
## Other Arithmetic Functions (n = 132910)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 53160 Total number of positive integers not greater than n that are coprime to n λ(n) 26580 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 12374 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 53,160 positive integers (less than 132,910) that are coprime with 132,910. And there are approximately 12,374 prime numbers less than or equal to 132,910.
## Divisibility of 132910
m n mod m 2 3 4 5 6 7 8 9 0 1 2 0 4 1 6 7
The number 132,910 is divisible by 2 and 5.
• Arithmetic
• Deficient
• Polite
• Square Free
• Sphenic
## Base conversion (132910)
Base System Value
2 Binary 100000011100101110
3 Ternary 20202022121
4 Quaternary 200130232
5 Quinary 13223120
6 Senary 2503154
8 Octal 403456
10 Decimal 132910
12 Duodecimal 64aba
20 Vigesimal gc5a
36 Base36 2ujy
## Basic calculations (n = 132910)
### Multiplication
n×y
n×2 265820 398730 531640 664550
### Division
n÷y
n÷2 66455 44303.3 33227.5 26582
### Exponentiation
ny
n2 17665068100 2347864201171000 312054630977637610000 41475181003237814745100000
### Nth Root
y√n
2√n 364.568 51.0332 19.0937 10.5855
## 132910 as geometric shapes
### Circle
Diameter 265820 835098 5.54964e+10
### Sphere
Volume 9.83471e+15 2.21986e+11 835098
### Square
Length = n
Perimeter 531640 1.76651e+10 187963
### Cube
Length = n
Surface area 1.0599e+11 2.34786e+15 230207
### Equilateral Triangle
Length = n
Perimeter 398730 7.6492e+09 115103
### Triangular Pyramid
Length = n
Surface area 3.05968e+10 2.76698e+14 108521
## Cryptographic Hash Functions
md5 738f3b918f4c0cd73b79ac93c3d5dfc5 355f9a8d012e4997aae8b243ec18c6f2b2ff3345 f51f77ef01be9083eda2205763c5cc135c04139f8445dbc6a47937c17f4ccaa5 7e99b9e2f8559fd1b61c9feb30a7a3dd92ad8845290d85ecb874ec5c0c00f3a35c3f31f515ebb83f02d5bd3a3e47c61310acc5680db290040ae2ce0c5b57bf43 86fb1bd6faafec1ab055f3c342646703ae81bebd
| 1,437
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Solving Inequalities: An Overview (page 2 of 3)
Sections: Linear inequalities, Quadratic inequalities, Other inequalities
The previous inequalities are called "linear" inequalities because we are dealing with linear expressions like "x – 2" ("x > 2" is just "x – 2 > 0", before you finished solving it). When we have an inequality with "x2" as the highest-degree term, it is called a "quadratic inequality". The method of solution is more complicated.
• Solve x2 – 3x + 2 > 0
First, I have to find the x-intercepts of the associated quadratic, because the intercepts are where y = x2 3x + 2 is equal to zero. Graphically, an inequality like this is asking me to find where the graph is above or below the x-axis. It is simplest to find where it actually crosses the x-axis, so I'll start there.
Factoring, I get x2 3x + 2 = (x 2)
(x
1) = 0, so x = 1 or x = 2. Then the graph crosses the x-axis at 1 and 2, and the number line is divided into the intervals (negative infinity, 1), (1, 2), and (2, positive infinity). Between the x-intercepts, the graph is either above the axis (and thus positive, or greater than zero), or else below the axis (and thus negative, or less than zero).
There are two different algebraic ways of checking for this positivity or negativity on the intervals. I'll show both.
1) Test-point method. The intervals between the x-intercepts are (negative infinity, 1), (1, 2), and (2, positive infinity). I will pick a point (any point) inside each interval. I will calculate the value of y at that point. Whatever the sign on that value is, that is the sign for that entire interval.
For (negative infinity, 1), let's say I choose x = 0; then y = 0 0 + 2 = 2, which is positive. This says that y is positive on the whole interval of (negative infinity, 1), and this interval is thus part of the solution (since I'm looking for a "greater than zero" solution).
For the interval (1, 2), I'll pick, say, x = 1.5; then y = (1.5)2 3(1.5) + 2 = 2.25 4.5 + 2 = 4.25 4.5 = 0.25, which is negative. Then y is negative on this entire interval, and this interval is then not part of the solution.
For the interval (2, positive infinity), I'll pick, say, x = 3; then y = (3)2 3(3) + 2 = 9 9 + 2 = 2, which is positive, and this interval is then part of the solution. Then the complete solution for the inequality is x < 1 and x > 2. This solution is stated variously as:
inequality notation interval, or set, notation number line with parentheses (brackets are used for closed intervals) number line with open dots (closed dots are used for closed intervals)
2) Factor method. Factoring, I get y = x2 3x + 2 = (x 2)(x 1). Now I will consider each of these factors separately.
The factor x 1 is positive for x > 1; similarly, x 2 is positive for x > 2. Thinking back to when I first learned about negative numbers, I know that (plus)×(plus) = (plus), (minus)×(minus) = (plus), and (minus)×(plus) = (minus). So, to compute the sign on y = x2 3x + 2, I only really need to know the signs on the factors. Then I can apply what I know about multiplying negatives.
First, I set up a grid, showing the factors and the number line. Now I mark the intervals where each factor is positive. Where the factors aren't positive, they must be negative. Now I multiply up the columns, to compute the sign of y on each interval.
Then the solution of x2 3x + 2 > 0 are the two intervals with the "plus" signs:
(negative infinity, 1) and (2, positive infinity).
• Solve –2x2 + 5x + 12 < 0.
First I find the zeroes, which are the endpoints of the intervals: y = –2x2 + 5x + 12 =
(–2x – 3)(x – 4) = 0
for x = –3/2 and x = 4. So the endpoints of the intervals will be at 3/2 and 4. The intervals are between the endpoints, so the intervals are (negative infinity, –3/2], [–3/2, 4], and [4, positive infinity). (Note that I use brackets for the endpoints in "or equal to" inequalities, instead of parentheses, because the endpoints will be included in the final solution.)
To find the intervals where y is negative by the Test-Point Method, I just pick a point in each interval. I can use points such as x = –2, x = 0, and x = 5.
To find the intervals where y is negative by the Factor Method, I just solve each factor: –2x – 3 is positive for –2x – 3 > 0, –3 > 2x, –3/2 > x, or x < –3/2; and x – 4 is positive for x – 4 > 0,
x > 4
. Then I fill out the grid:
Then the solution to this inequality is all x's in
(negative infinity, –3/2 ] and [4, positive infinity).
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Cite this article as: Stapel, Elizabeth. "Solving Inequalities: An Overview." Purplemath. Available from https://www.purplemath.com/modules/ineqsolv2.htm. Accessed [Date] [Month] 2016
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Topic: covariance matrix, correlation matrix, decomposition
Replies: 4 Last Post: Nov 24, 1999 5:37 AM
Messages: [ Previous | Next ]
Michael Pronath Posts: 14 Registered: 12/15/04
Re: covariance matrix, correlation matrix, decomposition
Posted: Nov 24, 1999 5:11 AM
Michael Pronath <[email protected]> writes:
> It is often necessary to decompose a given covariance matrix C
> (resp. correlation matrix R) as C=G*G'. Cholesky decomposition is
> commonly used here.
> But there are some others possible. Starting from the eigenvalue
> decomposition C=Q*L*Q', there is
>
> G1 = Q*sqrt(L), as G1*G1' = Q*sqrt(L)*sqrt(L)*Q'=Q*L*Q' = C
>
> or
>
> G2 = Q*sqrt(L)*Q', as G2*G2' = Q*sqrt(L)* Q'*Q *sqrt(L)*Q' =
> = Q*sqrt(L)* 1 *sqrt(L)*Q' = C
>
> ...
>
> I'd like to know if anybody has made some more profound analysis about
> this, and the pro's and con's of the various methods.
Sargis Dallakyan <[email protected]> writes:
> IMO if you have a covariance matrix and want to decompose it then sure
> enough Cholesky is the best. But if you want to generate a random
> vectors with a desired confidence ellipsoid then 2) and 3) gives you a
> direct way to do that.
Why do you think that Cholesky is the best? It is commonly used for
that task, and I wanted to know the reason.
Helmut Jarausch <[email protected]> writes:
> What about an SVD of G (not C !) Just computing C (not even
> decomposing) makes the condition number worse. Say, you have the
> SVD of G : G = U S V' where U and V are orthogonal matrices and S
> has nontrivial elements only on its diagonal. Then G*G' = U S S' U'
> where SS' has the diagonal (s_{ii}^2) and zeros elsewhere. You can
> extract all information about C from U and S and it should be much
> more.
Indeed, if T is the diagonal matrix of the standard deviations, and R
is the correlation matrix, then C=T*R*T'. Then, the decomposition
G*G'=C=T*R*T' and so inverse(T)*G*G'*inverse(T') = R = H*H', with
H=inverse(T)*G. So, it is probably better to decompose R=H*H' (and
then calculate G=T*H if necessary).
Nevertheless, for the decomposition R=H*H', the same options remain:
Cholesky or Eigenvalue I or II? Which is when the best and why?
At a first glance, I like Eigenvalue I best, because it generates
orthogonal grids. As C and R are symmetric, and the number of
statistical parameters is commonly < 100, the eigenvalue
decomp. should be fast and stable. Plus, the parameter ordering does
not influence the decomposition, as it does in Cholesky.
As the decomposition R=G*G' is frequently used when generating random
numbers, grids in statistical parameter spaces, etc., I assume that
somebody has already performed an analysis of this, but where?
Michael Pronath
--
Date Subject Author
11/23/99 Michael Pronath
11/23/99 Sargis Dallakyan
11/24/99 Helmut Jarausch
11/24/99 Michael Pronath
11/24/99 Emilio Lopes
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Although only 2 percent of drivers on Lalaland’s highways : GMAT Critical Reasoning (CR)
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# Although only 2 percent of drivers on Lalaland’s highways
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24 Jul 2013, 10:39
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Although only 2 percent of drivers on Lalaland’s highways drove sports cars, 25 percent of all vehicles ticketed for drunk driving in the past 90 days were sports cars. Clearly, sports car drivers on Lalaland highways are more likely to drive drunk than are drivers of other kinds of vehicles.
The conclusion drawn above depends on which of the following assumptions?
A)Drivers on Lalaland highways drive drunk more often than do drivers on highways not covered in the report.
B)Many of the vehicles ticketed for drunk driving were ticketed more than once during the time period covered by the report.
C)Drivers who are ticketed for drunk driving are more likely to drive drunk regularly than are drivers who are not ticketed.
D)The number of drivers ticketed for drunk driving was greater than the number of sportscars.
E)Drivers of sports cars are less likely to be ticketed for drunk driving than are drivers of other kinds of cars.
[Reveal] Spoiler: OA
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Last edited by Zarrolou on 24 Jul 2013, 13:31, edited 1 time in total.
Edited the question.
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Re: Although only 2 percent of drivers on Lalaland’s highways [#permalink]
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24 Jul 2013, 23:09
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This question uses a very classic logic assumption – Statistical Assumption. It is similar to hypothesis you see in Probability & Statistics of Quantitative.
The form is:
A is more likely to do X than B is
Assumption: Over the statical period, A does X more often than B does.
For example:
Team A has more chances to win this game than Team B does.
Assumption: Over the statical period, Team A wins more often than Team B
More example:
Mount A is more likely to have volcano than Mount B is
Assumption: Over the statical period, Mount A has volcano more often than Mount B
APPLY TO THIS QUESTION:
Fact: only 2 percent of drivers on Lalaland’s highways drove sports cars,
Fact: 25 percent of all vehicles ticketed for drunk driving in the past 90 days were sports cars.
Conclusion: Sports car drivers on Lalaland highways are more likely to drive drunk than are drivers of other kinds of vehicles.
Assumption: Over the statical period, sport car drivers drives drunk more often (more regularly) than drivers of other kind of vehicles do.
A)Drivers on Lalaland highways drive drunk more often than do drivers on highways not covered in the report.
Wrong. Out of scope. Nothing about “drivers on highways not covered…”. We just talk about Lalaland highways.
B)Many of the vehicles ticketed for drunk driving were ticketed more than once during the time period covered by the report.
Wrong. TEMPTING. Why? The wording is quite simple, but if you do not read carefull, you may assume “many of the vehicles ticked for drunk driving more than once” are “sport vehicles”. However, B is wrong because B does not say these vehicles are sport cars. If that was the case, B would be correct assumption. But what if the vehicles ticked for drunk more than once may NOT be sport cars.
C)Drivers who are ticketed for drunk driving are more likely to drive drunk regularly than are drivers who are not ticketed.
Correct. This is exactly assumption stated above.
D)The number of drivers ticketed for drunk driving was greater than the number of sportscars.
Wrong. We do not compare the number of drivers vs the number of sportcars.
E)Drivers of sports cars are less likely to be ticketed for drunk driving than are drivers of other kinds of cars.
Wrong. It’s a reverse answer. Hence, E is wrong.
Hope it helps.
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Re: Although only 2 percent of drivers on Lalaland’s highways [#permalink]
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25 Jul 2013, 09:31
this assumption question relies on the connection between 25 % and more likely. Assuming that if a higher percentage of people have something happen to them then that makes it more likely. The answer choices are designed to distract the reader with other, irrelevant issues.
Stiv wrote:
Although only 2 percent of drivers on Lalaland’s highways drove sports cars, 25 percent of all vehicles ticketed for drunk driving in the past 90 days were sports cars. Clearly, sports car drivers on Lalaland highways are more likely to drive drunk than are drivers of other kinds of vehicles.
The conclusion drawn above depends on which of the following assumptions?
A)Drivers on Lalaland highways drive drunk more often than do drivers on highways not covered in the report. the subject of other highways is irrelevant becuase the argument does not address other highways
B)Many of the vehicles ticketed for drunk driving were ticketed more than once during the time period covered by the report. Being ticketed more than once would acutally weaken the argument becuase it would show a problem with the statistics
C)Drivers who are ticketed for drunk driving are more likely to drive drunk regularly than are drivers who are not ticketed. this is correct because it directly connects the number of tickets to the likelyhood of driving drunk
D)The number of drivers ticketed for drunk driving was greater than the number of sportscars. This is irrelevant becuase the ratio of tickets to sportscars doesn't matter unless all of the sportscars were getting tickets
E)Drivers of sports cars are less likely to be ticketed for drunk driving than are drivers of other kinds of cars.While this may strengthen the argument, it is not an assumption becuase it is a new fact, not a connection of two facts in the argument itself.
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Re: Although only 2 percent of drivers on Lalaland’s highways [#permalink]
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08 Aug 2013, 01:50
pqhai wrote:
This question uses a very classic logic assumption – Statistical Assumption. It is similar to hypothesis you see in Probability & Statistics of Quantitative.
The form is:
A is more likely to do X than B is
Assumption: Over the statical period, A does X more often than B does.
For example:
Team A has more chances to win this game than Team B does.
Assumption: Over the statical period, Team A wins more often than Team B
More example:
Mount A is more likely to have volcano than Mount B is
Assumption: Over the statical period, Mount A has volcano more often than Mount B
APPLY TO THIS QUESTION:
Fact: only 2 percent of drivers on Lalaland’s highways drove sports cars,
Fact: 25 percent of all vehicles ticketed for drunk driving in the past 90 days were sports cars.
Conclusion: Sports car drivers on Lalaland highways are more likely to drive drunk than are drivers of other kinds of vehicles.
Assumption: Over the statical period, sport car drivers drives drunk more often (more regularly) than drivers of other kind of vehicles do.
A)Drivers on Lalaland highways drive drunk more often than do drivers on highways not covered in the report.
Wrong. Out of scope. Nothing about “drivers on highways not covered…”. We just talk about Lalaland highways.
B)Many of the vehicles ticketed for drunk driving were ticketed more than once during the time period covered by the report.
Wrong. TEMPTING. Why? The wording is quite simple, but if you do not read carefull, you may assume “many of the vehicles ticked for drunk driving more than once” are “sport vehicles”. However, B is wrong because B does not say these vehicles are sport cars. If that was the case, B would be correct assumption. But what if the vehicles ticked for drunk more than once may NOT be sport cars.
C)Drivers who are ticketed for drunk driving are more likely to drive drunk regularly than are drivers who are not ticketed.
Correct. This is exactly assumption stated above.
D)The number of drivers ticketed for drunk driving was greater than the number of sportscars.
Wrong. We do not compare the number of drivers vs the number of sportcars.
E)Drivers of sports cars are less likely to be ticketed for drunk driving than are drivers of other kinds of cars.
Wrong. It’s a reverse answer. Hence, E is wrong.
Hope it helps.
Hi, could you please elaborate a little more. I am not able to catch the exact point in 'C'.
The explanation you haven for negating 'B' applies to 'C' as well as in even 'C' doesn't talk about the sports cars. So, I am unable to figure out why 'C' is a better option than 'B'.
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Re: Although only 2 percent of drivers on Lalaland’s highways [#permalink]
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15 Aug 2013, 14:19
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mneeti wrote:
Hi, could you please elaborate a little more. I am not able to catch the exact point in 'C'.
The explanation you haven for negating 'B' applies to 'C' as well as in even 'C' doesn't talk about the sports cars. So, I am unable to figure out why 'C' is a better option than 'B'.
Hi mneeti
First of all, B and C are totally different in meaning. Let analyze.
B)Many of the vehicles ticketed for drunk driving were ticketed more than once during the time period covered by the report.
B is wrong because B only mentions that those who were ticketed for drunk driving will be ticketed again during the time period. But what if those who were ticketed more than one time are are drivers of other kinds of vehicles, NOT drivers of sport vehicles. If that’s the case, you cannot conclude drivers of sport vehicles drunk more regularly than drivers of other kind of vehicles do.
Hence, B is not the assumption.
C)Drivers who are ticketed for drunk driving are more likely to drive drunk regularly than are drivers who are not ticketed.
C means those who are ticketed fro drunk driving are more likely ticketed again. It does not matter drivers of sport cars or other kind of cars. This is the base for a comparison between the number of ticket/driver of sport cars AND the number of ticket/driver of other kind of cars.
Name “drivers of sport cars” is SET 1, name “drivers of other kind of cars is SET 2.
So, if the number of ticket/driver of SET 1 is greater than that of SET 2, the conclusion is correct.
Let analyze an example:
Assume we have 1000 drivers
SET 1 accounts for 2% = 20 drivers
SET 2 accounts for 98% = 980 drivers
Assume we have 100 tickets for drunk driving.
SET 1 accounts for 25% = 25 tickets
SET 2 accounts for 75% = 75 tickets.
Ticket / Driver:
20 drivers of sports cars got 25 tickets ==> 1.25 ticket / driver
980 drivers of other kind of cars got 75 tickets ==> 0.07 ticket / driver
Clearly, drivers of sport cars are more likely are more likely to drive drunk regularly than are drivers. Hence, C is the assumption.
Hope it helps.
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Re: Although only 2 percent of drivers on Lalaland’s highways [#permalink]
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Re: Although only 2 percent of drivers on Lalaland’s highways [#permalink]
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Re: Although only 2 percent of drivers on Lalaland’s highways [#permalink] 19 Feb 2016, 21:43
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Approaches to fractions and equivalence
In this month's maths article from Caroline Clissold, we'll be looking at fractions and in particular approaching equivalence.
We will be covering:
How to approach equivalence in Year 1
What children need to be able to understand about equivalence in years 2 and 3
Activities to help children understand the concept of equivalence
How Fluency with Fractions can help
Equivalence is a very important aspect of mathematics. It’s basically about finding relationships or aspects that are the same. In measurement we often ask the children to find equivalent units of measure, for example 12kg 500g, 12.5kg and 12 500g, one mile is the same as 1.61km (rounded to nearest two decimal places). In number we ask the children to make equivalent missing number statements, 15 x 3 = 60 - ? (15). Finding solutions to calculations is also all about equivalence. When they see the symbol =, children often see that symbol as an indication that the answer comes next. They need to be shown that equals means equivalent.
Equivalence in Year 1
In fractions, we often ask the children to find equivalences, for example 1/2 = 2/4, 8/10 = 4/5. Finding equivalences begins in Year 1 when children need to recognise, find and name a half as one of two equal parts of an object, shape or quantity and recognise, find and name a quarter as one of four equal parts of an object, shape or quantity. When they explore this, they will notice that two halves and four quarters are equivalent to one whole and two quarters are equivalent to one half. It is important to give the opportunity for the children to find this out for themselves. A simple way to do this is to give them strips of paper and ask them to keep one whole and then fold a second in half and a third into quarters:
If they do this they can clearly see that 2/4 = 1/2 and the 4/4 and 2/2 are equivalent to one whole. They can also clearly see that 1/4 is smaller than 1/2.
Try introducing the word equivalent to children in Year 1 and expect them to use this word when describing these types of factions.
Equivalence in Year 2 and 3
In Year 2 children need to recognise, find, name and write 1/3, 1/4, 2/4 and 3/4 of a length, shape, set of objects or quantity, write simple fractions for example, 1/2 of 6 = 3 and recognise the equivalence of 2/3 and 1/2.
In Year 3 the children begin adding and subtracting fractions with the same denominator and in Year 4 they do this for fractions with common equivalent fractions. In Years 5 and 6 they begin, first multiplying, and then dividing fractions. These operations all require an understanding of equivalence.
The NCETM has a great video of children in Year 2 adding and subtracting halves and quarters which demonstrates just what they are capable of if they are given appropriate visual representations. These children are able to use equivalence to do this. It’s worth watching!
All the way through, children need to be given opportunities to compare and order fractions. This often requires them to find equivalent fractions in order to do this accurately.
To develop a conceptual understanding, the children first need to be aware that a fraction is an area and not a shape. Many develop the misconception that if, for example, fractions of shapes look the same they are equivalent but if they are different shapes then they are not. Thus misconception often develops because of the way fractions are presented in textbooks or on worksheets. Here is a typical example:
Activities to help children understand the concept of equivalence
There are lots of nice activities to help the children understand the concept of equivalence.
Try presenting fractions so that they are not identical. For example, you could ask them to find out whether these are quarters:
By cutting the pieces out and folding them in half, they can see that each piece is one eighth. They should know that 2/8 is equivalent to 1/4, so each part is 1/4 and each quarter is equivalent.
You could give them two pieces of A4 paper. They tear one in half. Ask them to tell you what fraction the smaller piece is of the larger. Most children will tell you half. Ask them to stick the half in the middle of the large piece. Ask them what fraction the piece in the middle is and then ask them what fraction the part around the outside is of the whole. Many will give random guesses as to the latter because they may think that the parts should look the same and so that part can’t be the other half.
You could then ask the children to cut around the smaller piece so that there is one long piece of paper. Ask them what fraction this is of the whole again and they may well give you other random fractions that are different from their first guesses. You could then ask them to cut this long pieces into parts and stick them onto the smaller piece so proving that they are both halves.
You could write 1/4 on the board and ask the children to write a variety of ways to show this, drawing on their experiences of equivalence. For example:
Tangrams are also a great way to explore equivalence:
You could ask the children to explore what fraction each part is of the whole and then other fractions and equivalences. There are lots of ideas that you can adapt to help your children get to grips with the concept of equivalence.
How can Fluency with Fractions help?
The Fluency with Fractions series from Rising Stars provides great support for using a variety of visual representations of fractions to support understanding. For example, in each year level fraction bar images are used to introduce the concept of fractions as equal parts of a whole, equivalence, counting (linked to the number line) and calculating. Finding and identifying equivalent fractions paves the way for later understanding of equivalent ratios, so the Fluency with Fractions series moves from work on equivalent fractions through decimals and percentages to incorporate ratio and proportion in Year 6. Developing understanding of equivalence early on is essential to becoming confident with these concepts.
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Assessment, English and Literacy, Fluency with Fractions, Grammar, Spelling and Punctuation, Mathematics, Revision and Practice
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# 1.
(a) The force on the electron is
ˆ FB = qv × B = q vx ˆ + v y ˆ × Bx ˆ + By j = q ( vx By − v y Bx ) k i j i = −1.6 ×10−19 C ⎡ 2.0 × 106 m s ( −0.15 T ) − 3.0 × 106 m s ( 0.030 T ) ⎤ ⎣ ⎦ ˆ = 6.2 × 10−14 N k.
(
) (
)
( (
)( )
)
(
)
Thus, the magnitude of FB is 6.2 × 1014 N, and FB points in the positive z direction. (b) This amounts to repeating the above computation with a change in the sign in the charge. Thus, FB has the same magnitude but points in the negative z direction, namely, ˆ F = − 6.2 × 10−14 N k.
B
(
)
2. (a) We use Eq. 28-3: FB = |q| vB sin φ = (+ 3.2 × 10–19 C) (550 m/s) (0.045 T) (sin 52°) = 6.2 × 10–18 N. (b) a = FB/m = (6.2 × 10– 18 N) / (6.6 × 10– 27 kg) = 9.5 × 108 m/s2. (c) Since it is perpendicular to v , FB does not do any work on the particle. Thus from the work-energy theorem both the kinetic energy and the speed of the particle remain unchanged.
3. (a) Eq. 28-3 leads to v= 6.50 × 10−17 N FB = = 4.00 × 105 m s . −19 −3 eB sin φ 160 × 10 C 2.60 × 10 T sin 23.0° .
c
hc
h
(b) The kinetic energy of the proton is
K=
2 1 2 1 mv = (1.67 ×10−27 kg )( 4.00 × 105 m s ) = 1.34 ×10−16 J , 2 2
which is equivalent to K = (1.34 × 10– 16 J) / (1.60 × 10– 19 J/eV) = 835 eV.
4. The force associated with the magnetic field must point in the j direction in order to
cancel the force of gravity in the − j direction. By the right-hand rule, B points in the
− k direction (since i × − k = j ). Note that the charge is positive; also note that we need to assume By = 0. The magnitude |Bz| is given by Eq. 28-3 (with φ = 90°). Therefore, with m = 1.0 × 10−2 kg , v = 2.0 ×104 m/s and q = 8.0 × 10−5 C , we find ⎛ mg ⎞ ˆ ˆ ˆ B = Bz k = − ⎜ ⎟ k = (−0.061 T)k qv ⎠ ⎝
e j
5. Using Eq. 28-2 and Eq. 3-30, we obtain
F = q v x By − v y Bx k = q v x 3Bx − v y Bx k
d
i
d b g
i
where we use the fact that By = 3Bx. Since the force (at the instant considered) is Fz k where Fz = 6.4 × 10–19 N, then we are led to the condition
q ( 3vx − v y ) Bx = Fz ⇒ Bx = Fz . q ( 3vx − v y )
Substituting vx = 2.0 m/s, vy = 4.0 m/s and q = –1.6 × 10–19 C, we obtain Fz 6.4 × 10−19 N Bx = = = −2.0 T. q(3vx − v y ) (−1.6 ×10−19 C)[3(2.0 m/s) − 4.0 m]
6. The magnetic force on the proton is F = qv × B where q = +e . Using Eq. 3-30 this becomes (4 × 10−17 )i + (2 × 10−17)j = e[(0.03vy + 40)i + (20 – 0.03vx)j – (0.02vx + 0.01vy)k]
^ ^ ^ ^ ^
with SI units understood. Equating corresponding components, we find (a) vx = −3.5×103 m/s, and (b) vy = 7.0×103 m/s.
7. Straight line motion will result from zero net force acting on the system; we ignore gravity. Thus, F = q E + v × B = 0 . Note that v ⊥B so v × B = vB . Thus, obtaining the
d
i
speed from the formula for kinetic energy, we obtain
B= E E 100 V /(20 ×10−3 m) = = = 2.67 ×10−4 T. 3 −19 −31 v 2 K / me 2 (1.0 ×10 V ) (1.60 ×10 C ) / ( 9.11×10 kg )
ˆ In unit-vector notation, B = −(2.67 ×10−4 T)k .
8. Letting F = q E + v × B = 0 , we get vB sin φ = E . We note that (for given values of the fields) this gives a minimum value for speed whenever the sin φ factor is at its maximum value (which is 1, corresponding to φ = 90°). So
d
i
vmin =
E 1.50 ×103 V/m = = 3.75 ×103 m/s . B 0.400 T
9. We apply F = q E + v × B = me a to solve for E :
E= me a + B×v q
−31
d
i
c9.11 × 10 = e
kg 2.00 × 1012 m s i
2 −19
hd
−160 × 10 .
. = −114 i − 6.00 j + 4.80k V m .
j
C
i + b400µTgi × b12.0 km sg j + b15.0 km sgk
10. (a) The net force on the proton is given by ˆ j i F = FE + FB = qE + qv × B = (1.60 × 10−19 C ) ⎡( 4.00 V m ) k+ ( 2000 m s ) ˆ × ( −2.50 × 10−3 T ) ˆ ⎤ ⎣ ⎦ −18 ˆ = (1.44 × 10 N ) k. (b) In this case
F = FE + FB = qE + qv × B
ˆ j i = 1.60 × 10−19 C ⎡( −4.00 V m ) k + ( 2000 m s ) ˆ × ( −2.50 mT ) ˆ ⎤ ⎣ ⎦ ˆ = 1.60 × 10−19 N k. (c) In the final case,
( (
) )
F = FE + FB = qE + qv × B
i+ j i = 1.60 × 10−19 C ⎡( 4.00 V m ) ˆ ( 2000 m s ) ˆ × ( −2.50 mT ) ˆ ⎤ ⎣ ⎦ −19 −19 ˆ i+ = 6.41× 10 N ˆ 8.01× 10 N k.
( (
) ) (
)
11. Since the total force given by F = e E + v × B vanishes, the electric field E must be perpendicular to both the particle velocity v and the magnetic field B . The magnetic field is perpendicular to the velocity, so v × B has magnitude vB and the magnitude of the electric field is given by E = vB. Since the particle has charge e and is accelerated through a potential difference V , mv 2 / 2 = eV and v = 2eV m . Thus,
2 1.60 × 10−19 C 10 ×103 V 2eV E=B = (1.2 T ) = 6.8 × 105 V m. m 9.99 × 10−27 kg
d
i
(
(
)(
)
)
12. (a) The force due to the electric field ( F = qE ) is distinguished from that associated with the magnetic field ( F = qv × B ) in that the latter vanishes at the speed is zero and the former is independent of speed. The graph (Fig.28-36) shows that the force (ycomponent) is negative at v = 0 (specifically, its value is –2.0 × 10–19 N there) which (because q = –e) implies that the electric field points in the +y direction. Its magnitude is
2.0 × 10−19 N E= = = 1.25 N/C = 1.25 V/m . |q| 1.6 × 10−19 C Fnet, y
(b) We are told that the x and z components of the force remain zero throughout the motion, implying that the electron continues to move along the x axis, even though magnetic forces generally cause the paths of charged particles to curve (Fig. 28-11). The exception to this is discussed in section 28-3, where the forces due to the electric and magnetic fields cancel. This implies (Eq. 28-7) B = E/v = 2.50 × 10−2 T. For F = qv × B to be in the opposite direction of F = qE we must have v × B in the opposite direction from E which points in the +y direction, as discussed in part (a). Since the velocity is in the +x direction, then (using the right-hand rule) we conclude that ^ ^ ^ the magnetic field must point in the +z direction ( i × k = −j ). In unit-vector notation, we ˆ have B = (2.50 ×10−2 T)k .
13. We use Eq. 28-12 to solve for V:
V=
( 23A )( 0.65 T ) iB = = 7.4 × 10−6 V. 28 nle ( 8.47 ×10 m3 ) (150 µ m ) (1.6 ×10−19 C )
14. For a free charge q inside the metal strip with velocity v we have F = q E + v × B . We set this force equal to zero and use the relation between (uniform) electric field and potential difference. Thus, 3.90 × 10−9 V E Vx − Vy d xy = 0.382 m s . v= = = B B 120 × 10−3 T 0.850 × 10−2 m .
d
i
c
c
hc
h
h
15. (a) We seek the electrostatic field established by the separation of charges (brought on by the magnetic force). With Eq. 28-10, we define the magnitude of the electric field as | E | = v | B | = ( 20.0 m/s )( 0.030 T ) = 0.600 V/m . Its direction may be inferred from Figure 28-8; its direction is opposite to that defined by v × B . In summary, ˆ E = −(0.600 V m)k which insures that F = q E + v × B vanishes. (b) Eq. 28-9 yields V = Ed = (0.600 V/m)(2.00 m) = 1.20 V .
d
i
16. We note that B must be along the x axis because when the velocity is along that axis there is no induced voltage. Combining Eq. 28-7 and Eq. 28-9 leads to
d=
V V = E vB
where one must interpret the symbols carefully to ensure that d , v and B are mutually perpendicular. Thus, when the velocity if parallel to the y axis the absolute value of the voltage (which is considered in the same “direction” as d ) is 0.012 V, and
d = dz =
0.012 V = 0.20 m . (3.0 m/s)(0.020 T)
On the other hand, when the velocity if parallel to the z-axis the absolute value of the appropriate voltage is 0.018 V, and d = dy = Thus, our answers are (a) dx = 25 cm (which we arrive at “by elimination” – since we already have figured out dy and dz ), (b) dy = 30 cm, and (c) dz = 20 cm . 0.018 V = 0.30 m . (3.0 m/s)(0.020 T)
17. (a) From K =
1 me v 2 we get 2
. . 2 120 × 103 eV 160 × 10−19 eV J 2K v= = = 2.05 × 107 m s . −31 9.11 × 10 kg me
c
hc
h
(b) From r = me v / qB we get 9.11 × 10−31 kg 2.05 × 107 m s me v B= = = 4.67 × 10−4 T. −19 −2 qr 160 × 10 C 25.0 × 10 m .
c
c
hc hc
h
h
(c) The “orbital” frequency is f = 2.07 × 107 m s v . = = 131 × 107 Hz. −2 2 πr 2 π 25.0 × 10 m
c
h
(d) T = 1/f = (1.31 × 107 Hz)–1 = 7.63 × 10–8 s.
18. (a) The accelerating process may be seen as a conversion of potential energy eV into 1 kinetic energy. Since it starts from rest, me v 2 = eV and 2
2 1.60 ×10−19 C ( 350 V ) 2eV = = 1.11× 107 m s. v= −31 me 9.11×10 kg
(
)
(b) Eq. 28-16 gives
9.11× 10−31 kg 1.11× 107 m s me v = = 3.16 ×10−4 m. r= −19 −3 eB 1.60 × 10 C 200 ×10 T
(
(
)( )(
)
)
19. From Eq. 28-16, we find
9.11× 10−31 kg 1.30 × 106 m s me v = = 2.11× 10−5 T. B= −19 er 1.60 × 10 C ( 0.350 m )
(
(
)(
)
)
20. Using Eq. 28-16, the radius of the circular path is mv 2mK = qB qB
r=
where K = mv 2 / 2 is the kinetic energy of the particle. Thus, we see that K = (rqB)2/2m ∝ q2m–1. (a) Kα = ( qα q p ) ( m p mα ) K p = ( 2 ) (1 4 ) K p = K p = 1.0MeV;
2 2
(b) K d = ( qd q p ) ( m p md ) K p = (1) (1 2 ) K p = 1.0 MeV 2 = 0.50MeV.
2 2
21. (a) The frequency of revolution is 35.0 × 10−6 T 160 × 10−19 C . Bq f = = = 9.78 × 105 Hz. −31 2 πme 2 π 9.11 × 10 kg
c
c
hc
h
h
(b) Using Eq. 28-16, we obtain
2me K mv r= e = = qB qB . 2 9.11 × 10−31 kg 100 eV 160 × 10−19 J eV . c160 × 10 Chc35.0 × 10 Th
−19 −6
c
hb
gc
h = 0.964 m .
1 22. Combining Eq. 28-16 with energy conservation (eV = 2 mev2 in this particular application) leads to the expression me r=eB 2eV me
2me / eB 2 . From Fig. 2me / eB 2
which suggests that the slope of the r versus V graph should be and solving we find B = 6.7 × 10−2 T.
28-39, we estimate the slope to be 5 × 10−5 in SI units. Setting this equal to
23. Let ξ stand for the ratio ( m / | q | ) we wish to solve for. Then Eq. 28-17 can be written as T = 2πξ/B. Noting that the horizontal axis of the graph (Fig. 28-40) is inverse-field (1/B) then we conclude (from our previous expression) that the slope of the line in the graph must be equal to 2πξ. We estimate that slope as 7.5 × 10−9 T.s, which implies
ξ = m / | q | = 1.2 × 10−9 kg/C .
24. With the B pointing “out of the page,” we evaluate the force (using the right-hand rule) at, say, the dot shown on the left edge of the particle’s path, where its velocity is down. If the particle were positively charged, then the force at the dot would be toward the left, which is at odds with the figure (showing it being bent towards the right). Therefore, the particle is negatively charged; it is an electron. (a) Using Eq. 28-3 (with angle φ equal to 90°), we obtain
v= |F| = 4.99 ×106 m s. e| B|
(b) Using either Eq. 28-14 or Eq. 28-16, we find r = 0.00710 m. (c) Using Eq. 28-17 (in either its first or last form) readily yields T = 8.93 × 10–9 s.
25. (a) Using Eq. 28-16, we obtain . . 2 4.50 × 10−2 m 160 × 10−19 C 120 T 2eB rqB v= = = = 2.60 × 106 m s . −27 4.00 u mα . 4.00 u 166 × 10 kg u
c
b
gc
hc
hb h
g
(b) T = 2πr/v = 2π(4.50 × 10–2 m)/(2.60 × 106 m/s) = 1.09 × 10–7 s. (c) The kinetic energy of the alpha particle is
4.00 u 166 × 10 −27 kg u 2.60 × 106 m s . 1 2 K = mα v = 2 2 160 × 10−19 J eV .
b
gc
c
hc
h
h
2
= 140 × 105 eV . .
(d) ∆V = K/q = 1.40 × 105 eV/2e = 7.00 × 104 V.
26. Using F = mv 2 / r (for the centripetal force) and K = mv 2 / 2 , we can easily derive the relation 1 K = 2 Fr . With the values given in the problem, we thus obtain K = 2.09 × 10−22 J.
27. Reference to Fig. 28-11 is very useful for interpreting this problem. The distance → traveled parallel to B is d|| = v||T = v||(2πme /|q|B) using Eq. 28-17. Thus, v|| = d|| e B = 50.3 km/s 2 π me
using the values given in this problem. Also, since the magnetic force is |q|Bv⊥, then we find v⊥ = 41.7 km/s. The speed is therefore v = v⊥2 + v||2 = 65.3 km/s.
28. We consider the point at which it enters the field-filled region, velocity vector pointing downward. The field points out of the page so that v × B points leftward, which indeed seems to be the direction it is “pushed’’; therefore, q > 0 (it is a proton). (a) Eq. 28-17 becomes T = 2π mp / e | B | , or 2 (130 ×10 which yields B = 0.252 T . (b) Doubling the kinetic energy implies multiplying the speed by 2 . Since the period T does not depend on speed, then it remains the same (even though the radius increases by a factor of 2 ). Thus, t = T/2 = 130 ns.
−9
)=
(1.60 ×10 ) | B |
−19
2π (1.67 ×10−27 )
29. (a) If v is the speed of the positron then v sin φ is the component of its velocity in the plane that is perpendicular to the magnetic field. Here φ is the angle between the velocity and the field (89°). Newton’s second law yields eBv sin φ = me(v sin φ)2/r, where r is the radius of the orbit. Thus r = (mev/eB) sin φ. The period is given by 2π ( 9.11×10−31 kg ) 2πme 2πr = = = 3.58 ×10−10 s. T= −19 v sin φ eB (1.60 ×10 C ) ( 0.100T ) The equation for r is substituted to obtain the second expression for T. (b) The pitch is the distance traveled along the line of the magnetic field in a time interval 1 of one period. Thus p = vT cos φ. We use the kinetic energy to find the speed: K = 2 me v 2 means
2 ( 2.00 ×103 eV )(1.60 ×10−19 J eV ) 2K = = 2.65 ×107 m s . v= −31 9.11×10 kg me
Thus,
p = 2.65 × 107 m s 3.58 × 10−10 s cos 89° = 1.66 × 10−4 m . (c) The orbit radius is
9.11× 10−31 kg 2.65 × 107 m s sin 89° me v sin φ = = 1.51×10−3 m . R= −19 eB 1.60 × 10 C ( 0.100 T )
(
)(
)
(
(
)(
)
)
30. (a) Eq. 3-20 gives φ = cos−1(2/19) = 84°. (b) No, the magnetic field can only change the direction of motion of a free (unconstrained) particle, not its speed or its kinetic energy. (c) No, as reference to to Fig. 28-11 should make clear. (d) We find v⊥ = v sin φ = 61.3 m/s, so r = mv⊥ /eB = 5.7 nm.
31. (a) We solve for B from m = B2qx2/8V (see Sample Problem 28-3): B= 8Vm . qx 2
We evaluate this expression using x = 2.00 m:
B=
8 100 × 103 V 3.92 × 10−25 kg
c
c3.20 × 10 Chb2.00 mg
−19
hc
2
h = 0.495 T .
(b) Let N be the number of ions that are separated by the machine per unit time. The current is i = qN and the mass that is separated per unit time is M = mN, where m is the mass of a single ion. M has the value 100 × 10−6 kg M= = 2.78 × 10−8 kg s . 3600 s Since N = M/m we have 3.20 × 10−19 C 2.78 × 10−8 kg s qM i= = = 2.27 × 10−2 A . −25 3.92 × 10 kg m (c) Each ion deposits energy qV in the cup, so the energy deposited in time ∆t is given by E = NqV ∆t = For ∆t = 1.0 h, . E = 2.27 × 10−2 A 100 × 103 V 3600 s = 817 × 106 J . To obtain the second expression, i/q is substituted for N. iqV ∆t = iV ∆t . q
c
hc
h
c
hc
hb
g
32. Eq. 28-17 gives T = 2πme /eB. Thus, the total time is 1⎞ ⎛T ⎞ ⎛ T ⎞ πme ⎛ 1 ⎜ 2 ⎟ + tgap + ⎜ 2 ⎟ = e ⎜B + B ⎟ + tgap . ⎝ 1 ⎝ ⎠1 ⎝ ⎠2 2⎠ The time spent in the gap (which is where the electron is accelerating in accordance with Eq. 2-15) requires a few steps to figure out: letting t = tgap then we want to solve
2 K 0 1 ⎛ e∆V 1 d = v0t + at 2 ⇒ 0.25 m = t+ ⎜ 2 me 2 ⎝ me d ⎞ 2 ⎟t ⎠
for t. We find in this way that the time spent in the gap is t ≈ 6 ns. Thus, the total time is 8.7 ns.
33. Each of the two particles will move in the same circular path, initially going in the opposite direction. After traveling half of the circular path they will collide. Therefore, using Eq. 28-17, the time is given by t=
π ( 9.11×10−31 kg ) T πm = = = 5.07 ×10−9 s. −3 −19 2 Bq (3.53 ×10 T) (1.60 ×10 C )
34. Let v = v cos θ . The electron will proceed with a uniform speed v|| in the direction of B while undergoing uniform circular motion with frequency f in the direction perpendicular to B: f = eB/2πme. The distance d is then d = v||T = v|| f
( v cosθ ) 2πme = 2π (1.5 ×107 m s )( 9.11×10−31 kg ) ( cos10° ) = 0.53m. =
eB
(1.60 ×10
−19
C )(1.0 ×10−3 T )
35. We approximate the total distance by the number of revolutions times the circumference of the orbit corresponding to the average energy. This should be a good approximation since the deuteron receives the same energy each revolution and its period does not depend on its energy. The deuteron accelerates twice in each cycle, and each time it receives an energy of qV = 80 × 103 eV. Since its final energy is 16.6 MeV, the number of revolutions it makes is n= 16.6 × 106 eV = 104 . 2 80 × 103 eV
c
h
Its average energy during the accelerating process is 8.3 MeV. The radius of the orbit is given by r = mv/qB, where v is the deuteron’s speed. Since this is given by v = 2 K m , the radius is m 2K 1 r= = 2 Km . qB m qB For the average energy r= 2 8.3 × 106 eV 160 × 10−19 J eV 3.34 × 10−27 kg . . . c160 × 10 Chb157 Tg
−19
c
hc
hc
h = 0.375 m .
The total distance traveled is about n2πr = (104)(2π)(0.375) = 2.4 × 102 m.
36. (a) Using Eq. 28-23 and Eq. 28-18, we find
(1.60 ×10−19 C ) (1.20T ) = 1.83×107 Hz. qB = f osc = 2πm p 2π (1.67 ×10−27 kg )
(b) From r = mp v qB = 2mP k qB we have
( rqB ) K=
2m p
2
⎡( 0.500m ) (1.60 ×10−19 C ) (1.20T ) ⎤ ⎦ = 1.72 ×107 eV. =⎣ −27 −19 2 (1.67 ×10 kg )(1.60 ×10 J eV )
2
37. (a) By conservation of energy (using qV for the potential energy which is converted into kinetic form) the kinetic energy gained in each pass is 200 eV. (b) Multiplying the part (a) result by n = 100 gives ∆K = n(200 eV) = 20.0 keV. (c) Combining Eq. 28-16 with the kinetic energy relation (n(200 eV) = mpv2/2 in this particular application) leads to the expression mp r=eB 2n(200 eV) . mp
which shows that r is proportional to n . Thus, the percent increase defined in the problem in going from n = 100 to n = 101 is 101/100 – 1 = 0.00499 or 0.499%.
38. (a) The magnitude of the field required to achieve resonance is 2π fm p q 2π(12.0×106 Hz) (1.67 ×10−27 kg ) 1.60 ×10−19 C
B=
=
= 0.787T.
(b) The kinetic energy is given by 1 1 2 K = 1 mv 2 = m ( 2πRf ) = (1.67 ×10−27 kg ) 4π 2 (0.530 m) 2 (12.0 ×106 Hz) 2 2 2 2 6 −12 = 1.33 ×10 J = 8.34 ×10 eV. (c) The required frequency is
(1.60 ×10−19 C ) (1.57T ) = 2.39 ×107 Hz. qB f= = 2πm p 2π (1.67 ×10−27 kg )
(d) The kinetic energy is given by 1 1 2 K = 1 mv 2 = m ( 2πRf ) = (1.67 ×10−27 kg ) 4π 2 (0.530 m) 2 (2.39 ×107 Hz) 2 2 2 2 −12 = 5.3069 ×10 J = 3.32 ×107 eV.
39. (a) The magnetic force on the wire must be upward and have a magnitude equal to the gravitational force mg on the wire. Since the field and the current are perpendicular to each other the magnitude of the magnetic force is given by FB = iLB, where L is the length of the wire. Thus,
2 mg ( 0.0130 kg ) ( 9.8 m s ) iLB = mg ⇒ i = = = 0.467 A. LB ( 0.620 m )( 0.440 T )
(b) Applying the right-hand rule reveals that the current must be from left to right.
40. (a) From symmetry, we conclude that any x-component of force will vanish (evaluated over the entirety of the bent wire as shown). By the right-hand rule, a field in the k direction produces on each part of the bent wire a y-component of force pointing in the − j direction; each of these components has magnitude | Fy | = i | B | sin 30° = (2.0 A)(2.0 m)(4.0 T) sin 30° = 8 N. Therefore, the force on the wire shown in the figure is (−16ˆ N . j) (b) The force exerted on the left half of the bent wire points in the − k direction, by the right-hand rule, and the force exerted on the right half of the wire points in the + k direction. It is clear that the magnitude of each force is equal, so that the force (evaluated over the entirety of the bent wire as shown) must necessarily vanish.
41. (a) The magnitude of the magnetic force on the wire is given by FB = iLB sin φ, where i is the current in the wire, L is the length of the wire, B is the magnitude of the magnetic field, and φ is the angle between the current and the field. In this case φ = 70°. Thus, FB = 5000 A 100 m 60.0 × 10−6 T sin 70° = 28.2 N .
b
gb
gc
h
(b) We apply the right-hand rule to the vector product FB = iL × B to show that the force is to the west.
42. The magnetic force on the (straight) wire is
FB = iBL sin θ = (13.0A ) (1.50T ) (1.80m ) ( sin 35.0° ) = 20.1N.
43. The magnetic force on the wire is
ˆ ˆ i j j FB = iL × B = iLˆ × By ˆ + Bz k = iL − Bz ˆ + By k
(
)
(
)
ˆ j = ( 0.500A ) ( 0.500m ) ⎡ − ( 0.0100T ) ˆ + ( 0.00300T ) k ⎤ ⎣ ⎦ −3 ˆ −3 ˆ = −2.50 ×10 j + 0.750 ×10 k N.
(
)
44. (a) The magnetic force on the wire is FB = idB, pointing to the left. Thus
v = at =
FB t idBt (9.13×10−3 A)(2.56 ×10−2 m)(5.63×10−2 T)(0.0611s) = = 2.41×10−5 kg m m
= 3.34 ×10−2 m/s. (b) The direction is to the left (away from the generator).
45. (a) The magnetic force must push horizontally on the rod to overcome the force of friction, but it can be oriented so that it also pulls up on the rod and thereby reduces both the normal force and the force of friction. The forces acting on the rod are: F, the force of the magnetic field; mg, the magnitude of the (downward) force of gravity; FN , the normal force exerted by the stationary rails upward on the rod; and f , the (horizontal) force of friction. For definiteness, we assume the rod is on the verge of moving eastward, which means that f points westward (and is equal to its maximum possible value µsFN). Thus, F has an eastward component Fx and an upward component Fy, which can be related to the components of the magnetic field once we assume a direction for the current in the rod. Thus, again for definiteness, we assume the current flows northward. Then, by the right-hand rule, a downward component (Bd) of B will produce the eastward Fx, and a westward component (Bw) will produce the upward Fy. Specifically, Fx = iLBd , Fy = iLBw . Considering forces along a vertical axis, we find FN = mg − Fy = mg − iLBw so that
f = f s , max = µ s mg − iLBw .
b
g
It is on the verge of motion, so we set the horizontal acceleration to zero:
Fx − f = 0 ⇒ iLBd = µ s ( mg − iLBw ) .
The angle of the field components is adjustable, and we can minimize with respect to it. Defining the angle by Bw = B sinθ and Bd = B cosθ (which means θ is being measured from a vertical axis) and writing the above expression in these terms, we obtain
iLB cos θ = µ s ( mg − iLB sin θ ) ⇒ B =
µ s mg iL ( cos θ + µ s sin θ )
which we differentiate (with respect to θ) and set the result equal to zero. This provides a determination of the angle:
θ = tan −1 µ s = tan −1 0.60 = 31° .
Consequently,
Bmin = 0.60 (1.0 kg ) 9.8 m s 2
b g
b g
)
( 50 A )(1.0 m )( cos 31° + 0.60sin 31° )
(
= 0.10 T.
(b) As shown above, the angle is θ = tan −1 ( µ s ) = tan −1 ( 0.60 ) = 31°.
46. We use dFB = idL × B , where dL = dx i and B = Bx i + By j . Thus,
xf xf ˆ FB = ∫ idL × B = ∫ idxˆ × Bx ˆ + By ˆ = i ∫ By dxk i i j
= ( −5.0A )
(∫
xi
3.0
1.0
ˆ ˆ (8.0 x dx ) ( m ⋅ mT ) ) k = (−0.35N)k.
2
(
)
xi
47. The applied field has two components: Bx > 0 and Bz > 0. Considering each straightsegment of the rectangular coil, we note that Eq. 28-26 produces a non-zero force only for the component of B which is perpendicular to that segment; we also note that the equation is effectively multiplied by N = 20 due to the fact that this is a 20-turn coil. Since we wish to compute the torque about the hinge line, we can ignore the force acting on the straight-segment of the coil which lies along the y axis (forces acting at the axis of rotation produce no torque about that axis). The top and bottom straight-segments experience forces due to Eq. 28-26 (caused by the Bz component), but these forces are (by the right-hand rule) in the ±y directions and are thus unable to produce a torque about the y axis. Consequently, the torque derives completely from the force exerted on the straight-segment located at x = 0.050 m, which has length L = 0.10 m and is shown in Figure 28-47 carrying current in the –y direction. Now, the Bz component will produce a force on this straight-segment which points in the –x direction (back towards the hinge) and thus will exert no torque about the hinge. However, the Bx component (which is equal to B cosθ where B = 0.50 T and θ = 30°) produces a force equal to NiLBx which points (by the right-hand rule) in the +z direction. Since the action of this force is perpendicular to the plane of the coil, and is located a distance x away from the hinge, then the torque has magnitude
τ = ( NiLBx )( x ) = NiLxB cos θ = ( 20 )( 0.10 A )( 0.10 m )( 0.050 m )( 0.50 T ) cos 30°
= 0.0043 N ⋅ m . Since τ = r × F , the direction of the torque is –y. In unit-vector notation, the torque is ˆ τ = (−4.3×10−3 N ⋅ m)j An alternative way to do this problem is through the use of Eq. 28-37. We do not show those details here, but note that the magnetic moment vector (a necessary part of Eq. 2837) has magnitude µ = NiA = 20 010 A 0.0050 m2 .
b gb
gc
h
and points in the –z direction. At this point, Eq. 3-30 may be used to obtain the result for the torque vector.
48. We establish coordinates such that the two sides of the right triangle meet at the origin, and the y = 50 cm side runs along the +y axis, while the x = 120 cm side runs along the +x axis. The angle made by the hypotenuse (of length 130 cm) is
θ = tan–1 (50/120) = 22.6°,
relative to the 120 cm side. If one measures the angle counterclockwise from the +x direction, then the angle for the hypotenuse is 180° – 22.6° = +157°. Since we are only asked to find the magnitudes of the forces, we have the freedom to assume the current is flowing, say, counterclockwise in the triangular loop (as viewed by an observer on the +z axis. We take B to be in the same direction as that of the current flow in the hypotenuse. Then, with B = B = 0.0750 T, Bx = − B cos θ = −0.0692 T , By = B sin θ = 0.0288T.
(a) Eq. 28-26 produces zero force when L || B so there is no force exerted on the hypotenuse of length 130 cm. (b) On the 50 cm side, the Bx component produces a force i y Bx k, and there is no contribution from the By component. Using SI units, the magnitude of the force on the side is therefore . b4.00 Agb0.500 mgb0.0692 Tg = 0138 N.
y
(c) On the 120 cm side, the By component produces a force i x By k, and there is no contribution from the Bx component. The magnitude of the force on the . . b4.00 Agb120 mgb0.0288 Tg = 0138 N. (d) The net force is i y Bx k + i x By k = 0, keeping in mind that Bx < 0 due to our initial assumptions. If we had instead assumed B went the opposite direction of the current flow in the hypotenuse, then Bx > 0 but By < 0 and a zero net force would still be the result.
x
side is also
49. Consider an infinitesimal segment of the loop, of length ds. The magnetic field is perpendicular to the segment, so the magnetic force on it has magnitude dF = iB ds. The horizontal component of the force has magnitude dFh = (iB cos θ )ds and points inward toward the center of the loop. The vertical component has magnitude dFy = (iB sin θ )ds and points upward. Now, we sum the forces on all the segments of the loop. The horizontal component of the total force vanishes, since each segment of wire can be paired with another, diametrically opposite, segment. The horizontal components of these forces are both toward the center of the loop and thus in opposite directions. The vertical component of the total force is Fv = iB sin θ ∫ ds = 2πaiB sin θ = 2π (0.018 m)(4.6 ×10−3 A)(3.4 ×10−3 T) sin 20° = 6.0 ×10−7 N. We note that i, B, and θ have the same value for every segment and so can be factored from the integral.
50. The insight central to this problem is that for a given length of wire (formed into a rectangle of various possible aspect ratios), the maximum possible area is enclosed when the ratio of height to width is 1 (that is, when it is a square). The maximum possible value for the width, the problem says, is x = 4 cm (this is when the height is very close to zero, so the total length of wire is effectively 8 cm). Thus, when it takes the shape of a square the value of x must be ¼ of 8 cm; that is, x = 2 cm when it encloses maximum area (which leads to a maximum torque by Eq. 28-35 and Eq. 28-37) of A = (0.020 m)2 = 0.00040 m2. Since N = 1 and the torque in this case is given as 4.8 × 10−4 N.m, then the aforementioned equations lead immediately to i = 0.0030 A.
51. (a) The current in the galvanometer should be 1.62 mA when the potential difference across the resistor-galvanometer combination is 1.00 V. The potential difference across the galvanometer alone is iRg = 162 × 10−3 A 75.3 Ω = 0122 V, . . so the resistor must be in series with the galvanometer and the potential difference across it must be 1.00 V – 0.122 V = 0.878V. The resistance should be R = 0.878 V
c
hb
g
b
. g c162 × 10 Ah = 542 Ω.
−3
(b) As stated above, the resistor is in series with the galvanometer. (c) The current in the galvanometer should be 1.62 mA when the total current in the resistor and galvanometer combination is 50.0 mA. The resistor should be in parallel with the galvanometer, and the current through it should be 50.0 mA – 1.62 mA = 48.38 mA. The potential difference across the resistor is the same as that across the galvanometer, 0.122 V, so the resistance should be R = 0122 V 48.38 × 10−3 A = 2.52 Ω. .
b
gc
h
(d) As stated in (c), the resistor is in parallel with the galvanometer.
52. We use τ max = | µ × B |max = µ B = iπr 2 B, and note that i = qf = qv/2πr. So
τ max = ⎜
1 1 ⎛ qv ⎞ 2 −19 6 −11 −3 ⎟ πr B = qvrB = (1.60 ×10 C)(2.19 ×10 m/s)(5.29 ×10 m)(7.10 ×10 T) 2 2 ⎝ 2πr ⎠ −26 = 6.58 ×10 N ⋅ m.
53. We use Eq. 28-37 where µ is the magnetic dipole moment of the wire loop and B is the magnetic field, as well as Newton’s second law. Since the plane of the loop is parallel to the incline the dipole moment is normal to the incline. The forces acting on the cylinder are the force of gravity mg, acting downward from the center of mass, the normal force of the incline FN, acting perpendicularly to the incline through the center of mass, and the force of friction f, acting up the incline at the point of contact. We take the x axis to be positive down the incline. Then the x component of Newton’s second law for the center of mass yields mg sin θ − f = ma. For purposes of calculating the torque, we take the axis of the cylinder to be the axis of rotation. The magnetic field produces a torque with magnitude µB sinθ, and the force of friction produces a torque with magnitude fr, where r is the radius of the cylinder. The first tends to produce an angular acceleration in the counterclockwise direction, and the second tends to produce an angular acceleration in the clockwise direction. Newton’s second law for rotation about the center of the cylinder, τ = Iα, gives fr − µB sin θ = Iα . Since we want the current that holds the cylinder in place, we set a = 0 and α = 0, and use one equation to eliminate f from the other. The result is mgr = µ B. The loop is rectangular with two sides of length L and two of length 2r, so its area is A = 2rL and the dipole moment is µ = NiA = Ni (2rL). Thus, mgr = 2 NirLB and 0.250 kg 9.8 m s2 mg i= = = 2.45 A . 2 NLB 2 10.0 0100 m 0.500 T .
b gb
b
gc
gb
h
g
54. (a) µ = NAi = πr 2i = π 0.150 m (b) The torque is
b
. g b2.60 Ag = 0184A ⋅ m .
2 2
τ = µ × B = µ B sin θ = ( 0.184 A ⋅ m 2 ) (12.0 T ) sin 41.0° = 1.45 N ⋅ m.
55. (a) The magnitude of the magnetic dipole moment is given by µ = NiA , where N is the number of turns, i is the current in each turn, and A is the area of a loop. In this case the loops are circular, so A = πr2, where r is the radius of a turn. Thus
i=
µ
Nπr 2
=
b160gbπgb0.0190 mg
2.30 A ⋅ m2
2
= 12.7 A .
(b) The maximum torque occurs when the dipole moment is perpendicular to the field (or the plane of the loop is parallel to the field). It is given by
τ max = µ B = ( 2.30 A ⋅ m 2 )( 35.0 ×10−3 T ) = 8.05 ×10−2 N ⋅ m.
56. From µ = NiA = iπr2 we get
i=
µ
πr 2
=
8.00 × 1022 J T π 3500 × 10 m
c
3
h
2
= 2.08 × 109 A.
57. (a) The area of the loop is A =
b30 cmgb40 cmg = 6.0 × 10 cm , so µ = iA = b5.0 A gc6.0 × 10 m h = 0.30 A ⋅ m .
1 2 2 2
−2
2
2
(b) The torque on the loop is
τ = µB sin θ = 0.30 A ⋅ m2 80 × 103 T sin 90° = 2.4 × 10−2 N ⋅ m.
c
hc
h
58. (a) The kinetic energy gained is due to the potential energy decrease as the dipole swings from a position specified by angle θ to that of being aligned (zero angle) with the field. Thus, K = U i − U f = − µB cosθ − − µB cos 0° .
b
g
Therefore, using SI units, the angle is
θ = cos−1 1 −
FG H
K 0.00080 = cos−1 1 − 0.020 0.052 µB
IJ K
F GH b
gb
I = 77° . gJK
(b) Since we are making the assumption that no energy is dissipated in this process, then the dipole will continue its rotation (similar to a pendulum) until it reaches an angle θ = 77° on the other side of the alignment axis.
59. (a) The magnitude of the magnetic moment vector is
2 2 µ = ∑ in An = πr12i1 + πr22i2 = π ( 7.00A ) ⎡( 0.200m ) + ( 0.300m ) ⎤ = 2.86A ⋅ m 2 .
n
(b) Now,
2 2 µ = πr22i2 − πr12i1 = π ( 7.00A ) ⎡( 0.300m ) − ( 0.200m ) ⎤ = 1.10A ⋅ m 2 .
60. Eq. 28-39 gives U = − µ · B = −µB cos φ, so at φ = 0 (corresponding to the lowest point on the graph in Fig. 28-52) the mechanical energy is K + U = Ko + (−µB) = 6.7 × 10−4 J + (−5 × 10−4 J) = 1.7 × 10−4 J. The turning point occurs where K = 0, which implies Uturn = 1.7 × 10−4 J. So the angle where this takes place is given by
→ →
φ = − cos −1 ⎜
⎛ 1.7 × 10−4 J ⎞ ⎟ = 110° µB ⎝ ⎠
where we have used the fact (see above) that µB = 5 × 10−4 J.
61. The magnetic dipole moment is µ = µ 0.60 i − 0.80 j , where
e
j
µ = NiA = Niπr2 = 1(0.20 A)π(0.080 m)2 = 4.02 × 10–4 A·m2.
Here i is the current in the loop, N is the number of turns, A is the area of the loop, and r is its radius. (a) The torque is
τ = µ × B = µ 0.60i − 0.80 j × 0.25i + 0.30k
j e j = µ b0.60gb0.30ge i × k j − b0.80gb0.25ge j × i j − b0.80gb0.30ge j × k j
= µ −018 j + 0.20k − 0.24 i . .
e
Here i × k = − j, j × i = − k, and j × k = i are used. We also use i × i = 0 . Now, we substitute the value for µ to obtain
ˆ i j τ = −9.7 × 10−4 ˆ − 7.2 × 10−4 ˆ + 8.0 × 10−4 k N ⋅ m.
(
)
(b) The potential energy of the dipole is given by
U = − µ ⋅ B = − µ 0.60 i − 0.80 j ⋅ 0.25i + 0.30k = − µ 0.60 0.25 = −015µ = −6.0 × 10−4 J. .
b gb g
e
je
j
Here i ⋅ i = 1, i ⋅ k = 0, j ⋅ i = 0, and j ⋅ k = 0 are used.
62. Let a = 30.0 cm, b = 20.0 cm, and c = 10.0 cm. From the given hint, we write
ˆ ˆ j j ˆ j µ = µ1 + µ 2 = iab −k + iac ˆ = ia cˆ − bk = ( 5.00A )( 0.300m ) ⎡( 0.100m ) ˆ − ( 0.200m ) k ⎤ ⎣ ⎦
2
( ) () ( ˆ j = ( 0.150ˆ − 0.300k ) A ⋅ m .
)
63. If N closed loops are formed from the wire of length L, the circumference of each loop is L/N, the radius of each loop is R = L/2πN, and the area of each loop is 2 A = πR 2 = π L 2 πN = L2 4 πN 2 .
b
g
(a) For maximum torque, we orient the plane of the loops parallel to the magnetic field, so the dipole moment is perpendicular (i.e., at a 90° angle) to the field. (b) The magnitude of the torque is then
τ = NiAB = Ni
iL B b g FGH 4πLN IJK B = 4πN .
2 2 2
To maximize the torque, we take the number of turns N to have the smallest possible value, 1. Then τ = iL2B/4π. (c) The magnitude of the maximum torque is iL2 B (4.51×10−3 A)(0.250 m) 2 (5.71×10−3 T) τ= = = 1.28 ×10−7 N ⋅ m 4π 4π
64. Looking at the point in the graph (Fig. 28-54(b)) corresponding to i2 = 0 (which means that coil 2 has no magnetic moment) we are led to conclude that the magnetic moment of coil 1 must be µ1 = 2.0 ×10−5 A ⋅ m 2 . Looking at the point where the line crosses the axis (at i2 = 5.0 mA) we conclude (since the magnetic moments cancel there) that the magnitude of coil 2’s moment must also be µ2 = 2.0 × 10−5 A ⋅ m 2 when i2 = 0.0050 A which means (Eq. 28-35) N 2 A2 =
µ2
i2
=
2.0 × 10−5 A ⋅ m 2 = 4.0 × 10−3 m 2 . 0.0050 A
Now the problem has us consider the direction of coil 2’s current changed so that the net moment is the sum of two (positive) contributions – from coil 1 and coil 2 – specifically for the case that i2 = 0.007 A. We find that total moment is
µ = (2.0 × 10−5 A·m2) + (N2A2 i2) = 4.8 × 10−5 A·m2.
65. (a) Using Eq. 28-35 and Figure 28-23, we have j) j µ = ( NiA)(−ˆ = −(0.0240 A ⋅ m 2 )ˆ . Then, Eq. 28-38 gives U = − µ ⋅ B = −(−0.0240 A ⋅ m 2 )( − 3.00 × 10−3 T) = −7.20 × 10−5 J . (b) Using the fact that j × j = 0, Eq. 28-37 leads to
→ ^ ^
τ = µ × B = (–0.0240j) × (2.00 × 10−3 i) + (–0.0240j) × (–4.00 × 10−3 k) ^ ^ = (4.80 × 10−5 k + 9.60 × 10−5 i ) N·m.
^ ^ ^ ^
66. The unit vector associated with the current element (of magnitude d ) is − j. The (infinitesimal) force on this element is dF = i d − j × 0.3 y i + 0.4 yj
e j e
j
with SI units (and 3 significant figures) understood. Since j × i = − k and j × j = 0 , we obtain ˆ ˆ dF = 0.3iy d k = ( 6.00 × 10−4 N m 2 ) y d k . We integrate the force element found above, using the symbol ξ to stand for the coefficient 6.00 × 10–4 N/m2, and obtain
2 ˆ 0.25 ydy = ξ k ⎛ 0.25 ⎞ = (1.88 ×10−5 N)k . ˆ ˆ F = ∫ dF = ξ k ∫ ⎜ ⎟ 0 2 ⎠ ⎝
67. The period of revolution for the iodine ion is T = 2πr/v = 2πm/Bq, which gives 45.0 × 10−3 T 160 × 10−19 C 129 × 10−3 s . . BqT m= = = 127 u. 2π 7 2 π 166 × 10 −27 kg u .
c
hc b gb gc
hc
h
h
68. (a) The largest value of force occurs if the velocity vector is perpendicular to the field. Using Eq. 28-3, FB,max = |q| vB sin (90°) = ev B = (1.60 × 10– 19 C) (7.20 × 106 m/s) (83.0 × 10– 3 T) = 9.56 × 10– 14 N. (b) The smallest value occurs if they are parallel: FB,min = |q| vB sin (0) = 0. (c) By Newton’s second law, a = FB/me = |q| vB sin θ /me, so the angle θ between v and B is
θ = sin
−1
F m a I = sin GH q vB JK
e
−1
LM c9.11 × 10 kghd4.90 × 10 m s i . MN c160 × 10 Chc7.20 × 10 m shc83.0 × 10
−31
14 2
−16
6
−3
OP = 0.267° . Th P Q
69. (a) We use τ = µ × B, where µ points into the wall (since the current goes clockwise around the clock). Since B points towards the one-hour (or “5-minute’’) mark, and (by the properties of vector cross products) τ must be perpendicular to it, then (using the right-hand rule) we find τ points at the 20-minute mark. So the time interval is 20 min. (b) The torque is given by
τ =| µ × B |= µ B sin 90° = NiAB = πNir 2 B = 6π ( 2.0A )( 0.15m ) ( 70 ×10−3 T )
2
= 5.9 ×10−2 N ⋅ m.
70. (a) We use Eq. 28-10: vd = E/B = (10 × 10–6V/1.0 × 10–2 m)/(1.5 T) = 6.7 × 10–4 m/s. (b) We rewrite Eq. 28-12 in terms of the electric field:
n= Bi Bi Bi = = V e Ed e EAe
b g
which we use A = d . In this experiment, A = (0.010 m)(10 × 10–6 m) = 1.0 × 10–7 m2. By Eq. 28-10, vd equals the ratio of the fields (as noted in part (a)), so we are led to
n= Bi i 3.0 A = = = 2.8 × 1029 m3 . 2 −4 −7 −19 E Ae vd Ae 6.7 × 10 m s 1.0 ×10 m 1.6 × 10 C
(
)(
)(
)
(c) Since a drawing of an inherently 3-D situation can be misleading, we describe it in terms of horizontal north, south, east, west and vertical up and down directions. We assume B points up and the conductor’s width of 0.010 m is along an east-west line. We take the current going northward. The conduction electrons experience a westward magnetic force (by the right-hand rule), which results in the west side of the conductor being negative and the east side being positive (with reference to the Hall voltage which becomes established).
71. From m = B2qx2/8V we have ∆m = (B2q/8V)(2x∆x). Here x = 8Vm B 2 q , which we substitute into the expression for ∆m to obtain
F B q IJ 2 ∆m = G H 8V K
2
mq 8mV ∆x = B ∆x . 2 Bq 2V
Thus, the distance between the spots made on the photographic plate is
∆x = =
∆m 2V B mq
( 37 u − 35 u ) (1.66 ×10−27 kg
0.50 T
u
)
2 7.3 × 103 V
( 36 u ) (1.66 ×10
(
−27
kg u 1.60 × 10−19 C
)(
)
)
= 8.2 × 10−3 m.
72. (a) Equating the magnitude of the electric force (Fe = eE) with that of the magnetic force (Eq. 28-3), we obtain B = E / v sin φ. The field is smallest when the sin φ factor is at 1 its largest value; that is, when φ = 90°. Now, we use K = mv 2 to find the speed: 2
2K v= = me 2 2.5 × 103 eV 160 × 10−19 J eV . 9.11 × 10
−31
c
hc
kg
h = 2.96 × 10 m s.
7
Thus, B= E 10 × 103 V m = = 3.4 × 10−4 T. 7 v 2.96 × 10 m s
The direction of the magnetic field must be perpendicular to both the electric field ( −ˆ ) j j and the velocity of the electron ( + ˆ ). Since the electric force Fe = (−e) E points in the + ˆ i j direction, the magnetic force F = (−e)v × B points in the −ˆ direction. Hence, the
B
ˆ ˆ direction of the magnetic field is − k . In unit-vector notation, B = (−3.4 ×10−4 T)k.
73. The fact that the fields are uniform, with the feature that the charge moves in a straight line, implies the speed is constant (if it were not, then the magnetic force would vary while the electric force could not — causing it to deviate from straight-line motion). This is then the situation leading to Eq. 28-7, and we find | E | = v| B| = 500 V m. j Its direction (so that F = q E + v × B vanishes) is downward, or −ˆ , in the “page” ˆ coordinates. In unit-vector notation, E = (−500 V/m)j
d
i
74. (a) For the magnetic field to have an effect on the moving electrons, we need a nonnegligible component of B to be perpendicular to v (the electron velocity). It is most efficient, therefore, to orient the magnetic field so it is perpendicular to the plane of the page. The magnetic force on an electron has magnitude FB = evB, and the acceleration of the electron has magnitude a = v2/r. Newton’s second law yields evB = mev2/r, so the radius of the circle is given by r = mev/eB in agreement with Eq. 28-16. The kinetic 1 energy of the electron is K = 2 me v 2 , so v = 2 K me . Thus,
r= me 2 K = eB me
2me K . e2 B 2
This must be less than d, so
2me K 2me K ≤ d , or B ≥ . 2 2 e2d 2 e B
(b) If the electrons are to travel as shown in Fig. 28-57, the magnetic field must be out of the page. Then the magnetic force is toward the center of the circular path, as it must be (in order to make the circular motion possible).
75. (a) Since K = qV we have K p = 1 Kα ( as qα = 2 K p ) , or K p / Kα = 0.50. 2 (b) Similarly, qα = 2 K d , K d / Kα = 0.50. (c) Since r = 2mK qB ∝ mK q , we have
rd = md K d q p rp = m p K p qd
( 2.00u ) K p r = 10 (1.00u ) K p p
2cm = 14cm.
(d) Similarly, for the alpha particle, we have
rα = mα Kα q p rp = m p K p qα
( 4.00u ) Kα erp = 10 (1.00u ) ( Kα 2 ) 2e
2cm=14cm.
76. The equation of motion for the proton is
F = qv × B = q v x i + v y j + vz k × B i = qB vz j − v y k
x y z p p
e e j LF dv IJ i + FG dv IJ j + FG dv IJ kOP . = m a = m MG NH dt K H dt K H dt K Q
dv y dvx dvz = 0, = ω vz , = −ω v y , dt dt dt
j
Thus,
where ω = eB/m. The solution is vx = v0x, vy = v0y cos ωt and vz = –v0y sin ωt. In summary, we have v t = v0 x i + v0 y cos ωt j − v0 y sin ωt k .
bg
b g
b
g
77. By the right-hand rule, we see that v × B points along − k . From Eq. 28-2 ˆ F = qv × B , we find that for the force to point along + k , we must have q < 0. Now,
d
i
examining the magnitudes in Eq. 28-3, we find | F | = | q | v | B |sin φ , or
0.48 N = | q | ( 4000 m/s ) ( 0.0050 T ) sin 35°
which yields |q| = 0.040 C. In summary, then, q = –40 mC.
q v = . With the speed of the ion m B′r giving by v = E / B (using Eq. 28-7), the expression becomes 78. Using Eq. 28-16, the charge-to-mass ratio is q E/B E = = . m B′r BB′r
79. (a) We use Eq. 28-2 and Eq. 3-30:
F = qv × B = ( +e ) = 1.60 ×10−19 ˆ (( v B − v B ) ˆi + ( v B − v B ) ˆj + ( v B − v B ) k ) i+ ) ( ( ( 4 )( 0.008) − ( −6 ) ( −0.004 ) ) ˆ
y z z y z x x z x y y x
(
ˆ ( ( −6 )( 0.002 ) − ( −2 ) ( 0.008) ) ˆj + ( ( −2 )( −0.004 ) − ( 4 )( 0.002 ) ) k )
i j = (1.28 ×10−21 ) ˆ + ( 6.41×10−22 ) ˆ with SI units understood. (b) By definition of the cross product, v ⊥ F . This is easily verified by taking the dot (scalar) product of v with the result of part (a), yielding zero, provided care is taken not to introduce any round-off error. (c) There are several ways to proceed. It may be worthwhile to note, first, that if Bz were 6.00 mT instead of 8.00 mT then the two vectors would be exactly antiparallel. Hence, the angle θ between B and v is presumably “close” to 180°. Here, we use Eq. 3-20: ⎛ v⋅B ⎞ −68 ⎞ −1 ⎛ θ = cos −1 ⎜ ⎟ = cos ⎜ ⎟ = 173° ⎝ 56 84 ⎠ ⎝ | v || B | ⎠
80. (a) In Chapter 27, the electric field (called EC in this problem) which “drives” the current through the resistive material is given by Eq. 27-11, which (in magnitude) reads EC = ρJ. Combining this with Eq. 27-7, we obtain EC = ρnevd . Now, regarding the Hall effect, we use Eq. 28-10 to write E = vdB. Dividing one equation by the other, we get E/Ec = B/neρ. (b) Using the value of copper’s resistivity given in Chapter 26, we obtain
E B 0.65 T = = = 2.84 × 10−3. −19 −8 28 3 Ec ne ρ 8.47 × 10 m 1.60 ×10 C 1.69 × 10 Ω ⋅ m
(
)(
)(
)
81. (a) The textbook uses “geomagnetic north” to refer to Earth’s magnetic pole lying in the northern hemisphere. Thus, the electrons are traveling northward. The vertical component of the magnetic field is downward. The right-hand rule indicates that v × B is to the west, but since the electron is negatively charged (and F = qv × B ), the magnetic force on it is to the east. We combine F = mea with F = evB sin φ. Here, B sin φ represents the downward component of Earth’s field (given in the problem). Thus, a = evB / me. Now, the electron 1 speed can be found from its kinetic energy. Since K = mv 2 , 2
2 12.0 × 103 eV 160 × 10−19 J eV . 2K v= = = 6.49 × 107 m s . me 9.11 × 10−31 kg
c
hc
h
Therefore,
−19 7 −6 evB (1.60 ×10 C ) ( 6.49 ×10 m s ) ( 55.0 ×10 T ) 2 2 = = 6.27 × 1014 m s ≈ 6.3 × 1014 m s . a= −31 9.11× 10 kg me
(b) We ignore any vertical deflection of the beam which might arise due to the horizontal component of Earth’s field. Technically, then, the electron should follow a circular arc. However, the deflection is so small that many of the technicalities of circular geometry may be ignored, and a calculation along the lines of projectile motion analysis (see Chapter 4) provides an adequate approximation: ∆x = vt ⇒ t = ∆x 0.200m = = 3.08 × 10−9 s . 7 v 6.49 ×10 m s
Then, with our y axis oriented eastward, ∆y =
2 1 2 1 at = ( 6.27 × 1014 ) ( 3.08 × 10−9 ) = 0.00298m ≈ 0.0030 m. 2 2
ˆ 82. (a) We are given B = Bx ˆ = (6 × 10−5T)i , so that v × B = − v y Bx k where vy = 4×104 m/s. i
We note that the magnetic force on the electron is − e − v y Bx k and therefore points in the + k direction, at the instant the electron enters the field-filled region. In these terms, Eq. 28-16 becomes me v y r= = 0.0038 m. e Bx (b) One revolution takes T = 2πr/vy = 0.60 µs, and during that time the “drift” of the electron in the x direction (which is the pitch of the helix) is ∆x = vxT = 0.019 m where vx = 32 × 103 m/s. (c) Returning to our observation of force direction made in part (a), we consider how this is perceived by an observer at some point on the –x axis. As the electron moves away from him, he sees it enter the region with positive vy (which he might call “upward’’) but “pushed” in the +z direction (to his right). Hence, he describes the electron’s spiral as clockwise.
b ge
j
83. Using Eq. 28-16, the radius of the circular path is r= mv 2mK = qB qB
where K = mv 2 / 2 is the kinetic energy of the particle. Thus, we see that r ∞ mK qB . (a) rd md K d q p 2.0u e = = = 2 ≈ 1.4 , and 1.0u e rp m p K p qd rα mα Kα q p 4.0u e = = = 1.0. rp m p K p qα 1.0u 2e
(b)
84. Letting Bx = By = B1 and Bz = B2 and using Eq. 28-2 ( F = qv × B ) and Eq. 3-30, we obtain (with SI units understood) ˆ ˆ ˆ 4i − 20ˆ + 12k = 2 ( 4 B2 − 6 B1 ) ˆ + ( 6 B1 − 2 B2 ) ˆ + ( 2 B1 − 4 B1 ) k . j i j Equating like components, we find B1 = –3 and B2 = –4. In summary, ˆ ˆ B = −3.0i − 3.0ˆ − 4.0k T. j
(
)
(
)
ˆ 85. The contribution to the force by the magnetic field B = Bx ˆ = (−0.020 T)i i
(
) is given
by Eq. 28-2: ˆ ˆ i j i i FB = qv × B = q 17000i × Bx ˆ + −11000ˆ × Bx ˆ + 7000k × Bx ˆ ˆ j = q −220k − 140ˆ
(
((
) (
) (
) (
)) )
in SI units. And the contribution to the force by the electric field E = E y ˆ = 300ˆ V/m is j j given by Eq. 23-1: FE = qE y j . Using q = 5.0 × 10–6 C, the net force on the particle is ˆ F = (0.00080ˆ − 0.0011k) N. j
86. The current is in the + i direction. Thus, the i component of B has no effect, and (with x in meters) we evaluate
F = ( 3.00A ) ∫
1 3 ˆ × ˆ = ⎛ −1.80 1 A ⋅ T ⋅ m ⎞ k = (−0.600N)k. ˆ ˆ ( −0.600 T m ) x dx i j ⎜ ⎟ 3 ⎝ ⎠ 2 2
0
( )
87. We replace the current loop of arbitrary shape with an assembly of small adjacent rectangular loops filling the same area which was enclosed by the original loop (as nearly as possible). Each rectangular loop carries a current i flowing in the same sense as the original loop. As the sizes of these rectangles shrink to infinitesimally small values, the assembly gives a current distribution equivalent to that of the original loop. The magnitude of the torque ∆τ exerted by B on the nth rectangular loop of area ∆An is given by ∆τ n = NiB sin θ∆An . Thus, for the whole assembly
τ = ∑ ∆τ n = NiB ∑ ∆An = NiAB sin θ .
n n
| 18,011
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CC-MAIN-2018-17
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latest
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| 0.827408
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https://www.elitedigitalstudy.com/1120/an-object-is-placed-at-a-distance-of-10-cm-from-a-convex-mirror-of-focal-length-of-15-cm-find-the-positionand-nature-of-the-image
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| 10,771
|
An object is placed at a distance of 10 cm from a convex mirror of focal length of 15 cm. Find the position and nature of the image.
Asked by Vishal kumar | 2 years ago | 399
##### Solution :-
Focal length of convex mirror (f) = +15 cm
Object distance (u) = - 10 cm
According to the mirror formula,
$$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$$
$$\frac{1}{v}=\frac{1}{15}-\frac{1}{-10} =\frac{2+3}{30}$$
$$v = \frac{5}{30}= 6 cm$$
$$Magnification =\frac{-v}{u}=\frac{-6}{-10}=0.6$$
The image is located at a distance of 6 cm from the mirror on the other side of the mirror. The positive and a value less than 1 of magnification indicates that the image formed is virtual and erect and diminished.
Answered by Shivani Kumari | 2 years ago
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| 4.34375
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CC-MAIN-2023-40
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longest
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https://www.univerkov.com/for-the-reduction-of-240-g-of-magnesium-oxide-80-g-of-coal-were-consumed-mgo-c-mg/
| 1,713,429,054,000,000,000
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crawl-data/CC-MAIN-2024-18/segments/1712296817200.22/warc/CC-MAIN-20240418061950-20240418091950-00509.warc.gz
| 950,256,797
| 6,263
|
# For the reduction of 240 g of magnesium oxide, 80 g of coal were consumed: MgO + C = Mg
For the reduction of 240 g of magnesium oxide, 80 g of coal were consumed: MgO + C = Mg + CO What is the mass fraction of pure carbon in this coal?
1. Let’s write down the reaction equation:
MgO + C = Mg + CO.
2. Find the amount of magnesium oxide:
n (MgO) = m (MgO) / M (MgO) = 240 g / 40 g / mol = 6 mol.
3. Using the equation, we find the amount of pure coal and the mass:
np (C) = n (MgO) = 6 mol.
mh (C) = n (C) * M (C) = 6 mol * 12 g / mol = 72 g.
4. Let’s find the mass fraction of pure coal:
ω (C) = (mh (C) / mth (C)) * 100% = (72 g / 80 g) * 100% = 90%.
| 242
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| 3.84375
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CC-MAIN-2024-18
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latest
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| 0.886247
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https://stevecotler.com/2011/04/02/final-four-math-2011/
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crawl-data/CC-MAIN-2024-38/segments/1725700651420.25/warc/CC-MAIN-20240912011254-20240912041254-00005.warc.gz
| 518,349,408
| 16,407
|
# Final Four Math — 2011
This year’s March Madness has brought us a Final Four with no #1 or #2 seeds, unique in NCAA tournament history. But the absence of high-seed teams has not dulled enthusiasm for the last three games. In fact, some sports pundits are trumpeting the “Cinderella” factor: can a #11 seed, Virginia Commonwealth University, pull off the all-time, unexpected upset?
But no matter who matter who wins the game, the bookies win their money. The bookie’s odds always include a built-in percentage for the house. In 2009 I calculated the Las Vegas Final Four edge at 9.8%; in 2010, the edge was larger (14.3%). This year the edge, if you can actually get these published odds, is unbelievably small…only 2.2%!
First a repeat of my tutorial on how the house gets its take:
Among the simplest edges to compute is Las Vegas roulette. If your chips are on one of the numbers from 1 to 36, and you win, you get paid 35-1. That means that if you put \$1 on each of those 36 numbers, when the ball drops onto one of those numbers, you’ll get your winning bet back plus \$35; you’ll break even. Those are fair odds. But the House, as I said, always has an edge. Las Vegas wheels include two other numbers that also pay 35-1: 0 and 00. So to be sure you’ll win, you’d have to place 38 one-dollar bets, thus giving the House a \$2 profit on every \$38 you bet (a 5.3% margin).
This weekend’s Final Four games pit Connecticut, Butler, Kentucky, and VCU. The odds quoted by Cantor Gaming, which runs four sports books in Las Vegas, are:
+160 (8-5) Kentucky
+220 (11-5) Connecticut
+400 (4-1) Butler
+700 (7-1) VCU
What do these odds mean? How did I compute the House’s take at a skimpy 2.2%?
The math is easy.
Assume you bet on all four teams to win the national championship. One of them will definitely triumph, and you want a \$100 payout. For Kentucky, the favorite at 8-5, you need to bet \$38.46 (I’ll explain where that strange number comes from below) If Kentucky wins, you receive \$61.54 in winnings (which is 38.46 times 8/5) plus the return of your bet, totaling \$100. For Connecticut, the odds are longer, so you need to bet less…only \$31.25. At 11-5, if the Huskies win, you’ll win \$68.75…again totaling \$100. On Butler, a longer shot at 4-1, your bet has to be \$20.00 to get back \$80.00. And on VCU, the 7-1 long shot, you need bet only \$12.50 to win \$87.50. No matter who wins, you’ll end up with \$100.
But what did it cost you?
You made four bets: \$38.46, \$31.25, \$20.00, and \$12.50. That totals \$102.21, but you only got back \$100, so the House kept \$2.21, for a profit margin of 2.2%. Something’s wrong, Las Vegas never lets you get in with such thin overhead.
So I tried another site. The odds at vegasinsider.com are what you’d expect…
+150 (3-2) Kentucky
+200 (2-1) Connecticut
+300 (3-1) Butler
+400 (4-1) VCU
Vegas Insider’s take is a fat 15.5%.
Covers.com is even greedier. They skim 16.9% with their odds set at:
+115 (23-20) Kentucky
+200 (2-1) Connecticut
+330 (33-10) Butler
+480 (24-5) VCU
Moral of this story: The horses may be the same, but the tracks compute differently.
## * * * * *
Here’s how the \$38.46 bet on Kentucky is computed:
Let B be your bet and L be the line (the odds). If your team wins, you’re going to get your bet returned PLUS your winnings, so B + L * B = \$100. The kentucky odds are 8-5.
B + 8/5 * B = \$100
13/5 * B = \$100
B = \$100 / 2.6
B = \$38.46
Similarly for the other three teams.
| 1,036
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| 3.953125
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CC-MAIN-2024-38
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en
| 0.931365
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http://clay6.com/qa/52788/what-work-is-done-by-a-force-f-2xn-i-3n-j-with-x-in-meters-that-moves-a-par
| 1,534,615,880,000,000,000
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crawl-data/CC-MAIN-2018-34/segments/1534221213693.23/warc/CC-MAIN-20180818173743-20180818193743-00259.warc.gz
| 85,791,347
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# What work is done by a force $F=(2xN)i+(3N)j$,with $x$ in meters,that moves a particle from a position $r_i=(2m)i+(3m)j$ to a position $r_f=-(4m)i-(3m)j$?
$\begin{array}{1 1}-6J\\-5J\\6J\\-15J\end{array}$
We use the general definition of work (for a two-dimensional problem),
$W=\int\limits_{x_i}^{x_f}F_x(r)dx+\int\limits_{y_i}^{y_f} F_y(r)dy$
With $F_x=2x$ and $F_y=3$
$W=\int\limits_{2m}^{-4m}2xdx+\int\limits_{3m}^{-3m}3dy$
$\;\;\;\;=x^2\big|_{2m}^{-4m}+3x\big|_{3m}^{-3m}$
$\;\;\;\;=[(16)-(4)]J+[(-9)-(9)]J$
$\;\;\;\;=-6J$
| 257
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| 3.984375
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CC-MAIN-2018-34
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http://faculty.kfupm.edu.sa/MATH/afarahat/Math%20Courses/Kfupm/Math002/Homework%20and%20Cal%20Solutions/Quiz%208s/Quiz%208s.htm
| 1,558,670,126,000,000,000
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4.
____________O___________
Again, Mathcad will only give answers that lie in the range of the inverse sine function, i. e.,
Check using Mathcad
==>
==>
Divide by 2
or
==>
Use unit circle
Take sin(2x) as a common factor
Take as a common factor
Thus,
or
Use the zero product property
or
==>
or
Solve for t
or
Substitute cos(x) for t
==>
or
==>
Given
To find the other angle (in Quadrant IV) whose cosine is 1/3, we subtract 70.5 from 360:
Given that cos-1(1/3) = 70.5o , solve for x in the interval
Solution:
Let
Substitute t for cos(x)
==>
Factor
or
(Which property did we use?)
or
The solutions are:
If we substitute x = 180o in the original equation, we get . Similarly for the value
x = 270o. Thus, x = 180o and x = 270o are not solutions.
However, because we squared both sides in the process of the solution, one has to be careful about introducing "extraneous" solutions (solutions that will not satisfy the original equation).
or
or
or
Use zero product property
or
Divide by 2
Math 002-041
Major Quiz 8 Solution
Nov. 29, 2004
Answer all questions. Show all your work
Solve the following problems for x:
1.
Solution:
Square both sides
Expand
Use the identity
Using the fact that the sine function is odd
The sine function is negative in Quadrant III and IV. Using the method of the unit circle (see Lecture Notes web page), we have
or
Check using Mathcad:
Remark
Mathcad returned only one value because it used the inverse sine function (sin-1 ). Recall that the range of sin-1(x) is between and .
_____________O_____________
3.
Solution:
Group to take common factors
or
Remark
Mathcad can be used to check your answer using the "solve" command as follows:
The solution is given in radians which is the same as the one we obtained in degrees.
____________O_____________
2.
Solution:
Using
Where
and
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# Slope Calculator
Created by Mateusz Mucha and Julia Żuławińska
Reviewed by Bogna Szyk and Jack Bowater
Last updated: Jan 30, 2024
The slope calculator determines the slope or gradient between two points in the Cartesian coordinate system. The slope is basically the amount of slant a line has and can have a positive, negative, zero, or undefined value. Before using the calculator, it is probably worth learning how to find the slope using the slope formula. To find the equation of a line for any given two points that this line passes through, use our slope intercept form calculator.
## How to use this slope calculator
Here, we will walk you through how to use this calculator, along with an example calculation, to make it simpler for you. To calculate the slope of a line, you need to know any two points on it:
1. Enter the x and y coordinates of the first point on the line.
2. Enter the x and y coordinates of the second point on the line.
3. We instantly get the slope of the line. But the magic doesn't stop there, for you also get a bunch of extra results for good measure:
• The equation of your function (same as the equation of the line).
• The y-intercept of the line.
• The angle the line makes with respect to the x-axis (measure anti-clockwise).
• Slope as a percentage (percentage grade).
• The distance between the two points.
For example, say you have a line that passes through the points (1, 5) and (7, 6). Enter the x and y coordinates of the first point, followed by the x and y coordinates of the second one. Instantly, we learn that the line's slope is 0.166667. If we need the line's equation, we also have it now: y = 0.16667x + 4.83333.
You can use this calculator in reverse and find a missing x or y coordinate! For example, consider the line that passes through the point (9, 12) and has a 12% slope. To find the point where the line crosses the y-axis (i.e., x = 0), enter 12% in percent grade (9, 12) as the coordinate of the first point, and x2 = 0. Right away, the calculator tells us that y2 = 10.92.
The slope of a line has many significant uses in geometry and calculus. The article below is an excellent introduction to the fundamentals of this topic, and we insist that you give it a read.
## The slope formula
$\mathrm{slope} = \frac{y_2 - y_1}{x_2 - x_1}$
Notice that the slope of a line is easily calculated by hand using small, whole number coordinates. The formula becomes increasingly useful as the coordinates take on larger values or decimal values.
It is worth mentioning that any horizontal line has a gradient of zero because a horizontal line has the same y-coordinates. This will result in a zero in the numerator of the slope formula. On the other hand, a vertical line will have an undefined slope since the x-coordinates will always be the same. This will result the division by zero error when using the formula.
## How to find slope
1. Identify the coordinates $(x_1, y_1)$ and $(x_2, y_2)$. We will use the formula to calculate the slope of the line passing through the points $(3, 8)$ and $(-2, 10)$.
2. Input the values into the formula. This gives us $(10 - 8)/(-2 - 3)$.
3. Subtract the values in parentheses to get $2/(-5)$.
4. Simplify the fraction to get the slope of $-2/5$.
5. Check your result using the slope calculator.
To find the slope of a line, we need two coordinates on the line. Any two coordinates will suffice. We are basically measuring the amount of change of the y-coordinate, often known as the rise, divided by the change of the x-coordinate, known as the run. The calculations in finding the slope are simple and involve nothing more than basic subtraction and division.
🙋 To find the gradient of non-linear functions, you can use the average rate of change calculator.
Just as slope can be calculated using the endpoints of a segment, the midpoint can also be calculated. The midpoint is an important concept in geometry, particularly when inscribing a polygon inside another polygon with its vertices touching the midpoint of the sides of the larger polygon. This can be obtained using the midpoint calculator or by simply taking the average of each x-coordinates and the average of the y-coordinates to form a new coordinate.
The slopes of lines are important in determining whether or not a triangle is a right triangle. If any two sides of a triangle have slopes that multiply to equal -1, then the triangle is a right triangle. The computations for this can be done by hand or by using the right triangle calculator. You can also use the distance calculator to compute which side of a triangle is the longest, which helps determine which sides must form a right angle if the triangle is right.
The sign in front of the gradient provided by the slope calculator indicates whether the line is increasing, decreasing, constant or undefined. If the graph of the line moves from lower left to upper right it is increasing and is therefore positive. If it decreases when moving from the upper left to lower right, then the gradient is negative.
## Making this slope calculator
The slope calculator is one of the oldest at Omni Calculator, built by our veterans Mateusz and Julia, who make creating accurate scientific tools look easy. The idea for this calculator was born when the two were crunching data analytics and trends and realized how a slope calculator would make their job easier. Even today, you can find them occasionally using this tool for reliable calculations.
We put extra care into the quality of our content so that it is as accurate and dependable as possible. Each tool is peer-reviewed by a trained expert and then proofread by a native speaker. You can learn more about our standards in our page.
## FAQ
### How to find slope from an equation?
The method for finding the slope from an equation will vary depending on the form of the equation in front of you. If the form of the equation is y = mx + c, then the slope (or gradient) is just m. If the equation is not in this form, try to rearrange the equation. To find the gradient of other polynomials, you will need to differentiate the function with respect to x.
### How do you calculate the slope of a hill?
1. Use a map to determine the distance between the top and bottom of the hill as the crow flies.
2. Using the same map, or GPS, find the altitude between the top and bottom of the hill. Make sure that the points you measure from are the same as step 1.
3. Convert both measurements into the same units.
4. Divide the difference in altitude by the distance between the two points.
5. This number is the gradient of the hill if it increases linearly. If it does not, repeat the steps but at where there is a noticeable change in slope.
### How do you calculate the length of a slope?
1. Measure the difference between the top and bottom of the slope in relation to both the x and y axis.
2. If you can only measure the change in x, multiply this value by the gradient to find the change in the y axis.
3. Make sure the units for both values are the same.
4. Use the Pythagorean theorem to find the length of the slope. Square both the change in x and the change in y.
5. Add the two values together.
6. Find the square root of the summation.
7. This new value is the length of the slope.
### What is a 1 in 20 slope?
A 1/20 slope is one that rises by 1 unit for every 20 units traversed horizontally. So, for example, a ramp that was 200 ft long and 10 ft tall would have a 1/20 slope. A 1/20 slope is equivalent to a gradient of 1/20 (strangely enough) and forms an angle of 2.86° between itself and the x-axis.
### How do you find the slope of a curve?
As the slope of a curve changes at each point, you can find the slope of a curve by differentiating the equation with respect to x and, in the resulting equation, substituting x for the point at which you’d like to find the gradient.
### Is rate of change the same as slope?
The rate of change of a graph is also its slope, which are also the same as gradient. Rate of change can be found by dividing the change in the y (vertical) direction by the change in the x (horizontal) direction, if both numbers are in the same units, of course. Rate of change is particularly useful if you want to predict the future of previous value of something, as, by changing the x variable, the corresponding y value will be present (and vice versa).
### Where do you use slope in everyday life?
Slopes (or gradients) have a number of uses in everyday life. There are some obvious physical examples - every hill has a slope, and the steeper the hill, the greater its gradient. This can be useful if you are looking at a map and want to find the best hill to cycle down. You also probably sleep under a slope, a roof that is. The slope of a roof will change depending on the style and where you live. But, more importantly, if you ever want to know how something changes with time, you will end up plotting a graph with a slope.
### What is a 10% slope?
A 10% slope is one that rises by 1 unit for every 10 units travelled horizontally (10 %). For example, a roof with a 10% slope that is 20 m across will be 2 m high. This is the same as a gradient of 1/10, and an angle of 5.71° is formed between the line and the x-axis.
### How do you find the area under a slope?
To find the area under a slope given by the equation y = mx + c, follow these steps:
1. Define the lower and upper bounds of x to get a value for Δx.
2. Multiply Δx by the slope (m) to obtain Δy.
3. Multiply Δx by Δy.
4. Divide by 2 to give you the area under the slope.
### What degree is a 1 to 5 slope?
A 1 to 5 slope is one that, for every increase of 5 units horizontally, rises by 1 unit. The number of degrees between a 1 to 5 slope and the x-axis is 11.3°. This can be found by first calculating the slope, by dividing the change in the y direction by the change in the x direction, and then finding the inverse tangent of the slope.
Mateusz Mucha and Julia Żuławińska
First point coordinates
x₁
y₁
Second point coordinates
x₂
y₂
Result
Slope (m)
Related numbers
Y - intercept
Angle (θ)
deg
%
Distance (d)
Distance between x's (Δx)
Distance between y's (Δy)
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http://www.glassdoor.com/Interview/Given-a-2D-rectangular-matrix-of-boolean-values-write-a-function-which-returns-whether-or-not-the-matrix-is-the-same-when-QTN_472612.htm
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Engaged Employer
## Interview Question
Interview New York, NY
# Given a 2D rectangular matrix of boolean values, write a
function which returns whether or not the matrix is the same when rotated 180 degrees. Additionally verify that every boolean true is accessible from every other boolean true if a traversal can be made to an adjacent cell in the matrix, excluding diagonal cells. That is , (x , y ) can access the set [ ( x + 1 , y ) , ( x - 1 , y ) , (x , y - 1 ) , (x , y + 1 ) ] For example, the matrix { { true , false } , { false , true } } should not pass this test.
1
if the matrix A is a11, a12 a21, a22 after 180 rotation a22, a21 a12, a11 so a11 == a22 and a12 == a21 function is BOOL isSame = (a11==a22) && (a12==a21) done.
David C on Jun 17, 2013
0
public static boolean isMatrixEqualToFlip(boolean[][] matrix) { if (matrix==null || matrix.length == 0 || matrix[0].length == 0) { return true; } int rowlen = matrix[0].length; int highInd = matrix.length/2; int lowInd = highInd - 1 + (matrix.length % 2); System.out.println("rowlen: " + rowlen + " high: "+ highInd + " low: " + lowInd); while (lowInd >= 0) { System.out.println("high: " + highInd + " lowInd: " + lowInd); for(int i=0; i < rowlen; i++) { System.out.println("Compare " + matrix[highInd][i] + " to " + matrix[lowInd][rowlen - 1 - i]); if (matrix[highInd][i] != matrix[lowInd][rowlen - 1-i]) { return false; } } lowInd--; highInd++; } return true; }
Ben H on Jul 26, 2013
0
def rotate180(mtx): col=mtx col.reverse() for row in col: row.reverse() print col
Anon on Jul 28, 2013
0
If the matrix is: true, false false, true 90 degree rotation would be: false, true true, false 180 degree rotation would be: true, false false, true If we define the matrix as: a11, a12 a21, a22 Then the solution would be: boolean isMatch = !(a11 && a12) && !(a11 && a21) && !(a12 && a22) && !(a21 && a22);
kaetem on May 9, 2014
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How do I calculate the wallpaper requirement for wallpapers with repeat?
First divide the wall width by the roll width of the wallpaper. This will give you the number of strips of wallpaper you need.
In the next step, to find out how many strips you can get from one roll of wallpaper, divide the length of the wallpaper roll by the ceiling height and the repeat. If you replace the numbers in our example with your numbers, this is easy to calculate with a calculator.
Example:
Room height: 2.65 m
Wall width: 5.00 m
Roll width of wallpaper: 0.53 m
roll length: 10.05 m
Repeat of the wallpaper: 0.72 m
How many strips of wallpaper do I need?
5,00 m (wall width) : 0,53 m (roll width) = 9,43 strips >> rounded up 10 strips
What length do I need?
1. determine the number of recurring patterns:
2.65m (room height) : 0.72 m (repeat) = 3.68 patterns >> 4 patterns.
2. calculate the length of the panels:
4 patterns x 0.72 m (repeat) = 2.88 m
How many panels do I get from one roll?
10,05 m : 2,88 = 3,49 >> rounded down 3 panels
How many rolls do I need?
10 : 3 = 3,33 >> rounded up 4 rolls
Result:
From a 10-metre roll of wallpaper with a repeat of 0.72 m, you get 3 strips with a ceiling height of 2.65 metres and a wall width of 5 metres. You therefore need 4 rolls of wallpaper to cover the wall.
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https://www.vedantu.com/question-answer/find-the-number-of-sides-of-a-polygon-having-35-class-8-maths-cbse-5ed97be8251062488817e26b
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Question
# Find the number of sides of a polygon having 35 diagonals.
Hint- The Number of diagonals are given here .So, we use the formula of finding the Number of diagonals of a polygon having n sides $= \dfrac{{n\left( {n - 3} \right)}}{2}$
As we know that the number of diagonals of polygon having n sides $= \dfrac{{n\left( {n - 3} \right)}}{2}$
Now it is given that polygons have 35 diagonals.
$\therefore 35 = \dfrac{{n\left( {n - 3} \right)}}{2}$
$\begin{gathered} \Rightarrow {n^2} - 3n = 70 \\ \Rightarrow {n^2} - 3n - 70 = 0 \\ \end{gathered}$
Now factorize the equation we have
$\begin{gathered} \Rightarrow {n^2} - 10n + 7n - 70 = 0 \\ \Rightarrow n\left( {n - 10} \right) + 7\left( {n - 10} \right) = 0 \\ \Rightarrow \left( {n - 10} \right)\left( {n + 7} \right) = 0 \\ \Rightarrow \left( {n - 10} \right) = 0{\text{ \& }}\left( {n + 7} \right) = 0 \\ \therefore n = 10,{\text{ - 7}} \\ \end{gathered}$
But the number of sides of a polygon cannot be negative.
So, the number of sides of a polygon having 35 diagonals is 10.
Note- In such types of questions the key concept we have to remember is that always recall the formula of number of diagonals of a polygon having n sides, then according to given condition substitute the value and simplify, we will get the required number of sides having 35 dia
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https://en.wikipedia.org/wiki/Spectral_risk_measure
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# Spectral risk measure
Jump to: navigation, search
A Spectral risk measure is a risk measure given as a weighted average of outcomes where bad outcomes are, typically, included with larger weights. A spectral risk measure is a function of portfolio returns and outputs the amount of the numeraire (typically a currency) to be kept in reserve. A spectral risk measure is always a coherent risk measure, but the converse does not always hold. An advantage of spectral measures is the way in which they can be related to risk aversion, and particularly to a utility function, through the weights given to the possible portfolio returns.[1]
## Definition
Consider a portfolio ${\displaystyle X}$ (denoting the portfolio payoff). Then a spectral risk measure ${\displaystyle M_{\phi }:{\mathcal {L}}\to \mathbb {R} }$ where ${\displaystyle \phi }$ is non-negative, non-increasing, right-continuous, integrable function defined on ${\displaystyle [0,1]}$ such that ${\displaystyle \int _{0}^{1}\phi (p)dp=1}$ is defined by
${\displaystyle M_{\phi }(X)=-\int _{0}^{1}\phi (p)F_{X}^{-1}(p)dp}$
where ${\displaystyle F_{X}}$ is the cumulative distribution function for X.[2][3]
If there are ${\displaystyle S}$ equiprobable outcomes with the corresponding payoffs given by the order statistics ${\displaystyle X_{1:S},...X_{S:S}}$. Let ${\displaystyle \phi \in \mathbb {R} ^{S}}$. The measure ${\displaystyle M_{\phi }:\mathbb {R} ^{S}\rightarrow \mathbb {R} }$ defined by ${\displaystyle M_{\phi }(X)=-\delta \sum _{s=1}^{S}\phi _{s}X_{s:S}}$ is a spectral measure of risk if ${\displaystyle \phi \in \mathbb {R} ^{S}}$ satisfies the conditions
1. Nonnegativity: ${\displaystyle \phi _{s}\geq 0}$ for all ${\displaystyle s=1,\dots ,S}$,
2. Normalization: ${\displaystyle \sum _{s=1}^{S}\phi _{s}=1}$,
3. Monotonicity : ${\displaystyle \phi _{s}}$ is non-increasing, that is ${\displaystyle \phi _{s_{1}}\geq \phi _{s_{2}}}$ if ${\displaystyle {s_{1}}<{s_{2}}}$ and ${\displaystyle {s_{1}},{s_{2}}\in \{1,\dots ,S\}}$.[4]
## Properties
Spectral risk measures are also coherent. Every spectral risk measure ${\displaystyle \rho :{\mathcal {L}}\to \mathbb {R} }$ satisfies:
1. Positive Homogeneity: for every portfolio X and positive value ${\displaystyle \lambda >0}$, ${\displaystyle \rho (\lambda X)=\lambda \rho (X)}$;
2. Translation-Invariance: for every portfolio X and ${\displaystyle \alpha \in \mathbb {R} }$, ${\displaystyle \rho (X+a)=\rho (X)-a}$;
3. Monotonicity: for all portfolios X and Y such that ${\displaystyle X\geq Y}$, ${\displaystyle \rho (X)\leq \rho (Y)}$;
4. Sub-additivity: for all portfolios X and Y, ${\displaystyle \rho (X+Y)\leq \rho (X)+\rho (Y)}$;
5. Law-Invariance: for all portfolios X and Y with cumulative distribution functions ${\displaystyle F_{X}}$ and ${\displaystyle F_{Y}}$ respectively, if ${\displaystyle F_{X}=F_{Y}}$ then ${\displaystyle \rho (X)=\rho (Y)}$;
6. Comonotonic Additivity: for every comonotonic random variables X and Y, ${\displaystyle \rho (X+Y)=\rho (X)+\rho (Y)}$. Note that X and Y are comonotonic if for every ${\displaystyle \omega _{1},\omega _{2}\in \Omega :\;(X(\omega _{2})-X(\omega _{1}))(Y(\omega _{2})-Y(\omega _{1}))\geq 0}$.[2]
In some texts[which?] the input X is interpreted as losses rather than payoff of a portfolio. In this case the translation-invariance property would be given by ${\displaystyle \rho (X+a)=\rho (X)+a}$ instead of the above.
## References
1. ^ Cotter, John; Dowd, Kevin (December 2006). "Extreme spectral risk measures: An application to futures clearinghouse margin requirements". Journal of Banking & Finance. 30 (12): 3469–3485. doi:10.1016/j.jbankfin.2006.01.008.
2. ^ a b Adam, Alexandre; Houkari, Mohamed; Laurent, Jean-Paul (2007). "Spectral risk measures and portfolio selection" (pdf). Retrieved October 11, 2011.
3. ^ Dowd, Kevin; Cotter, John; Sorwar, Ghulam (2008). "Spectral Risk Measures: Properties and Limitations" (pdf). CRIS Discussion Paper Series (2). Retrieved October 13, 2011.
4. ^ Acerbi, Carlo (2002), "Spectral measures of risk: A coherent representation of subjective risk aversion", Journal of Banking and Finance, Elsevier, 26 (7), pp. 1505–1518, doi:10.1016/S0378-4266(02)00281-9
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# How do you do long division?
Very difficultly. No, really, it's not difficult. It's a pattern. Once you find the pattern, it's easy. It is hard to describe in words. So, I have a problem worked out for you as I describe it.
First, the problem we will do is 79 "divided by" 3. Notice the setup. That's one thing students do incorrect a lot. The first number goes inside; the second goes outside.
So, now, remember, it's a pattern. There's about 5-7 steps to it. So, be patient, especially at the start.
You start to go through the numbers one by one. What I mean is, "How many times does 3 go into 7?" If the answer was none, you would "include" the next number (here, 79). But, for us, 3 goes into 7 twice. So, 2 goes above the 7.
Then, do 3 times that number you found, 2. 3*2 = 6. That goes under the 7.
Then, you do 7-6 = 1. Put the 1 underneath all, just like a subtraction problem.
Then, you drop the next number that is "inside" the long division symbol. For us, that's the "9". So, you write the 9 beside the 1, giving "19". As it looks, the "9" dropped straight down beside the 1.
Then, you repeat these steps. As in. . .
"How many times does 3 go into 19?" That is 6. So, 6 goes above the 9, next to the 2 that is above the 7, on top, giving us "26" on top.
Then, 3 times that number, 6, = 18. Put that under the 19, then subtract those two numbers. 19-18 = 1.
There are no other numbers to drop. So, 1 is considered our "remainder". So, on top, after the 26, we would put "r1". So, our entire answer would be "26 r1".
Hopefully not making things more complicated, depending where you are in your math, you may or may not need to know this yet. You will get it at some point in time. I will try to keep it simple. If you are looking for a "decimal" for your answer, you would repeat these steps still. "But, what number do I drop?" You would drop "0". Keep dropping "0", until you get a remainder of "0", or you start to get a pattern, or you are told to stop.
If you need a fraction for your answer, you have the "26". We need a fraction after that, a top number and bottom number. The top number is the remainder always; here "1". The bottom number is always the "outside number" at the beginning here, "3". So, the fraction would be 1/3, and the answer would be "26 1/3".
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# 11821 (number)
11,821 (eleven thousand eight hundred twenty-one) is an odd five-digits prime number following 11820 and preceding 11822. In scientific notation, it is written as 1.1821 × 104. The sum of its digits is 13. It has a total of 1 prime factor and 2 positive divisors. There are 11,820 positive integers (up to 11821) that are relatively prime to 11821.
## Basic properties
• Is Prime? Yes
• Number parity Odd
• Number length 5
• Sum of Digits 13
• Digital Root 4
## Name
Short name 11 thousand 821 eleven thousand eight hundred twenty-one
## Notation
Scientific notation 1.1821 × 104 11.821 × 103
## Prime Factorization of 11821
Prime Factorization 11821
Prime number
Distinct Factors Total Factors Radical ω(n) 1 Total number of distinct prime factors Ω(n) 1 Total number of prime factors rad(n) 11821 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 9.37763 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 11,821 is 11821. Since it has a total of 1 prime factor, 11,821 is a prime number.
## Divisors of 11821
2 divisors
Even divisors 0 2 2 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 2 Total number of the positive divisors of n σ(n) 11822 Sum of all the positive divisors of n s(n) 1 Sum of the proper positive divisors of n A(n) 5911 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 108.724 Returns the nth root of the product of n divisors H(n) 1.99983 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 11,821 can be divided by 2 positive divisors (out of which 0 are even, and 2 are odd). The sum of these divisors (counting 11,821) is 11,822, the average is 5,911.
## Other Arithmetic Functions (n = 11821)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 11820 Total number of positive integers not greater than n that are coprime to n λ(n) 11820 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1425 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares
There are 11,820 positive integers (less than 11,821) that are coprime with 11,821. And there are approximately 1,425 prime numbers less than or equal to 11,821.
## Divisibility of 11821
m n mod m 2 3 4 5 6 7 8 9 1 1 1 1 1 5 5 4
11,821 is not divisible by any number less than or equal to 9.
• Arithmetic
• Prime
• Deficient
• Polite
• Prime Power
• Square Free
## Base conversion (11821)
Base System Value
2 Binary 10111000101101
3 Ternary 121012211
4 Quaternary 2320231
5 Quinary 334241
6 Senary 130421
8 Octal 27055
10 Decimal 11821
12 Duodecimal 6a11
20 Vigesimal 19b1
36 Base36 94d
## Basic calculations (n = 11821)
### Multiplication
n×y
n×2 23642 35463 47284 59105
### Division
n÷y
n÷2 5910.5 3940.33 2955.25 2364.2
### Exponentiation
ny
n2 139736041 1651819740661 19526161154353681 230818751005614863101
### Nth Root
y√n
2√n 108.724 22.7799 10.4271 6.52425
## 11821 as geometric shapes
### Circle
Diameter 23642 74273.5 4.38994e+08
### Sphere
Volume 6.91913e+12 1.75597e+09 74273.5
### Square
Length = n
Perimeter 47284 1.39736e+08 16717.4
### Cube
Length = n
Surface area 8.38416e+08 1.65182e+12 20474.6
### Equilateral Triangle
Length = n
Perimeter 35463 6.05075e+07 10237.3
### Triangular Pyramid
Length = n
Surface area 2.4203e+08 1.94669e+11 9651.81
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# Elements of a ring problem
## Homework Statement
A) If ab+ba = 1 and a^3 = a in a ring, show that a^2 = 1
none
## The Attempt at a Solution
Little confused. If we know that a^3 = a, can't we just multiply each on the right or left side by a^-1 to get a^2 = 1? Or could we only do that if the ring is said to be commutative?
edit: I realized that to show that a^2 = 1 in the way I mentioned above that a would have to be a division ring. Normal rings aren't guaranteed to have multiplicative inverses.
Last edited:
RUber
Homework Helper
I would start by multiplying the first equation by an a on the left, then do the same with an a on the right. Compare the resulting equations. Then there should only be one reasonable conclusion that satisfies ##a^3=a## as well.
So I get b + aba = a and aba + b = a, (considering a^2=1) is there something obvious that I'm missing here? The two equations are the same thing are they not?
Last edited:
RUber
Homework Helper
## ab+ba = 1,## so left multiplying gives ## a^2 b + aba = a## and right multiplying gives ## aba+ba^2=a##. Moving ##aba## to the other side shows that ##a^2b = a-aba = ba^2##.
Similarly, you could right/left multiply ##a^2## to show that ## a^3b+a^2ba=a^2 = aba^2 + ba^3 ##. Substituting the fact that ## a^2b=ba^2 ## should allow you to cancel out some (edit) ##a^3##s.
Last edited:
RUber
Homework Helper
So I get b + aba = a and aba + b = a, (considering a^2=1) is there something obvious that I'm missing here? The two equations are the same thing are they not?
I would not assume what you are trying to show at this point.
PsychonautQQ
Fredrik
Staff Emeritus
Gold Member
There aren't many many ways to use the two given equations. Let's start with ##ab+ba=1##. What can we do with this other than to insert the assumption ##a=a^3##? This gives us ##1=ab+ba=a^3b+ba^3##. How can we rewrite the right-hand side? Replacing ##a## with ##a^3## again doesn't look very promising, and replacing ##a^3## with ##a## just takes us back to where we started. So we're almost certainly going to have to use ##ab+ba=1## (possibly more than once) to rewrite the right-hand side.
Update:
I now have four (seven?) equations I'm moving around.
1) a = (a^2)*b + a*b*a
2) a = aba + b(a^2)
3) 1 = (a^3)b + b(a^3)
4) 1 = ab + ba
Hence...
5) ab + ba = (a^3)b + b(a^3)
6) aba + b(a^2) = (a^2)*b + a*b*a
I even multiplied 5 and 6 together to try to get something. Got messy.
Hmmm.
RUber
Homework Helper
From 1 and 2 you have ##a^2b=ba^2##.
Multiplying 4 by ##a^2## gives either ## a^2 = a^3b+a^2ba ## or ## a^2 = aba^2+ba^3##.
Using the facts that ##a^3 = a## and ##a^2b=b^2a## you can reduce this down to something equivalent to ##a^2 = 1##.
Fredrik
Staff Emeritus
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# Next level of resistor divider
I know how to calculate proportions of resistor values in resistor divider, but dont know how to jump into "next level" and calculate the precise values of resistors from proportion :
For example to devide 10V to obtain 5V you can use: - 10 ohm R1 / 10 ohm R2 And as well you can use - 100kohm R1 / 100kohm R2
Proportion is the same 1/2 but it for sure is different.
What I have to look for, what I have to know to calculate this?
Is time also critical? I want to divide pulsing signal, want change only voltage range, but still I want to keep form/shape of this vawe.
• This answer discusses how to calculate resistors when the current drawn, minimum and maximum voltages at the middle node are known. electronics.stackexchange.com/a/83659/8627 Jan 6, 2015 at 21:01
To determine the values of resistors in a divider, you need to consider how much power you are willing to "waste" throught the divider, and how much current you will draw from the divider mid-point. Don't forget that any load you connect to the divider mid-point will appear as another resistor in parallel with the "bottom" resistor, and so must be included when calculating the resistor values.
Two main factors are how much current can you draw (that is the voltage applied divided by the total resistance - sum of the two resistors. The second factor is what output impedance you want (or can tolerate) from the divider point. That value is the parallel combination of the two resistors if it's being fed from a stiff voltage source.
So in your example, assume 10V is applied, the 100K + 100K divider draws 50uA and has a source impedance of 100K||100K = 50K ohms (so if you load it with 50M ohms it will drop by about 0.1%)
In the case of the 10R + 10R, it will draw 500mA and have a source impedance of 5 ohms, so if you load it with 5K ohms it will drop about 0.1%.
The other practical factor if you still have some range in the above two constraints is that it's nice to be able to use commercially available resistor values. Maybe you can change the total resistance a bit to get it to fit with values that you can buy in the tolerance and power rating you need. If you can't buy the exact values you calculate and can't seem to make it fit, another approach is to pick one of the two values that you can buy and then find a parallel or series combination of standard values that makes the ratio correct (since the ratio is usually much more important than the total resistance).
For example, suppose you want to divide 12V down to 2.5V so you can measure it with a microcontroller, and the microcontroller can only tolerate 10K source impedance before errors increase.
So you might pick 38K and 10K, which has a source impedance of 7.92K, but 38K isn't a standard E96 value- the closest is 38.3K, which is an error of about 0.6%. You could also use 5.62K and 21.3K which would have a ratio error of only about 0.2% but would draw more current from the 12V input (and would still meet the 10K max source impedance limit, of course).
Edit: Given your added comment about AC signals, if you match the resistance (actually you match conductance- reciprocal of resistance) with capacitance across the divider you can maintain the same shape to the signals. Since resistors are not designed to have any specific capacitance, only stray and trace capacitance, you might have to adjust the capacitance across one of the resistors. That's essentially what you do when you adjust your 10:1 oscilloscope probe compensation.
I assume the question is how to go from proportions to explicit resistor values.
The main difference is the amount of current that will flow through your divider, and that is dictated by the application. As an example, if you're using a divider to measure a battery voltage in a uC design, where your ADC can see values of 0-3.3V, and your battery is 9-12.6V (as it would be in a 3S LiPo), you would want the highest resistor values that still provide a measurable amount of current to the uC to minimize the amount of current through it.
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# Differenciate
• Aug 8th 2013, 05:56 AM
srirahulan
Differenciate
How can i differenciate $\displaystyle e^\frac{y}{x}=x$ respect to x>>>>>
• Aug 8th 2013, 06:09 AM
HallsofIvy
Re: Differenciate
Pretty straightforward, using the "chain rule". First, of course, we integrate functions, not equations, so what you really mean is "how do I differentiate each side of this equation?"
To differentiate $\displaystyle e^{\frac{y}{x}}$, let $\displaystyle u=\frac{y}{x}= yx^{-1}$. Then $\displaystyle \frac{de^u}{dx}= \frac{de^u}{du}\frac{du}{dx}$. $\displaystyle \frac{de^u}{du}= e^u$ and $\displaystyle \frac{dyx^{-1}}{dx}= (-1)yx^{-2}= -\frac{y}{x^2}$. So the derivative of $\displaystyle e^{\frac{y}{x}}$, with respect to x is $\displaystyle -\frac{y}{x^2}e^{\frac{y}{x}}$.
Now, what is the derivative of the right side of the equation?
• Aug 8th 2013, 06:33 AM
Soroban
Re: Differenciate
Hello, srirahulan!
Quote:
$\displaystyle \text{Differentiate: }\:e^\frac{y}{x}\:=\:x$
$\displaystyle \text{Take logs: }\:\ln\left(e^{\frac{y}{x}}\right) \:=\:\ln(x) \quad\Rightarrow\quad \frac{y}{x}\underbrace{\ln(e)}_{\text{This is 1}} \:=\:\ln(x)$
. . . . . . . . $\displaystyle \frac{y}{x} \:=\:\ln(x) \quad\Rightarrow\quad y \:=\:x\!\cdot\!\ln(x)$
$\displaystyle \text{Differentiate: }\:y' \;=\;x\!\cdot\!\frac{1}{x} + 1\!\cdot\!\ln(x)$
. .$\displaystyle \text{Therefore: }\:y' \;=\;1 + \ln(x)$
• Aug 8th 2013, 07:48 PM
ibdutt
Re: Differenciate
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Vol. Control Systems
Chapter Closed-loop Control Systems
# Integral (Reset) Control
### Integral control, also called ‘reset’ control, accumulates error over time, eliminating even the slightest offset errors with a carefully calculated integral gain value.
Imagine a liquid-level control system for a vessel, where the position of a level-sensing float sets the position of a potentiometer, which then sets the speed of a motor-actuated control valve. If the liquid level is above the setpoint, the valve continually opens up; if below the setpoint, the valve continually closes off:
Unlike the proportional control system where valve position was a direct function of float position, this control system sets the speed of the motor-driven valve according to the float position. The further away from the setpoint the liquid level is, the faster the valve moves open or closed. In fact, the only time the valve will ever halt its motion is when the liquid level is precisely at the setpoint; otherwise, the control valve will be in constant motion.
This control system does its job in a very different manner than the all-mechanical float-based proportional control system illustrated previously. Both systems are capable of regulating liquid level inside the vessel, but they take very different approaches to doing so. One of the most significant differences in control behavior is how the proportional system would inevitably suffer from offset (a persistent error between PV and SP), whereas this control system actively works at all times to eliminate offset. The motor-driven control valve literally does not rest until all error has been eliminated!
Instead of characterizing this control system as proportional, we call it integral in honor of the calculus principle (“integration”) whereby small quantities are accumulated over some span to form a total. Don’t let the word “calculus” scare you! You are probably already familiar with the concept of numerical integration even though you may have never heard of the term before.
### What is Integration?
Calculus is a form of mathematics dealing with changing variables, and how rates of change relate between different variables. When we “integrate” a variable with respect to time, what we are doing is accumulating that variable’s value as time progresses. Perhaps the simplest example of this is a vehicle odometer, accumulating the total distance traveled by the vehicle over a certain time period. This stands in contrast to a speedometer, indicating the rate of distance traveled per unit of time.
Imagine a car moving along at exactly 30 miles per hour. How far will this vehicle travel after 1 hour of driving this speed? Obviously, it will travel 30 miles. Now, how far will this vehicle travel if it continues for another 2 hours at the exact same speed? Obviously, it will travel 60 more miles, for a total distance of 90 miles since it began moving. If the car’s speed is a constant, calculating total distance traveled is a simple matter - just multiply the speed by the travel time.
The odometer mechanism that keeps track of the mileage traveled by the car may be thought of as integrating the speed of the car with respect to time. In essence, it is multiplying speed times time continuously to keep a running total of how far the car has gone. When the car is traveling at a high speed, the odometer “integrates” at a faster rate. When the car is traveling slowly, the odometer “integrates” slowly.
If the car travels in reverse, the odometer may be expected to decrement (count down) rather than increment (count up) because it sees a negative quantity for speed (in reality, it may not actually do this). The rate at which the odometer decrements depends on how fast the car travels in reverse. When the car is stopped (zero speed), the odometer holds its reading and neither increments nor decrements.
Now let us return to the context of an automated process to see how this calculus principle works inside a process controller. Integration is provided either by a pneumatic mechanism, an electronic opamp circuit, or by a microprocessor executing a digital integration algorithm. The variable being integrated is error (the difference between PV and SP) over time. Thus the integral mode of the controller ramps the output either up or down over time in response to the amount of error existing between PV and SP, and the sign of that error.
We saw this “ramping” action in the behavior of the liquid level control system using a motor-driven control valve commanded by a float-positioned potentiometer: the valve stem continuously moves so long as the liquid level deviates from setpoint. The reason for this ramping action is to increase or decrease the output as far as it is necessary in order to completely eliminate any error and force the process variable to precisely equal setpoint. Unlike proportional action, which simply moves the output an amount proportional to any change in PV or SP, integral control action never stops moving the output until all error is eliminated.
#### The Effect of Integral Gain
If proportional action is defined by the error telling the output how far to move, integral action is defined by the error telling the output how fast to move. One might think of integral as being how “impatient” the controller is, with integral action constantly ramping the output as far as it needs to go in order to eliminate error. Once the error is zero (PV = SP), of course, the integral action stops ramping, leaving the controller output (valve position) at its last value just like a stopped car’s odometer holds a constant value.
#### Integral Gain Algorithm
If we add an integral term to the controller equation, we get something that looks like this:
$m = K_p e + {1 \over \tau_i} \int e \> dt + b$
Where,
$$m$$ = Controller output
$$e$$ = Error (difference between PV and SP)
$$K_p$$ = Proportional gain
$$\tau_i$$ = Integral time constant (minutes)
$$t$$ = Time
$$b$$ = Bias
The most confusing portion of this equation for those new to calculus is the part that says “$$\int e \> dt$$”. The integration symbol (looks like an elongated letter “S”) tells us the controller will accumulate (“sum”) multiple products of error ($$e$$) over tiny slices of time ($$dt$$). Quite literally, the controller multiplies error by time (for very short segments of time, $$dt$$) and continuously adds up all those products to contribute to the output signal which then drives the control valve (or other final control element). The integral time constant ($$\tau_i$$) is a value set by the technician or engineer configuring the controller, proportioning this cumulative action to make it more or less aggressive over time.
### Proportional Integral Controller (PI Controller) Benefit
To see how this works in a practical sense, let’s imagine how a proportional + integral controller would respond to the scenario of a heat exchanger whose inlet temperature suddenly dropped. As we saw with proportional-only control, an inevitable offset occurs between PV and SP with changes in load, because an error must develop if the controller is to generate the different output signal value necessary to halt further change in PV. We called this effect proportional-only offset.
Once this error develops, though, integral action begins to work. Over time, a larger and larger quantity accumulates in the integral mechanism (or register) of the controller due to the persistent error between PV and SP. That accumulated value adds to the controller’s output, driving the steam control valve further and further open. This, of course, adds heat at a faster rate to the heat exchanger, which causes the outlet temperature to rise. As the temperature re-approaches setpoint, the error becomes smaller and thus the integral action proceeds at a slower rate (like a car’s odometer incrementing at a slower rate as the car’s speed decreases). So long as the PV is below SP (the outlet temperature is still too cool), the controller will continue to integrate upwards, driving the control valve further and further open. Only when the PV rises to exactly meet SP does integral action finally rest, holding the valve at a steady position. Integral action tirelessly works to eliminate any offset between PV and SP, thus neatly eliminating the offset problem experienced with proportional-only control action.
As with proportional action, there are (unfortunately) two completely opposite ways to specify the degree of integral action offered by a controller. One way is to specify integral action in terms of minutes or minutes per repeat. A large value of “minutes” for a controller’s integral action means a less aggressive integral action over time, just as a large value for proportional band means a less aggressive proportional action. The other way to specify integral action is the inverse: how many repeats per minute, equivalent to specifying proportional action in terms of gain (large value means aggressive action). For this reason, you will sometimes see the integral term of a PID equation written differently:
$\tau_i$ = minutes per repeat $K_i$ = repeats per minute
${1 \over \tau_i} \int e \> dt$ $K_i \int e \> dt$
Many modern digital electronic controllers allow the user to select the unit they wish to use for integral action, just as they allow a choice between specifying proportional action as gain or as proportional band.
Integral is a highly effective mode of process control. In fact, some processes respond so well to integral controller action that it is possible to operate the control loop on integral action alone, without proportional. Typically, though, process controllers implement some form of proportional plus integral (“PI”) control.
### Oscillation from High Integral Gain
Just as too much proportional gain will cause a process control system to oscillate, too much integral action (i.e. an integral time constant that is too short) will also cause oscillation. If the integration happens at too fast a rate, the controller’s output will “saturate” either high or low before the process variable can make it back to setpoint. Once this happens, the only condition that will “unwind” the accumulated integral quantity is for an error to develop of the opposite sign, and remain that way long enough for a canceling quantity to accumulate. Thus, the PV must cross over the SP, guaranteeing at least another half-cycle of oscillation.
### Integral Windup or Reset Windup
A similar problem called reset windup (or integral windup) happens when external conditions make it impossible for the controller to achieve setpoint. Imagine what would happen in the heat exchanger system if the steam boiler suddenly stopped producing steam. As outlet temperature dropped, the controller’s proportional action would open up the control valve in a futile effort to raise temperature. If and when steam service is restored, proportional action would just move the valve back to its original position as the process variable returned to its original value (before the boiler died). This is how a proportional-only controller would respond to a steam “outage”: nice and predictably. If the controller had integral action, however, a much worse condition would result. All the time spent with the outlet temperature below setpoint causes the controller’s integral term to “wind up” in a futile attempt to admit more steam to the heat exchanger. This accumulated quantity can only be undone by the process variable rising above the setpoint for an equal error-time product, which means when the steam supply resumes, the temperature will rise well above the setpoint until the integral action finally “unwinds” and brings the control valve back to the same position again.
#### Solving Integral Windup
Various techniques exist to manage integral windup. Controllers may be built with limits to restrict how far the integral term can accumulate under adverse conditions. In some controllers, integral action may be turned off completely if the error exceeds a certain value. The surest fix for integral windup is human operator intervention, by placing the controller in manual mode. This typically resets the integral accumulator to a value of zero and loads a new value into the bias term of the equation to set the valve position wherever the operator decides. Operators usually wait until the process variable has returned at or near the setpoint before releasing the controller into automatic mode again.
While it might appear that operator intervention is again a problem to be avoided (as it was in the case of having to correct for proportional-only offset), it is noteworthy to consider that the conditions leading to integral windup usually occur only during shut-down conditions. It is customary for human operators to run the process manually anyway during a shutdown, and so the switch to manual mode is something they would do anyway and the potential problem of windup often never manifests itself.
Integral control action has the unfortunate tendency to create loop oscillations (“cycling”) if the final control element exhibits hysteresis, such as the case with a “sticky” control valve. Imagine for a moment our steam-heated heat exchanger system where the steam control valve possesses excessive packing friction and therefore refuses to move until the applied air pressure changes far enough to overcome that friction, at which point the valve “jumps” to a new position and then “sticks” in that new position. If the valve happens to stick at a stem position resulting in the product temperature settling slightly below setpoint, the controller’s integral action will continually increase the output signal going to the valve in an effort to correct this error (as it should). However, when that output signal has risen far enough to overcome valve friction and move the stem further open, it is very likely the stem will once again “stick” but this time do so at a position making the product temperature settle above setpoint. The controller’s integral action will then ramp downward in an effort to correct this new error, but due to the valve’s friction making precise positioning impossible, the controller can never achieve setpoint and therefore it cyclically “hunts” above and below setpoint.
The best solution to this “reset cycling” phenomenon, of course, is to correct the hysteresis in the final control element. Eliminating friction in the control valve will permit precise positioning and allow the controller’s integral action to achieve setpoint as designed. Since it is practically impossible to eliminate all friction from a control valve, however, other solutions to this problem exist. One of them is to program the controller to stop integrating whenever the error is less than some pre-configured value (sometimes referred to as the “integral deadband” or “reset deadband” of the controller). By activating reset control action only for significant error values, the controller ignores small errors rather than “compulsively” trying to correct for any detected error no matter how small.
REVIEW:
• ##### Manually resetting the accumulated error on startup is a safe method of ensuring the reduction of windup error.
Interested in more PID and control system topics?
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#### Lessons in Industrial Automation
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# Why not think of derivatives as fractions?
Back in high school—back in the 1900s, as my sons say—when our calculus teacher was introducing the chain rule...
$$\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$$
...he made a special point of saying that we shouldn't think of the $$dt$$ as cancelling out in this equation. He said it was just happened to look that way. We were supposed to consider $$\frac{\partial}{dx}$$ as an indivisible operator. (That was probably not his term.)
So I was confused when my son's calculus book introduced differentials, and explicitly encouraged algebraic manipulation of expressions like $$\frac{dy}{dx}$$ — multiplying both sides of an equation by $$dx$$, and things like that. I've looked into it somewhat, and as far as I can tell this is regarded as mathematically legitimate. (Feel free to correct me on that.)
I'm mostly just curious now why our calculus teacher warned us off from thinking of Leibniz notation in this way. I could imagine that when students think about $$\frac{dy}{dx}$$, then there could be a lot of algebraic rabbit trails. Are there other reasons for discouraging high school students from thinking about differentials?
• What book is this? Most of the popular textbooks still discourage students from thinking of the derivative as a quotient, and they'll mention that cancelling differentials doesn't give a valid proof of the chain rule. But there's usually a later section on "differentials" in which they'll define the differential of $y=f(x)$ to be $dy=f'(x)dx$, and they'll use differentials to do linear approximation. This paper argues that the reason for this is to stop students from thinking of differentials as infinitesimals. Oct 12 at 0:16
• I edited with what I hope is a more informative title -- OP, if you think it's not good, feel free to edit or revert. Oct 12 at 1:16
• I took a look, and Saxon Calculus does show some informal arguments involving cancelling differentials and referring to them as infinitesimals. It doesn’t go as far as what some would consider “teaching with differentials.” This answer goes into some of the pros and cons of teaching with differentials. Oct 12 at 10:50
• @JustinHancock Aren't they the same questions anyway, regardless of title? And adam.baker: Is $\partial$ a typo and meant to be a $d$? The context you provide makes this seem a single-variable calculus question and not a multivariable calculus. Which is it? Oct 12 at 15:46
• Is this a duplicate of math.stackexchange.com/questions/21199/… ? Oct 16 at 15:44
If you have a function $$f(x,y)$$ where $$x=x(t)$$ and $$y=y(t)$$ are themselves functions of a parameter $$t,$$ and you blindly cancel out differentials, then you can get to incorrect statements like
$$\require{cancel}\dfrac{\partial f}{\partial t} = \dfrac{\partial f}{\cancel{\partial x}} \cdot \dfrac{\cancel{\partial x}}{\partial t} = \dfrac{\partial f}{\cancel{\partial y}} \cdot \dfrac{\cancel{\partial y}}{\partial t}, \quad {\color{red}\times}$$
whereas what's actually true is
$$\dfrac{\partial f}{\partial t} = \dfrac{\partial f}{\partial x} \cdot \dfrac{\partial x}{\partial t} + \dfrac{\partial f}{\partial y} \cdot \dfrac{\partial y}{\partial t}. \quad {\color{green}\checkmark}$$
You can't cancel because the $$\partial f$$'s in the numerators of $$\dfrac{\partial f}{\partial t},$$ $$\dfrac{\partial f}{\partial x},$$ $$\dfrac{\partial f}{\partial y}$$ all mean different things.
• The $$\partial f$$ in the numerator of $$\dfrac{\partial f}{\partial t},$$ represents the change in $$f$$ attributed to the change in $$t.$$
• The $$\partial f$$ in the numerator of $$\dfrac{\partial f}{\partial x}$$ represents the change in $$f$$ attributed to the change in $$x.$$
• The $$\partial f$$ in the numerator of $$\dfrac{\partial f}{\partial y}$$ represents the change in $$f$$ attributed to the change in $$y.$$
But in single-variable calculus (which is likely the focus of your son's textbook), you're working exclusively with functions that have only one input variable. And if you have a function $$f(x)$$ where $$x=x(t)$$ is itself a function of a parameter $$t,$$ then it's true that
$$\dfrac{\partial f}{\partial t} = \dfrac{\partial f}{\partial x} \cdot \dfrac{\partial x}{\partial t}.$$
The above is conventionally written with "total" derivative symbols ($$\mathrm d$$ means "total", $$\partial$$ means "partial") since the change attributed to the single variable is the same as the total change of the function.
$$\dfrac{\mathrm df}{\mathrm dt} = \dfrac{\mathrm df}{\mathrm dx} \cdot \dfrac{\mathrm dx}{\mathrm dt}$$
So in general, you can manipulate total derivatives ($$\mathrm d$$) like fractions, but you can't do the same with partial derivatives ($$\partial$$).
\begin{align*} \require{cancel} \textrm{valid:} \quad &\dfrac{\mathrm df}{\mathrm dt} = \dfrac{\mathrm df}{\cancel{\mathrm dx}} \cdot \dfrac{\cancel{\mathrm dx}}{\mathrm dt} \quad \color{green}\checkmark \\[5pt] \textrm{NOT valid:} \quad &\dfrac{\partial f}{\partial t} = \dfrac{\partial f}{\cancel{\partial x}} \cdot \dfrac{\cancel{\partial x}}{\partial t} \quad \color{red}\times \end{align*}
Response to comments about notation: The streamlined notation makes this answer is friendly for non-experts. Yes, it relies on some intuitive inference as to what is meant -- for instance, given $$f(x,y),$$ $$x(t),$$ $$y(t),$$ the streamlined notation $$\dfrac{\partial f}{\partial t}$$ refers to $$\dfrac{\partial f(x(t),y(t))}{\partial t}.$$ But non-experts generally find it easier to make reasonably intuitive inferences like this than to work with technically complete notation that overwhelms them. (Those who disagree: maybe it was different for you when you were a student, but that's how it is for 99% of students.)
If anyone has a suggested improvement that will make the notation more technically unambiguous without making the answer less friendly for non-experts, then feel free to make an edit suggestion.
• I think that you want to avoid the non-standard notation of the lone $\partial f$, $\partial x$, and $\partial y$ entirely and instead use the usual notation for the total differential $df = \frac{\partial f}{\partial x} dx+ \frac{\partial f}{\partial y} dy$. (i.e. the dual to the gradient.)
Oct 11 at 18:47
• You could argue that $\frac{\partial f}{\partial t}$ is also abuse of notation. The partial derivative means that we change one of $f$'s arguments and keep all the other arguments constant. Which argument of $f$ is $t$? None is! $f$ is a function of two variables: $f: (x,y)\mapsto f(x,y)$. We have secretly composed $f(x,y)$ and $x(t), y(t)$ together. So better would be to say $\frac{\mathrm df}{\mathrm dt}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial t}$. Usually I would not care about this, but in this case it is relevant Oct 12 at 14:33
• @ReinstateMonica $u$ is there so that it's proper to use $\frac{\partial f}{\partial t},$ not $\frac{\mathrm df}{\mathrm dt}.$ And the reason why I wanted to use only partials in the incorrect example is that it lets us bypass a discussion of $\mathrm d$ vs $\partial.$ I wanted to avoid getting caught up in that nuance at the beginning. One thing at a time. Oct 12 at 15:47
• Being for the moment a highly technical reader :), I would pick out $df$ to comment on, not whether $\partial$ is permitted on single-variable functions. As for $df$, I don't like treating functions as scalar variables. I write $z = f(x,y)$ and use $dz$. Or $df(x,y)$ — maybe, maybe not. I write ${df(x)\over dx}$ in the single-variable case. I think of Leibniz notation as applying to expressions such as $f(x)$ or $f(x,y)$ and not to functions such as $f$. Oct 13 at 22:09
There is a nice write up at the included link with an application to thermodynamics. I've included two of the images (subsequently transcribed) that give the core of the argument. Single variable functions and differentials (derivatives with $$d$$) are safe to use as fractions, but multivariate functions and partial derivatives (with a $$\partial$$) aren't.
https://johncarlosbaez.wordpress.com/2021/09/13/the-cyclic-identity-for-partial-derivatives/
### WHEN YOU REALIZE, SADLY, THAT DERIVATIVES DON'T ALWAYS ACT LIKE FRACTIONS
Suppose $$u$$, $$v$$, $$w$$ are functions on the plane and you can take any two as coordinates and write the third as a smooth function of those two. Then
$$\frac{\partial u}{\partial v}\Bigr|_w \frac{\partial v}{\partial w}\Bigr|_u \frac{\partial w}{\partial u}\Bigr|_v = \color{red}{-1}$$
### AN EXAMPLE
Say $$u$$, $$v$$, $$w$$ are these functions on the plane:
$$u = x \quad v = y \quad w = x + y$$
Then each is a function of the other two:
$$u = w - v \quad v = w - u \quad w = u + v$$
and we get
$$\underbrace{\frac{\partial u}{\partial v}\Bigr|_w}_{= -1} \underbrace{\frac{\partial v}{\partial w}\Bigr|_u}_{= 1} \underbrace{\frac{\partial w}{\partial u}\Bigr|_v}_{= 1} = \color{red}{-1}$$
• So, we can only cancel differentials in even-dimensional contexts. Just kidding. This is a good example, Sir Roger Penrose gives it as one of the fundamental confusions of calculus in his "Road to Reality" book if memory serves me correctly. Oct 14 at 16:30
First: $${\partial}\over{\partial x}$$ $$\neq$$ $${d}\over{d x}$$
The difference being the concepts in multivariable calculus vs single variable calculus are different and for different purposes. But by the context of your question I'm pretty sure you meant $${d}\over{d x}$$, so let's go with that.
Second: The cancel is actual a pretty helpful tool when solving differential equations, so I do love it. Buuuut, I 100% understand why you should be taught it last and be dissuaded from using it.
The reason why he said not to think of it like that is because its lazy, without conceptual meaning, and an abuse of notation. However, it's ironically right unless you go further with the thought. No seriously here's a simple proof.
$$f(x+dx)=f(x)+df(x)$$ So the chain rule comes naturally from: $$df(g(x))+f(g(x))=$$ $$f(g(x+dx))=f(g(x)+dg(x))$$ And the cancel: $$\frac{1}{dg(x)}\frac{dg(x)}{dx}=\frac{1}{dx}$$ And all together is the chain rule as you know and love: $$\frac{df}{dx}=\frac{df}{dg}\frac{dg}{dx}$$
So from this you may tell that, yes, the cancel is really there, but also an erroneously incomplete perception on the subject. In fact, you can do calculus without the fractions at all using lie algebra or using the techniques I used above.
$$[D,x]=Dx-xD=1 \implies [D,f(x)]=f'(x)$$
But probably the most horrendous abuse of notation is with the second derivative.
$$\frac{d^2f}{dx^2} \neq \frac{d^2f}{dg^2}\frac{dg^2}{dx^2}=\frac{d^2f}{dg^2}\left(\frac{dg}{dx}\right)^2$$
Nor is it anything remotely similar because if you start treating it as a linear algebra you should understand that the algebra is non-commutative which means $$ab \neq ba$$ and implies $$(ab)^c \neq a^c b^c$$. The product rule makes it more complicated:
$$\left(g'\frac{d}{dg}\right)^2=$$ $$g'\frac{d}{dg}g'\frac{d}{dg}=$$ $$g'\left(\frac{d}{dg} \cdot g'\right)\frac{d}{dg}=$$ $$g'\left(g''+g'\frac{d}{dg}\right)\frac{d}{dg}=$$ $$g'g''\frac{d}{dg}+(g')^2\frac{d^2}{dg^2}=$$
Note that this, as expected from the product rule, doesn't equal the equation that is wrongly created using just the cancel.
• That horrendous equation with the second derivative is true if you interpret the second derivative notation according to it's original meaning. See my answer. Oct 24 at 4:31
The reason given in the accepted answer is valid, but I'd like to point out something that Ben Crowell (user507) said in a comment several years back: that problem has got nothing to do with calculus, infinitesimals, or treating derivatives as fractions, but more with an incomplete notation used to denote the fractions involved. Because the same problem arises if we look at a linear function $$z=f(x,y)$$, with $$x,y$$ themselves being linear functions of $$t$$, and ask for $$\frac{\Delta z}{\Delta t}$$ (no calculus, no infinitesimals, true fractions). In that case we also have (with incomplete notation) $$\Delta z = \frac{\Delta z}{\Delta x}\Delta x + \frac{\Delta z}{\Delta y}\Delta y$$ and dividing by $$\Delta t$$ gives the 'chain rule'. Here we could literally cancel $$\Delta x$$ in the first summand and $$\Delta y$$ in the second, but then we arrive at the nonsense $$\Delta z= 2\Delta z$$. How come? The reason is that in the first summand $$\Delta z$$ denotes the change of $$z$$ when $$y$$ is held constant, while in the second it denotes the change of $$z$$ while $$x$$ is held constant. If the notation made explicit that those two $$\Delta z$$ are different, we could in principle avoid the problem. For instance writing $$\Delta z = \frac{\Delta z|_{y=\text{const.}}}{\Delta x}\Delta x + \frac{\Delta z|_{x=\text{const.}}}{\Delta y}\Delta y$$
If we now cancel we arrive at
$$\Delta z = \Delta z|_{y=\text{const.}}+ \Delta z|_{x=\text{const.}}$$
which might seem unusual but at least we wouldn't reduce it to $$\Delta z = 2 \Delta z$$. Similarly we could explain to a student that
$$\frac{\Delta z}{\Delta t} \neq \frac{\Delta z|_{y=\text{const.}}}{\Delta x}\frac{\Delta x}{\Delta t}$$
not because we are not allowed to cancel, but simply because in general $$\Delta z\neq \Delta z|_{y=\text{const.}}$$.
Returning to partial derivatives $$\frac{\partial z}{\partial x}$$ etc. the same problem arises there: the notation does not show explicitly which variables are held fixed when we take the derivative with respect to $$x$$. Jacobi was the first to draw attention to that problem, but his proposed solution was silly and its adoption probably caused more confusion. (It's likely the reason for the discussion under the accepted answer). If instead we use a more cumbersome but explicit notation like $$\left.\frac{\partial z}{\partial x}\right|_{y=\text{const.}}$$, which physicist sometimes write as $$\left(\frac{\partial z}{\partial x}\right)_y$$, then we could again explain why that false chain rule in the accepted answer is wrong: it becomes $$\left.\frac{\partial z}{\partial t}\right|_{u=\text{const.}} \neq \left.\frac{\partial z}{\partial x}\right|_{y=\text{const.}} \left.\frac{\partial x}{\partial t}\right|_{u=\text{const.}}$$ and we are not allowed to cancel the changes in $$x$$ (i.e. $$\partial x$$), since in the first factor $$\partial x$$ is considered under the assumption that $$y$$ is constant while in the second factor it is considered under the assumption that $$u$$ is constant. (I added the additional variable $$u$$, like Justin did in a comment, to avoid a discussion about the difference between $$\partial$$ and $$d$$. So I'm assuming $$x,y$$ are both functions of $$t,u$$.)
The other problem, mentioned in davdan angelo's answer, relating to second derivative notation also disappears, if we understand where that notation for second derivatives comes from: the equality $$\frac{d\frac{dy}{dx}}{dx}=\frac{d^2y}{dx^2}$$ was historically only considered correct, under the assumption that $$dx=\text{const.}$$ See here. Under that assumption (and further assumptions necessary to make sense of the involved operations), the chain rule-like equation $$\frac{d^2f}{dx^2}= \frac{d^2f}{dg^2}\left(\frac{dg}{dx}\right)^2$$ in davdan angelo's answer is true: we assume $$dx=\text{const.}$$, $$dg=\text{const.}$$ and that $$f$$ is a function of $$g$$ which is a function of $$x$$. These assumptions imply that $$g$$ is a linear function of $$x$$ and hence the equations holds.
(For people still worrying about what the differentials actually denote: one could make them precise by interpreting variable quantities $$z,x,y,f,g$$ etc. as real valued smooth functions on a manifold $$M$$ and differentials $$dx$$ etc. as differentials in the sense of differential geometry. The operation $$d^2 x$$ would additionally require a connection on the manifold in order to talk about the differential of a differential.)
Mathematics is a formal language, which people often use in an informal way. Think of the symbol ×. It is a binary operator - it needs two numbers to 'work':
3 × 2 = 6
If we wrote × with only one number ("3 ×" or "× 2") it doesn't make sense.
Differentials are the same. The symbol represents the relative change of one quantity when compared to the change in another quantity. $$\frac {dy}{dx}$$ means "how much does y change when compared to a change in x". Writing just "dy" is writing half a statement, and in formal understanding is meaningless.
I've had students say "but we use 'dx' alone in integral calculus". No we don't. Every integral that ends with a '.dx' starts with a $$\int$$. They are effectively the two ends of a set of brackets, and writing either part alone doesn't make sense.
That doesn't stop people from doing it, however. Derivatives look like fractions, but are not fractions. People can manipulate them in a way similar to fractions, and in many cases will get the right answer, but what they are doing and what they think they are doing are actually two different things. When you use the Chain Rule to make an expression simpler, you are not "cancelling out".
• In higher math, there actually is a fully rigorous meaning to these loose differential terms and it comes from the theory of Differential Forms. It just requires a bachelors degree level of mathematics to understand the formal definitions involved, so we don't teach it that way :+) Oct 11 at 23:30
There are two reasons, and they both stem from problematic notations being used in calculus.
The first reason is higher-order derivatives. The common notation for the second derivative, $$\frac{d^2y}{dx^2}$$, cannot be used as a fraction. However, this is because it is problematic notationally. If you consider $$\frac{dy}{dx}$$ a fraction, and then use the quotient rule to take its derivative, the second derivative of $$y$$ with respect to $$x$$ is $$\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$$, which does allow for straightforward algebraic manipulations of differentials.
The second reason is partial differentials. Unfortunately, partial differentials are individually completely ambiguous as to what they refer to. If you clean up the notation, however, and make each different unambiguous, the problem goes away.
You can check out more details about both of these in the paper "Total and Partial Differentials as Algebraically Manipulable Entities" by Maria Isabelle Fite and myself.
• Side note - I wrote my own calculus book, "Calculus from the Ground Up", specifically because I was annoyed at how Saxon Calculus teaches it. Oct 25 at 11:44
• The notation in the paper introduces more problems and I suspect that a mixup between the notion of function in the modern sense and in the original sense of "function of" is the origin. You write $\partial(f,x_1)=f(x_1+dx_1,x_2,\ldots)-f(x_1,x_2,\ldots)$: The term on the right is clearly an infinitesimal number, not a function, while the term on the left is probably supposed to be a function. You cannot equate a function with a number. But independent of that, the right hand side depends on $x_2,\ldots$ so these variables should also appear in the notation on the left. Oct 26 at 14:35
Remember that $$\frac{\partial y}{\partial t}$$ is actually $$\lim\limits_{h \to 0} \frac{y(t+h) - y(t)}{h} = \lim\limits_{h \to 0} \frac{y' - y}{h}$$ (I'm using $$y'$$ to make it less busy). It's not a fraction but the limit of a fraction.
If you rewrite the original:
$$\lim\limits_{h \to 0} \frac{y' - y}{x' - x} = \lim\limits_{h \to 0} \frac{y' - y}{t' - t} \cdot \lim\limits_{h \to 0} \frac{t' - t}{x' - x}$$
The $$t'-t$$ is inside each limit, so you can't cancel them out.
• Sure, you don't have a convergent limit for $1/(x'-x)$ and so trying to define the fraction $1/dx$ this way doesn't really work. (the bit with $y$ is fine, but zero) However, there are other ways to define $dx$ to give a sensible definition. e.g. differential forms or non-standard analysis.
Oct 13 at 14:36
• +1: I like the start of this answer and but I think it didn't get much attention because it leaves the reader confused: "If you can't cancel, then why do calculus textbooks manipulate $\dfrac{\mathrm dy}{\mathrm dx}$ as though it were a fraction? Is there a concrete example where canceling will lead you to an incorrect result? Couldn't you just assume the $h$'s are the same in each limit in the product, so that you can combine the limits and then cancel?" Just some things I imagine going on in readers' minds. Oct 14 at 20:56
• That's the modern mainstream definition. But there are other approaches like smooth infinitesimal analysis where $f'(a)$ it is defined as the unique number $m$ such that $f(a+\epsilon)=f(a)+m\epsilon$ for every infinitesimal $\epsilon$. In this definition it is quite naturally seen as a fraction. Oct 20 at 11:27
• @MichaelBächtold A slope $m=f'(a)$ is naturally seen as a fraction, too, with or without using infinitesimals. But this answer gives the same reason that I was given as to why we should not think of the $dt$s in the OP's chain rule as cancelling. This seems to answer "why our calculus teacher warned us off from thinking of Leibniz notation in this way." Oct 20 at 19:59
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# Finding an unknown force at an unknown angle
## Homework Statement
So we have a force of unknown magnitude acting on these struts at an angle θ measured from strut AB.
The component of the force acting along AB is 600lb, and the magnitude of the force acting along BC is 500lb.
If Φ = 60°, what is the magnitude of F and the angle θ?
Fcos(θ) = 600lb
## The Attempt at a Solution
Ok. So, it'll probably help if I knew the third angle of the triangle formed.
180° = ɣ + (60° + 45°)
180° - 105° = ɣ
75° = ɣ
Great. So, I know that Fcos(θ) = 600lb, and Fcos(75° - θ) = 500lb
hm. Fcos(θ)/600 = 1 = Fcos(75 - θ)/500
500Fcos(θ) = 600Fcos(75° - θ)
5Fcos(θ) = 6Fcos(75° - θ)
5cos(θ) = 6cos(75° - θ)
0 = 5cos(θ) - 6cos(75° - θ)
Originally I tried finding where z = 5cos(θ) - 6cos(75° - θ) intersected with z = θ + η where η = 75° + θ, but I couldn't get Wolfram Alpha to understand what I was talking about. Here, I see I should have just left η as
75° - θ, but even still, I have to *ask* Wolfram Alpha what θ works for 0 = 5cos(θ) - 6cos(75° - θ) when
0<=θ<=75° (it gives me an angle of ~30.7°).
Worse, since I couldn't figure it out, I figures if I gave in on the magnitude of F, I could still find the angle. Mastering Engineering told me F = 870lb. So, if Fcos(θ) = 600, θ=arccos(600/F) and arccos(600/870) ≈ 46.4°.
Which was wrong. The θ it wanted was ~34°
Which means everything I did was wrong.
So what triangle magic do I do to get from the initial problem, to the final F=870 θ=34°, without doing something so complicated I need Wolfram Alpha to crunch it out?
Svein
A hint: cos(x-y) = cos(x)cos(y) + sin(x)sin(y). Apply that to cos(75° - θ).
Doc Al
Mentor
Great. So, I know that Fcos(θ) = 600lb, and Fcos(75° - θ) = 500lb
Good. I suggest using a trig identity to simplify the expression Fcos(75° - θ).
While that does get everything in simpler terms of θ, that doesn't address that 870cos(34) = 721.3 and not 600.
My error is a lot farther up.
Doc Al
Mentor
While that does get everything in simpler terms of θ, that doesn't address that 870cos(34) = 721.3 and not 600.
My error is a lot farther up.
Start over with the two simpler equations. You'll get a different value for θ and F.
what do you mean by that? Because F=870 and θ≈34° are set in stone Those are the answers for this problem.
Doc Al
Mentor
what do you mean by that? Because F=870 and θ≈34° are set in stone
And yet you know that:
870cos(34) = 721.3 and not 600.
Those are the answers for this problem.
Says who?
Why not just solve it using that trig substitution. You'll solve it easily without needing Wolfram.
Says who?
Says The homework. Because I've already gotten this question wrong, and the answers were displayed. I'm asking about this question here so I know what to do when this sort of problem comes up again.
Doc Al
Mentor
Says The homework. Because I've already gotten this question wrong, and the answers were displayed.
Do you not agree that the given answers do not work? That they contradict the problem statement?
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Quick Answer: Do The Two Variables Have A Linear Relationship?
What is a linear relationship between two variables?
A linear relationship (or linear association) is a statistical term used to describe a straight-line relationship between two variables.
Linear relationships can be expressed either in a graphical format or as a mathematical equation of the form y = mx + b.
Linear relationships are fairly common in daily life..
Is there a linear relationship?
There are only three criteria an equation must meet to qualify as a linear relationship: It can have up to two variables. The variables must be to the first power and not in the denominator. It must graph to a straight line.
What is a perfect linear relationship?
A perfect linear relationship (r=-1 or r=1) means that one of the variables can be perfectly explained by a linear function of the other.
What is the difference between linear and directly proportional?
If a relationship is linear, then a change in one variable will cause a change in another variable by a fixed amount. … A directly proportional relationship is a special type of linear relationship. When one variable is equal to 0, the second variable will also have a value of 0.
Do all proportional relationships go through the origin?
Directly proportional relationships always pass through the origin (0,0). There are other linear relationships that do not pass through the origin.
How do you know if two variables have a linear relationship?
When two variables are perfectly linearly related, the points of a scatterplot fall on a straight line as shown below. … The more the points tend to fall along a straight line the stronger the linear relationship. The figure below shows two variables (X and Y) that have a strong but not a perfect linear relationship.
How do you know if a linear relation exists?
If the absolute value of the correlation coefficient is greater than the critical value, we say a linear relation exists between the two variables. Otherwise, no linear relation exists.
How do you know if a linear correlation exists?
Linear correlation : A correlation is linear when two variables change at constant rate and satisfy the equation Y = aX + b (i.e., the relationship must graph as a straight line). Non-Linear correlation : A correlation is non-linear when two variables don’t change at a constant rate.
What is the difference between linear and nonlinear equations?
While a linear equation has one basic form, nonlinear equations can take many different forms. … Literally, it’s not linear. If the equation doesn’t meet the criteria above for a linear equation, it’s nonlinear.
Which pairs of variables have a linear relationship?
Answer: a linear pairs are x and y.
What are three variables that have a linear relationship?
If a, b, c and r are real numbers (and if a, b, and c are not all equal to 0) then ax + by + cz = r is called a linear equation in three variables. (The “three variables” are the x, the y, and the z.) The numbers a, b, and c are called the coefficients of the equation.
Does a linear relationship go through the origin?
Similar descriptions can be used for the curved graphs that show a ‘decrease of y with x’. The formal term to describe a straight line graph is linear, whether or not it goes through the origin, and the relationship between the two variables is called a linear relationship.
Does a linear relationship have to be proportional?
The relationship between variables can be linear, non-linear, proportional or non-proportional. A proportional relationship is a special kind of linear relationship, but while all proportional relationships are linear relationships, not all linear relationships are proportional.
How do you tell if an equation is linear or nonlinear?
Using an Equation Simplify the equation as closely as possible to the form of y = mx + b. Check to see if your equation has exponents. If it has exponents, it is nonlinear. If your equation has no exponents, it is linear.
Does correlation mean linear relationship?
Correlation can tell if two variables have a linear relationship, and the strength of that relationship.
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You are here:
Question
good morning, I studying to take my exam, have a math question that is fooling me some. Question as follows,
A belt filter press receives a feed sludge at 3% total solids and produces a cake that is 20% total solids. If influent flow rate to the press is 50 GPM, what will the volume of cake produce be if the press runs for 10 hours?
I thank you very much for your time and help, again thank you.
The way I figure it is that the filter press is able to squeeze out enough water to produce a 20% solid component flow (cake) from a 3% flow (sludge). This assumes that the press captures all the waste solids.
Converting percents to fractions, i.e., 3% -> 0.03 and 20% -> 0.20, this means that the cake has a volume (0.03/0.20) = 0.15 as much as the sludge. That is, the same amount of solids make up the 2 fractions of solids, which means the volume for the 20% must be less than for the 3%.
Using this fraction, we have the production of cake = (0.15)(50 GPM) = 7.5 GPM.
Now, 10 hours converts to (10 hours)(60 min/hour) = 600 minutes.
Thus the total amount of cake produced in 10 hours is (600 min)(7.5 GPM) = 4500 gal.
Questioner's Rating
Rating(1-10) Knowledgeability = 10 Clarity of Response = 10 Politeness = 10 Comment thank you very much,
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# Stretched Spring
#### 12boone
1. Homework Statement
A spring with 58 hangs vertically next to a ruler. The end of the spring is next to the 15- mark on the ruler. If a 2.0- mass is now attached to the end of the spring, where will the end of the spring line up with the ruler marks?
2. Homework Equations
I used F=KX should it be mgy=kx? or mgy=1/2kx^2
3. The Attempt at a Solution
My solution was 32 centimeters and 34 centimeters and those were both wrong. Must use two sig figs.
Related Introductory Physics Homework Help News on Phys.org
#### LowlyPion
Homework Helper
1. Homework Statement
A spring with 58 hangs vertically next to a ruler. The end of the spring is next to the 15- mark on the ruler. If a 2.0- mass is now attached to the end of the spring, where will the end of the spring line up with the ruler marks?
2. Homework Equations
I used F=KX should it be mgy=kx? or mgy=1/2kx^2
3. The Attempt at a Solution
My solution was 32 centimeters and 34 centimeters and those were both wrong. Must use two sig figs.
With F=-kx you can know k by the original measurement - assuming that 0 is the 0 force point. This implies that each unit represents 58*g/15 units of force. Adding another 2*g units of force means that you will get an additional 2*g*15/58*g units. This suggest then it will lengthen to 15+(30/58) units.
I'm curious where you got cm out of your description. Or is there a picture you aren't sharing and other units you haven't mentioned?
#### 12boone
oh the ruler was 15 cm
#### 12boone
sorry i guess my copy paste didnt work. The 58 is 58 N/m and is the K and it hangs next to a spring that is next to a ruler at 15 cm. If you add 2 kg to it what is the new measured distance?
#### LowlyPion
Homework Helper
sorry i guess my copy paste didnt work. The 58 is 58 N/m and is the K and it hangs next to a spring that is next to a ruler at 15 cm. If you add 2 kg to it what is the new measured distance?
I still have no idea where the origin is. But 2 kg will move it 2*9.8/58 m.
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# How to solve the following exercise?
17 views (last 30 days)
Ellen De Jonghe on 14 Jan 2020
Edited: Andrei Bobrov on 14 Jan 2020
Hye!
I have to solve the following problem:
Consider a matrix M with only the numbers 1 to 9 as elements.
M =
2 9 3 2 4
8 6 4 8 5
5 7 1 6 4
9 8 9 5 1
Consider one of the elements M(i,j) that's not on the edge of the matrix. Such element always has 8 neighbours. If M(i,j) > 1 and each number from 1 to M(i,j)-1 is one of the 8 neighbours, we say that element is neighboring. If M(i,j) = 1, the element is automatically neighboring.
For example, M(2,2) is neighboring because 1,2,3,4 and 5 are one of the element's neighbours. M(3,4) on the other hand isn't neighboring because 2 and 3 don't occur around the element.
Now, I have to write a function that has 3 inputs: a matrix M and a row- and column index. The function has to control whether de element is neighboring or not and has a logical 0 or 1 as output.
Bob Nbob on 14 Jan 2020
That's definitely a really good start. I have a couple of thoughts to add.
1) How does your code respond to the selected element being 1? I think you may have a problem there.
2) You can always just take the entire 3x3 around the selected element. Even if it contains the element that shouldn't be a problem with your logic check.
n = matrix(row-1:row+1,column-1:column+1);
Guillaume on 14 Jan 2020
@Ellen, you've got the correct algorithm. As you suspec and Bob pointed out, the construction of n can be done in just one line with simple indexing.
@Bob, no the code also works for 1. m would be empty, so ismember will return empty. sum(empty) is 0 which is also the length of empty.
For the record, the one-liner I was talking about is
res = all(ismember(1:matrix(row, col)-1, matrix(row-1:row+1, col-1:col+1)))
Ellen De Jonghe on 14 Jan 2020
Thanks!
I indeed forgot to check the case where M(i,j) = 1, I will try to figure that out.
Andrei Bobrov on 14 Jan 2020
Edited: Andrei Bobrov on 14 Jan 2020
function out = find_neighbor(M,i,j)
out = all(ismember(1:M(i,j)-1,M(i-1:i+1,j-1:j+1)));
end
Guillaume on 14 Jan 2020
The if is not needed, out is true anyway, if M(i,j) is 1, since all([]) is true.
Andrei Bobrov on 14 Jan 2020
Thanks Guillaume! I'm fixed.
Ellen De Jonghe on 14 Jan 2020
That's a very easy one! Thanks.
I seem to always make my scripts much longer than it has to.
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# Binary Search Tree
My professor posted some review questions for the final exam. And I can't seem to find the answers for it. Any help will be greatly appreciated!
Consider a binary tree of n nodes:
a. What is the minimal and maximal number of leaf nodes?
b. What is the minimal and maximal value of the height?
c. How many pointers are used by the tree (not counting the null pointers, and assuming we do not keep a field that stores the parent)?
*d. What is the worst care running time for inserting n nodes into a (initially empty) binary search tree?
• You should go talk to the professor and see if you can get what he was leading you to understand from these questions. Maybe he can shed some light on the overall concept he's looking for. – Heath Hunnicutt May 24 '11 at 17:31
• Do you understand what a binary tree is? If so try putting some numbers to check if you can figure out answers like n=3, 4 etc – d-live May 24 '11 at 17:32
• The Prof did not say "Balanced Binary Tree", and the worst-case Binary Tree degenerates to a.... Bueler? Bueler? Anybody? – Heath Hunnicutt May 24 '11 at 17:33
• I know this is a really old question, but just in case anyone is still looking for an answer... you may want to take a look at his post on how to implement AVL trees from scratch. medium.com/amiralles/… – Alejandro Miralles Nov 22 '18 at 17:55
• The maximum number of leaves is ceil(n / 2). The minimum number is 1
• The maximum height is n. The minimum is floor(log_2(n))
Try drawing various trees on a paper and see what you get. Remember that a binary tree is defined as a tree where each node may have 0 (in which case it is a leaf), 1 or 2 children. For your question you should examine the very unbalanced case of 1 child per node.
Consider:
If you're trying to maximize the number of leaves, you want as few internal nodes as possible (and the reverse if you're trying to minimize the number of leaves). How can you accomplish that?
To get a tree of maximal height, you'll put as few nodes in each level as possible. How can you do that? Conversely, for the minimum height, what is the maximum number of nodes you can put at each level?
How many ways are there to get to each node of a tree? Thus, how many pointers do you need?
I'm assuming you're either coding in C or C++.
a. A node, if the structure is defined like this: struct node { struct node *left, *right; }; You can observe that the structure can either have 0, 1, or 2 leaves. So, the max is 2, min is 0 leaves.
b.Minimal height is zero, in which would only contain the root node. Note that the root node does not count as a height of 1. It's also called depth at times. Here is an algorithm for the height:
`````` int height(struct node *tree)
{
if (tree == NULL) return 0;
return 1 + max (height (tree->left), height (tree->right));
}
``````
c. Pardon me if I take this the worng way, but I'm assuming if we mapped this out on a piece of paper, we'd be trying to find the number of "links" that we would use? In that case, it'd simply be the number of nodes in the tree -1 for root node. This algorithm found on this page http://forums.techarena.in/software-development/1147688.htm can help you: check if root is null, then pass the left and right nodes as parameters into the function.
``````int countnodes(Node* root)
{
if (root == null || k<=0)
{
return 0;
} else {
return 1 + count(root.left,k-1) + count(root.right,k-1);
}
}
// remember to subtract one at the end.
int totalnodes = countnodes(root) - 1;
``````
d. The time complexity for best case is O(nlogn) where n is the number of nodes to insert. The worst case, is O(n). It is directly linear.
If you have any other questions just google it, there's plenty of things to know about binary search trees. But most of it is simply recursion that you can learn in 30 seconds.
I hope this helps. Good luck on your exam! I had mine a few months ago. ;)
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Home » Posts tagged 'ratio'
# Tag Archives: ratio
## A Question from Current Electricity
The resistances of 2 bulbs are in the ratio 1:2.if they are joined in series,find the ratio of the energy consumed by the 2 bulbs?
When connected in series, the same current flows through each of the resistors. Therefore, the energy dissipated is directly proportional to their resistances. That is, the ratio of energy dissipated will nbe same as the ratio of resistances. (1:2)
## de Broglie Wavelength
Alpha particle and a proton are accelerated from rest by the same potential. Find the ratio of their de- broglie wavelength
Charge of alpha particle = 2e
Mass of alpha particle = 4 u
Charge of proton = e
mass of proton = u
The energy acquired by proton when accelerated through a pd of V,
E=eV
The momentum acquired by proton=${\sqrt{2ueV}}$
The de Broglie wavelength is given by $\lambda =\frac{h}{mv}$
Therefore, de Broglie wavelength of Proton, $\lambda _{proton}=\frac{h}{\sqrt{2ueV}}$
Similarly,$\lambda _{alpha}=\frac{h}{\sqrt{2\times 4u\times 2e\times V}}$
$\frac{\lambda _{alpha}}{\lambda _{proton}}=\frac{\sqrt{2ueV}}{\sqrt{2\times 4u\times 2e\times V}}=\frac{1}{2\sqrt{2}}$
## A numerical problem on Drift velocity
Two conducting wires X and Y of same diameter but different materials are joined in series across a battery . if the number density of electrons in X is twice that in Y . Find the ratio of drift velocity of electrons in the two wires.
## Power factor in AC circuits
“what is meant by power factor?”
Ans:
In AC circuits, Power factor is defined as the ratio of true power to apparent power.
$Power Factor =\frac{True Power}{Apparent Power}$
If φ is the phase difference between E and I, then
True Power = EvIv cos φ and the Apparent Power is EvIv
Therefore,
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Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
# What is the Smallest Positive Integer N for Which ( 1 + I ) 2 N = ( 1 − I ) 2 N ? - Mathematics
What is the smallest positive integer n for which $\left( 1 + i \right)^{2n} = \left( 1 - i \right)^{2n}$ ?
#### Solution
$\left( 1 + i \right)^{2n} = \left( 1 - i \right)^{2n}$
$\Rightarrow \left[ \left( 1 + i \right)^2 \right]^n = \left[ \left( 1 - i \right)^2 \right]^n$
$\Rightarrow \left( 1^2 + i^2 + 2i \right)^n = \left( 1^2 + i^2 - 2i \right)^n$
$\Rightarrow \left( 1 - 1 + 2i \right)^n = \left( 1 - 1 - 2i \right)^n [ \because i^2 = - 1]$
$\Rightarrow \left( 2i \right)^n = \left( - 2i \right)^n$
$\Rightarrow \left( 2i \right)^n = \left( - 1 )^n (2i \right)^n$
$\Rightarrow ( - 1 )^n = 1$
$\Rightarrow \text { n is a multiple of } 2$
Thus, the smallest positive integer n for which
$\left( 1 + i \right)^{2n} = \left( 1 - i \right)^{2n}$ is 2.
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 11 Mathematics Textbook
Chapter 13 Complex Numbers
Exercise 13.2 | Q 24 | Page 33
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Careers and Employability Service
# Verbal Logic Test
Two things are infinite: the universe and human stupidity; and I'm not sure about the universe.
Albert Einstein
This practice test involves verbal logic puzzles, some of which have a numerical element. They test your ability to think logically, analytically and numerically, and also to extract meaning from complex information.. Some employers use similar tests as part of their selection procedures and this test will give you some idea of what to expect.
The test has 21 questions and you will have 40 minutes to do them. At the end of the test (when 40 minutes have elapsed), you will be given a score. You are allowed to use scrap paper and a calculator for working out answers.
The test will start with three easy example questions which will not be marked or timed. Now click on the "First Question" button below to begin. Click on the button beneath your answer to answer it. You can change your answer by clicking on another button.You can also go back to the previous question, by clicking on the previous question button. When you have finished the test click on "Get Your Score"
Time Left : Test developed by Bruce Woodcock
For further example tests and help with tests, see our Aptitude Tests page with tips on how to pass tests.
## Score
• 16 or above. This is an above average score for graduates.
• 12 - 15. This is in the average group compared to other graduates.
• 11 or below. This is below the typical score of graduates but remember that a number of factors may have distorted your score - see above for some of these factors. You may wish to discuss your results with a careers adviser.
Question
1 2nd
In a horse race Hill Royal came in ahead of Trigger. Hill Royal finished after Black Beauty. Copenhagen beat Black Beauty but finished after Bucephalus
Where did Copenhagen finish? 2nd
Bucephalus, Copenhagen, Black Beauty, Hill Royal, Trigger
2 4th Where did Hill Royal finish? 4th
3 66
Forty two is seven times a particular number. What is eleven times that number?
7 x 6 = 42
1x6 = 66
Question
1. Kimi Debbie, Kimi and Michael have Ferraris. Michael also has a Reliant Robin. Jensen has a Mercedes and a Model T. Rubens also has a Mercedes. Debbie also has a Bugatti Veyron. Rubens has just bought a Toyota Prius. Who has the fewest cars?
Debbie: Ferrari, Bugatti Veyron
Kimi: Ferrari
Michael: Ferrari, Reliant Robin,
Jensen: Mercedes, Model T,
Rubens: Mercedes, Toyota Prius
2 66
One third of a number is four times eleven. What is half of that number?
1/3 =44
Number = 132
Half 132 = 66
3. Jensen doesn't qualify
Jensen, Lewis and Mika need to be able to run 100m in under 12.5 seconds to qualify for a championship. Lewis and Mika run faster than Jensen. Jensen's best time for the 100m is 13.1 seconds. Which of these 5 options MUST be true:
• Only Lewis qualifies
• Jensen doesn't qualify
• Lewis and Mika both qualify
• Jensen qualifies
• No one qualifies
4. 7
Wayne is double the age of Fernando and one third as old as Didier who will be 48 years old in 6 years. How old is Fernando?
Didier is 48-6 =42 years old
Wayne is 1/3 of 42 years old = 14
Fernando is half 14 years old = 7 years old
5. HILARY
Hanif, Horace, Hilary and Hannah are students. Hanif and Horace speak Chinese, whereas the others speak Arabic. Horace and Hannah speak Albanian. Everyone except Hanif speaks Esperanto.
Hanif speaks Chinese;
Horace: Chinese, Albanian and Esperanto;
Hilary: Arabic and Esperanto;
Hannah: Arabic, Albanian, and Esperanto
Who only speaks Arabic and Esperanto? HILARY
6. HORACE
Who speaks more than one language but not Arabic? HORACE
7. 13
Josh the postman has eleven red rubber bands; he gives Sunita three bands. Sunita now has twice the number of bands Josh has left. How many bands did Sunita have at the beginning?
11-3 = 8
Sunita = 2x8=16
16-3 = 13
8. 4.15
Simon, Cheryl and Dannii are all going by train to London to watch a singing competition. Cheryl gets the 2.15 pm train. Simon's train journey takes 50% longer than Dannii's. Simon catches the 3.00 train. Dannii leaves 20 minutes after Cheryl and arrives at 3.25 pm. When will Simon arrive?
Dannii leaves at 2.35 arrives 3.25 therefore 50m journey
Simon's journey takes 75m therefore arrives at 4.15
9. 10
5 bricklayers can lay a total of 50 bricks in 30 minutes. How many bricklayers will be required to lay a total of 60 bricks in 18 minutes?
1 bricklayer lays 10 bricks in 30 minutes = 1 brick every 3 minutes
60 bricks would take 1 bricklayer 180 minutes
Therefore 180/18 bricklayers are required to lay these in 18 minutes =10
10. W An old treasure map has the following instructions:
Stand next to the black rock and face West. Walk 20 yards and then turn 90 degrees clockwise
Walk another 10 yards and then turn 45 degrees anticlockwise. Walk another 15 yards, reverse your direction and walk 5 yards back. Turn 135 degrees clockwise and walk another 10 yards. In which direction are you now facing? N
Ignore the distances (these are red herrings) , the direction you face is all that matters!
W, N, NW, SE, W
11.
Skoda and BMW You are the head of purchasing for a company and have to buy the following 9 cars for the company pool in the next three months. The company is very bureaucratic and has a rule that exactly £60,000 must be spent in each of the three months and you are only allowed to buy 3 cars each month. In the first month you buy the Fiat Uno. Which other two cars must you buy that month?
You must spend £60,000 each month The Fiat accounts for £8,000 of this leaving £52,000 to be spent on two more cars. The only way of achieving this is to buy the BMW and Skoda.
• BMW Series 3 £40,000
• Lexus £36,000
• Volvo Estate £32,000
• Ford Focus £14,000
• Ford Focus £14,000
• Ford Focus £14,000
• Skoda Octavia £12,000
• Ford Car £10,000
• [Fiat Uno £8,000]
12. Athos
Athos, Portos and Aramis live in three adjoining houses. Aramis has a black cat called d'Artagnan, Portos has a white dog whereas Athos has a red herring. Portos has a neighbour with a red door. The owner of a four legged animal has a blue door. Either a feline or fishy owner has a green door. Aramis and Portos are not neighbours. Whose door is red?
Portos, white dog, blue door
Athos, red herring, red door
Aramis, black cat, green door
13.
Mr Bumpem and Mr Parker Mrs Krashem, a driving instructor has to arrange booking for a number of her pupils. She has 8 new pupils who wish to book either a morning or afternoon of a particular day. As these are two hour introductory lessons, she only sees one pupil each morning or afternoon.
• Miss Banger is only available Tuesday mornings
• Mr Bumpem can make any time on a Wednesday
• Mrs Exhaust is free on Tuesdays all day
• Mr Hilstart is only free Wednesday afternoons
• Miss Boot is only available Friday mornings
• Miss Bonnet can only make Saturday afternoons
• Mrs Speed is available all day Fridays
• Mr Parker can make any time on a Saturday
Which must both have morning appointments?
• Mr Bumpem AM
• Mrs Exhaust PM
• Mr Parker AM
• Mrs Speed PM
Mr Bumpem and Mr Parker AM
14. Mrs Exhaust & Mrs Speed
Which must both have morning appointments?
Mrs Exhaust & Mrs Speed
15. 15 degrees
How many degrees are there between clock hands at 6.30 am?
Minute hand is at 180 degrees
Hour hand is at 180 + 360/24 degrees as it has moved half an hour further = 180 +15 degrees.
Therefore difference in angle = 15 degrees
16. 3
It is 4.30 am and your alarm goes off. You crawl sleepily out of bed. You have a job interview to go to that day and you must leave by 5 am to catch the train as the interview is being held 150 miles away. Suddenly all the lights go off. There has been a power failure. It is still pitch dark outside and you have no other light available. You have 6 brown socks and 6 pink socks in your drawer, all mixed up and identical in all but colour.
How many socks must you take from the drawer to be sure of taking a matching pair of the same colour?
Once you have taken two socks from the drawer (the possible combinations are BB, PP or BP), the third sock you remove must give you a pair of either brown or pink.
17. 8
How many socks would you need to take from the drawer to be certain of getting two pink socks?
The first six socks you remove could all be brown, but the seventh and eighth socks would then have to be pink.
By the way, you didn't get the job.
18.
Magician and Ball games
You are holding a children's party for 7 children and have asked the children what activities they would like at the party. Because of time constraints, you will only have time for two activities, but want to make sure that everyone gets either their first or second choice. The children and activity preferences are as follows:
Rachel Face painting Magician Bouncy castle Ball games Disco Debbie Bouncy castle Ball games Face painting Magician Disco Sunita Magician Face painting Magician Disco Ball games Ben Ball games Face painting Disco Magician Bouncy castle Mia Disco Magician Face painting, Bouncy castle Ball games Jo Magician Bouncy castle Disco Face painting Ball games Amel Face painting Ball games Bouncy castle Magician Disco
Which two activities should you choose? Magician and Ball games.
You only need to look at the second and third columns to make sure that everyone gets their first or second choice.
19. £16
You spend 56 pounds in total on a cactus, a stuffed porcupine and a puncture repair kit. The cactus costs twice as much as the puncture repair kit and the stuffed porcupine double the price of the cactus. How much did the cactus cost?
Using simple algebra. Say cost of puncture repair kit = x, then cost of cactus is 2x and cost of porcupine is 4x
Therefore total cost = x + 2x + 4x = 7x
7x = £56
Therefore x = £8
Cactus costs 2x which is £16
20. Kitten
You take seven children to a toy shop to buy each a soft toy. The toy shop only has one of each type so you ask the children their preferences. You decide to give each child one of their preferred toys.
James Wombat Panda Gorilla Kitten Rat Josh Panda Donkey Dog Jezebel Panda Donkey Kitten Dog Jamelia Panda Donkey Dog Janine Wombat Gorilla Donkey Kitten Rat Jasbeet Panda Dog Jason Wombat Gorilla Dog Rat Kitten
Which animal will you give Jezebel? Kitten
The trick here is to realise that you can ignore the children with more than four choices.
Josh, Jamelia and Jasbeet all have only two or three choices and all want a panda, donkey or dog, so they must each get one of these.
Jezebel also wants a panda, donkey or dog, but she would also accept a kitten, so she must be given this, if each child is to be allowed a preferred animal.
21. Run
A crofter has to get to his herd of sheep quickly as he has been told they are being attacked by a dog. His sheep are on the other side of a steep hill. He can run over the hill (3 miles) at 4 miles an hour, or take his tractor via an old dirt track which is 5 miles at an average of 6 miles an hour or he can drive his car along a very narrow winding road but this is 14 miles and he can only go at 18 miles an hour on average.
Which method should he choose?
Running 3 miles at 4 mph takes 45 minutes
Driving the tractor for 5 miles at 6 mph takes 50 minutes
Driving the car for 14 miles at 18 mph takes 47 minutes
Therefore he should run.
Careers and Employability Service - © University of Kent
The University of Kent, Canterbury, Kent, CT2 7ND, T: +44 (0)1227 764000 ext. 3299
Last Updated: 05/04/2018
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# Polygons with equal area and perimeter but different number of sides?
Let's say we have two polygons with different numbers of sides. They can be any sort of shape, but they have to have the same area, and perimeter.
There could be such possibilities, but can someone show me with pictures? I just need visualize it.
Sometimes in life you just have to know it, and sometimes we need a picture shown in our faces :).
┌─┐ ┌┐
│ │ │└─┐
│ └─┐ │ └┐
└───┘ └───┘
Edit: I like the above figures because they're easy to generalize to many sides. But if it's unclear that they have the same area, here's another pair: the L and T tetrominoes.
You can imagine sliding the square on the right side up and down relative to the 1x3 bar on the left side; this operation preserves both area and perimeter. Explicitly, both tetrominoes have area 4 and perimeter 10. The L has 6 sides, and the T has 8 sides.
• How do I know those have the same area? Commented Jul 25, 2014 at 16:51
• @oconnor0 Chop a 2x2 square from the top of the first figure and put in into the "notch" that remains. You can do the same with the 2x1 rectangle from the top of the second figure. Commented Jul 25, 2014 at 17:01
• The Tetris pieces make it much more obvious. And they're prettier. :)
– cHao
Commented Jul 27, 2014 at 4:46
For any triangle there is a rectangle with the same area and perimeter.
proof: by Heron's formula the area of the triangle with semiperimeter $S$ and triangle sides $a,b,c$ is $$\sqrt{S(S-a)(S-b)(S-c)}\leq \sqrt{\frac{8S^4}{27}}$$
A square with semiperimeter $S$ has area $\frac{S^2}{4}\geq\sqrt{\frac{8S^4}{27}}$.
We can then proceed to make the square into a rectangle, making the area smaller and smaller preserving the perimeter. Until the area is as small as the triangle's.
• In fact I am sure for any n<m and any n-gon we can find an m-gon with the same area and perimeter as the n-gon, but I can't prove this Commented Jul 25, 2014 at 0:37
• See my answer "cut and flip" Commented Aug 4, 2014 at 20:01
• are you sure that works for any arbitrary n,n and an y arbitrary n-gon? Commented Aug 5, 2014 at 2:58
• actually, upon thinking about it, there are some cases I am not sure about. I think it deserves its own question. math.stackexchange.com/questions/888011/… Commented Aug 5, 2014 at 10:58
Just worked out a quick example, so the numbers may not be optimal: take a triangle with side lengths 2,3,4 - this has perimeter 9 and area $3\sqrt{15}/4$. It's easy enough to construct a rectangle with this data as well, by solving the equations $st = 3\sqrt{15}/4$ and $2s+2t = 9$. In fact, the sides lengths $s,t$ of the rectangle work out to be $\frac{1}{4}(9 \pm \sqrt{81-12\sqrt{15}})$.
• A 3-4-5 triangle has perimeter 12 and area 6, the corresponding rectangle satisfies $x(6-x)=6$ or $x^2 - 6x + 6 = 0$. By the quadratic formula, $x = 3 \pm \sqrt{3}$, so a $3 + \sqrt{3}$ by $3 - \sqrt{3}$ rectangle will work. Commented Jul 25, 2014 at 4:53
• love the way you solve it! Commented Jul 27, 2014 at 12:39
This the first example that came to my mind, using four identical $45-45-90$ triangles.
• Might be better to clone the first shape, but flip one of the sub-triangles (horizontally in this case) so that the dotted-line sides still entirely touch. That should make it more obvious that the perimeter is exactly the same, without requiring any math.
– cHao
Commented Jul 27, 2014 at 4:38
• @chao actually, that was my original construction, but I somehow convinced myself they had the same number of sides! Thanks for rescuing the idea from my lapse :) Commented Jul 27, 2014 at 12:51
• @chao rolled back to revision 1. It's minimal and requires basically no explanation. Commented Aug 9, 2014 at 1:56
• You already had my upvote :) But yeah, that does seem a lot cleaner.
– cHao
Commented Aug 9, 2014 at 3:23
Cut-(flip)-join method
Take a polygon and choose a line segment between two points on the perimeter that is fully contained inside the polygon.
Separate the polygon along this line segment to produce two polygons. Be sure to distinguish the "exposed cut edge".
Choose one of the two parts and join it to the other so that the exposed cut edge is touching an edge of the other part along its full length. It does not have to be joined to the other exposed edge, as long as it is touching some edge along its full length. You may choose to flip one of the parts before rejoining.
This will produce a polygon with the same area and same perimeter as the original polygon.
rschwieb's answer is an example of this method.
Special case: cut-flip-join-in-same-place
Cut the polygon in two as described above. Flip one part. Join the two exposed cut edges.
Unless the part flipped was symmetric, you will have made a new polygon (as opposed to simply restoring the original polygon). If at least one end of the line segment is not at a vertex, then you will also change the number of sides.
The simplest example is an isosceles triangle and a parallelogram:
However you can start with any triangle and progressively make shapes with more and more sides:
• ...and this one looks like an example of your 'flipping' method in my question, iterated repeatedly. You really like these two problems :) Commented Aug 4, 2014 at 11:22
• @Semiclassical I'm a mathematician; sometimes I find it hard to let things go. :) Commented Aug 4, 2014 at 19:45
• @DavidButlerUofA very nice "video", may I ask you which programm you use to make such things? Commented Jul 26, 2019 at 18:55
• @Siegfried.V It was Geogebra, which can export on of your geometry constructions as a gif. Commented Jul 26, 2019 at 20:04
• @DavidButlerUofA thanks for answer, will lookfor it :) Commented Jul 26, 2019 at 21:15
Here are Gordon–Webb–Wolpert polygons, they have more than just common area and perimeter, they are isospectral! Drums shaped as them would sound the same. It's a fun exercise to construct more examples using tangram pieces.
More isospectral polygons can be found here, including ones with different numbers of sides.
• I wonder if the common area & perimeter property is sufficient as well as necessary. Probably not, but a counter-example to that would be interesting as well. Commented Jul 25, 2014 at 0:28
• Unfortunately this does not answer the original question, as the two polygons given both have 8 sides. Commented Jul 25, 2014 at 0:29
You can make a hexagon and a pentagon with the same area and perimeter by gluing triangles to a square:
The area of both is $6$ units$^2$ and the perimeter of both is $4 + 4\sqrt2$ units. It's particularly pleasant that both shapes are convex.
(Note also that these two are related by a "cut and flip" as described in my other answer.)
• That's also nice as an example in the spirit of the bump case you did for my problem. Commented Aug 3, 2014 at 23:30
• The idea of adding "bumps" inspired both of them actually -- good example of multiple problems feeding into each other. Commented Aug 3, 2014 at 23:32
This may be cheating a bit, but why not just create two polygons that "look the same" but have a different graph? For example, we may take a triangle and insert a new vertex somewhere in one of its edges. This produces a new polygon that looks like that triangle, but is actually a rectangle quadrilateral with three collinear vertices.
• that's not a rectangle, and not even a 4-sided polygon because its sides must be non-adjacent Commented Jul 27, 2014 at 12:48
• @mattecapu According to which definition? Commented Jul 27, 2014 at 12:55
• A rectangle has four right interior angles. Your figure is not a rectangle.
– JRN
Commented Jul 17, 2019 at 5:44
• @JoelReyesNoche Yes, it's a quadrilateral, not a rectangle. That was a slip or mistranslation. My question concerning the definition was about the claim that it's "not even a 4-sided polygon because its sides must be non-adjacent", not about the definition of rectangles. Sorry for the confusion. Commented Jul 18, 2019 at 10:08
For the first shape, take a rectangle with sides $1,x$. For the second shape, take a right triangle with sides $4,\frac{x}{2}, \sqrt{(\frac{x}{2})^2+4}$. They both have area $x$, for every $x$. Making them have the same perimeter reduces to solving a quadratic equation; a solution is $x=\frac{3+\sqrt{33}}{2}$. Then both shapes have perimeter $5+\sqrt{33}$.
Let $A$ be a triangle with corners at the points $(0,0)$, $(2,0)$, and $(0,1)$, with area $1$ and perimeter a little larger than 5. Let $B$ be a parallellogram with corners at $(0,0)$, $(1,0)$, $(x,1)$, and $(x+1,1)$. For any $x$, the areas of $A$ and $B$ are the same, namely $1$. If $x=0$, $A$ has largest perimeter, while if $x$ is large then $B$ has largest perimeter. So by continuity, there is a value of $x$ such that the perimeters are equal.
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# Talk:85: Paths
This is the kind of thing that comes up in story problems in Calculus often. If you can travel in/over one medium at one speed, and in/over another medium at a different speed, what is the optimum path to minimize your travel time.
An example of this problem would be if there is a drowning swimmer 100 meters offshore, you are 300 meters from the point on the shoreline closest to the swimmer, and you can run at 15mph and swim at 2mph, how far do you run along the shoreline before going into the water to get to the swimmer as quickly as possible?
The fact that Randall shows two different paths over the "grass" makes me think that he was thinking more along the line of obsessively optimizing his path rather than about whether it might be acceptable or not to walk over the grass. -- mwburden 70.91.188.49 21:23, 13 December 2012 (UTC)
Along similar lines, this mathematician's dog uses Calculus (albeit at an intuitive, rather than mathematical level) to optimize the path that it takes to retrieve the ball from the water. -- mwburden 70.91.188.49 21:27, 13 December 2012 (UTC)
This particular situation is less interesting, since the walker's speed is the same for all three paths! This is seen by the times being directly proportional to the distances. Normally, the off-normal-path is at a lower speed, but some shorter path still gives the smallest time.DrMath 08:22, 14 October 2013 (UTC)
Where do the equations come from to figure out #2 & #3 - can anybody derive it? 108.162.219.185 (talk) (please sign your comments with ~~~~)
The equation #2 comes from the second route. t(1+√2)/3 is how far the second path takes the guy. If each block is a unit square, the diagonal to the corner is √2 while the next part is 1. The t/3 part is making it comparable to the first one (the first one is t despite it being 3 unit squares). Equation #3 is t√(5)/3. Plugging 1, 2, and √5 into wolfram|alpha for triangle side lengths makes it a right triange, so the √5 comes from the side length (assuming unit squares) while the t/3 makes it comparable to the first one.Mulan15262 (talk) 02:36, 1 December 2014 (UTC)
Interestingly enough, if the three sides are equal in time taken (20 seconds each), the time it would take for path #2 would be 20rt2 + 20, and path three would be roughly 40, which comes out to 60, 48.28, and 40 seconds by using very simple geometry. 108.162.237.179 (talk) (please sign your comments with ~~~~)
Based on his times, it is two squares with the same side length. Base on that geometry, Path #2 will be the hypotenuse of a 45 degree right triangle. t = √(20^2 + 20^2) = 20 * √2 = 48.28. Path #3 would be t = √(20^2 + 40^2) = 20 * √5 = 44.72. Not sure where you got roughly 40 from. Were you thinking of the sin of 30 degree rule where the hypotenuse is double the opposite side? In this case, the adjacent is double the opposite which puts the hypotenuse at √5 times the opposite.Flewk (talk) 17:10, 24 December 2015 (UTC)
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# NCERT Solutions for Class 11 Maths Chapter 7 - Binomial Theorem Miscellaneous Exercise
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## NCERT Solutions for Maths Class 11 Binomial Theorem Miscellaneous Exercise - Free PDF Download
Chapter 7 of Class 11 Maths focuses on the Binomial Theorem, which is a useful technique for multiplying formulas with any power. Because it makes complex algebraic calculations easier this theorem has significance for solving difficult mathematical problems. This chapter's Miscellaneous Exercise offers various kinds of difficult tasks that help in the understanding and successful application of the binomial theorem for students.
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1. NCERT Solutions for Maths Class 11 Binomial Theorem Miscellaneous Exercise - Free PDF Download
2. Access NCERTSolutions Class 11 Maths Chapter 7 Binomial Theorem
2.1Miscellaneous Exercise
3. Class 11 Maths Chapter 7: Exercises Breakdown
4. CBSE Class 11 Maths Chapter 7 Other Study Materials
5. Chapter-Specific NCERT Solutions for Class 11 Maths
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Students can get a thorough understanding of binomial coefficients and the general shape of the binomial expansion by completing these exercises. With the detailed clarifications and step-by-step instructions provided in Class 11 Maths NCERT Solutions, students may confidently solve any related problem and improve their overall mathematical skills. Get the latest CBSE Class 11 Maths Syllabus here.
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## Access NCERTSolutions Class 11 Maths Chapter 7 Binomial Theorem
### Miscellaneous Exercise
1. If a and b are distinct integers, prove that a-b is a factor of ${a^n} - {b^n}$, whenever n is a positive integer
(hint: ${a^n} = {(a - b + b)^n}$)
Ans: To prove to prove that(a-b) is a factor of (${a^n} - {b^n}$), it must be proved that
${a^n} - {b^n}$= $k(a - b)$, where k is some natural number
It can be written that, $a = a - b + b$
${((a - b) + b)^n}{ = ^n}{C_0}{(a - b)^n}{ + ^n}{C_1}{(a - b)^{n - 1}}b{ + ^n}{C_2}{(a - b)^{n - 2}}{b^2} + \ldots { + ^n}{C_{n - 1}}(a - b){b^{n - 1}}{ + ^n}{C_n}{b^n}$
=${(a - b)^n}{ + ^n}{C_1}{(a - b)^{n - 1}}b{ + ^n}{C_2}{(a - b)^{n - 2}}{b^2} + \ldots { + ^n}{C_{n - 1}}(a - b){b^{n - 1}} + {b^n}$
=${a^n} - {b^n} = (a - b)$$[{(a - b)^{n - 1}}{ + ^n}{C_1}{(a - b)^{n - 2}}b{ + ^n}{C_2}{(a - b)^{n - 3}}{b^2} + \ldots { + ^n}{C_{n - 1}}{b^{n - 1}}]$
$\Rightarrow {a^n} - {b^n} = k(a - b)$
Where k =$[{(a - b)^{n - 1}}{ + ^n}{C_1}{(a - b)^{n - 2}}b{ + ^n}{C_2}{(a - b)^{n - 3}}{b^2} + \ldots { + ^n}{C_{n - 1}}{b^{n - 1}}]$ is a natural number this shows that $(a - b)$is a factor of $({a^n} - {b^n})$,
Where n is a positive integer.
2. Evaluate $\left( {\sqrt 3 } \right. + {\left. {\sqrt 2 } \right)^6} - \left( {\sqrt 3 } \right. - {\left. {\sqrt 2 } \right)^6}$
Ans: Firstly, the expression
${\left( {a + b} \right)^6} - {\left( {a - b} \right)^6}$ is simplified by using Binomial Theorem. This can be done as
${\left( {a + b} \right)^6}$=$^6{C_0}{(a)^6}{ + ^6}{C_1}{(a)^5}b{ + ^6}{C_2}{(a)^4}{b^2}{ + ^6}{C_3}{(a)^3}{b^3}{ + ^6}{C_4}{(a)^2}{b^4}{ + ^6}{C_5}(a){b^5}{ + ^6}{C_6}{b^6}$
=${(a)^6} + 6{(a)^5}b + 15{(a)^4}{b^2} + 20{(a)^3}{b^3} + 15{(a)^2}{b^4} + 6a{b^5} + {b^6}$
Putting a=$\sqrt 3 \,and\,$b=$\sqrt 2$, we obtain
$\left( {\sqrt 3 } \right. + {\left. {\sqrt 2 } \right)^6} - \left( {\sqrt 3 } \right. - {\left. {\sqrt 2 } \right)^6}$
$= 2\left( {6 \cdot {{(\sqrt 3 )}^5}(\sqrt 2 ) + 20 \cdot {{(\sqrt 3 )}^3}{{(\sqrt 2 )}^3} + 6 \cdot (\sqrt 3 ){{(\sqrt 2 )}^5}} \right)$
=$2 \times 198\sqrt 6$
$= 396\sqrt 6$
3. Find the values of${\left( {{a^2} + \sqrt {{a^2} - 1} } \right)^4} + {\left( {{a^2} - \sqrt {{a^2} - 1} } \right)^4}$
Ans: Firstly, the expression is simplified by using Binomial Theorem.
${\left( {x + y} \right)^4} + {\left( {x - y} \right)^4}$
This can be done as
${\left( {x + y} \right)^4}$=$^4{C_0}{(x)^4}{ + ^4}{C_1}{(x)^3}y{ + ^4}{C_2}{(x)^2}{y^2}{ + ^4}{C_3}x\,{y^3}{ + ^4}{C_4}{y^4}$
=${(x)^4} + 4{(x)^3}y + 6{(x)^2}{y^2} + 4x\,{y^3} + {y^4}$
${\left( {x - y} \right)^4}$=$^4{C_0}{(x)^4}{ - ^4}{C_1}{(x)^3}y{ - ^4}{C_2}{(x)^2}{y^2}{ - ^4}{C_3}x\,{y^3}{ - ^4}{C_4}{y^4}$
=${(x)^4} - 4{(x)^3}y - 6{(x)^2}{y^2} - 4x\,{y^3} - {y^4}$
Putting x=${a^2}$ and $y = \sqrt {{a^2} - 1} ,$ We obtain
${\left( {{a^2} + \sqrt {{a^2} - 1} } \right)^4} + {\left( {{a^2} - \sqrt {{a^2} - 1} } \right)^4}$
$= 2\left[ {{{\left( {{a^2}} \right)}^4} + 6{{\left( {{a^2}} \right)}^2}{{\left( {\sqrt {{a^2} - 1} } \right)}^2} + {{\left( {\sqrt {{a^2} - 1} } \right)}^4}} \right]$
$= 2\left[ {\left( {{a^8}} \right) + 6\left( {{a^4}} \right)\left( {{a^2} - 1} \right) + {{\left( {{a^2} - 1} \right)}^2}} \right]$
$= 2\left[ {{a^8} + 6{a^6} - 6{a^4} + {a^4} - 2{a^2} + 1} \right]$
$= 2\left[ {{a^8} + 6{a^6} - 5{a^4} - 2{a^2} + 1} \right]$
$= 2{a^8} + 12{a^6} - 10{a^4} - 4{a^2} + 2$
4. Find an approximation of ${\left( {0.99} \right)^5}$using the first three terms of its expansion.
Ans: $0.99\, = 1 - 0.01$
${\left( {0.99} \right)^5} = {\left( {1 - 0.01} \right)^5}$
$^5{C_0}{(1)^5}{ - ^5}{C_1}{(1)^4}\left( {0.01} \right){ - ^5}{C_2}{(1)^3}{\left( {0.01} \right)^2}$
(Approximately)
$= 1 - 0.05 + 0.001$
$= 1.001 - 0.05$
=$= 0.951$
Thus, the value of ${\left( {0.99} \right)^5}$is approximately 0.951
5. Expand using Binomial Theorem ${\left( {1 + \dfrac{x}{2} - \dfrac{2}{x}} \right)^4},\,x \ne 0$
Ans: ${\left( {1 + \dfrac{x}{2} - \dfrac{2}{x}} \right)^4}$
${ = ^n}{C_0}\left( {1 + {{\dfrac{x}{2}}^4}} \right){ - ^n}{C_1}{\left( {1 + {{\dfrac{x}{2}}^4}} \right)^3}\left( {\dfrac{2}{x}} \right) - {\,^n}{C_2}{\left( {1 + {{\dfrac{x}{2}}^4}} \right)^2}{\left( {\dfrac{2}{x}} \right)^2}{ - ^n}{C_3}\left( {1 + {{\dfrac{x}{2}}^4}} \right){\left( {\dfrac{2}{x}} \right)^3}{ - ^n}{C_4}{\left( {\dfrac{2}{x}} \right)^4}$
$= \left( {1 + {{\dfrac{x}{2}}^4}} \right) - 4{\left( {1 + {{\dfrac{x}{2}}^4}} \right)^3}\left( {\dfrac{2}{x}} \right) + \,6\left( {1 + x + {{\dfrac{x}{4}}^2}} \right)\left( {\dfrac{4}{{{x^2}}}} \right) - 4\left( {1 + \dfrac{x}{2}} \right)\left( {\dfrac{8}{{{x^3}}}} \right) + \left( {\dfrac{{16}}{{{x^4}}}} \right)$
$= \left( {1 + {{\dfrac{x}{2}}^4}} \right) - {\left( {1 + {{\dfrac{x}{2}}^4}} \right)^3}\left( {\dfrac{8}{x}} \right) + \,\left( {\dfrac{8}{{{x^2}}}} \right) + \dfrac{{24}}{x} + 6 - \left( {\dfrac{{32}}{{{x^3}}}} \right) + \left( {\dfrac{{16}}{{{x^4}}}} \right)$…..(1)
Again, by using the Binomial Theorem, we obtain
${\left( {1 + \dfrac{x}{2}} \right)^4}{ = ^4}{C_0}{(1)^4}{ + ^4}{C_1}{(1)^3}\left( {\dfrac{x}{2}} \right){ + ^4}{C_2}{(1)^2}{\left( {\dfrac{x}{2}} \right)^2}{ + ^4}{C_3}\,{\left( {\dfrac{x}{2}} \right)^3}{ + ^4}{C_4}{\left( {\dfrac{x}{2}} \right)^4}$
$= 1 + 4 \times \dfrac{x}{2} + 6 \times \dfrac{{{x^4}}}{4} + 4 \times \dfrac{{{x^3}}}{8} + \dfrac{{{x^3}}}{{16}}$
$= 1 + 2x + \dfrac{{3{x^2}}}{2} + \dfrac{{{x^3}}}{2} + \dfrac{{{x^4}}}{{16}}$…..(2)
${\left( {1 + \dfrac{x}{2}} \right)^3}{ = ^3}{C_0}{(1)^3}{ + ^3}{C_1}{(1)^2}\left( {\dfrac{x}{2}} \right){ + ^3}{C_2}(1){\left( {\dfrac{x}{2}} \right)^2}{ + ^3}{C_3}\,{\left( {\dfrac{x}{2}} \right)^3}$
$= \,1 + \dfrac{{3x}}{2} + \dfrac{{3{x^2}}}{4} + \dfrac{{{x^3}}}{8} + \dfrac{{{x^3}}}{8}$…… (3)
From (1), (2), and (3) we obtain
${\left( {\left( {1 + \dfrac{x}{2}} \right) - \dfrac{2}{x}} \right)^4}$
$= 1 + 2x + \dfrac{{3{x^2}}}{2} + \dfrac{{{x^3}}}{2} + \dfrac{{{x^4}}}{{16}} - \left( {\dfrac{8}{x}} \right)\left( {1 + \dfrac{{3x}}{2} + \dfrac{{3{x^2}}}{4} + \dfrac{{{x^3}}}{8}} \right) + \dfrac{8}{{{x^2}}} + \dfrac{{24}}{x} + 6 - \dfrac{{32}}{{{x^3}}} + \dfrac{{16}}{{{x^4}}}$
$= 1 + 2x + \dfrac{{3{x^2}}}{2} + \dfrac{{{x^3}}}{2} + \dfrac{{{x^4}}}{{16}} - \dfrac{8}{x} - 12 - 6x - {x^2} - \dfrac{8}{{{x^2}}} + \dfrac{{24}}{x} + 6 - \dfrac{{32}}{{{x^3}}} + \dfrac{{16}}{{{x^4}}}$
$= \dfrac{{16}}{x} + \dfrac{8}{{{x^2}}} - \dfrac{{32}}{{{x^3}}} + \dfrac{{16}}{{{x^4}}} - 4x + \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{2} + \dfrac{{{x^4}}}{{16}} - 5$.
6. Find the expansion of ${\left( {3{x^2} - 2ax + 3{a^2}} \right)^3}$using binomial theorem.
Ans: Using the Binomial Theorem, the given expression
${\left( {3{x^2} - 2ax + 3{a^2}} \right)^3}$Can be expanded as
${\left( {3{x^2} - 2ax + 3{a^2}} \right)^3}$
${ = ^3}{C_0}{\left( {3{x^2} - 2ax} \right)^3}{ - ^3}{C_1}{\left( {3{x^2} - 2ax} \right)^2}\left( {3{a^2}} \right){ + ^3}{C_2}\left( {3{x^2} - 2ax} \right){\left( {3{a^2}} \right)^2}{ - ^3}{C_3}{\left( {3{a^2}} \right)^3}$
$= {\left( {3{x^2} - 2ax} \right)^3} + 3\left( {9{x^4} - 12a{x^3} + 4{a^2}{x^2}} \right)\left( {3{a^2}} \right) + 3\left( {3{x^2} - 2ax} \right)\left( {9{a^4}} \right) + \left( {2{a^6}} \right)$
$= {\left( {3{x^2} - 2ax} \right)^3} + 81{a^2}{x^4} - 108{a^3}{x^3} + 36{a^4}{x^2} + 81{a^4}{x^2} - 54{a^5}x + 27{a^6}$
$= {\left( {3{x^2} - 2ax} \right)^3} + 81{a^2}{x^4} - 108{a^3}{x^3} + 117{a^4}{x^2} - 54{a^5}x + 27{a^6}$…. (1)
Again, by using the Binomial Theorem, we obtain
${\left( {3{x^2} - 2ax} \right)^3}$
${ = ^3}{C_0}{\left( {3{x^2}} \right)^3}{ - ^3}{C_1}{\left( {3{x^2}} \right)^2}\left( {2ax} \right){ + ^3}{C_2}\left( {3{x^2}} \right){\left( {2ax} \right)^2}{ - ^3}{C_3}{\left( {2ax} \right)^3}$
$= \left( {27{x^6}} \right) - 3\left( {9{x^4}} \right)\left( {2ax} \right) + 3\left( {3{x^2}} \right)\left( {4{a^2}{x^2}} \right) - 8{a^3}{x^3}$
$= 27{x^6} - 54a{x^5} + 36{a^2}{x^4} - 8{a^3}{x^3}$……… (2)
From (1) and (2), we obtain
${\left( {3{x^2} - 2ax + 3{a^2}} \right)^3}$
$= 27{x^6} - 54a{x^5} + 36{a^2}{x^4} - 8{a^3}{x^3} + 81{a^2}{x^4} - 108{a^3}{x^3} + 117{a^4}{x^2} - 54{a^5}x + 27{a^6}$
$= 27{x^6} - 54a{x^5} + 117{a^2}{x^4} - 116{a^3}{x^3} + 117{a^4}{x^2} - 54{a^5}x + 27{a^6}$.
## Conclusion
NCERT Solutions for Class 11 Maths Miscellaneous Exercise in Chapter 7 on the Binomial Theorem are essential for scoring this key concept. These solutions offer clear explanations, helping students solve complex problems easily. By regularly practising these exercises, students can improve their understanding and use of the binomial theorem. This practice improves their confidence and prepares them well for exams and future math challenges.
## Class 11 Maths Chapter 7: Exercises Breakdown
Exercise Number of Questions Exercise 7.1 14 Questions & Solutions
## Chapter-Specific NCERT Solutions for Class 11 Maths
Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.
## FAQs on NCERT Solutions for Class 11 Maths Chapter 7 - Binomial Theorem Miscellaneous Exercise
1. What is the Binomial Theorem in NCERT solutions of Miscellaneous Exercise Class 11 Chapter 7?
The Binomial Theorem of Miscellaneous Exercise Class 11 Chapter 7 provides a formula for expanding expressions raised to any positive integer power. It is essential for understanding polynomial expansions. This theorem is expressed as a sum involving terms of the form. Each term represents a specific combination of the variables.
2. Why is the Binomial Theorem important in NCERT solutions of Miscellaneous Exercise Class 11 Chapter 7?
In NCERT Solutions of Miscellaneous Exercise Class 11 Chapter 7 the binomial theorem simplifies the expansion of binomials. It is widely used in various fields of mathematics such as algebra, calculus, and probability theory. This makes it a critical tool for solving complex mathematical problems. Understanding this theorem helps in efficiently handling polynomial expansions and related calculations.
3. What are binomial coefficients in NCERT solutions of Binomial Theorem Class 11 Miscellaneous Exercise?
In the Binomial Theorem Class 11 Miscellaneous Exercise, binomial coefficients are the numerical factors in the binomial expansion. They are calculated using combinations. These coefficients determine the weight of each term in the expansion. They play a crucial role in the precise calculation of terms in the binomial expression.
4. What is Pascal's Triangle in NCERT solutions of Binomial Theorem Class 11 Miscellaneous Exercise?
Pascal's Triangle of Binomial Theorem Class 11 Miscellaneous Exercise is a triangular array of numbers where each number is the sum of the two directly above it. This pattern is used to find binomial coefficients. Pascal's Triangle simplifies the process of identifying coefficients in binomial expansions.
5. What is the importance of the NCERT solutions of class 11 maths chapter 7 miscellaneous exercise?
The NCERT solutions of class 11 maths chapter 7 miscellaneous exercise include a variety of problems. It helps reinforce the understanding and application of the Binomial Theorem. By practising diverse questions, students can grasp the concepts more deeply. This exercise ensures comprehensive preparation for exams by covering all aspects of the chapter.
6. What is the main topic of the Miscellaneous Exercise on Chapter 7 Class 11?
Miscellaneous Exercise on Chapter 7 Class 11 focuses on Permutations and Combinations, which are essential concepts in counting and probability. These concepts help in determining the number of ways to arrange or select items.
7. How do NCERT Solutions help with the Miscellaneous Exercise of Chapter 7?
Miscellaneous Exercise on Chapter 7 Class 11 provides step-by-step explanations and detailed solutions for the Miscellaneous Exercise. They help students understand complex problems and learn the correct methods to solve them, ensuring thorough preparation.
8. What types of problems are included in the Binomial Theorem Miscellaneous Exercise?
The Binomial Theorem Miscellaneous Exercise includes problems on permutations, combinations, factorials, and the application of these concepts in various situations. These problems range from simple to complex, providing comprehensive practice.
9. How can students effectively solve the problems in the Binomial Theorem Miscellaneous Exercise?
Students should carefully read each problem, understand the concepts involved, and apply the correct formulas. Practising regularly and referring to NCERT Solutions for guidance can help in solving problems accurately.
10. What are the benefits of using NCERT Solutions for Binomial Theorem Miscellaneous Exercise?
NCERT Solutions Binomial Theorem Miscellaneous Exercise helps students understand difficult concepts with ease. They provide clear explanations and solve problems step-by-step, which improves comprehension and problem-solving abilities. Using these solutions ensures better preparation for exams.
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https://www.shaalaa.com/question-bank-solutions/a-consumer-consumes-only-two-goods-explain-consumer-s-equilibrium-help-utility-analysis-conditions-consumer-s-equilibrium-using-marginal-utility-analysis_2319
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# A Consumer Consumes Only Two Goods. Explain Consumer'S Equilibrium with the Help of Utility Analysis. - Economics
A consumer consumes only two goods. Explain consumer's equilibrium with the help of utility analysis.
A consumer consumes only two goods X and Y. Explain the conditions of consumer’s equilibrium using Marginal Utility Analysis.
A consumer consumes only two goods. Explain the conditions of consumers’ equilibrium using utility analysis.
#### Solution
Conditions of consumer’s equilibrium using marginal utility analysis:
When a consumer buys both Goods X and Y, the consumer’s equilibrium condition is expressed through the equation:
(MU_x)/P_x=(MU_y)/P_y=(MU_m)/P_n=MU_m
Consider the following numerical example to understand the consumer’s equilibrium using marginal utility. A consumer Marginal Utility of Money (MUm) is 16 utils and two Goods X and Y whose prices are Rs 1 (Px) and Rs 1 (Py) per unit, respectively. Consider the following schedule to analyse the marginal utility of good x (MUx) and good y (MUy).
Units of x MU x (Utils) MU y3 (Utils) 1 28 32 2 24 29 3 21 27 4 20 23 5 16 20 6 13 18 7 9 17 8 5 16 9 3 12 10 1 9
Based on the given schedule, the consumer is in equilibrium at the consumption of 5 units of commodity x and 8 units of commodity y. At such a consumption combination, the
marginal utility of a rupee spent on the commodity x((MU_x)/P_x) is equal to the marginal utility of a rupee spent on the commodity y((MU_y)/P_y)and also equal to the marginal utility of money (MUm).
Marginal utility of a rupee spent on commodity x = marginal utility of a rupee spent on commodity y = Marginal utility of money
(MU_x)/P_x=(MU_y)/P_y=MU_m
(MU_x)/P_x=(MU_y)/P_y=16/1=16=MU_m
In the diagram, OO1 is the total income of a consumer. MUx and MUy are the marginal utility curves of commodity x and commodity y, respectively.
The consumer does not attain equilibrium at Point L because the point at L is
(MU_x)/P_x>(MU_y)/P_y
The consumer does not attain equilibrium at Point R because the point at R is
(MU_x)/P_x<(MU_y)/P_y
So, when OF amount of income is spent on commodity x and FO1 amount is spent on commodity y, the consumer is in equilibrium at Point E. Hence, at this point
(MU_x)/P_x=(MU_y)/P_y=MU_m
Concept: Conditions of Consumer's Equilibrium Using Marginal Utility Analysis
Is there an error in this question or solution?
2013-2014 (March) Delhi Set 1
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1. ## Perpendicular Lines
Find all a for which the lines represented by the equations 2y+3+x=0 and 3y+ax+2=0 are perpendicular
2. Originally Posted by Rimas
Find all a for which the lines represented by the equations 2y+3+x=0 and 3y+ax+2=0 are perpendicular
Put the first line equation in slope-intercept form:
2y + 3 + x = 0
y = -(1/2)x - (3/2)
The second line needs to have a slope of +2 to be perpendicular to this line.
Thus:
3y + ax + 2 = 0
y = -(a/3)x - (2/3)
Thus -a/3 = 2. Thus a = -6.
-Dan
3. Originally Posted by Rimas
Find all a for which the lines represented by the equations 2y+3+x=0 and 3y+ax+2=0 are perpendicular
If line l_1 has slope m_1 and line l_2 has slope m_2 then if neither line is vertical then the two line are perpendicular if (m_1)(m_2)=-1.
2y+3+x=0 has slope (-1/2); 3y+ax+2=0 has slope (-a/3).
Now solve for a.
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# 5. (Optional) AI Move
Workshop Resources
### Let’s make the computer think
Previously, we used a Random object to randomly generate a move for the computer. As a result, the competitiveness of the computer is fairly low.
In this exercise, we want to increase the difficulty of the game by making optimal decisions for the computer.
We will do so by adding artificial intelligence into our program with the use of the Minimax algorithm (a well-defined procedure that allows computers to solve problems).
### Minimax Algorithm
Minimax is an algorithm that is used in two-player games to make optimal decisions for a player.
• The two players are labeled as maximizer and minimizer respectedly. While the maximizer maximizes its chance to win, the minimizer tries to minimize its loss.
• The algorithm examines all the possible future game states based on the current game board assuming both the maximizer and minimizer will pick the move that benefits them the most.
• In our case, we pick the computer to be the maximizer, and the player to be the minimizer. We will attempt to make optimal decision for the computer by maximize its chance to win!
### How will Minimax Algorithm work in TicTacToe?
• We examine all the possible moves of "X" and "O" and give a score to a game board if there is a winner or tie.
• Since we want the computer to win with the fewest steps possible, we design the score for the game boards as the following:
• If computer wins, calculate the score with the formula 1 * (number of available spots on the board + 1).
• If the player wins, calculate the score with the formula -1 * (number of available spots on the board + 1).
• If there is a tie, the score is 0.
• Note that by giving larger scores to game states where the computer can win with fewer steps, we are teaching our code to pick the optimal move for the computer.
Let’s look at an example below:
1. In the first row, we considered the 3 possible moves for the computer "O", who is the maximizer.
2. We examine all the game states until all the moves end will a computer win, player win, or a tie. We then give them their corresponding score.
For example, in the second board in row 1, the computer wins by placing "O" at position 8. That state will then be given a score of 1 * (number of available spots on the board + 1) = 1 * (2+1) = 3.
3. At game states that don’t have a winner or a tie, we pick the smallest score during minimizing rounds (when "X" make a move), and the largest score during maximizing rounds (when "O" make a move).
4. If you follow along through the maximizing/minimizing rounds on the picture above, you should note that the optimal move for the computer is to place "O" at position 8, allowing the computer to win with 1 move from the starting game board.
### Code Structure
In activity-3, you wrote the method int getComputerMove(String[] curBoard) to randomly generate a spot for the computer. Let’s write another method called getComputerMoveAI(String[] curBoard) that returns the optimal move for the computer by calling the int minimax(String[] curBoard, boolean isMaximizing) method.
int getComputerMove(String[] curBoard){
// 1. this method calls minimax() on all the possible moves the computer can pick
// 2. it takes the maximum out of all of them
// 3. return the optimal move
}
int minimax(String[] curBoard, boolean isMaximizing){
// 1. In the maximizing round, it calls minimax() on all the possible moves for the computer, "O", return the maximum score
// 2. In the minimizing round, it calls minimax() on all the possible moves for the player, "X", return the minimum score
}
### Write the Method getComputerMoveAI()
1. For each available spot on the game board, place "O" at that spot and get the score for that board by calling minimax().
1. Track the largest score and its corresponding position of the board at each iteration. Return the position with the largest score.
### Write the Method minimax()
As we discussed above, the minimax() method has the header int minimax(String[] curBoard, boolean isMaximizing).
1. Call getWinner() on the board to check if there is a winner. If so, return the corresponding score.
(Score: computer wins (1 * number of available spots on the board + 1), player wins (-1 * number of available spots on the board + 1), tie (0)).
2. Get Minimax Score
• If it’s maximizer’s turn ("O"), for each available spot on the game board, place "O" at that spot and get the score for that board by calling minimax().
• If it’s minimizer’s turn ("X"), for each available spot on the game board, place "X" at that spot and get the score for that board by calling minimax().
1. Return Largest Score
• If it’s maximizer’s turn ("O"), track the largest score and the corresponding position at each iteration, and return that score.
• If it’s minimizer’s turn ("X"), track the smallest score and the corresponding position at each iteration, and return that score.
Copy and paste your two methods below main().
Click Run to test around your methods. We provided a given test of the example of the picture above.
You can plug in different boards to test if you get a desirable output.
### Update the Program
After testing the methods, update all the method calls of getComputerMove() to getComputerMoveAI() in your TicTacToe program.
It should be way more challenging to win this time around😀!
You did it!
Workshop completed!
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## Do You Want a Map or just Directions?
### Richard E. Little
Department of Mathematics
121 Ayres Hall
University of Tennessee, Knoxville
Knoxville, TN 37996
[email protected]
### 1. INTRODUCTION
One role of a calculator in mathematics is to ease numerical computations. In many problems a calculator can be used in more than one way, for example, consider finding the solutions to x^2+71x-12=0. If you want to find them by completing the square having a calculator do the arithmetic for you would be convenient, which is also the case if you're going to use the quadratic formula. One method which requires minimal effort on your part is using one of today's more powerful calculators which can find the real and complex roots of a polynomial given just its coefficients.
Completing the square is a technique which can be used for more than just finding roots of quadratic polynomials, for example placing conic sections in standard form. In contrast, the calculator finding the roots directly can be used for nothing else. Perhaps the roots are put to other uses, but this is distinct from the finding of the roots. The quadratic formula falls somewhere between these two positions --- it is used in other ways and yet most are involved with finding roots.
An image that illustrates this contrast between one use and many uses is the following. Think of the calculator as a vehicle to be used to reach an address (e.g., the roots of a quadratic equation). You can either follow an explicit set of directions that you have been given (the calculator's root finding feature) or you can look at a map of the area (complete the square) and put together the route yourself (have the calculator do the arithmetic). Directions may be of limited use in reaching more than one address. Maps can often be used to reach locations that lie at some distance from each other.
None of us would forsake teaching completing the square and/or the quadratic formula in favor of just showing the students how to use their calculator. On the other hand, almost all students favor explicit algorithms over abstract techniques. Since algorithms are easier to teach than abstractions this student preference, coupled with time pressures, can lead to situations in which we give solutions to problems which may be more "directionlike" than we intend or even realize. One goal of this paper will be to give some ways of identifying when solutions are directions. In addition an approach to using calculators which can lead more naturally to using maps to solve problems is described.
### 2. EXAMPLE
The example of solving a quadratic equation is a rather simple one. I want to now consider a problem whose solution does not have any explicit equations, such as the quadratic formula, associated with it . The problem will be to find a complete graph of an arbitrary polynomial function p(x), with degree p(x)> 1. Here complete means the graph should display all the real roots, it should be continuous, and should display correct behavior as x increases and decreases without bound. I give two solutions below, one a set of directions and the other a map. The set of directions is written with a Texas Instruments TI-85 in mind, the recommended graphing calculator for my department.
#### Method I (Directions)
Step 1: After entering the degree and coefficients of p(x) in the `POLY` menu use `POLY/SOLVE` to find all the roots of p(x). Write down the largest and smallest real roots.
Step 2: Using the functions `evalF`, `fMin`, and `fMax` located under the `CALC` menu compute the maximum and minimum values of p(x) on the interval between the two roots written down in step 1.
Step 3: Graph p(x) with the viewing window (slightly) wider than is needed to contain the roots written down in step 1 and tall enough to contain the values computed in step 2.
(Note: This can all be made precise and put into a program that asks only for the degree and coefficients of p(x) and then puts up the graph.)
#### Method II (Map)
Step 1: If possible, factor p(x) to obtain roots.
Step 2: Graph p(x) relative to the default viewing window. Increase the width of the viewing window to include all roots, if any, found in step 1.
Step 3: Examine the behavior of the graph on the screen as x increases and decreases without bound. If behavior is incorrect increase width of window on appropriate side(s) until it is correct.
Step 4: Adjust the height of the viewing window to obtain a continuous graph; this can be done by guesswork or using `TRACE` to get values of p(x).
When Method I is applied to p(x)=.02x^3-0.4x^2-.5x+10 the graph shown in Figure 1 below is obtained ("slightly" in step 3 was taken to be 25%). Method II yields a similar graph. It is clear that Figure 1 is a complete graph so both methods work, yet neither one is without flaw. Step 3 of Method II is not guaranteed to bring all the roots in view if a subset of the graph mimics the total graph's behavior as x increases/decreases without bound. There is also the problem of what to do if the default graph in step 2 is blank. Method I has a much more serious problem, namely what to do if p(x) has fewer than two distinct real roots.
• Fig. 1: Graph of p(x)=.02x^3-0.4x^2-.5x+10 for -6.25 <= x <= 25, -20.52 <= y <= 10.15.
Aside from the issue of when the methods work and when they don't there is one major difference between the two solutions. Method I replaces general calculations by the student (e.g., factoring p(x) and resizing the window of the graph to obtain correct behavior) with specific calculator functions (e.g., directly computing roots and using `fmin` and `fmax`). This replacement leads to the shortcoming observed above in addition to making the entire method a "black box": students enter data according to certain directions and the graph is displayed.
Method II is anything but a black box. While it is true that the calculator replaces student work such as graphing p(x) it does not replace student insight. Each of the steps of the method requires the student to interpret information and to decide how to act; it is this distinction, that of black box versus thought, that makes Method I a set of directions and Method II a map.
### 3. TELLTALE SIGNS OF DIRECTIONS
Using the calculator as a black box is a worst case example of directions and none of us would give out Method I to our students. However, it does exemplify some of the common themes present in directions, which is nice since in general the division between directions and maps is not as clear as it is in the case of Methods I and II. Take the quadratic formula solution from the example in the introduction. A program could be written that asks for the coefficients of the polynomial and then applies the quadratic formula to give the roots. This program would be a black box set of directions, possibly with limitations (say complex roots). Yet the quadratic formula itself is very much a map, one which leads to the three types of solutions depending on the sign of b^2-4ac. The map/directions distinction here depend entirely on how the formula is presented and used.
Mathematical formulas in solutions abound and they can always be programmed for a calculator (given suitable constraints). This doesn't mean that the program has to be written or if it is written that it must be given out to the students. (Though somewhere, someone will, and nothing may prove too simple to write a program for. A colleague of mine had a student of hers who'd been given a program which, when you gave it a, b, and c, computed the vertex of the parabola y=ax^2+bx+c.) I'm not saying here that programs should never be given out, as often they are very useful. Just be aware of the potential for students to treat programs as black boxes.
In addition to programs watch out for the use of one step problem solving features of a calculator. These include finding roots of polynomials (as in Method I), finding coordinates of intersections of curves, evaluating derivatives at a point, evaluation of definite integrals, finding points of inflection, solving systems of equations, and so on. There are courses in which each of these features would prove useful and others in which the students should demonstrate their ability to do the work on their own. A danger with all of these features is the student coming to rely on the calculator exclusively and not knowing what to do when it cannot do a problem. Worse yet, they may never develop any sense of intuition and not be able to recognize when the calculator gives them a wrong answer.
In general, a test to apply to a calculator solution technique to see if it's more a set of directions than a map is to ask yourself the following questions. How broad a class of problems can it be applied to? Are there any major restrictions (as with Method I) on how it can be applied? If there are restrictions, will they be ones that the students remember, and what are the students to do in the cases where the method doesn't apply? If given many methods to solve many separate cases, will the students be able to keep track of them all in other courses in the future?
Also look and see if the solution applies general principles. Method II is based on basic concepts concerning polynomials such as finding roots and limiting behavior. The calculator's main use is to relieve the tedium of computing values of p(x) as well as the actual graphing of p(x). If the student had to graph a polynomial without the aid of a calculator, but remembered this method, they would possess almost all the conceptual knowledge needed to put a solution on the page.
A final test to apply is to see how much the method in question relies on the student using a particular model of calculator. For example, Method I relies on the student having a TI-85. If they happen to own a TI-81 they are out of luck. This certainly implicates this solution as a set of directions, a set which says you can get there from here with these directions only if you have a vehicle capable of offroad travel. Another difficulty with relying on a specific model of calculator is that there is no guarantee that the student will always be using that model: he may borrow a friends; she may lose hers and when replacing it choose a cheaper one; or in the future she may buy a more powerful model which replaces the feature you rely upon with a different one.
### 4. A DIFFERENT OUTLOOK
The Mathematics Department at UTK requires the use of a graphing calculator in it's science and engineering pre-calculus and calculus courses. It does not require that all students buy the same model (though it does recommend one). As such, the classes I taught were filled with up to five different models of calculators at once. To be fair to all I tried to teach from a common ground. Since all of the calculators shared a common set of features when I wished to use a calculator to help solve a problem this set of features is what I would attempt to work with and nothing more.
I began by giving handouts to let the students know what features I wanted them to be familiar with. In the case of the recommended model I could go further and actually give solutions to sample problems, sometimes keystroke by keystroke ([1,2]). Students with other models used their manuals or we worked together in my office. I'll admit that what I was doing was giving explicit directions as solutions to problems. The problems were basic ones: graphing a function, zooming, tracing, some scientific functions. My goal was for the students to know how to get to certain landmarks so that I would later be able in class to ask them to graph a function or to zoom in on an intersection point and not have to stop and answer questions on how to do it.
One immediate, but small, benefit to the handouts was saving some class time. The main advantage was that I later could present solutions to problems in a form like Method II above. These solutions would be maps that relied on the landmarks that the students already knew how to reach. Many components non-calculator based (e.g., factoring polynomials and writing expressions in certain forms) so that student insight was never sacrificed.
Some departments and/or courses require that all of the students have the same model of calculator in the interests of uniformity in both teaching and student capability. Even in this situation it is still possible to act as if the students don't all have the same model. Class time should still be devoted to mathematics and the use of the calculator can stay on the general level with specifics left to handouts for the students to work through on their own time (often in advance).
The only drawback I encountered with my methodology was in distributing actual calculator programs. There are times when explicit programs are almost a necessity, an example I encountered in one class being Simpson's rule. It's too tedious to do manually and so a program is extremely useful. However, in that class distributing a program was out of the question since I had four models of calculator and didn't even know how to program two of the four. The next time I am faced with a similar problem I may try to give the students a handout on the basics of writing programs. Then they can either continue to do Simpson's rule (or whatever) manually or, if they are industrious enough, they can write a program themselves. If they write the program it won't nearly be so much of a black box as if the program was just handed to them to use.
### 5. CONCLUSION
The answer to the question of maps versus directions is clear now: maps, not directions. The main exception concerns basic features of the calculator. In this instance giving directions establishes landmarks that can later be used when setting up maps for solutions to mathematical problems.
In closing I would also like to note that another way of looking at the entire issue of maps versus directions involves remembering that the goal of teaching mathematics is for the students to learn mathematics, not how to punch keys and run programs on a calculator. I am not implying that calculators should not be used at all in mathematics courses as they often can play a positive role. It is just important to remember that the role should be to enhance the learning of the subject material rather than outright replacing the material.
### REFERENCES
[1] Richard E. Little, Basic Guide to the TI-85 Graphics Calculator, UTK Department of Mathematics, August 1994.
[2] Richard E. Little, TI-85 Math 130 Supplement, UTK Department of Mathematics, August 1994.
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# Oblique projection
This article discusses the imaging of 3D objects. For an abstract mathematical discussion, see Projection (linear algebra).
Oblique projection is a simple type of technical drawing of graphical projection used for producing two-dimensional images of three-dimensional objects. The objects are not in perspective, so they do not correspond to any view of an object that can be obtained in practice, but the technique does yield somewhat convincing and useful images.
Oblique projection is commonly used in technical drawing. The cavalier projection was used by French military artists in the 18th century to depict fortifications. Oblique projection was used almost universally by Chinese artists from the first or second centuries to the 18th century, especially when depicting rectilinear objects such as houses.[1]
## Overview
Comparison of several types of graphical projection. The presence of one or more 90° angles is usually a good indication that the perspective is oblique.
Oblique projection with foreshortening by half, seen from the side.
Oblique projection is a type of parallel projection:
• it projects an image by intersecting parallel rays (projectors)
• from the three-dimensional source object with the drawing surface (projection plane).
In both oblique projection and orthographic projection, parallel lines of the source object produce parallel lines in the projected image. The projectors in oblique projection intersect the projection plane at an oblique angle to produce the projected image, as opposed to the perpendicular angle used in orthographic projection.
Mathematically, the parallel projection of the point (x, y, z) on the xy-plane gives (x + az, y + bz, 0). The constants a and b uniquely specify a parallel projection. When a = b = 0, the projection is said to be "orthographic" or "orthogonal". Otherwise, it is "oblique". The constants a and b are not necessarily less than 1, and as a consequence lengths measured on an oblique projection may be either larger or shorter than they were in space. In a general oblique projection, spheres of the space are projected as ellipses on the drawing plane, and not as circles as you would expect them from an orthogonal projection.
Oblique drawing is also the crudest "3D" drawing method but the easiest to master. Oblique is not really a 3D system but a two-dimensional view of an object with 'forced depth'. One way to draw using an oblique view is to draw the side of the object you are looking at in two dimensions, i.e. flat, and then draw the other sides at an angle of 45°, but instead of drawing the sides full size they are only drawn with half the depth creating 'forced depth' – adding an element of realism to the object. Even with this 'forced depth', oblique drawings look very unconvincing to the eye. For this reason oblique is rarely used by professional designers and engineers.
## Oblique pictorial
In an oblique pictorial drawing, the angles displayed among the axis, as well as the foreshortening factors (scale) are arbitrary. More precisely, any given set of three coplanar segments originating from the same point may be construed as forming some oblique perspective of three sides of a cube. This result is known as Pohlke's theorem, from the German mathematician Pohlke, who published it in the early 19th century.[2]
The resulting distortions make the technique unsuitable for formal, working drawings. Nevertheless, the distortions are partially overcome by aligning one plane of the image parallel to the plane of projection. Doing so creates a true shape image of the chosen plane. This specific category of oblique projections, whereby lengths along the directions x and y are preserved, but lengths along direction z are drawn at angle using a reduction factor is very much in use for industrial drawings.
• Cavalier projection is the name of such a projection, where the length along the z axis remains unscaled.[3]
• Cabinet projection, popular in furniture illustrations, is an example of such a technique, wherein the receding axis is scaled to half-size[3] (sometimes instead two thirds the original).[4]
## Cavalier projection
Further information: Mathematics and art
In cavalier projection (sometimes cavalier perspective or high view point) a point of the object is represented by three coordinates, x, y and z. On the drawing, it is represented by only two coordinates, x″ and y″. On the flat drawing, two axes, x and z on the figure, are perpendicular and the length on these axes are drawn with a 1:1 scale; it is thus similar to the dimetric projections, although it is not an axonometric projection, as the third axis, here y, is drawn in diagonal, making an arbitrary angle with the x″ axis, usually 30 or 45°. The length of the third axis is not scaled.[5][6]
It is very easy to draw, especially with pen and paper. It is thus often used when a figure must be drawn by hand, e.g. on a black board (lesson, oral examination).
The representation was initially used for military fortifications. In French, the « cavalier » (literally rider, horseman, see Cavalry) is an artificial hill behind the walls that allows to see the enemy above the walls.[7] The cavalier perspective was the way the things were seen from this high point. Some also explain the name by the fact that it was the way a rider could see a small object on the ground from his horseback.[8]
## Cabinet projection
The term cabinet projection (sometimes cabinet perspective) stems from its use in illustrations by the furniture industry.[9] Like cavalier perspective, one face of the projected object is parallel to the viewing plane, and the third axis is projected as going off at an angle (typically 63.4°). Unlike cavalier projection, where the third axis keeps its length, with cabinet projection the length of the receding lines is cut in half.
### Mathematical formula
As a formula, if the plane facing the viewer is xy, and the receding axis is z, then a point P is projected like this:
Where is the mentioned angle.
The transformation matrix is:
Alternatively you could remove one third from the leading arm projected off the starting face, thus giving the same
## Military projection
In the military projection, the angles of the x- and z-axes are at 45°, meaning that the angle between the x-axis and the z-axis is 90°. That is, the xz-plane is not skewed. It is rotated over 45°, though.[10]
## Examples
Besides technical drawing and illustrations, video games (especially those preceding the advent of 3D games) also often use a form of oblique projection. Examples include SimCity, Ultima VII, Ultima Online, EarthBound, Paperboy and, more recently, Tibia.
## References
1. Cucker, Felix (2013). Manifold Mirrors: The Crossing Paths of the Arts and Mathematics. Cambridge University Press. pp. 269–278. ISBN 978-0-521-72876-8.
2. Weisstein, Eric W. "Pohlke's Theorem". From MathWorld—A Wolfram Web Resource.
3. Parallel Projections from PlaneView3D Online
4. Bolton, William (1995), Basic Engineering, Butterworth-Heinemann GNVQ Engineering Series, BH Newnes, p. 140, ISBN 9780750625845.
5. ["Archived copy". Archived from the original on 22 August 2010. Retrieved 22 August 2010./content/draftsman/14276/css/14276_307.htm Illustrator Draftsman 3 & 2 – Volume 2 Standard Practices and Theory, page 67] from "Archived copy". Archived from the original on 22 August 2010. Retrieved 22 August 2010.
6. Ingrid Carlbom, Joseph Paciorek, Planar Geometric Projections and Viewing Transformations, ACM Computing Surveys, v.10 n.4, pp. 465–502, Dec. 1978
7. Etymologie des maths, letter C (French)
8. DES QUESTIONS D’ORIGINES (French)
9. Ching, Francis D. K.; Juroszek, Steven P. (2011), Design Drawing (2nd ed.), John Wiley & Sons, p. 205, ISBN 9781118007372.
10. "The Geometry of Perspective Drawing on the Computer". Retrieved 24 April 2015.
Wikimedia Commons has media related to Oblique projection.
Wikimedia Commons has media related to Cabinet projection.
Wikimedia Commons has media related to Cavalier perspective.
• Foley, James (1997). Computer Graphics. Boston: Addison-Wesley. ISBN 0-201-84840-6.
• Ingrid Carlbom, Joseph Paciorek, Planar Geometric Projections and Viewing Transformations, ACM Computing Surveys, v.10 n.4, p. 465–502, Dec. 1978
This article is issued from Wikipedia - version of the 10/25/2016. The text is available under the Creative Commons Attribution/Share Alike but additional terms may apply for the media files.
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# Algorithm to find the path with minimum bending points on a square grid board
Let's suppose we have a square grid board like the one shown in the picture below:
I'm wondering how I can find the path with minimum number of "bending" points (like the ones shown in red) between two given squares ("Start" and "Finish"). The board has some impediments or broken squares on it (Black squares).
I tried to run a BFS algorithm and find all of the minimum cost paths and then pick up the one with minimum bending points, but this won't give us the optimal answer.
Is there a way to efficiently solve this problem?
Motivation: I was trying to solve a computer game using A* algorithm, and I figured out I can use the solution to this problem as a heuristic function to solve the game.
Create an undirected weighted graph $$G$$ with two vertices $$h_c$$, $$v_c$$ for each (non-obstacle) cell $$c$$ of the grid. Intuitively $$h_c$$ encodes entering $$c$$ from the left or right and $$v_c$$ encodes entering $$c$$ from the top or bottom.
For each pair of horizontally adjacent cells $$c, c'$$ add the edge $$(h_c, h_{c'})$$ with weight $$0$$. For each pair of vertically adjacent cells $$c, c'$$ add the edge $$(v_c, v_{c'})$$ with weight $$0$$. For each cell $$c$$, add the edge $$(v_c, h_c)$$ of cost $$1$$ (this encodes changing direction).
If the starting cell is $$s$$, add a new vertex $$s^*$$ along with the edges $$(s^*, h_s)$$ and $$(s^*, v_s)$$ of cost $$0$$. Similarly, if the target cell is $$t$$, add a new vertex $$t^*$$ along with the edges $$(h_t, t^*)$$ and $$(v_t, t^*)$$ of cost $$0$$.
The path with the minimum number of bends in the grid is the one corresponding to the shortest path between $$s^*$$ and $$t^*$$ in $$G$$. Such a shortest path can be found in time $$O(n)$$ where $$n$$ is the number of cells (since weights are non-negative and bounded by a constant, there are $$2n+2$$ vertices, and each vertex has degree at most $$4$$).
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Anda di halaman 1dari 6
# Triangle Area = bh b = base h = vertical height Rectangle Area = bh b = breadth h = height Trapezoid (US) Trapezium (UK) Area = (a+b)h
h = vertical height
## Square Area = a2 a = length of side
Parallelogram Area = bh b = breadth h = height Circle Area = r2 Circumference=2r r = radius Sector Area = r2 r = radius = angle in radians
Ellipse Area = ab
## Area of Plane Shapes
Area Calculation Tool
Circumference of a Circle
## Keep your students safe online!
Unit
A circle is a shape with all points the same distance from the center. It is named by the center. The circle to the left is called circle A since the center is at point A. If you measure the distance around a circle and divide it by the distance across the circle through the center, you will always come close to a particular value, depending upon the accuracy of your measurement. This value is approximately 3.14159265358979323846... We use the Greek letter (pronounced Pi) to represent this value. The number goes on forever. However, using computers, has been calculated to over 1 trillion digits past the decimal point.
The distance around a circle is called the circumference. The distance across a circle through the center is called the diameter. is the ratio of the circumference of a circle to the diameter. Thus, for any circle, if you divide the circumference by the diameter, you get a value close to . This relationship is expressed in the following formula:
where is circumference and is diameter. You can test this formula at home with a round dinner plate. If you measure the circumference and the diameter of the plate and then divide by , your quotient should come close to . Another way to write this formula is: where means multiply. This second formula is commonly used in problems where the diameter is given and the circumference is not known (see the examples below).
The radius of a circle is the distance from the center of a circle to any point on the circle. If you place two radii end-to-end in a circle, you would have the same length as one diameter. Thus, the diameter of a circle is twice as long as the radius. This relationship is expressed in the following formula: , where is the diameter and is the radius. Circumference, diameter and radii are measured in linear units, such as inches and centimeters. A circle has many different radii and many different diameters, each passing through the center. A real-life example of a radius is the spoke of a bicycle wheel. A 9inch pizza is an example of a diameter: when one makes the first cut to slice a round pizza pie in half, this cut is the diameter of the pizza. So a 9-inch pizza has a 9-inch diameter. Let's look at some examples of finding the circumference of a circle. In these examples, we will use = 3.14 to simplify our calculations.
Example 1: Solution:
## The radius of a circle is 2 inches. What is the diameter? = 2 (2 in) = 4 in
Example 2: The diameter of a circle is 3 centimeters. What is the circumference? Solution: = 3.14 (3 cm) = 9.42 cm Example 3: Solution: = 2 (2 in) = 4 in = 3.14 (4 in) = 12.56 in Example 4: Solution: 15.7 cm = 3.14 15.7 cm 3.14 = = 15.7 cm 3.14 = 5 cm Summary: The number is the ratio of the circumference of a circle to the diameter. The value of is approximately 3.14159265358979323846...The diameter of a circle is twice the radius. Given the diameter or radius of a circle, we can find the circumference. We can also find the diameter (and radius) of a circle given the circumference. The formulas for diameter and circumference of a circle are listed below. We round to 3.14 in order to simplify our calculations. The circumference of a circle is 15.7 centimeters. What is the diameter? The radius of a circle is 2 inches. What is the circumference?
Circle Formulas
Circumference =
=
Circle Area =
=
4 r
=
Sphere Volume =
d)/6
d 4/3 r
## *************************** Cylinder Formulas
Surface Area =
Where (2 (2
(2 r) + (2 r height)
=
Volume =
r height
height
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### lewin's blog
By lewin, 17 months ago, ,
Short solutions:
• Div2A: How many times do we need to do a cyclic shift to consider all possible ones? Afterwards, what data structure allows us to easily check number of distinct elements?
• Div2B: Imagine the process in reverse. What types of identical shapes can I get if I cut a rectangle into two pieces? Remember, pieces cannot be rotated or flipped.
• Div2C / Div1A: How do we handle components with special nodes? What do we do with the ones without special nodes?
• Div2D / Div1B: We don't get many questions, so is there a way to "parallelize" questions? Another approach, can we split up the condition i != j somehow using bits?
• Div2E / Div1C: First, somehow reduce it so r_i,b_i <= n. Now, we can bound the excess tokens by a small number, so we can do bitmask dp from here.
• Div1D: The optimal circle must touch a blue point. Now, either consider the inversion, or do a binary search
• Div1E: What makes a list good? How fast can we do this check, and how many times do we need to do this check?
Long solutions:
## Hongcow Learns the Cyclic Shift
We only need to consider at most |s| cyclic shifts (since |s| cyclic shifts returns us back to the original string).
So, we can put these all in a set, and return the size of the set.
code
## Hongcow Solves A Puzzle
I really apologize for the ambiguity of this problem. We worked hard to make it concise and accurate, but we left out too many details.
Basically, the idea is we want to overlay two of these pieces together so that no square has more than 1 X, and the region of X's forms a rectangle.
Now for the solution:
First, let's look at it backwards. I have a rectangle, and I cut it in two pieces. These two pieces have the same exact shape. What shapes can I form?
A necessary and sufficient condition is that the piece itself is a rectangle itself! There are a few ways to check this. One is, find the min/max x/y coordinates, and make sure the number of X's match the bounding box of all the points.
code
## Hongcow Builds a Nation
First, let's make all connected components cliques. This graph is still stable.
Now, there are some components without special nodes. Where should we connect them?
If there is a component with size A and a component with size B, we can add A*B edges if we connect these two components. So, it makes sense to choose the largest component.
code
## Hongcow's Game
For the bits solution: We want to create 20 questions where for every i != j, there exists a question
that contains j and not i, and also a qusetion that contains i and not j. If we can do this, we can find the min for each row.
Note that i != j implies that there exists a bit index where i and j differ.
So, let's ask 2 questions for each bit position, one where all indices have a value of 0 in that position, and one where all indices have a value of 1 in that position. This is a total of at most 20 questions, and we can show that this satisfies the condition above, so this solves the problem.
Parallelization will basically reduce to the above solution, but is another way of looking at the problem.
First, let's ask {1,2,...,n/2} and {n/2+1,...,n} This handles the case where the min lies on the opposite half.
[
OOOOXXXX
OOOOXXXX
OOOOXXXX
OOOOXXXX
XXXXOOOO
XXXXOOOO
XXXXOOOO
XXXXOOOO
]
For example, this handles the case where the min lies in the X part of the matrix, and we split it into two identical problems of size n/2 within the O matrix.
Now, we can ask questions for each submatrix, but we can notice that these two don't interact so we can combine all the questions at this level.
However, we should ask the questions in parallel, as we don't have that many questions For example, for n=8, we should ask
First level:
[1,2,3,4]
[5,6,7,8]
Second level
[1,2],[5,6] (i.e. ask 1,2,5,6 all together, but this is actually two different subproblems, one for the top left, and one for the bottom right).
[3,4],[7,8]
Third level
[1],[3],[5],[7]
[2],[4],[6],[8]
As you can see, this reduces to the bit approach above if N is a power of 2.
code
## Hongcow Buys a Deck of Cards
Also note that if r_i or b_i >= n, we need to collect tokens no matter what since those costs can't be offset. So, we can assume that r_i, b_i <= n.
Let's only buy tokens when we need them. Note that after buying a card, you will have either 0 red tokens or 0 blue tokens, so our dp state can be described by [mask][which one is zero][how many of the other] The dimensions of this dp table are 2^n * 2 * (n^2) (n^2 because the costs to buy cards is at most n).
See the code for more details on how to update this dp.
code
## Hongcow Draws a Circle
First to check if an answer can be arbitrarily large, we can see if there is any red point that is on the convex hull of all our points. So from now on, we can assume the answer is finite.
We can show that the optimal circle must touch a blue point. To see this, consider any optimal circle that doesn't touch a blue point. We can make it slightly bigger so that it does touch one.
So, let's binary search for the answer. However, you have to very careful and notice that the binary search isn't monotonic if we only consider circles touching blue points. However, if we consider circles that touch either a red or blue point, then the binary search is monontonic, so everything works out.
To check if a radius works, we can do a angle sweep around our center point. We have a fixed radius and fixed center, so each other point has at most two angles where it enters and exits the circle as we rotate it about the center point. We can keep track of these events and find an interval where the circle only contains red points.
code for binary search
For the inversion solution, let's fix the blue point that our circle touches. Then, let's take the inversion around this point (i.e. https://en.wikipedia.org/wiki/Inversive_geometry). Now, circles that pass through our center points become lines, and the interior of those circles are in the halfplane not containing the center point. The radius of the circle is inversely proportional to the distance between our center point to the line after inversion.
So, we can say we want to solve the following problem after inversion. Find the closest line that contains no blue points in the halfplane facing away from our center point and at least one red point. We can notice that we only need to check lines that contain a blue point on the convex hull after inversion.
To make implementation easier, you can make the additional observation that the sum of all convex hull sizes will be linear through the process of the algorithm. Some intuition behind this observation is that only adjacent nodes in a delaunay triangluation can appear on the convex hull after inversion, so the sum is bounded by the number of edges in such a triangulation (of course, we do not need to explicitly find the triangulation).
code for inversion
## Hongcow Masters the Cyclic Shift
Let M denote the total number of characters across all strings.
Consider how long it takes to compute f(L) for a single list.
Consider a graph where nodes are suffixes of strings. This means we already spelled out the prefix, and still need to spell out the suffix.
There are at most M nodes in this graph. Now, draw at most N edges connecting suffixes to each other. We can find the edges efficiently by doing suffix arrays or z algorithm or hashes.
Now, we claim is the list is good if and only if there is no cycle in this graph. You can notice that a cycle exists => we can construct a bad word. Also, if a bad word exists => we can form a cycle. So, we can check if there is a cycle, which takes O(N*M) time.
Next step is to notice that extending a bad list will never make it good. So we can do two pointers to find all good intervals, which requires O(n) calls to the check function. So, overall this runs in O(N^2*M) time.
You might be wondering why this problem asks for sublists rather than the entire list. To be honest, it's just to make tests slightly stronger (i.e. I get ~30^2x the number of tests in the same amount of space).
code
•
• +88
•
» 17 months ago, # | +10 For Div1C, why is the size of the dimension "how many of the other" n^2?
• » » 17 months ago, # ^ | 0 "First, somehow reduce it so r_i,b_i <= n"
• » » 17 months ago, # ^ | 0 That dimension represents how much of the payment is done by cards and not by tokens. Since for ith card at max (i — 1) payment can be done by cards, hence its size is n^2.I had the same idea but couldn't get it AC in contest.:)
• » » » 17 months ago, # ^ | 0 You don't need extra dimensions for that, you can directly derive it from the bitmask of which cards you've already visited (at least, I assume that's what the 2n represents).
• » » » » 17 months ago, # ^ | ← Rev. 2 → 0 The amount of payment done in cards depends on the order of the cards also, not just the mask. Eg. if there are 3 cards {(2, 0), (1, 0), (1, 0)}, then : If order is {1, 2, 3}, then payment done in cards = 1 + 1 = 2 If order is {2, 3, 1}, then payment done in cards = 1 + 2 = 3 Payment for getting a card can be done in 2 ways — token or using previous cards. Each card can assist in payment of all subsequent cards. ith card can be used in payment of n-i subsequent cards. This sum is obviously n^2.
• » » 17 months ago, # ^ | ← Rev. 2 → 0 Yeah I don't buy that either. For the following input you ended up with > n2 tokens of one color as intermediate value: 4 R 100 0 R 100 0 B 0 100 B 0 100
• » » » 17 months ago, # ^ | +15 For the token R 100 0 at some moment you pay max(100-r,0), where r is the current amount of red tokens collected. Note that it will always be 100-r as r<=4 here. You can transform it to 96+max(4-r,0) which means you buy 96 red tokens at some other moment and change the token description to R 4 0. By doing this kind of transformation to all of the tokens (in any case, not just this example) you can end up with every token price <=n and some additional price to pay separately.
• » » » » 17 months ago, # ^ | 0 If i transform it to 96+max(4-r,0),i should buy 96 red cards at some other moment,but when i buy 96 red cards,i also have another 96 blue cards.These bule cards can also contribute to the answer,so i doubt how to handle with these 96 blue cards?
• » » » » » 17 months ago, # ^ | 0 My idea is that you can shift the payments to any moment in time (you may be in debt at some point but have to pay it at the end). I would shift all the additions (e.g. this 96 red) to the end and have a total of R and B to add after the dp is done. This seems complicated though, I wonder if the author had something simpler in mind.
• » » » » » » 17 months ago, # ^ | ← Rev. 2 → 0 I am not quite convinced on the point you can shift the payment if you can buy it earlier, wouldn't it contributes to the discount caused by owning that card incorrectly?Edit: I think I've got it
» 17 months ago, # | -11 For Div1D, if any nice circle of radius r does not touch any blue point, I assumed there exist a circle of radius r which has center on one of the red point, is it wrong? since I'm getting wa for this idea :(
» 17 months ago, # | -24 Too bad I solved Div2D but I did not count properly the number of indexes which I was asking as a question.
» 17 months ago, # | -6 wow, this Editorial is even faster than my Internet speed
» 17 months ago, # | +7 the description of Div2 B may be hard to many contestants.
• » » 17 months ago, # ^ | +47
• » » » 17 months ago, # ^ | 0 lol!same here!
» 17 months ago, # | +15 How to use Binary Search to solve D? Anyone could explain it in more detail?
• » » 17 months ago, # ^ | +6 First, the circle has to touch some point red or blue.Try each of these points, and call this the center point. Now, we can rotate the circle about the center point. Notice that every other point forms an interval on the angles, so this reduces to a pretty standard problem. Note you count the center point as part of the condition (i.e. when we rotate around a red point, we don't necessarily have to include a different red point).
• » » » 17 months ago, # ^ | ← Rev. 4 → -10 For Div2 Problem B Hangcow And Puzzle This is also a valid test case 5 5X X X X X. . . . .X X X X X. . . . .X X X X Xbecause final rectangle made just by shifting second rectangle one unit down and five units left both will fit together (without rotating or overlap or picking up some part)X X X X XX X X X X X X X X XX X X X XX X X X XX X X X X
• » » » » 17 months ago, # ^ | +11 It is guaranteed that the puzzle pieces are one 4-connected piece.
• » » » » » 17 months ago, # ^ | +10 So sir you mean such a test case is not possible ????????
• » » » 17 months ago, # ^ | 0 I had a problem with understanding it too.After reading editorial I've concluded that need binary search by angle too :)
» 17 months ago, # | ← Rev. 3 → 0 ...XXXin Div2B i can't understand why this is wrong ! as it's mentioned in the problem statement that i'm allowed to move any piece so if i moved the bottom left piece 3 timesi can get this .X.X.X and then i can make a rectangle ,, what is the thing that i misunderstood ?
• » » 17 months ago, # ^ | +1 I made the same mistake because of the ambiguity of the question. The pieces (or the Xs) are all connected together (in one 4-connected piece). So you either move them together as a whole, or don't move any of them.
• » » 17 months ago, # ^ | +1 The whole thing is a piece, you cannot move individual parts of the piece.
• » » 17 months ago, # ^ | 0 actually what I understood is, the given input is a single piece in a whole..it does not cosists of pieces,i.e. 'X' blocks are not single pieces.
» 17 months ago, # | 0 Editorial posted 13 hours ago? :/
• » » » 17 months ago, # ^ | +12 It was saved as a draft 13 hours ago
» 17 months ago, # | +13 This contest is excellent! Thank you authors(lewin) and MikeMirzayanov!!!
» 17 months ago, # | 0 My friends and I have a better solution which is easier understood for Div2E/Div1C. We can make the problem reversed. The solution is also bitmask dp. We can start from the state (0,0), which means we cost no red tokens and blue tokens at first. And then we can buy one cards each time, whose cost is ( max( red_cost[i]-red_left, 0 ), max( blue_cost[i]-blue_left, 0 ) ), while i is the card we are ready to buy and red_left means how many red cards we haven't bought except i. We say a state (a,b) is better than another state (a2,b2) when max(a,b) < max(a2,b2). But my program has a little bug and I could not find it until 2 minutes after the contest. QAQ.
» 17 months ago, # | ← Rev. 2 → +1 In DIV2B why the answer for the below test case would be NO.2 5.XXX..X.X.We can move the X at (2,2) to (1,1) and the X at (2,4) to (1,5). So in this way we can form a rectangle. But the answer for the above test case is NO. Why ???
• » » 17 months ago, # ^ | 0 i have the same problem ! i can't get what in the statement that make the answer for this testcase is no !
• » » 17 months ago, # ^ | +3 You can only move all Xs (all the component) to any direction.
• » » 17 months ago, # ^ | ← Rev. 3 → 0 I think the whole structure can be moved not a single part XXX X.X We can move this whole structure not a single X. I am not sure though. But I hacked a solution using the similar test case. UPD : My solution for B failed system test. So, at this moment I have no clue what the fuck is the problem with problem B :(
• » » 17 months ago, # ^ | +1 I came up with exact same thing and passed pretest. Few mins later.. Unfortunately your solution for B is Hacked :'(
• » » 17 months ago, # ^ | 0 The idea of the problem is to check if you can make a rectangle with 2 exactly pieces as given in the input. You are allowed to move the whole piece, not part of the piece. So if there are any way to make a rectangle with the 2 exactly pieces, then you have print "YES", otherwise "NO".
» 17 months ago, # | 0 Can someone help me about problem D. My ten solutions dont work on pretest 1, but i have been using any flushes and endl's. Plz explain me my fault :http://codeforces.com/contest/745/submission/23071323.
• » » 17 months ago, # ^ | 0 Maybe it is because you only need to write fflush(stdout) after you write a complete line instead of writing it after every number on your list.
• » » » 17 months ago, # ^ | 0 i had written every variant... Idk.
» 17 months ago, # | +13 Div2 D is such a fun problem :D
» 17 months ago, # | ← Rev. 2 → 0 In div1E, I've tried to do almost the same thing. But I don't know how to "check the match of the suffix with the end." So we start from a suffix adding words and going into new suffixes and achieve different suffixes. But we are not looking for a cycle in this graph, we are looking for an achievable suffix that, after adding at its end the prefix of the initial state, could get to a string that can be partitioned into more strings. So how do we do that? Wouldn't it be M^2*N?(fixing an initial state and then moving through the edges and checking whether we could have achieved to certain nodes that could be combined with the prefix of this initial state)
• » » 17 months ago, # ^ | +10 I'll add more description above, but as a quick reply.The part you're quoting is more for creating the edges rather than checking the condition of the problem. What you're saying is correct, but you have to notice that "need to spell out suffix starting at position i" and "already spelled out letters up to position i" are the same state, so it is sufficient to check if we have a cycle in our graph.
• » » » 17 months ago, # ^ | 0 Ohh... Yeah, I think I got it now. So close:))Anyway, very nice problemset:)Congrats!
» 17 months ago, # | ← Rev. 2 → +1 Could anyone explain Div2.D more detail please?
• » » 17 months ago, # ^ | -10 For row i, you need to cover all all j!=i. This means for some j!=i, there is a question where we did not take i but j was present. So write i and j in their binary expression(<=10 bits are needed), there will exist a bit where they differ. Acc to editorial solution, you would have asked a question where either i was present or either j. This is exactly what we needed.
» 17 months ago, # | 0 Could someone explain more about Div2E.
» 17 months ago, # | 0 When the problem can be submit at the problem set? I want to try my solution for Div2D. :)
» 17 months ago, # | 0 Problem B Div2 was extremely ambiguous !
» 17 months ago, # | 0 problem statement of div2B was really bad,many people misunderstood the question and got hacked.
» 17 months ago, # | +1 In Div 2 bif the input is X.X.Xthen it should print yes as moving one piece to one step right would make a rectangle. But according to editorial as count =3 the output should be no. can anyone explain where am i wrong?
• » » 17 months ago, # ^ | 0 This input is not valid because it is not a 4-connected piece: "The region is 4-connected if for any two cells X there is a path between them that uses only X cells and any two neighbouring cells of this path share a side."
» 17 months ago, # | 0 Has anybody solved D2D/D1B using the "parallelization solution"?
• » » 17 months ago, # ^ | +13 I don't know if my solution was the expected "parallelization" solution.My solution is:First, work with a matrix of dimension 2M where 2M is the smallest power of two that is bigger or equal than N.Now, in order to guess the matrix, I make questions over the upper triangular and lower triangular matrix.For example, given n = 8Every entry of this matrix represent a question. Every entry that has the same value is asked in the same question (except the diagonal). The same is done with the lower triangularNow, using this method only log2(n) questions are needed for every triangular. If n = 1000, then 20 questions will be made!
• » » » 17 months ago, # ^ | +7 Yes, this is the intended way of parallelization that I was thinking of.
• » » » 17 months ago, # ^ | +3 I used almost the same approach, and came up with asking questions like: 1, 3, 5, 7, 9... 2, 4, 6, 8, 10... 1, 2, 5, 6, 9... 3, 4, 7, 8, 11... 1, 2, 3, 4, 9... 5, 6, 7, 8, 13... But its failing on test 7 don't know why :(
• » » » » 17 months ago, # ^ | 0 i made the same thing and it worked
• » » » 17 months ago, # ^ | 0 How are you bounding the number of questions by 2 * log2(n)
» 17 months ago, # | ← Rev. 2 → +5 I think something is wrong here one code with 2 different verdict23067626 hack23072618 acceptEDIT : PS for more detail see my blog
» 17 months ago, # | ← Rev. 2 → 0 My solution on div 1 c Number of final pairs isnt exceed 16^4 Also for every a1
» 17 months ago, # | 0 We can notice that we only need to check lines on the convex hull of all blue points. How to check them? And why we shouldn't check tangents from red points to this polygon?
• » » 17 months ago, # ^ | ← Rev. 2 → 0 To check that if a line has a red point on the opposite side, you can find the convex hull of all points, and only blue-blue edges that aren't on the entire convex hull will be covered by a red point, so you only need to check those. Checking a line is checking the distance from the line to our center point (the radius of the circle is equal to the reciprocal of the distance from the center point to the line).Also, an easier way to implement, but harder to notice is to see that the sizes of all the convex hulls across all inversions will be linear (the intuition is that the points on the convex hull after inversion are the ones that are closest to the center point). So, you can also just do this by brute force also.
• » » » 17 months ago, # ^ | 0 Also, an easier way to implement, but harder to notice is to see that the sizes of all the convex hulls across all inversions will be linear (the intuition is that the points on the convex hull after inversion are the ones that are closest to the center point) Wow, can you elaborate on this?
• » » » » 17 months ago, # ^ | 0 I realize I misspoke in an earlier post, you do need to check those tangents, but what I said to solve the original problem is still relevant (i.e. you can either find the convex hull of all points or you make this observation and do it by brute force).I don't have a formal proof for this, but some intuition is to look at the delaunay triangulation. After inverting about a point P, only points that are connected to P in the delaunay triangulation will lie on the convex hull, so this shows that the sum of all convex hull sizes will be linear.
» 17 months ago, # | ← Rev. 5 → 0 I think that following solution to Div2B is really simple and easy to code. All we have to do is check every "square" (four fields). For every (1 ≤ i < n, 1 ≤ j < m) we count how many letters 'X' are in the square formed by a[i][j], a[i][j + 1], a[i + 1][j], a[i + 1][j + 1]. If there is at least one "square" with number of 'X's equal to 3 then the answer is NO, if not the answer is YES.
» 17 months ago, # | 0 For the div1C/div2E solution of setter, there is int ans = base + n; Why should we add an n there? I think base should be enough (which is incorrect).
• » » 17 months ago, # ^ | 0 It takes a day to buy something. So if you buy n things there are n extra days.
• » » » 17 months ago, # ^ | 0 I see, I was confused by the number of tokens and failed to consider the number turns. Thanks.
» 17 months ago, # | 0 For div2 C / div1 A I cant see the testcases and i failed at test 15, can someone tell whats wrong in this code .here is my code https://codeshare.io/21j0z5
• » » 17 months ago, # ^ | 0 maybe problem in size of array as n or n+1 0 indexed or 1 indexed
• » » » 17 months ago, # ^ | 0 No, I found the problem, thanks anyway.
• » » » » 17 months ago, # ^ | ← Rev. 2 → 0 Stuck on the same testcase What was the problem that you identified?This is my submission 23256857
• » » » » » 17 months ago, # ^ | 0 I was not adding edges to the components with less number of edges, for example if we have three components 1-2-3-4, 5-6-7 and 8-9, where 1 and 5 are government cities, then I was combining 1-2-3-4-8-9 but forgot to add the edges in 5-6-7 component.
• » » » » » » 17 months ago, # ^ | 0 i thought the same thing when I looked at your code, did you try fixing it and resubmitting?
• » » » » » » » 17 months ago, # ^ | 0 No, I didn't try my exams are going on but I know that's the problem, someone had commented a testcase, my code was failing with previous approach, with the one which I told you, it will give correct output, the code given in the editorial is pretty easy to understand, check it once.
• » » » » 17 months ago, # ^ | 0 had the same problem. thanks :)
» 17 months ago, # | ← Rev. 2 → 0 Hi !The simplest solution for 744C - Коровоконг покупает колоду карт is here : 23076523. Because the number of dp states are O(2n·n2), this code workes in ( because of map).
» 17 months ago, # | 0 Div 2C, what should be the output for this test case? 10 13 2 1 6 1 2 1 3 1 4 1 5 2 3 2 4 2 5 3 4 3 5 4 5 6 7 6 8 6 9 I saw an ac code produced 8 as output, while another ac code produced 7 as output.Which is the correct one? Is the test case valid? I tried to solve it manually and got 7 as output.
• » » 17 months ago, # ^ | +1 It should be 8. You can put edges 1,2,3,4,5-10 and edges 7-8, 8-9, 7-9
• » » » 17 months ago, # ^ | 0 Oh I see. I forgot to consider completing the other government components. Thanks!
» 17 months ago, # | 0 Guys, what is special about testcase #15 in DIV2 C? My approach: find available nodes or components to add. then result is (n * (n — 1)) / 2 — currentEdges where n is the number of vertices
» 17 months ago, # | 0 In Div1E, This means we already spelled out the prefix, and still need to spell out the suffix. What does this sentence mean? And what's the condition of connecting two nodes to each other?
» 17 months ago, # | +1 I am having problem understanding Div 2D. Could someone please suggest more problems like it? What is the topic that it falls under?
» 17 months ago, # | ← Rev. 2 → 0 I tried to solve Div2-D by first asking the first half of the elements, and then the second half. This would leave us with two smaller recursive problems which could be solved together as they don't intersect. Hence o(n) = 2 + o(n/2). Here is an ac using this solution http://codeforces.com/contest/744/submission/23080098
» 17 months ago, # | 0 Hi Help me please understand why I get RUNTIME_ERROR http://codeforces.com/contest/745/submission/23080649 Because code works good on my machine
• » » 17 months ago, # ^ | 0 oops. I see.
» 17 months ago, # | +10 I think your binary search for Div1D runs in time O(n^2 log n log W)? Probably there can be some hard cases to make it TLE.... During the contest I basically implemented the same algorithm, but with a random_shuffle among blue points and red points... That can bring the complexity down to something like O(n^2 log n + n log n log W)...
• » » 17 months ago, # ^ | 0 But if you do binary search individually for each point,the answer may not be right since it's no longer monotonic,did I just miss something?
• » » » 17 months ago, # ^ | 0 The very key thing of this to work is that you should also include the red points and calculate their individual answers first...After that, suppose the best radius for touching those red points is R, you can somehow prove that for the remaining blue points, either its individual answer is smaller than R, or it is monotonic after R...
» 17 months ago, # | ← Rev. 2 → +11 I'm sorry but the solution for Div 1 C/ Div 2 E has stumped me...Could someone please elaborate I've tried reading a a few codes but it was to no avail...:)UPD: I got it now if someone wants to know how to solve it my submission might help (I've filled it with comments to help make things easier)C++ code:23346668.:)
» 17 months ago, # | 0 Div.1 C could be solved by Simulated Annealing....
» 17 months ago, # | 0 If anyone has tried DIV2- C Build a Nation using Union-Find algo, please comment. Im stuck with the understanding using this implementation.
• » » 17 months ago, # ^ | 0 I did. Here's the code: 23060103.While building the DSU, we need to update components' sizes and count of edges in them. Then, turn each component into a clique. Find the largest component that contains a capital. Finally, merge it with components that do not contain a capital. Be sure to perform the final step only once for each component.
• » » » 17 months ago, # ^ | 0 Thanks. Pretty neat coding style. I messed up the entire DSU code for computing size and got lost. Im still not sure where my code is failing though.
» 17 months ago, # | ← Rev. 4 → 0 For Div1 E, I don't quite understand. It seems you're connecting suffix u of word a and suffix v of word b whenever there's a word c such that c completes word a and goes through a prefix of word b up to suffix v. Is this right? If so, why are there "at most N edges" for each suffix? I would think that a word can lead a suffix to different ones.Consider for example words:aaaaa aaaa aaaand a suffix aaa of aaaaa. Then, word aaaa takes you to prefix aa, which is a prefix of every word and so is represented by three different suffix nodes. What to do with this?EDIT: It seems AC solutions have assumed that being at a suffix of word A and then using word B necessarily takes you to a suffix of word B. Why is this?EDIT2: I understand now. The edge isn't using a complete word to jump, it's deciding to put that word after the current one, and using the current suffix as a prefix to it.
» 17 months ago, # | +5 When will the test cases be made public... Or is it just me who can't view the test cases?
» 17 months ago, # | 0 How do we count the number of edges in a graph component? Is there any optimal way ?
• » » 17 months ago, # ^ | ← Rev. 2 → 0 If you know the vertices that belong to the component, it is enough to take the number of neighbors for each vertex, add them up, and halve the result (every edge was counted twice, once for every endpoint).
• » » » 17 months ago, # ^ | 0 Then I will have to use extra memory to store vertices of each component, and then sum the indegrees. Will it not be too costly?
• » » » » 17 months ago, # ^ | 0 No extra memory is needed, you can compute the sum during the traversal, something similar to this: sum = 0 dfs(v): sum = sum + degree(v) [dfs stuff] And even if you did have to store vertices separately, it is not that big of a cost.
• » » » » » 17 months ago, # ^ | 0 Did I get it right?Suppose there are 3 cc's of a graph. Each time I run dfs on each component, sum is initialized to zero. for( int i= 0; i< V; ++i ){ if( !visited[i] ){ sum= 0; dfs( i ); // num sum/2 is the number of edges in the this graph component } }
• » » » » » » 17 months ago, # ^ | 0 Looks all right.
• » » » » » » » 16 months ago, # ^ | 0 hey! I implemented the code , need some help on test case 15! 23722340
» 17 months ago, # | 0 can somebody please explain easily why asking 2 questions for each position works in Div2D? i tried it by hand and it works, but i want to understand it...
• » » 17 months ago, # ^ | +3 As each position must have different bitmask representation, by asking 2 questions for each position, every two integers are guaranteed not to be grouped into the same question for at least once, therefore avoiding the query being diluted by the [0] at position i on each row.
» 17 months ago, # | ← Rev. 2 → 0 Can anyone help me find why 23268578 got WA for Div1 C ?
» 17 months ago, # | ← Rev. 2 → 0 Regarding div2c, stuck on testcase 15 23256857,any ideas?also is there a difference between a clique and completely connected component for a given graph?
• » » 17 months ago, # ^ | ← Rev. 3 → 0 Got It Your solution is failing for this test case. It's answer is 18 10 6 2 2 7 1 2 2 3 5 6 7 8 8 9 9 10
• » » » 17 months ago, # ^ | 0 Thanks for the help.
» 17 months ago, # | 0 Can anyone please give the code for DIV 2 Problem C (Hongcow builds a nation) in C++?
• » » 17 months ago, # ^ | 0 You can see all solutions here Here is my C++ AC solution 23417630
» 15 months ago, # | 0 In div2 E Hongcow buys a deck of cards, is it possible to use a map instead of static array, then we can solve without the first optimization. Has anyone done that?
» 12 months ago, # | 0 For problem 745B.. Following two solution is accepted.27559899 and 27559890.but for following case bith give different ans. Why? 4 4 .... .X.X .X.X ....I think its ambiguous.
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# Multiplying Polynomials. 1. Use FOIL method if you have 2 Binomials. ◦ F (first) O (outer) I (inner) L (last) 2. Use Distribution otherwise. Remember.
## Presentation on theme: "Multiplying Polynomials. 1. Use FOIL method if you have 2 Binomials. ◦ F (first) O (outer) I (inner) L (last) 2. Use Distribution otherwise. Remember."— Presentation transcript:
Multiplying Polynomials
1. Use FOIL method if you have 2 Binomials. ◦ F (first) O (outer) I (inner) L (last) 2. Use Distribution otherwise. Remember to combine any like terms and write answer in descending order.
1. 2x 3 (4x 2 – 3x + 2) 2. (6x + 1)(4x+ 7) 3. (x – 9)(4x + 9) 4. (2x + 6) 2 5. (2x – 1)(x 3 – 3x 2 + 4x – 2)
Two binomials are conjugates if they are the sum and difference of the same two terms. Find the conjugate of each: 1. -x + 5 2. 3x – 9 3. 2x 2 + 10y
(a + b)(a – b) = (a + b)(a – b) = a 2 - b 2 1) (2x + 4)(2x – 4) = 2) [(y + 3) + z][(y + 3) – z] =
1. Find (f·g)(x) 2. Find (f·g)(3) 3. Find (f·g)(-1)
Download ppt "Multiplying Polynomials. 1. Use FOIL method if you have 2 Binomials. ◦ F (first) O (outer) I (inner) L (last) 2. Use Distribution otherwise. Remember."
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# Lesson 8
Percentage Situations
Let's find unknown percentages.
### 8.1: Tax, Tip, and Discount
What percentage of the car price is the tax?
What percentage of the food cost is the tip?
What percentage of the shirt cost is the discount?
### 8.4: Info Gap: Sporting Goods
Your teacher will give you either a problem card or a data card. Do not show or read your card to your partner.
If your teacher gives you the problem card:
3. Explain how you are using the information to solve the problem.
Continue to ask questions until you have enough information to solve the problem.
4. Share the problem card and solve the problem independently.
If your teacher gives you the data card:
2. Ask your partner “What specific information do you need?” and wait for them to ask for information.
If your partner asks for information that is not on the card, do not do the calculations for them. Tell them you don’t have that information.
3. Before sharing the information, ask “Why do you need that information?” Listen to your partner’s reasoning and ask clarifying questions.
4. Read the problem card and solve the problem independently.
5. Share the data card and discuss your reasoning.
### 8.5: Card Sort: Percentage Situations
Your teacher will give you a set of cards. Take turns with your partner matching a situation with a descriptor. For each match, explain your reasoning to your partner. If you disagree, work to reach an agreement.
### Summary
There are many everyday situations in which a percentage of an amount of money is added to or subtracted from that amount, in order to be paid to some other person or organization:
goes to how it works
sales tax the government added to the price of the item
gratuity (tip) the server added to the cost of the meal
interest the lender (or account holder) added to the balance of the loan, credit card, or bank account
markup the seller added to the price of an item so the seller can make a profit
markdown (discount) the customer subtracted from the price of an item to encourage the customer to buy it
commission the salesperson subtracted from the payment that is collected
For example, if a restaurant bill is $34 and the customer pays$40, they left 6 dollars as a tip for the server. That is 18% of \$34, so they left an 18% tip. From the customer's perspective, we can think of this as an 18% increase of the restaurant bill. If we know the initial amount and the final amount, we can also find the percent increase or percent decrease. For example, a plant was 12 inches tall and grew to be 15 inches tall. What percent increase is this? Here are two ways to solve this problem:
The plant grew 3 inches, because $$15-12=3$$. We can divide this growth by the original height, $$3\div 12 = 0.25$$. So the height of the plant increased by 25%.
The plant's new height is 125% of the original height, because $$15 \div 12 = 1.25$$. This means the height increased by 25%, because $$1.25 - 1 = 0.25$$.
Here are two ways to solve the problem: A rope was 2.4 meters long. Someone cut it down to 1.9 meters. What percent decrease is this?
The rope is now $$2.4 - 1.9$$, or 0.5 meters shorter. We can divide this decrease by the original length, $$0.5 \div 2.4 = 0.208\overline{3}$$. So the length of the rope decreased by approximately 20.8%.
The rope's new length is about 79.2% of the original length, because $$1.9 \div 2.4 = 0.791\overline{6}$$. The length decreased by approximately 20.8%, because $$1 - 0.792 = 0.208$$.
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# Math help
I need help solving each equation by the substitution method.
1. X=5y-1
X+2y=13
2. -2x+3y=11
X=4y-3
3. X=-9y-1
2x+4y=5
I need help on these so please guide me in the right direction because I'm struggling on this topic!!
1. 👍 0
2. 👎 0
3. 👁 85
1. 1)
use 5 y - 1 for x
so
5 y-1 + 2y = 13
or
7 y - 1 = 13
7 y = 14
y = 2
now go back and get x
x = 5(2)-1
x = 9
1. 👍 0
2. 👎 0
posted by Damon
2. 2)
-2x+3y=11
X=4y-3
-2 (4y-3) + 3 y = 11
-8 y + 6 + 3 y = 11
- 5 y + 6 = 11
-5y = 5
y=-1
etc
1. 👍 0
2. 👎 0
posted by Damon
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2. ### Math
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Chapter 13 - Review - Page 960: 3
$(x - 5)^{2} + y^{2} = 25$
Work Step by Step
The radius is the diameter divided by two, or $\frac{10}{2} = 5$. The general form of a circle is $r^{2} = (x - h)^{2} + (y - k)^{2}$ where $r$ is the radius of the circle and the ordered pair $(h, k)$ is the center of the circle.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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## pattern block tiles.docx - Section 2: Warm up
pattern block tiles.docx
# Breaking down the design
Unit 5: Writing and comparing ratios
Lesson 3 of 14
## Big Idea: Reproducing the same pattern repeatedly results in a constant ratio.
Print Lesson
Standards:
Subject(s):
Math, modeling, Number Sense and Operations, cooperative learning activity, ratio, simplifying ratio, scaling, ratios, pattern, student ownership, cognitive demand
49 minutes
### Erica Burnison
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# Tag Archives: Keynesian Policy
Currently covering Keynes vs Monetarist in the A2 course. Here is a powerpoint on the theory that I use for revision purposes. I have found that the graphs are particularly useful in explaining the theory. The powerpoint includes explanations of:
C+I+G+(X-M)
• 45˚line
• Circular Flow and the Multiplier
• Diagrammatic Representation of Multiplier and Accelerator
• Quantity Theory of Money
• Demand for Money – Liquidity Preference
• Defaltionary and Inflationary Gap
• Extreme Monetarist and Extreme Keynesian
• Summary Table of “Keynesian and Monetarist”
• Essay Questions with suggested answers.
Keynes v Monetarist Keynote
# A2 Economics Multiple-Choice: Two ways of calculating the equilibrium level of income
With the CIE multiple-choice paper next week, I went through an A2 multiple-choice question with some students on calculating the equilibrium level of income. There are two ways that it can be worked out. Here is the question:
In a closed economy with no government C = 30 + 0.8 Y and I = 50, where C is consumption, Y is income and I is investment.
What is the equilibrium level of income?
A 64 B 80 C 250 D 400
Below is the most common way of working the question out:
Y = C + I
Y = 30 + 0.8Y + 50
0.2Y = 80
Y = 400
Here is the other way that you should be able to work out the equilibrium
Remember: Savings = Income – Consumption
S = Y – C
S = Y – (a + cY)
S = Y – a – cY
S = -a + (1-c) Y
So if we put the figures into the equation you get:
50 = -30 + (1-0.8) Y
50 = -30 + 0.2Y
80 = 0.2Y
Y = 400
# A2 Economics – Keynes 45˚ line
Just completed the Keynes 45˚ line (still in the CIE A2 syllabus) with my A2 class and find this graph useful to explain it. A popular multi-choice question and usually in one part of an essay. Make sure that you are aware of the following;
Common Errors:
1. C and S are NOT parallel
2. The income level at which Y=C is NOT the equilibrium level of Y which occurs where AMD crosses the 45˚ line.
To Remember:
1. OA is autonomous consumption.
2. Any consumption up to C=Y must be financed.
3. At OX1 all income is spent
4. At OB consumption = BQ and saving= PQ
5. Equilibrium level of Y shown in 2 ways
a) where AMD crosses 45˚ line
b) Planned S = Planned I – point D
Remember the following equilibriums:
2 sector – S=I
With Govt – S+T = I+G
With Govt and Trade – S+T+M = I+G+X
# A2 Revision – New Classical to Extreme Keynesian
The main competing views of macroeconomics (Keynesian vs Monetarist) is part of Unit 5 in the A2 syllabus and is a popular topic in the essay and multiple-choice papers. Begg covers this area very well in his textbook. In looking at different schools of thought it is important to remember the following:
Aggregate Demand – the demand for domestic output. The sum of consumer spending, investment spending, government purchases, and net exports
Demand Management – Using monetary and fiscal policy to try to stabilise aggregate demand near potential output.
Potential Output – The output firms wish to supply at full employment after all markets clear
Full Employment – The level of employment when all markets, particularly the labour market, are in equilibrium. All unemployment is then voluntary.
Supply-side policies – Policies to raise potential output. These include investment and work incentives, union reform and retraining grants to raise effective labour supply at any real wage; and some deregulation to stimulate effort and enterprise. Lower inflation is also a kind of supply-side policy if high inflation has real economic costs.
Hysteresis – The view that temporary shocks have permanent effects on long-run equilibrium.
There are 4 most prominent schools of macroeconomics thought today.
New Classical – assumes market clearing is almost instant and there is a close to continuous level of full employment. Also they believe in rational expectations which implies predetermined variables reflect the best guess at the time about their required equilibrium value. With the economy constantly near potential output demand management is pointless. Policy should pursue price stability and supply-side policies to raise potential output.
Gradualist Monetarists – believe that restoring potential output will not happen over night but only after a few years. A big rise in interest rates could induce a deep albeit temporary recession and should be avoided. Demand management is not appropriate if the economy is already recovering by the time a recession is diagnosed. The government should not fine-tune aggregate demand but concentrate on long-run policies to keep inflation down and promote supply-side policies to raise potential output.
Moderate Keynesians – believe full employment can take many years but will happen eventually. Although demand management cannot raise output without limit, active stabilisation policy is worth undertaking to prevent booms and slumps that could last several years and therefore are diagnosed relatively easily. In the long run, supply-side policies are still important, but eliminating big slumps is important if hysteresis has permanent effects on long-run equilibrium. New Keynesians provide microeconomics foundations for Keynesian macroeconomics. Menu costs may explain nominal rigidities in the labour market.
Extreme Keynesians – believe that departures from full employment can be long-lasting. Keynesian unemployment does not make real wage fall, and may not even reduce nominal wages and prices. The first responsibility of government is not supply-side policies to raise potential output that is not attained anyway, but restoration of the economy to potential output by expansionary fiscal and monetary policy, especially the former.
# Keynes v Hayek with Chinese characteristics.
You will no doubt have heard about the battle of ideas – Keynes v Hayek. In the 1930’s this was probably the most famous debate in the history of economics – the battle of ideas -government v markets.
Now there is Chinese version of the debate:
Justin Lin (Keynes) versus Zhang Weiying (Hayek) – both are Professors at Peking University. Lin is on the right of the image below.
Their latest debate is about industrial policy and the concept that the government can set the example of how to run successful industries – in the 1980’s textiles and today renewable energy. Although China’s growth record would seem to justify this some have seen these state run industries produce little innovation. Lin believes that countries that have a comparative advantage should receive help from the government whether it be in the form of tax cuts or improved infrastructure. Furthermore, because resources are limited the government should help in identifying industries which have earning potential. This assistance includes subsidies, tax breaks and financial incentives — aimed at supporting specific industries considered crucial for the nation’s economic growth.
Zhang sees this industrial policy as a failure in that he believes government officials don’t know enough about new technologies. He uses the example in the 1990s, when the Chinese government spent significant money on the television industry only for the cathode ray tubes to become outdated. He is also concerned about industrial inertia with local officials following the central government’s direction which tends to lead to an overcapacity. Zhang, however, credited the free market — not politically motivated government subsidies — with game-changing innovations that benefit society eg. James Watt and the steam engine, George Stephenson’s intercity railway, and Jack Ma’s innovative online marketplaces under Alibaba.
China’s ongoing transition to a market-based economy has relied on labour, capital and resource-intensive industries. But the transition’s negative side effects have included structural imbalances and excess capacity in certain sectors. Moreover, some state-owned enterprises such as telecoms have been challenged by disruptive innovators, such as social networks.
Zhang said industrial policy can foster greed. For example, companies may collude with government officials to win special favours. And policymakers can make mistakes, given that even the most well-informed intellectual cannot always predict market trends. Other economists have contributed to the debate stating that a lot of the most successful companies have not had any government assistance in their early years.
However the debate is sure to continue – what works best ‘Markets or Governments’?
Sources:
The Economist – 5th November 2016
CaixinOnline
# A2 Revision – Keynesian and Post-Keynesian Period
I have discussed with my A2 class the end of the Gold Standard and the new era of self-regulating markets that started in the 1980’s under Reagan (US) and Thatcher (UK). This relates to Unit 5 in the A2 syllabusMain schools of thought on how the macroeconomy functions – Keynesian and monetarist.
Robert Skidelsky, in his book “Keynes – The Return of the Master”, outlined the Keynesian and Post-Keynesian periods. The Keynesian period was the Bretton Woods system whilst the “New Classical” Washington consensus system succeeded it. Both are outlined below:
The Bretton Woods system was designed to improve the rules and practices of the liberal world economy which had grown up sporadically in the 19th century. However in 1971 the fixed exchange rate system collapsed (see post Fixed exchange rates and the end of the Gold Standard) and the full employment objective was cast aside. Futhermore controls on capital were removed in the 1990’s. The new system introduced was more free market based and took the name of the Washingotn Consensus System.
According to Skidelsky the two regimes were shaped by two different philosophies. The Bretton Woods system broadly reflected the Keynesian view that an international economy needed strong political and institutional supports if it was to be acceptably stable. The Washington consensus was driven by free market principles of self-regulation and limited government intervention.
# The Influence of Economists – Seven Bad Ideas.
Alan Blinder wrote a review of Jeff Madrick’s book “Seven Bad Ideas: How Mainstream Economists Have Damaged America and the World” – in the December edition of The New York Review of Books. The basis of the book is that ‘economists’ most fundamental ideas contributed centrally to the financial crisis of 2008 and the Great recession that followed.” Blinder quoted George Stigler’s contrary verdict “that economists exert a minor and scarcely detectable influence on the societies in which they live.” He comes up with a test that asks you which of the statements below comes closer to the truth.
The dominant academic thinking, research, and writing on economic policy issues exert a profound, if not dispositive, influence on decisions made by politicians.
Politicians use research findings the way a drunk uses a lamppost: for support, not for illumination.
Most people chose the second statement but Madrick’s answer seems closer to the first.
1. The Influence of Economists – they don’t have as much influence on economic policy as Madrick suggests.
Blinder quotes his idea of Murphy’s Law of Economic Policy:
Economists have the least influence on policy where they know the most and are most agreed; they have the most influence in policy where they know the least and disagree the most vehemently.
However when you consider who has the most influence on the election of politicians it is the general public and not economic experts. According to Binder the GFC of 2008 has in part been the fault of economists. Most graduate take at least one economics paper but professors have failed to convince the public of even the most obvious lessons, like the virtues of international trade and the success of expansionary fiscal policy in a slump. It’s a pedagogical failure on a grand scale. Many economists teach and praise the efficient market hypothesis.
2. Mainstream economics is right wing.
Milton Friedman (University of Chicago) is targeted by Madrick with regard to right wing doctrine. It is quoted in the book that Keynesian economics is not part of what anybody has taught graduate students since the 1960’s. Keynesian ideas are fairly tales that have been proved false. Blinder refutes these statements with the latter being farcical. One can argue over the macroeconomic policies of the Chicago School but it’s clear that their views are far from the mainstream.
The success of Keynesian policy since the GFC has been well documented. Experts were asked whether they agreed or disagreed with the two statements about fiscal stimulus.
1. Because of the American Recovery and Reinvestment Act of 2009, the US unemployment rate was lower at the end of 2010 that it have been without the stimulus bill. 82% agreed 2% disagreed
2. Taking into account all the ARRA’s economic consequences – including the economic costs of raising taxes to pay for the spending, its effects on future spending, and any other likely future effects – the benefits of the stimulus will end up exceeding its costs. 56% agreed 5% disagreed and 23% uncertain.
So from this the mainstream is overwhelmingly Keynesian.
Madrick’s first bad idea was Adam Smith’s invisible hand. Blinder sees this as a great idea as throughout history, there has never been a serious practical alternative to free competitive markets as a mechanism for delivering the right people at the lowest possible costs. So it is essential that students learn about the virtues of the invisible hand in their first economics course.
Blinder also challenges another of Madrick’s Bad Ideas – Say’s Law, which states that supply creates its own demand, means that an economy can never have a generalized insufficiency of demand (and hence mass unemployment) because people always spend what they earn. Therefore: no recessions, no depressions. The Great Depression and then Keynes put an end to Say’s Law.
A Bad Idea – Efficient Market Hypothesis (EMH)
This Bad Idea became destructive as the EMH gave Wall Street managers the tools with which to build monstrosities like Collateralized Debt Obligations and Credit Default Swaps on top of the rickety foundation of subprime mortgages. This was further backed up by the credit rating agencies who gave AAA ratings to risky investments. EMH also handed the conservative regulators a rationale for minimal financial regulation.
According to Blinder, Madrick is an important and eloquent voice for what’s left of the American left – at least in economic matters.
# Syriza’s rescue programme for Greece “ pure Keynesian policies”
A number of articles from The New Yorker magazine have outlined the problems facing Greece’s anti-austerity party Syriza. The party came to power on the election promise of reducing Greece’s debt burden and to liberate Greece from the Troika – the ECB, the IMF and the European Commission. However the extension recently granted to Greece will take place only within the framework of the existing arrangement. The budgetary targets for 2015 and 2016 have kept the economy stuck in recession.
* the Greek economy has contracted by 30% since 2008.
* 25% of the workforce are officially unemployed
* 50% of those under 24 years of age are unemployed
* 40% of Greek children live below the poverty line.
Money has been flowing out of the economy leaving the banking system on the verge of collapse see graphic from The Economist.
As with the Keynesian doctrine, Syriza’s solution in to create effective demand by pumping money into the system. One economics professor at the University of Athens called it “pure Keynesian policies. The big question is where will the money come from although some seem to think that it can raise revenue from tackling corruption and tax evasion. The latter is widespread in Greece amongst the upper-middle class and the very rich – the top-most bracket of households and businesses are responsible for 80% of the total tax debt owed to the government.
Greece’s creditors were mostly European banks, which had, in part, used public bailout money following the 2008 credit crunch to scoop up Greek bonds. For example, French and German banks were on the books for thirty-one and twenty-three billion euros, respectively. The troika stepped in during the spring of 2010, and again in 2012, to orchestrate bailouts of the Greek government, offering two hundred and forty billion euros in loans in exchange for a drastic reduction in government spending and other measures to make the Greek economy more competitive. Source: New Yorker
Grexit
The conventional wisdom is that returning to the drachma would be a catastrophe for Greece. There are pros and cons to this decision – the following would be concerns about returning to the drachma:
* An immediate devaluation;
* The value of savings would tumble;
* The price of imported goods would soar.
However on the positive side of things you would get the following:
* Greek exports would become cheaper
* Labour costs even more competitive.
* Tourism would likely boom.
* Regaining control of its monetary and fiscal policy for the first time since 2001
It would give Greece the chance to deal with its economic woes. Other countries that have endured sudden devaluations have often found that long-term gain outweighs short-term pain. When Argentina defaulted and devalued the peso, in 2001, months of economic chaos were followed by years of rapid growth. Iceland had a similar experience after the financial crisis. The Greek situation would entail an entirely new currency rather than just a devaluation.
This conflict is as much about the ideology of austerity and whether smaller countries will have a meaningful say in their own economic fate. However one needs look back in history to remember that in debt-saddled Weimar German, humiliation and dispossession festered until it a gave rise to the Nazi party. Greece’s neo-nazi party won the third greatest number of parliament seats in the last election.
# Why are people risk adverse?
One of the major concerns with regard to economic growth, especially in the US, the is lack of risk taking i.e. Entrepreneurial decision-making, self-employment etc. Ultimately risk-taking is the fulcrum of the economic cycle of booms and slumps which John Maynard Keynes referred to Animal Spirits. However research in Behavioural Economics has shown that there is a reason for this behaviour. A recent Free Exchange article in The Economist discussed this area.
A simple test that has been carried out is as follows. Would you prefer:
1. A gift of \$50, or
2. To play a game with a 50% chance of winning \$120
It might seem logical to chose the second option since the average return in \$60. However most people chose the more secure option rather than take the risk. But risks do vary according to the individual and the time period.
Individual as risk taker
The following are seen as influences in risk behaviour:
1. Education – more educated likely to take risks
2. Income – the higher the income the more likely to take risks
3. Gender – Men more likely to take risks
Historically
Research shows that those people who experienced high returns on the stockmarket in their early years were likely to tolerate more risk to own shares as a higher proportion of their income. However a bad experience can dampen risk-taking and this would be prevalent today with the impact of the GFC. Also after the Wall Street crash of 1929 the US economy came to standstill as people were very tentative of taking any risks after what they had been through.
It is not just a financial crisis which affects risk-taking. The tsunami that hit South-East Asia in 2004 and Japan in 2011 can put an end to risk-taking as people rebuild their lives. Furthermore military conflicts such as the Korean war can cause their victims more cautious for long periods of time.
Horror Story = Risk Aversion
One group of universtiy students watched a horror story which contained a gruesome torture scene whilst another group didn’t. On completition of the horror scene both were asked some questions about their propensity for risk. Students who had seen the horror story reported increased aversion to risk by roughly the same amount that the GFC had on people. With this in mind the GFC in likely to restrict consumers from taking the sort of risks that help propel the economy for decades to come. It could be that the lack of entrepreneurial spirit will hinder growth and governments rather than worry about debt will have a lack of risk-takers to deal with.
# Greenspan the Keynesian?
Recently John Cassidy in the New Yorker wrote a piece about Alan Greenspan’s new book “The Map and the Territory: Risk, Human Nature, and the Future of Forecasting” in which Greenspan admits to flaws in the neo-classical model that people and financial institutions act in their own self-interest. He has finally come round to admitting that people’s actions are often driven by “Animal Spirits” (originates from Keynes) and rather than behaving like calculators they respond to fear, greed, euphoria, and impatience. He identified 10 traits which he calls “Inbred Propensities”. It seems that during his tenure as Fed Chairman Greeenspan rarely mentioned Keynes although he studied it at Columbia University in the 1950’s. As John Cassidy points out by the time he met Ayan Rand he was of the neo-classical orthodox. The article is worth a look – plenty of good debate.
Alan Greenspan Rediscovers Keynes – Sort Of
Remember the difference between the two schools of thought
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# Quick Answer: How Many Days Old Is 30 Years?
## How many years we sleep in a lifetime?
1.
Sleeping.
A good night’s sleep is vital for every human being to survive.
Given that an average a person sleeps for 8 hours in a day, that means that an average person will sleep for 229,961 hours in their lifetime or basically one third of their life..
## How long does the average person live after they retire?
Here’s another – the later you retire, the earlier you will die. A variation on this theme is the “fact” that, in some jobs, average life expectancy after retirement is just 18 months.
## How old are you when your 10000 days old?
As of the 17th of February, 2017, I am 27 years, 4 months and 15 days old. Or in other words, 10,000 days old.
## How many years is 1996 to now?
The number of years from 1996 to 2020 is 24 years .
## How many days are in a lifetime calculator?
To find out how much time you have left if you are an average man or woman, take your age and multiply it by 365. Then, subtract it from 27,375 days. For example, if you are 25 years old, you have 18,250 days to live (27,375 days – 9,125 (because 365 x 25 = 9,125)).
## What will happen in 2050?
Stabilization in the population will happen in the second half of the century. It is calculated there will be 601,000 centenarians (people at least a hundred years old – born before 1950) in the United States by 2050. … According to this study, 9.075 billion people will inhabit Earth in 2050, against 7 billion today.
## How old will I be in 2020 if I was born in 2004?
The number of years from 2004 to 2020 is 16 years .
## What is my age if I was born in 1991?
How old am I if I was born in 1991? – If you were born in 1991, you are 29 years old. If your date of birth is after September 22, then your age is 28 years old.
## How many days is a 12 year old?
4382.91 DaysYears to Days Conversion TableYearsDays9 Years3287.1825 Days10 Years3652.425 Days11 Years4017.6675 Days12 Years4382.91 Days8 more rows
## How old is 5000days?
13.7 years5000 days equals 13.7 years. Convert 5000 days into minutes, hours, seconds, months, weeks, miliseconds, microseconds, nanoseconds, etc…
## What was 10000 days ago?
To get exactly ten thousand weekdays before now, you actually need to count 14,000 total days (including weekend days). That means that 10000 weekdays before today would be May 24, 1982.
## Why is the year 2020 special?
Being alive in 2020 is special because that is the only year you are likely to live through wherein the first two digits will match the second two digits. The next year that follows this pattern is 2121. A person alive now would have to be at least 101 to see that year. While that is possible, it is unlikely.
## How old would you be if born in 1997?
What is my age now if i was born in 1997?What is my age if I was born in 1997?Month of 1997AgeSeptember23 yearsOctober22 years 11 monthsNovember22 years 10 months9 more rows
## How many days old is a 50 year old?
18250 days oldAt 50 years old, you’ll be 18250 days old!
## How many seconds does a person live?
22,075,000 secondsThere are approximately 22,075,000 seconds in a lifetime.
## How old is 20000days?
20,000 days old is Tuesday, June 25. 2075. 21,000 days old is Monday, March 21.
## How old are you if you are born in 1994?
1994 to 2019 How Many Years?How old was I if I was born in 1994?Years RangeNo of year agoAge1994 to 20173 years ago23 years1994 to 20155 years ago21 years1994 to 20137 years ago19 years3 more rows
## How old will I be in 2040?
how old will you be in 2019, 2020, 2025, 2030, 2040 & 2050?…How Old Am I? – Age & Birthday Calculator.How Old will I be?Same Day onAgeDay203030 years 11 months 10 daysSunday204040 years 11 months 10 daysSaturday5 more rows
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