Outcome:
Model and solve situational questions using linear equations of the form:
where a, b, c, d, e, and f are rational numbers.
Indicator:
i. Solve a linear equation symbolically.
Feedback:
Solving Equations Lesson 9 Feedback (1)
Solving Equations Lesson 9 Feedback (2)
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]]>I have always been a fan of the IPL, right from its inception in 2008. Though its been in the news for all the wrong reasons recently, its popularity has not diminished. It was during one of the matches involving my favourite team (RCB) this year that me and some of my friends had an argument on WhatsApp over the most successful teams in IPL’s history. I had created a very primitive ranking system for IPL back in 2009 and had a fair idea of how a team in IPL should be ranked. Just counting the silverware a team has won will not give a fair picture of where each team stands. I lost the debate because I could not convince them. What then does a programmer with a love of statistics, mathematics and cricket do? Create a custom ranking algorithm of course!
The Groundwork
Before I started with the new algorithm, I tried to recall what my primitive algorithm looked like. I knew it took into account the number of seasons a team played, the number of teams participating in a season and the team rankings in a season. The sad part was I forgot how these parameters were connected. I knew I had to start from scratch. So I laid down some ground rules for the algorithm:
1. The rankings would be based on a team’s performance in the IPL and will not take into account its performance in the Champions League T20.
2. The primary input to the algorithm would be a team’s ranking in a particular season.
3. For the sake of uniformity and continuity, if a team were to be renamed but remained from the same city (as was the case with DC/SRH) , they would be regarded as one team only.
4. Any team that has completed one full season in the IPL would be included in the rankings.
These criteria were comprehensive and, if used correctly in the algorithm, would make it independent of the number of seasons a team has played, the number of teams participating in different seasons and other flaws that might creep in due to the fact that there have been different number of teams playing in different seasons in IPL. Once these criteria were defined, the only work left was the creation of the algorithm itself.
The Algorithm
The first step in creating an algorithm is to take a sample subset of the data you are going to process and create a prototype algorithm which you can then extrapolate to include the entire data. In this case, I took RCB’s record in six seasons. Their record is not spectacular for they have not won a single season. But it is not too bad either for they have been runners-up twice and semi-finalists once. They were above-average and ideal as a sample for the algorithm.
First, I collected their rankings in each season of the IPL: 7th/8, 2nd/8, 3rd/8, 2nd/10, 5th/9 and 5th/9. I collated these figures appropriately, summed up the ratios of each season and divided the figure by the number of seasons RCB have played. This gave me a progressive figure between 0 and 1 which accurately described the team’s performance over the years. Further modifying this algorithm so that it would give a number between 0 and 100, I arrived at the following formula:
X = 100 * [1 - (Σx)/S]
where, x = (R-1)/(N-1)
Here, X – Score; R – Rank of the team in a particular season; N – Number of teams participating in that season; S – Number of seasons played by that team.
This gave RCB a score of 60.17. I applied the algorithm to all the teams and here are the results:
1. CSK – 86.33
2. MI – 63.83
3. RCB – 60.17
4. RR – 48
5. DC/SRH – 44.17
6. KKR – 41.5
7. KXIP – 40.83
8. DD – 31.5
9. KTK – 22.22
10. PWI – 7.83
Conclusion
The algorithm was tested thoroughly for every scenario possible and it passed all the tests. Although there might be minor bugs in the algorithm, they would even out as the tournament progresses. It is safe to say that it should hold good for quite some time. The rankings prove my initial assertion that having silverware does not make a team great as there are three teams in the rankings below RCB that have won an IPL season. Time to call up my friends and share this with them!
]]>The two types of trapezoid are shown in blue and green. There are twenty-four blue ones (in eights set of three, surrounding each triangle) and twenty-four green ones (in twelve sets of two, with each set in “bowtie” formation).
This symmetrohedron follows logically from one that was already known, and pictured at http://www.cgl.uwaterloo.ca/~csk/projects/symmetrohedra/, with the name “bowtie cube.” Here’s a rotating version, which you may enlarge with a click, if you wish:
(Images created with Stella 4d — software you can try yourself at http://www.software3d.com/Stella.php.)
]]>The regular octagons are of the same size, but of two different types, when one considers the pattern of other faces surrounding them. This is why six of them are yellow, and twelve are red.
If the hexagons and isosceles trapezoids were closer to regularity, this would qualify as a near-miss to the Johnson solids, but it falls short on this test. Is is, instead, a “near-near-miss” — and not the first such polyhedron to appear on this blog, either.
(Image created with Stella 4d — software you can try yourself at http://www.software3d.com/Stella.php.)
]]>This contains twelve octagons, six squares, and eight triangles. The “holes” in it keep it from being a true polyhedron, but it is my hope than further study of this arrangement may lead to the discovery of new, interesting, and symmetrical polyhedra.
(Image created with Stella 4d — software you can try yourself at http://www.software3d.com/Stella.php.)
]]>There are sixty of the irregaular, pentagonal gaps. Also, the hexagons themselves are of three types, two of which are sixty in number, and one of which is thirty in number.
If the gaps are filled, and the color scheme changed to make each of the four polygon-types into its own color-group, this looks, instead, like this (click on it if you wish to see it enlarged). It has 210 faces.
(Images created with Stella 4d — software you can try yourself at http://www.software3d.com/Stella.php.)
]]>This zonohedron has fifty faces:
(Image created with Stella 4d — software you can try yourself at http://www.software3d.com/Stella.php.)
]]>It’s like the snub dodecahedron’s big brother.
(Image created with Stella 4d — software you can try yourself at http://www.software3d.com/Stella.php.)
]]>But how to prove your love? Is there a way to provide sufficient proofs to make your partner believe in your love? Even then, when they believed you, you will still need to present the evidence every now and then. I don’t think there is a way for you to say “I love you, and the proofs are this and this and this,” and then confidently write “q.e.d” at the end of your description.
Does that mean that love doesn’t exist? Or is the science of mathematics are not universal since it can’t describe love?
I think they are just two different universes, coexist in two different dimensions.
Q.E.D
]]>(Image created with Stella 4d — software you can try yourself at http://www.software3d.com/Stella.php.)
]]>Graphed with μ=0, σ ranging from 21 to 50.
]]>
In April 2000 NCTM released its Principles and Standards for School Mathematics, which are guidelines for excellence in pre-K
In representing the interests of its members in the debate of public issues, NCTM’s government relations activities are dedicated to ongoing dialogue and constructive discussion with all stakeholders about what is best for students.
The general aim of the PhD in Mathematics is to prepare eligible candidates to become productive workers in the industry or government. For the academia, it helps the research scholars to be able to communicate their knowledge and expertise to students and other members of the mathematical society. The doctorate program is designed to develop the students to be able to have a fundamental grasp of certain basics in the fields of mathematics as well a deep understanding of a major field of interest. It nourishes the ability to formulate and recognize significant research problems, postulate solu
Maintaining a certain grade point average in certain written preliminary and qualifying examinations is paramount in obtaining admission as well. Exemption from such examinations which covers core areas such as complex analysis, real analysis, algebra, applied mathematics and topology is subject to meeting certain qualifications as well.
Mathematics requires efficiency and accuracy but most of the students face problem in this, and they think they can hire a tutor with the help of tutor they can solve their problems, but in modern age it is totally time consuming and useless pattern. Now they need to fast medium and best way to resolve their problems that
]]>“Well, you’ll be able to understand it with an example. Think of mathematics. What kind of images do figures contain? Or the plus and minus signs? What kind of images does an equation contain? None. When you solve a problem in arithmetic or algebra, no image will help you solve it, you execute a formal task within the codes of thought that you have learned.”
“That’s right, Narcissus. If you give me a row of figures and symbols, I can work through them without using my imagination, I can let myself be guided by plus and minus, square roots, and so on… But I can’t imagine that solving such a formal problem can have any other value than exercising a student’s brain. It’s all right to learn how to count. But I’d find it meaningless and childish if a man spent his whole life counting and covering paper with rows of figures.”
“You are wrong, Goldmund. You assume that this zealous problem-solver continuously solves problems a teacher poses for him. But he can also ask himself questions; they can arise within him as compelling forces. A man must have measured and puzzled over much real and much fictitious space mathematically before he can risk facing the problem of space itself.”
“Well, yes. But attacking the problem of space with pure thought does not strike me as an occupation on which a man should waste his work and years. The word ‘space’ means nothing to me and is not worth thinking about unless I can imagine real space, say the space between stars; now, studying and measuring star space does not seem an unworthy task to me.”
Smilingly, Narcissus interrupted: “You are actually saying that you have a rather low opinion of thinking, but a rather high one of the application of thought to the practical, visible world. I can answer you: we lack no opportunities to apply our thinking, nor are we unwilling to do so. The thinker Narcissus has, for instance, applied the results of his thinking a hundred times to his friend Goldmund… and does so at every instant. But how would be be able to ‘apply’ something if he had not learned and practiced it before? And the artist also constantly exercises his eye and imagination, and we recognize this training, even if it finds realization only in a few good works. You cannot dismiss thinking as such and sanction only its ‘application’!… So let me go on thinking and judge my thoughts by their results, as I shall judge your art by your works…”
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This is another concise and clearly written book by Professor Susskind and a ‘student’ collaborator. This book, as was The Theoretical Minimum, is based on (video) lectures given by Professor Susskind at Stanford.
The book goes through the mathematical under-pinnings (linear algebra and linear operators). The mathematical concepts are related to the physics through examples starting with spin. The importance of Hermitian matrices, Unitary matrices, commutator relations and the relation to classical mechanics are explored clearly. The book explores what generalized uncertainty is, what entanglement is and is not. It deals with the areas of confusion in a pragmatic way.
I think this book is a useful accompaniment of the video lectures and book by Professor Binney.
I look forward to more Susskind books.
]]>
In representing the interests of its members in the debate of public issues, NCTM’s government relations activities are dedicated to ongoing dialogue and constructive discussion with all stakeholders about what is best for students.
Maintaining a certain grade point average in certain written preliminary and qualifying examinations is paramount in obtaining admission as well. Exemption from such examinations which covers core areas such as complex analysis, real analysis, algebra, applied mathematics and topology is subject to meeting certain qualifications as well.
In broader sense mathematics is a study of numerical, interpretation, calculation, logic, formula etc. that is the main part of mathematics where we can touch all the topics. Mathematics is a huge subject and it requires lot of practical work and as well as theoretical knowledge if we make any sum we have to clear our fundamental or our base should be absolutely clear. And now education system has been totally changed because now 1st standard students got lots of home work in mathematics subject than other subjects and mathematics concept is also has been changed.
Have one amongst our tutors add up of it! Mathematics and science are structured subjects and every one it takes is that the proper of teaching to assist a student very grasp them. With educational Tutoring at Prodigy school assignment, we have a tendency to follow the information of the student’s course and teach new topics prior their category schedule. This implies that once the subject comes up at school, the coed already is aware of it! This methodology permits the coed to master the fabric and ultimately get a high grade within the course. Such students participate additional and stand o
]]>Are you are sitting in your class, studying mathematics, or whatever equivalent is taught to you in your branch, and wondering why this subject is so challenging and whatever could be its use? Well this past century saw many complex mathematical problems being solved and our understanding of science improve. But the case with humans is that we always want more. We like to push the boundaries till none remain.
On May 24^{th} 2000, in a meeting in Paris, in Collège de France, The Clay Mathematics Institute announced a cash prize of one million dollars (for each problem) to anyone who would solve seven designated problems. This was done in an effort to push forward the boundaries of mathematics in the new Millennium.
The problems are:
Birch and Swinnerton-Dyer Conjecture
Hodge Conjecture
Navier-Stokes Equation
P vs NP Problem
Poincaré Conjecture
Riemann Hypothesis
Yang-Mills and Mass Gap
The first of these problems to be solved was the Poincaré Conjecture. The Clay Institute awarded the prize to Dr. Grigoriy Perelman of St. Petersburg, Russia on March 18, 2010. This leaves six unsolved problems.
Many people have dedicated their lives to this endeavor and are still in the pursuit of an answer. The solution of these problems will have huge impacts on the way we think about mathematics, and their real world applications are also profound. P vs NP is the holy grail of computer science. Its solution will help us create better security systems and better computers. Navier-Stokes is the basis of fluid dynamics. Yang-Mills and Mass Gap is the basis to particle physics.
As we have uncovered many mysteries of the universe, new, more complex ones have emerged. The Millennium Problems represent the latest and greatest ones in the field of mathematics.
]]>
Answer key says B.
Challenge:
1. If there is no typo, do you dare challenge the answer key? :)
2. Spot and fix the typo so that B is the correct answer.
Comment to answer! :)
]]>Something’s wrong with this question. Spot the typo, fix it and give the correct answer. :)
Comment to answer! :)
]]>Won’t that be convenient?
From the previous posts, we’ve easily determined that it is impossible for all the distances between all the points on either a flat Earth or a spherical Earth to be the same.
This should be obvious to anybody who understands the First Basic Law of Shapes, and who can easily spot that circles and spheres are not congruent.
This simple fact torpedoes the notion held by many flat Earth proponents, that there would be no difference between the distances involved in the 2 models, or our perceptions of the 2 models.
Now it’s time to take some measurements and show how they will prove that the Earth is not flat.
We’ll use North-West coordinates, so that anyone can find them using Google maps. All distances are rounded to 1 decimal place.
We’re going to begin by choosing 4 points in the USA:
Point I: North County Rd 125 East, Indianapolis, Indiana (39.8° N, 86.5° W).
Point S: NorthEast River Ridge Rd, St Joseph, Missouri (39.8° N, 94.75° W).
Point L: Outside the Southside Bank, off West Marshall Ave, Longview, Texas (32.5° N, 94.75° W).
Point M: Breakfast Creek Rd, Montgomery, Alabama (32.5° N, 86.5° W).
Firstly, we’ll calculate the circumference of the latitudes for the 2 models of the Earth.
Points L and M are roughly 6,404.7 km from the central point (the center of the flat Earth and the North Pole of the Spherical one). Traveling from point L to point I and from point M to point S, we find that these are 810 km North of these points, making them around 5,594.6 km from the central point.
The circumferences for the flat Earth are easy. These distances are our radii for the concentric circles, and the circumference is just 2pi x r.
Calculating the circumferences on an oblate spheroid, however, is a little bit harder.
First, we need to calculate the distance from any point on these latitudes to the center of the Earth; then we need to use a bit of trigonometry to calculate the distance between any point on these latitudes and a point on the Polar Axis, where the vectors would describe a right angle – and this will be our radius for the circumference at that latitude.
To calculate the distance to the center of the Earth, we shall need to use this equation (supplied here – http://www.summitpost.org/distance-to-the-center-of-the-earth/849764):
D = f(Φ,Z,G) = {(a^{4}cos^{2}Φ+ b^{4}sin^{2}Φ)/(a^{2}cos^{2}Φ + b^{2}sin^{2}Φ) + 2(Z + G)(a^{2}cos^{2}Φ + b^{2}sin^{2}Φ)^{½} + (Z + G)^{2}}^{½}
^{ }
Where:
D is the distance to the Center of the Earth.
Φ is the latitude (the angle measured north or south from the equator).
Z is the height above sea level (i.e., the elevation of the point on the Earth, such as a summit).
G is the geoid height above or below the ellipsoid.
a is the semimajor axis (the equatorial diameter of the Earth) .
b is the semiminor axis (the pole-to-pole diameter).
We’re looking for the circumference at the latitudes, so Z and G are negligible for our concerns.
Plug in the numbers and we find that the distance from the center of the Earth for the latitude at points S & I is 6,369.4 km; and for the latitude at points L & M it is 6,372 km.
Finding the length of the vector from these points that is perpendicular to the Polar Axis is simple trigonometry.
We know these points lie on vectors describing 39.8° & 32.5° angles from the equatorial plane, and that the equatorial plane is parallel to the vectors from these points that are perpendicular to the Polar Axis. This means the vectors centered on these points that intersect the Polar Axis at 90°, and the vectors from these points to the center of the Earth, describe the alternate interior angles to the angles above – which means they are the same. As we have 2 angles for each triangle, we can find the third – which for points S & I is 50.2°, and for L & M is 57.5° – these angles are also complimentary to the angles of latitude, so we can find them just by subtracting the relevant angle of latitude from 90°.
We now have, for both triangles, 3 angles (one of which is a right angle) and the hypotenuse – so we can figure out the adjacent lengths, which will be our radii, using the law of sines:
a/sin(A)=b/sin(B)=c/sin(C)
So, the equation we shall use to find the radii is:
a = (sin(A) x b)/sin(B)
Where:
a is the length of the vector we want.
b is the distance to the center of the Earth.
A is the angle opposite a (which is 90° minus the angle of latitude).
B is 90°.
We can now find that the radius for the circumference at the latitude of points S & I is 4,893.5 km; and for points L & M it is 5,374.1 km.
Plugging in the numbers for all the radii now, into the equation 2pi x r, we can find that the circumference that points I and S lie on is 35,152.2 km long on a flat Earth; and 30,746.9 km long on a spherical Earth.
The circumference that points L and M lie on is 40,241.7 km long on a flat Earth; and 33,766.4 km long on a spherical Earth.
Well, our 2 circumferences have differences of 4,405.3 km and 6,475.3 km, respectively. I’m betting we’ll notice those differences when we measure an arc of those circumferences.
The difference in our longitude coordinates tells us the angle described by the vectors from the central point to points I & S, or L & M.
Using this angle we can calculate what distances we can expect to cover, depending on which model is correct.
Here the simple equation is:
(d/360) x C
d is the difference between the angles of longitude, and C is the circumference.
On a flat Earth, the distance between points S & I on this circumference is 805.6 km; and between points L & M it is 922.2 km.
On the spherical Earth, the distances are 704.6 km and 773.8 km, respectively.
That’s a difference of 101 km for points S & I; and 148.4 km for points L & M.
And that’s going to be easily noticed when we measure them.
Now, many people will say that we’re just assuming the distances to the central point here, but that’s OK. We don’t need to.
All we need to do is to travel due East (or due West if you prefer) along either vector SI or LM, and measure the distance. We can do this by constantly triangulating our position and using reverse triangulation to measure distances. Again, I’ll go through these simple methods in a later post.
Having traveled from point S to point I, say, we find the distance to indeed be 704.6 km.
Given that we know we’ve covered an arc, and that that arc describes an angle of 8.25°, we can calculate what we’d expect the distance between points L & M to be, if the Earth is flat.
If we’ve walked 8.25° worth of a circumference and measured it as 704.6 km, then we know we can multiply this by 360/8.25 to find the entire circumference – which comes to 30,747.1 km.
Rearranging the equation for determining the circumference of a circle, to find its radius, we get:
r = C/(2pi)
This gives us a distance from points S & I to the central point, of 4,893.6 km.
We can now walk from either point S to point L, or point I to point M, measuring as we go to find the distance of 810 km. Adding this on now gives us a distance from here to the central point of 5,703.6 km. The circumference at this distance is now 35,836.6 km. Multiply this by 8.25/360 and we can expect, should we travel due East (or West, depending on our starting point), to measure the distance between points L & M to be 821.3 km.
So, we’d notice a difference of around 47.4 km from the round Earth prediction.
Equally, if we’d started off measuring the distance between points L & M to be 773.8 km, then we can perform the same calculations to see what we would expect the distance between points S & I to be.
Again, we multiply 773.8 by 360/8.25, which gives us a circumference at the distance from the central point of 33,766.3 km. Dividing this by 2 x pi, we find the radius to be 5,374.1 km. Measuring our distance walking North from either point L to point S, or point M to point I, to be 810 km, we take this away from 5,374.1 km to get our new radius of 4,564 km. We can now calculate that the circumference of the circle that points S & I stand on is 28,676.7 km. Multiplying this again by 8.25/360, and we find the distance we should measure between the points to be 657.2 km.
Again, this is a difference of 47.4 km between the flat Earth and the oblate spheroid Earth.
But wait, I can hear the protestations from flat Earth proponents already- we’ve just assumed the distances we’ve traveled have covered 8.25° of arc!
Well, that’s OK. We don’t have to assume that either.
We know the distances we’ve covered, and how far apart those arcs are. So we know part of the radius, and we can calculate the ratio between the arcs, using this ratio to find out the whole radius, and thus the angle.
773.81/704.62 = 1.0982. Divide this by 0.0982, and we find that we have to multiply the distance between the 2 arcs (810 km) by 11.18384.
This gives us a radius for our points L & M of 9,059.2 km, which gives that circle a circumference of 56,920.5 km. Using the equation:
A=LM x (360/C)
Where A is the angle, LM is the length of the arc and C is the circumference, we find that we’ve covered 4.894° of arc.
Checking this against the other arc at points S & I by taking 810 from 9,059.2, giving us a radius of 8,249.2 km and a circumference of 51,831 km, we again find that we’ve covered 4.894° of arc.
Well, that’s great then, surely? That proves that we can reconcile a flat Earth with a spherical one, doesn’t it?
I’m afraid not.
As you might have noticed, the circumferences are now a lot longer than our Spherical Earth’s circumferences – 21,084.1 km and 23,154.1 km longer, in fact.
Indeed, not only are the differences over half the length of the spherical Earth’s equator in themselves, but both circumferences are longer than the equatorial circumference!
Certainly, these differences would not go unnoticed, unless all those people on round the world cruises are being shipped about at supersonic speeds without their realising.
Besides, our little experiment isn’t over yet.
We need to do another little measurement, to prove whether the central point really is where we now think it is in relation to everything else. We’ll pick some new points within the area we’ve already measured:
Point A: A spot just off Martin Trail, Northeast of Upper Ouachita National Wildlife reserve, Louisiana (33N, 92W).
Point B: A spot off Willie Ray Dr, North of Cabot, Arkansas (35N, 92W).
Point C: Lastrada St, South Haven, Memphis, Tennessee (35N 90W).
Point D: A spot South of Newport Rd, Near Holmes County State Park, Mississippi (33N 90W).
We measure that the points A & D lie on a circle set 55.6 km North of our previous circle for points L & M. And we can measure that Points B & C lie on a circle 222.4 km North of the one that points A & D lie on.
This should make points A & D lie 9,003.6 km away from the central point, if our last attempts to reconcile the flat Earth with observation is to succeed.
We measure the distance between points A & D to be 186.5 km.
Now we measure the distance between points B & C to be 182.2 km.
Again, knowing the distance between these arcs to be 222.4 km, we calculate the ratio between the 2 arcs and use this ratio to find the whole radius, to determine if we’re correct about where the central point is.
186.5/182.2 = 1.0236.
1.0236/0.236 = 43.37.
Multiplying 222.4 by 43.37 gives us a radius of 9,645.5 km.
Wow.
Even though we’ve traveled North of the arc LM and thus should be closer to the central point, the central point has somehow jumped away from us by 641.9 km!
Clearly our attempts to reconcile what we’ve measured with a flat Earth have completely failed, unless flat Earth proponents can come up with some mechanism by which the center of a circle can jump around inside that circle.
The reason that the arcs don’t increase and decrease at a uniform rate as we measure them, as they would on a flat Earth is because, on a round Earth, the radii of the concentric circles don’t emanate from a central point on a flat plane.
Instead they emanate from a central axis, which means they curve as they move down the axis.
Now, let’s finish up.
The coordinates indicate that we have actually traveled along an arc of 2°. Using the formulae from the beginning of this post to find the circumferences, we find the circumference for AD to be 33576.8 km; and for BC to be 32791.8 km. Multiplying both of these by 2/360, and we discover that the round Earth model correctly predicted that we should measure the length of AD to be 186.5; and BC to be 182.2.
So, having taken these measurements, we can easily determine that only an arc of angle 8.25°, traveled over an oblate spheroid, at the given distances from the North Pole, can account for the distances we measured between points L & M and S & I. And only an arc of 2°, also traveled over an oblate spheroid, can account for the second set of measurements and produce consistent measurements for our distances from the North Pole.
Indeed, we can conclude without doubt, that it is impossible for these measurements to exist on a flat Earth, proving conclusively that the Earth cannot be flat.
The results of this experiment get even more interesting if you live in the Southern Hemisphere, as flat Earth proponents don’t believe in the South Pole – so you need to use the North Pole as your central point still. As we’ve seen before, this does amazing things to the circumnavigational lengths.
Next time, we’ll have some fun with something that Fernieboy100 has difficulty believing is possible, and we’ll demonstrate a simple way to measure the size of – and distance to – far away objects, without using a tape measure, or even having to approach the objects themselves.
]]>