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http://vtmathcoalition.org/talent-search/oct-1999/index.html | 1,632,804,123,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060201.9/warc/CC-MAIN-20210928032425-20210928062425-00033.warc.gz | 68,566,365 | 2,606 | # VERMONT STATE MATHEMATICS COALITION TALENT SEARCH
### October 4, 1999
Test l of the 1999-2000 school year
Student Name _______________________ School ______________________
Directions: Solve as many as you can of the problems and list your solutions on this sheet of paper. On separate sheets, in an organized way, show how you solved the problems. You will be awarded full credit for a complete correct answer that is adequately supported by mathematical reasoning. You can receive half credit for correct answers that are the result of guesses, conjectures or incomplete solutions. Included as incomplete solutions are solutions that list some, but not all, solutions when the problem asks for solutions of equations. The decisions of the graders are final. You may earn bonus points for "commendable solutions"- solutions that display creativity, ingenuity and clarity. Your answers and solutions must be postmarked by November 1, 1999 and submitted to Tony Trono, Vermont St Mathematics Coalition, 419 Colchester Avenue, Burlington, VT 05401.
(For Coalition Information and test: http://www.vermontinstitutes.org/vsmc)
1. The buyer for Green Mountain Books, trying to anticipate demand, was planning to order 21 copies of a biography of Bill Clinton and 21 copies of a biography of Hillary Clinton. The store owner recommended an order of 28 copies of a biography of George W. Bush. The two of them then decided to purchase an equal number of copies of each of the three biographies. In all of the three cases, the books ordered would have cost the same thing. How many books were ordered?
2. The array of numbers contains 13 positive integers with four of them known and nine of them unknown. Each of the four known numbers is the sum of the four unknown numbers that form a square around it (for example, e + f + i + h = 83). Furthermore, e, g, b, f are each 19 greater than a, h, i, d respectively. Evaluate the sum a + e + f + g.Answer: ______________________
3. A cylinder with circular cross section has length 16 inches and circumference 3 inches. A string is wound around the cylinder four times with the ends of the string at the top and bottom of the cylinder and one end of the string directly under the other end. Find the length of the string. Answer: ______________________
4. The function f(x) satisfies the equation f(2 - x) = f(2 +x) for all real numbers x. The equation f(x) = O has four distinct real roots. Find the sum of these roots.
5. Find the smallest real number x satisfying the equation (log3 x)3 -log3(x3)=(log3x)2 - log3(x2). Round off your answer to two decimal places.
6. The slope of AB is m, where m>1. BCAB, CDBC, and DECD. If the length of AE is three times the length of AC, then find m.Answer:______
7. The roots of x2 +ax + b=0 are the squares of the roots of x2 + 2x +3=0. The roots of x2 +cx + d = 0 are the cubes of the roots of x2 +ax + b=0. Find the numerical value of c + d. | 712 | 2,939 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2021-39 | latest | en | 0.935673 |
http://engineering.myindialist.com/2009/engineering-mechanics-equilibrium-of-bodies/ | 1,556,009,034,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578596541.52/warc/CC-MAIN-20190423074936-20190423100936-00131.warc.gz | 61,132,399 | 11,030 | # Engineering Mechanics | Equilibrium of Bodies
Equilibrium of Bodies
In the static part when we say that a body is in equilibrium, what we mean is that the body is not moving at all even though there may be forces acting on it. (In general equilibrium means that there is no acceleration i.e., the body is moving with constant velocity but in this special case we take this constant to be zero).
Let us start by observing what all can a force do to a body? One obvious thing it does is to accelerate a body. So if we take a point particle P and apply a force on it, it will accelerate. Thus if we want its acceleration to be zero, the sum of all forces applied on it must vanish. This is the condition for equilibrium of a point particle. So for a point particle the equilibrium condition is
where are the forces applied on the point particle (see figure 13)
That is all there is to the equilibrium of a point particle. But in engineering problems we deal not with point particles but with extended objects. An example is a beam holding a load as shown in figure 2. The beam is equilibrium under its own weight W , the load L and the forces that the supports S1and S2 apply on it.
To consider equilibrium of such extended bodies, we need to see the other effects that a force produces on them. In these bodies, in addition to providing acceleration to the body, an applied force has two more effects. One it tends to rotate the body and two it deforms the body. Thus a beam put on two supports S1 and S2 tends to rotate clockwise about S2 when a force F is applied downwards (figure 3).
The strength or ability of a force to rotate the body about a point O is given by the torque generated by it. The torque is defined as the vector product of the displacement vector from O to the point where the force is applied. Thus
This is also known as the moment of the force. Thus in figure 3 above, the torque about S2 will be given by the distance from the support times the force and its direction will be into the plane of the paper. From the way that the torque is defined, the torque in a given direction tends to rotate the body on which it is applied in the plane perpendicular to the direction of the torque. Further, the direction of rotation is obtained by aligning the thumb of one’s right hand with the direction of the torque; the fingers then show the way that the body tends to rotate (see figure 4). Notice that the torque due to a force will vanish if the force is parallel to .
We now make a subtle point about the tendency of force to rotate a body. It is that even if the net force applied on a body is zero, the torque generated by them may vanish i.e. the forces will not give any acceleration to the body but would tend to rotate it. For example if we apply equal and opposite forces at two ends of a rod, as shown in figure 5, the net force is zero but the rod still has a tendency to rotate. So in considering equilibrium of bodies, we not only have to make sure that the net force is zero but can also that the net torque is also zero.
A third possibility of the action by a force, which we have ignored above, and which is highly explicit in the case of a mass on top of a spring, is that the force also deforms bodies. Thus in the case of a beam under a force, the beam may deform in various ways: it may get compressed, it may get elongated or may bend. A load on top of a spring obviously deforms it by a large amount. In the first case we assume the deformation to be small and therefore negligible i.e., we assume that the internal forces are so strong that they adjust so that there is no deformation by the applied external force. This is known as treating the body as a rigid body. In this course, we are going to assume that all bodies are rigid. So the third kind of action is not considered at all.
So now focus strictly on the equilibrium of rigid bodies: As stated, we are going to assume that internal forces are so great that the body does not deform. The only conditions for equilibrium in them are:
(1) The body should not accelerate/ should not move which, as discussed earlier, is ensured if that is the sum of all forces acting on it must be zero no matter at what points on the body they are applied. For example consider the beam in figure 2. Let the forces applied by the supports S1 and S2 be F1 and F2, respectively. Then for equilibrium, it is required that
Assuming the direction towards the top of the page to be y-direction, this translates to
The condition is sufficient to make sure that the net force on the rod is zero. But as we learned earlier, and also our everyday experience tells us that even a zero net force can give rise to a turning of the rod. So F1 and F2 must be applied at such points that the net torque on the beam is also zero. This is given below as the second rule for equilibrium.
(2) Summation of moment of forces about any point in the body is zero i.e. , where is the torque due to the force about point O. One may ask at this point whether should be taken about many different points or is it sufficient to take it about any one convenient point. The answer is that any one convenient point is sufficient because if condition (1) above is satisfied, i.e. net force on the body is zero then the torque as is independent of point about which it is taken. We will prove it later.
These two conditions are both necessary and sufficient condition for equilibrium. That is all we need to do to achieve equilibrium so in principle solving for equilibrium is quite easy and what we should learn is how to apply these condition efficiently in different engineering situations. We are therefore going to spend time on these topics individually. | 1,236 | 5,732 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2019-18 | latest | en | 0.949907 |
http://infohost.nmt.edu/~shipman/soft/homcoord/web/intro.html | 1,531,993,618,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590794.69/warc/CC-MAIN-20180719090301-20180719110301-00121.warc.gz | 170,966,554 | 2,617 | ## 1. Homogeneous coordinates make life so much easier
Much of graphics programming revolves around coordinate transformations. Objects in the modeled scene are specified by coordinates in scene space. In order to display these objects, we must transform their coordinates into display space.
The contribution of this document is an implementation of this method using the Python language, and the extremely useful NumPy extensions to Python that make vector and matrix operations easy.
The author has found this reference to be extremely valuable in graphics work:
Foley, James D., Andries van Dam, Steven K. Feiner, and John F. Hughes. Computer graphics: principles and practice. Second edition. Addison-Wesley, 1990, ISBN 0-201-12110-7.
In the chapter on geometrical transformations, the authors bring up homogeneous coordinates early on. This ingenous mathematical artifice makes it possible to specify any coordinate transformation using a small matrix: 3×3 for 2-space, 4×4 for 3-space. Points are carried as 3-element vectors for 2-space, or 4-element vectors for 3-space, and a trivial operation converts these vectors back to `(x,y)` or `(x,y,z)` form.
Moreover, you can combine these transform matrices to model the effect of multiple transforms. For example, suppose you have a model of some small rigid object like a cup, positioned in a larger scene. The coordinates of the cup are described in some standard orientation; when the cup is placed into the scene, a single transform matrix describes the transformation of those coordinates into coordinates relative to the scene. The net transformation of the cup to display coordinates can be represented as a matrix which happens to be the product of the two transform matrices: the transform that describes the cup's placement in the scene, and the transform that desribes how scene coordinates become display coordinates.
Transform matrices are easily reversible. For example, suppose the user clicks on the display, and you want to know where that click is in scene space. To convert a display coordinate to a scene coordinate requires a single matrix multiplication.
This document is structured in three general parts. | 435 | 2,192 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2018-30 | latest | en | 0.847799 |
https://www.khanacademy.org/math/ap-calculus-bc/bc-series-new/bc-10-14/v/function-as-a-geometric-series | 1,721,716,247,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518014.29/warc/CC-MAIN-20240723041947-20240723071947-00783.warc.gz | 725,447,386 | 142,401 | If you're seeing this message, it means we're having trouble loading external resources on our website.
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Main content
# Function as a geometric series
Expressions of the form a/(1-r) represent the infinite sum of a geometric series whose initial term is a and constant ratio is r, which is written as Σa(r)ⁿ. Since geometric series are a class of power series, we obtained the power series representation of a/(1-r) very quickly.
## Want to join the conversation?
• At , the sum of geometric sequence is a/(1-r), but this is only true for |r|<1, isn't it?
Applying this to f(x) seems like a bit of a stretch to me.
(34 votes)
• Yes, you are correct. This series converges when |-x³| < 1 ⇔ |x| < 1.
(12 votes)
• How can we say that Geometric series are Maclaurin series? or Maclaurin series are Power series? I think there's an information gap there. Can you explain to me?
(10 votes)
• I don't like asking these type of questions but What is the point of this ?
what the point of representing a nice rational function with a series that only works when x is between (-1,1)
is this trick used in some proofs or when solving some questions if so can someone give me an example
second question is what if we wanted to have a taylor series and expand it at x=3 for example what would I do ?
(8 votes)
• How is this a Maclaurin series? Doesn't the Maclaurin series have a factorial denominator in every term?
(4 votes)
• Good question. If you take the first, second and third derivative of f(u), you can then evaluate f f' f'' f''' at u=0. That will give you the first four coefficients of the Maclaurin Series. Next, put each of those coefficients to their corresponding terms of the M. series using u as your variable - corresponding terms meaning u^0/0! + u^1/1! + u^2/2! + ... . You can then substitute every u back to x^3. Then simplify. The process will take only a few minutes - using u in place of x^3 - and you will quickly see that it reduces to the same equation as above, in Sal's video.
(5 votes)
• How do you get rid of that autoplay button at the end..it doesn't let me rewind the video..so annoying..
(5 votes)
• You can push the counter clockwise button in the down left corner to rewatch the video.
(2 votes)
• When Sal makes this connection with the geometric series formula, how can we be sure it will be the Maclaurin Series? Is there only one unique Maclaurin Series for every function?
(2 votes)
• Yes, there is only one unique Maclaurin series for every function. Maclaurin series are always constructed around the function where x=0. To check that this is the Maclaurin series for the function, plug x=0 into any partial sum of the Maclaurin expansion, and you will find that it is equal to the exact function.
(2 votes)
• at around , how do we know a Maclaurin series is a power series?
(2 votes)
• Well, think about how a power series is defined. If I have a power series centred at c, it'd be the sum from n = 0 (or 1) to infinity of (a_n)(x-c)^n, where a_n is a real number.
See that this is exactly how a Maclaurin series looks like (just that the coefficients themselves need to be found out using a different formula). So, a Maclaurin (and in general, a Taylor) series is a kind of power series.
(2 votes)
• Why is the power series not over n!
Are geometric series a special circumstance that look different from a normal Maclaurin series?
(2 votes)
• The serie is p(x)=6 -6x^3 +6x^6......
the funtion is f(x)= 6/(1+x^3)
if x=1 f(1)=3 and p(1)=0 or 6
why f and p not equal in a infinite maclaurin's serie the error is zero?
(2 votes)
• I don't think what you did was allowed I mean isn't this an oscillating sequence that diverges.
I think that you are not allowed to make x=1 in p(x) because we have a constriction that the ratio in a geometric sequence must be less than 1 so we have |x^3| < 1 meaning that
-1<x<1
a lot of contradiction happen when you allow p(1) = 3
I mean notice that I can simply can add p(1) = 6-6+6-6... and it wouldn't change the seires
2p(1)=p(1)
implying that 2=1 which is absurd.
maybe someone who is an expert can shed some more light on this topic.
(1 vote)
• at around , f'(x)=-18x^2/(1+x^3)^2.
substitute x=0, we get f'(0)=0.
according to the formula of the Maclaurin series (f^n(0)*x^n/n!), we get the 2nd term of the Maclaurin series to be 0, which is different from -6x^3.
Can anyone point out where I did wrong? Thanks.
(1 vote)
• You're not wrong. The second term will be 0. However, if you go ahead and find the third derivative, you get -36. So, the third term becomes -36 (x)^(3)/3! which is -6x^3.
Essentially, see that each term of the power series Sal derived has only powers which are multiples of 3. So, even when finding the Maclaurin series with your traditional method, you'll only get a term every third time (You'll get a term on the 3rd, 6th, 9th... derivatives and every other derivative will be 0). So, just to get 5 terms like Sal did, you'll need to take 12 derivatives, which is definitely not something you'd want to do. Hence, he proposed this method.
(2 votes)
## Video transcript
- [Instructor] We're asked to find a power series for f, and they've given us f of x is equal to six over one plus x to the third power. Now, since they're letting us pick which power series, you might say, well, let me just find the Maclaurin series because the Maclaurin series tends to be the simplest to find 'cause we're centered at zero. And so you might immediately go out and say all right, well, let me evaluate this function at zero, evaluate its derivative at zero, its second derivative at zero, so on and so forth. And then I can use the formula for the Maclaurin series to just expand it out. But very quickly, you will run into roadblocks because the first, evaluating this at f, at x equals zero is pretty straightforward. Evaluating the first derivative is pretty straightforward. But then once you start taking the second and third derivatives, it gets very hairy, very fast. You could do a simplification, where you could say, well, let me find the Maclaurin series for f of u is equal to six over one plus u, where u is equal to x to the third. So you find this Maclaurin expansion, in terms of u, and then you substitute for x to the third. And actually, that makes it a good bit simpler, so that is another way to approach it. But the simplest way to approach it is to say, hey, you know what, this form right over here, this rational expression, it looks similar. It looks like the sum of a geometric series. Let's just remind ourselves what the sum of a geometric series looks like. If I have a plus a times r, so a is my first term, r is my common ratio, plus, I'm gonna multiply it times r again, plus a times r squared, plus a times r to the third power, and I keep going on, and on, and on forever. We know that this is going to be equal to a, our first term, over one minus our common ratio, and this just comes from a, the sum of a geometric series. And notice that what we have here, our f of x, our definition of f of x, and the sum of a geometric series look very, very similar. If we say that this, right over here, is a, so a is equal to six. And if negative r is equal to x to the third, or we could say, let me rewrite this. I could write this denominator as one minus negative x to the third. And so now, you could say, okay, well, r could be equal to negative x to the third. And just like that, we can expand it out. Well, if a is equal to six, and r is equal to negative x to the third, well, then we could just write this out as a geometric series, which is very straightforward. So let's do that. And I will do this in, I'll do this in this nice pink color. So the first term would be six, plus six times our common ratio, six times negative x to the third. And so actually, let me just write that as negative six x to the third, and then we're gonna multiply by negative x to the third again. So that's going to be, if I multiply this times negative x to the third, that's gonna be positive six times x to the sixth power. And then I'm gonna multiply it by times negative x to the third again, so it's gonna be minus six times x to the ninth power, and I'm gonna go on, and on, and on. So it's gonna be, and then I could keep going, I multiply it times negative x to the third. I will get six x to the twelfth power. Now, we can go on, and on, and on, and on forever. And so the key here was, and this is the Maclaurin series expansion for our f of x, but the key is to not have to go through all of this business and just to recognize that hey, the way this function was defined is it looks a lot like the sum of a geometric series. And it can be considered the sum of a geometric series, and we can use that to find the power series expansion for our function. This is a very, very, very useful trick. | 2,324 | 8,942 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-30 | latest | en | 0.94792 |
https://www.coursehero.com/file/203766/rev-test2/ | 1,490,874,709,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218193716.70/warc/CC-MAIN-20170322212953-00476-ip-10-233-31-227.ec2.internal.warc.gz | 870,458,958 | 21,615 | # Rev_test2 - Physics 1 East Los Angeles College Review Sheet TEST#2 Mr V Kiledjian 1 An object’s original position is 30 meters at an angle of 150
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Unformatted text preview: Physics 1, East Los Angeles College Review Sheet, TEST #2 Mr. V. Kiledjian 1) An object’s original position is 30 meters at an angle of 150 degrees and final position is 30 meters at 70 degrees. The object is traveling at constant speed in a circular arc clockwise and makes the journey in 12 seconds. a) Find the Displacement vector. b) Find the total Distance traveled. c) Find the avg Velocity vector, avg Speed, and instantaneous Speed. d) Find the avg Acceleration vector and instantaneous Acceleration vector. Show that the avg Acceleration vector points towards the center when placed at the midpoint of the arc!!! 2) A particle is traveling along the graph of the equation y = x 3 from (0,0) to (2,8) at a rate of dx/dt = 1 m/s. It makes this path in 2 seconds. Assume x and y are in meters. a) Find the total distance it travels and the magnitude and direction of its displacement. b) Find its average speed and the magnitude and direction of its average velocity. c) Find the instantaneous velocity vector and acceleration vector at t = 2 seconds and the instantaneous speed at t = 2 seconds. d) Find the magnitude and direction of its average acceleration 3) A fish in a river is originally at a location of 5i + 7j. Its original velocity is 2i + 3j and in 2 seconds, its velocity is –1i + 5j. velocity is –1i + 5j....
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## This note was uploaded on 06/27/2008 for the course PHYSICS 001 taught by Professor Kiledjian during the Spring '08 term at East Los Angeles College.
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Rev_test2 - Physics 1 East Los Angeles College Review Sheet TEST#2 Mr V Kiledjian 1 An object’s original position is 30 meters at an angle of 150
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Ask a homework question - tutors are online | 548 | 2,225 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2017-13 | longest | en | 0.883495 |
http://www.utdallas.edu/~nrr150130/gmbook/mrfs.html | 1,529,570,046,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864110.40/warc/CC-MAIN-20180621075105-20180621095105-00147.warc.gz | 507,637,008 | 5,759 | A Brief Introduction to Graphical Models
# Undirected Graphical Models
Similar to Bayesian networks, undirected graphical models also provide a visual representation of the structure of a joint probability distribution from which conditional independence relationships can be inferred. However, there are significant representational and computational differences between these models.
## Markov Random Fields (MRFs)
A Markov random field is an undirected graph $G = (V,E)$ together with a collection of potential functions defined over cliques of the graph $G$. Undirected graphs are the same as directed graphs except that the edges no longer have an associated direction. A clique in an undirected graph is a completely connected subgraph (i.e., there is an edge between every pair of vertices in the sub graph). See Figure 1 for an example of a clique.
Figure 1. An undirected graph $G$ with four vertices $A,B,C,$ and $D$. The subgraph over vertices $A, B,$ and $C$ is a clique.
As before, each vertex in a MRF corresponds to a random variable. Each clique, $C\subseteq V$, in the graph $G$ is associated with a nonnegative, real-valued potential function, $\psi_C$. The potential functions need not correspond to conditional probability distributions of any kind. A joint probability distribution, $p$, factorizes with respect to an undircted graph $G$ if there exist potential functions defined on the cliques of the graph $G$ such that $p$ can be expressed as follows.
$p(x_1,\ldots, x_{|V|}) = \frac{1}{Z} \prod_{C\in \mathcal{C}(G)} \psi_{C}(x_C)$
Here, $\mathcal{C}(G)$ is the set of all cliques in the graph $G$. The normalizing constant or partition function, $Z$, is chosen so that the product of potential functions defines a proper probability distribution.
$Z = \sum_{x_1,\ldots,x_{|V|}} \prod_{C\in \mathcal{C}(G)} \psi_{C}(x_C)$
If such a factorization exists, then the random variables are said to form a Markov random field with respect to the graph $G$. In general, we can restrict ourselves to models with only potential functions over the maximal cliques of $G$. A clique is maximal if it is not contained as a subgraph inside of a larger clique. As an example, in Figure 1, both sets $\{A,B\}$ and $\{A,B,C\}$ are cliques of the graph. However, as $\{A,B\} \subset \{A,B,C\}$, the clique $\{A,B\}$ is not a maximal clique. We can restrict our attention to maximal cliques because if we are given a factorization of a joint probability distribution that contains a potential function $\psi_{AB}$ and a potential function $\psi_{ABC}$, we could create a new potential function $\psi'_{ABC} \triangleq \psi_{ABC}\cdot\psi_{AB}$ and and remove the old potential functions yielding a new factorization of the joint probability distribution over only the maximal cliques.
Figure 2. An undirected graph $G$ with four vertices $A,B,C,$ and $D$. A joint distribution $p(x_A, x_B, x_C, x_D)$ factorizes over $G$ if there exist potential functions such that $p(x_A, x_B, x_C, x_D) = \frac{1}{Z} \psi_{AB}(x_A,x_B)\psi_{BC}(x_B,x_C)\psi_{CD}(x_C,x_D)$ where $Z$ is the normalization constant and the $\psi$'s are nonnegative real-valued functions.
To see how the MRF framework can be used to easily express certain types of probability distributions that would have been more challening to express in Bayesian networks, consider the problem of constructing a probability distribution over the independent sets of a given graph $H = (V_H, E_H)$. Recall that an independent set in $H$ is a subset of $V_H$ such that no two elements of the chosen subset are joined by an edge in $E_H$. Number the vertices of $H$ from $1$ to $|V_H|$. We will represent each possible subset of $V_H$ as a binary vector, $x$, whose $i^{th}$ component is equal to $1$ if the $i^{th}$ vertex of $H$ is selected to be in the subset and zero otherwise. We can express the uniform distribution over independent sets of $H$ as
$p(x_1,\ldots, x_{|V|}) = \frac{1}{Z} \prod_{(i,j)\in E_H} 1_{x_i + x_j \leq 1}.$
Here, $1_{x_i + x_j \leq 1}$ is an indicator function for the constraint that $x_i + x_j \leq 1$. That is, $1_{x_i + x_j \leq 1} = 1$ if $x_i + x_j \leq 1$ and zero otherwise. As the goal is to construct the uniform probability distribution over independent sets, the probability of any vector $x'$ whose corresponding subset is not an independent set should be zero. The potential functions are enforcing exactly this constraint: if two adjacent nodes in $H$, say $i$ and $j$, are both selected, i.e., $x'_i = x'_j = 1$, then $1_{x'_i + x'_j \leq 1} = 0$ and thus the whole joint probability will be zero for the vector $x'$. However, if $x'$ defines an independent set, then the joint probability will be $\frac{1}{Z}$. Finally, $Z = \sum_{x_1,\ldots,x_n\in\{0,1\}} \prod_{(i,j)\in E_H} 1_{x_i + x_j \leq 1}$ is simply the total number of independent sets in the graph. Which means that, with this choice of potential functions, $p$ is the uniform probabiltiy distribution over independent sets of $H$.
### Conditional Independence and Graph Separation
As in the case of Bayesian networks, independence relationships that are a consequence of the factorization can be inferred from the undirected graph itself. However, as there are no directions on the arrows, the separation conditions are somewhat easier to verify. Consider three sets of mutually disjoint sets $S_1,S_2,S_3\subseteq V$. The set of vertices $S_1 \subseteq V$ is graph separated from a set of vertices $S_2\subseteq V$ given a set $S_3\subseteq V$ if every path from a vertex $v_1\in S_1$ to a vertex $v_2\in S_2$ contains at least one vertex from the set $S_3$. Equivalently, $S_1$ and $S_2$ are graph separated given $S_3$ if removing the vertices in the set $S_3$ and all of their incident edges from the graph $G$ results in a new graph such that no connected component simultaneously contains a subset of $S_1$ and a subset of $S_2$.
Figure 3. An undirected graph on six vertices.
Is $D$ graph separated from $A$ given $B$ and $E$ in the graph in Figure 3?
Yes, as every path from $D$ to $A$ must pass through the neighbors of $A$, namely either $B$ or $E$.
Is $C$ graph separated from $A$ given $B$ and $D$ in the graph in Figure 3?
No, the path $C - E - A$ does not contain $B$ or $D$.
Theorem 1: For mutually disjoint sets $S_1,S_2,S_3\subseteq V$, if $S_1$ is graph separated from $S_2$ given $S_3$ in an undirected graph $G = (V,E)$, then for any probability distribution that factorizes over the maximal cliques of $G$, $X_{S_1} \perp X_{S_2} | X_{S_3}$.
To begin, we have that the joint distribution $p$ can be expressed as
$p(x_1,\ldots, x_{|V|}) = \frac{1}{Z} \prod_{C\in \mathcal{C}(G)} \psi_{C}(x_C).$
By the graph separation assumption, there can be no edges between vertices in the set $S_1$ and vertices in the set $S_2$. Therefore, no clique $C\in\mathcal{C}(G)$ can contain both vertices of $S_1$ and vertices of $S_2$. Now, let's divide the cliques into two sets: those cliques, $\mathcal{C}_1$, that contain at least one vertex of $S_1$ and those cliques, $\mathcal{C}_2$ that do not contain a vertex of $S_1$. With the above definition
$p(x_1,\ldots, x_{|V|}) = \frac{1}{Z} \left[\prod_{C\in \mathcal{C}_1} \psi_{C}(x_C)\right] \left[\prod_{C\in \mathcal{C}_2} \psi_{C}(x_C)\right].$
Again, by graph separation, if we remove the vertices in $S_3$ and their corresponding edges from the graph $G$ to form a new graph $G'$, the vertices in $V \setminus (S_1\cup S_2\cup S_3)$ must belong to connected components of $G'$ that contain either vertices of $S_1$ or vertices of $S_2$ but not both vertices of $S_1$ and $S_2$. Hence, we can partition the vertices of $V \setminus (S_1\cup S_2\cup S_3)$ into two sets: the set $T_1$ all of those vertices that are in a connected component of $G'$ with some vertex in $S_1$ and the set $T_2$ of those vertices that are not in such a connected component of $G'$. Note that if $C\in\mathcal{C}_1$, then $C$ can only contain vertices in $S_1$, $T_1$, and $S_3$ as $C$ must contain a vertex of $S_1$ by construction and, as a result, it cannot contain a vertex of $S_2$ or $T_2$. Similarly, if $C\in\mathcal{C}_2$, then $C$ can only contain vertices in $S_2$, $T_2$, and $S_3$. Using these observations, define
$f(x_{S_1}, x_{T_1}, x_{S_3}) \triangleq \prod_{C\in \mathcal{C}_1} \psi_{C}(x_C)$
and
$g(x_{S_2}, x_{T_2}, x_{S_3}) \triangleq \prod_{C\in \mathcal{C}_2} \psi_{C}(x_C).$
This allows us to write
$p(x_1,\ldots, x_{|V|}) = \frac{1}{Z} f(x_{S_1}, x_{T_1}, x_{S_3})\cdot g(x_{S_2}, x_{T_2}, x_{S_3}).$
Now,
\begin{align*} p(x_{S_1}, x_{S_2} | x_{S_3}) &= \frac{\sum_{x''_{T_1}, x''_{T_2}}\frac{1}{Z} f(x_{S_1}, x''_{T_1}, x_{S_3})\cdot g(x_{S_2}, x''_{T_2}, x_{S_3})}{\sum_{x'_{T_1}, x'_{T_2}, x'_{S_1}, x'_{S_2}}\frac{1}{Z} f(x'_{S_1}, x'_{T_1}, x_{S_3})\cdot g(x'_{S_2}, x'_{T_2}, x_{S_3})}\\ &= \left[\frac{\sum_{x''_{T_1}} f(x_{S_1}, x''_{T_1}, x_{S_3})}{\sum_{x'_{T_1}, x'_{S_1}}f(x'_{S_1}, x'_{T_1}, x_{S_3})}\right]\cdot\left[\frac{\sum_{x''_{T_2}}g(x_{S_2}, x''_{T_2}, x_{S_3})}{\sum_{x'_{T_2},x'_{S_2}}g(x'_{S_2}, x'_{T_2}, x_{S_3})}\right]. \end{align*}
Further,
\begin{align*} p(x_{S_1} | x_{S_3}) &= \frac{\sum_{x''_{T_1}, x''_{T_2}, x''_{S_2}}\frac{1}{Z} f(x_{S_1}, x''_{T_1}, x_{S_3})\cdot g(x''_{S_2}, x''_{T_2}, x_{S_3})}{\sum_{x'_{T_1}, x'_{T_2}, x'_{S_1}, x'_{S_2}}\frac{1}{Z} f(x'_{S_1}, x'_{T_1}, x_{S_3})\cdot g(x'_{S_2}, x'_{T_2}, x_{S_3})}\\ &= \left[\frac{\sum_{x''_{T_1}} f(x_{S_1}, x''_{T_1}, x_{S_3})}{\sum_{x'_{T_1}, x'_{S_1}}f(x'_{S_1}, x'_{T_1}, x_{S_3})}\right]\cdot\left[\frac{\sum_{x''_{T_2}, x''_{S_2}}g(x''_{S_2}, x''_{T_2}, x_{S_3})}{\sum_{x'_{T_2},x'_{S_2}}g(x'_{S_2}, x'_{T_2}, x_{S_3})}\right]\\ &= \frac{\sum_{x''_{T_1}} f(x_{S_1}, x''_{T_1}, x_{S_3})}{\sum_{x'_{T_1}, x'_{S_1}}f(x'_{S_1}, x'_{T_1}, x_{S_3})} \end{align*}
and
\begin{align*} p(x_{S_2} | x_{S_3}) &= \frac{\sum_{x''_{T_1}, x''_{T_2}, x''_{S_1}}\frac{1}{Z} f(x''_{S_1}, x''_{T_1}, x_{S_3})\cdot g(x_{S_2}, x''_{T_2}, x_{S_3})}{\sum_{x'_{T_1}, x'_{T_2}, x'_{S_1}, x'_{S_2}}\frac{1}{Z} f(x'_{S_1}, x'_{T_1}, x_{S_3})\cdot g(x'_{S_2}, x'_{T_2}, x_{S_3})}\\ &= \left[\frac{\sum_{x''_{T_1},x''_{S_1} } f(x''_{S_1}, x''_{T_1}, x_{S_3})}{\sum_{x'_{T_1}, x'_{S_1}}f(x'_{S_1}, x'_{T_1}, x_{S_3})}\right]\cdot\left[\frac{\sum_{x''_{T_2}}g(x_{S_2}, x''_{T_2}, x_{S_3})}{\sum_{x'_{T_2},x'_{S_2}}g(x'_{S_2}, x'_{T_2}, x_{S_3})}\right]\\ &= \frac{\sum_{x''_{T_2}}g(x_{S_2}, x''_{T_2}, x_{S_3})}{\sum_{x'_{T_2},x'_{S_2}}g(x'_{S_2}, x'_{T_2}, x_{S_3})}, \end{align*}
which demostrates that $p(x_{S_1}, x_{S_2} | x_{S_3}) = p(x_{S_1} | x_{S_3})\cdot p(x_{S_2} | x_{S_3})$. In other words, $X_{S_1} \perp X_{S_2} | X_{S_3}$.
As a simple example of the kinds of independence relationships that are implied by graph separation, consider the local Markov independence assumptions: for each $i\in V$ the random variable $x_i$ is independent of its nonneighbors given its neighbors in the graph $G$. A vertex $i\in V$ is a neighbor of a vertex $j\in V$ if there is an edge $(i,j)\in E$. | 3,901 | 11,036 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2018-26 | latest | en | 0.908524 |
http://www.math-problems-solved.com/q272_tq7x_pt5x3 | 1,511,184,382,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806066.5/warc/CC-MAIN-20171120130647-20171120150647-00499.warc.gz | 444,137,010 | 4,098 | Guide :
# PT=5x+3 and TQ=7x-9
for exceecises 20-22,use the figure below. Find the value of PT
## Research, Knowledge and Information :
### Find the value of PT PT = 5x + 3 and TQ 7x - 9 - Brainly.com
Find the value of PT PT = 5x + 3 and TQ 7x - 9 - 1847363
### SOLUTION: solve PT PT=5x+3 and TQ=7x-9
solve PT PT=5x+3 and TQ=7x-9-----PT = 5x + 3 ...
### PT=5x+3 and TQ=7x-9? - Weknowtheanswer - #1 Questions ...
PT= 5x+3 TQ=7x9 G Find PT. 8) ... 3 September 18, 2013 ... Read more. Positive: 15 %. Show more results recent questions recent answers. what has ...
### PT = 5x + 3 and TQ = 7X - 9 - OpenStudy
PT = 5x + 3 and TQ = 7X - 9what are you solving for?. ... Using the figure below to find the value of PT and graph has P-----T-----Q. anonymous
### If PT=5x+3 and TQ=7x-9, what is x? - weknowtheanswer.com
If T is the midpoint of PQ, PT = 5x + 3, TQ = 7x – 9, find x, PT, TQ, and PQ. Draw ... TQ = _____ PQ = _____ 10. If AE = 12 and AC = 4x – 36, find x ...
### PT=5x+3 and TQ=7x-9 plzz help and fastt - Brainly.com
PT=5x+3 and TQ=7x-9 plzz help and fastt - 1655814
### Help. PT=5x+3 and TQ=7x-9 Find the value of PT - jiskha.com
PT=5x+3 and TQ=7x-9 Find the value of PT. 94,967 results. Math ... Can someone help me to solve these questions please? 1) 3^x+1=9^x. Find the value of x. 2) ...
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• Use proper spelling and grammar | 622 | 1,717 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2017-47 | longest | en | 0.803309 |
http://mathhelpforum.com/calculus/76775-taylor-series-power-series.html | 1,519,440,601,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815034.13/warc/CC-MAIN-20180224013638-20180224033638-00743.warc.gz | 252,982,348 | 11,502 | # Thread: Taylor Series / Power Series
1. ## Taylor Series / Power Series
Alright, I'm just a little confused on what my work is asking me. The problem says "Use the definition to find the Taylor series, centered at c, for the function."
The problem is:
$\cos(x), c = \pi/4$
So I have an answer book that has the answer for this problem. I understand Taylor series / power series, but I'm just a bit confused on what is going on.
The answer book goes on to do a few derivatives, plugs in the c, etc. Then it forms the sum which is:
$\frac{\sqrt2}{2}\sum{\frac{(-1)^{n(n+1)/2}(x-\frac{\pi}{4})^n}{n!}}$
Now I understand this is in the form of:
$\sum{\frac{f^n(c)(x-c)^n}{n!}}$
Yet... Why does it not follow the "basic list" of power series? Which has:
$\cos(x) = \sum{\frac{(-1)^nx^{2n}}{(2n)!}}$
Why is there only n powers in the answer, and no 2n to be found, etc.? Can someone please explain? Maybe I am mixing two concepts...
2. Originally Posted by Fire Mage
Alright, I'm just a little confused on what my work is asking me. The problem says "Use the definition to find the Taylor series, centered at c, for the function."
The problem is:
$\cos(x), c = \pi/4$
So I have an answer book that has the answer for this problem. I understand Taylor series / power series, but I'm just a bit confused on what is going on.
The answer book goes on to do a few derivatives, plugs in the c, etc. Then it forms the sum which is:
$\frac{\sqrt2}{2}\sum{\frac{(-1)^{n(n+1)/2}(x-\frac{\pi}{4})^n}{n!}}$
Now I understand this is in the form of:
$\sum{\frac{f^n(c)(x-c)^n}{n!}}$
Yet... Why does it not follow the "basic list" of power series? Which has:
$\cos(x) = \sum{\frac{(-1)^nx^{2n}}{(2n)!}}$
Why is there only n powers in the answer, and no 2n to be found, etc.? Can someone please explain? Maybe I am mixing two concepts...
the one from the "basic list" is centered at c = 0 ... the odd powered terms cancel out because the odd degree derivatives are of the form $\pm \sin(0)$
3. Yes, I also understand that.
But what I'm trying to get at is: why does it become just n and not 2n? The only thing that changed it was the c. So why does changing the c automatically change the 2n's?
4. Originally Posted by Fire Mage
Yes, I also understand that.
But what I'm trying to get at is: why does it become just n and not 2n? The only thing that changed it was the c. So why does changing the c automatically change the 2n's?
Because when you center at c=0, you have every other term evaluated at sin 0=0. BUT in this case all the derivatives of cosine are either plus and minus cosine or sine. And these four functions evaluated at $\pi/4$ you'll never get zero. Instead you'll get plus and minus $\sqrt{2}/2$.
5. Originally Posted by matheagle
Because when you center at c=0, you have every other term evaluated at sin 0=0. BUT in this case all the derivatives of cosine are either plus and minus cosine or sine. And these four functions evaluated at $\pi/4$ you'll never get zero. Instead you'll get plus and minus $\sqrt{2}/2$.
Ah thank you for explaining it out. Like I comprehended the problem, understood everything, I can work the problem myself... I just didn't understand WHY.
But now that I read it, it's like a slap in the face. I totally ignored that fact. Thanks.
6. Sorry that I had to get physical with you.
7. Haha, of course. | 921 | 3,359 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2018-09 | longest | en | 0.933073 |
https://www.gradesaver.com/textbooks/math/algebra/college-algebra-6th-edition/chapter-4-exponential-and-logarithmic-functions-exercise-set-4-4-page-490/100 | 1,560,719,325,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998298.91/warc/CC-MAIN-20190616202813-20190616224813-00118.warc.gz | 769,657,788 | 13,083 | ## College Algebra (6th Edition)
Since $x \ne \frac{-5-\sqrt {37}}{2}$, then $x = \frac{-5+\sqrt {37}}{2}$
$\ln 3 - \ln(x+5) - \ln x = 0$ $\ln \frac{3}{x+5} = \ln x$ $\frac{3}{x+5} = x$ $3 = x(x+5)$ $3 = x^{2} + 5x$ $x^{2} + 5x - 3 = 0$ $x = \frac{-b±\sqrt {b^{2}-4ac}}{2a}$ $x = \frac{-(5)±\sqrt {5^{2}-4(1)(-3)}}{2(1)}$ $x = \frac{-5±\sqrt {25+12}}{2}$ $x = \frac{-5±\sqrt {37}}{2}$ Since $x \ne \frac{-5-\sqrt {37}}{2}$, then $x = \frac{-5+\sqrt {37}}{2}$ | 247 | 459 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2019-26 | latest | en | 0.307922 |
https://fr.mathworks.com/matlabcentral/answers/493549-how-can-i-change-zeros-matrix-to-1-by-specific-points | 1,579,699,753,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250607118.51/warc/CC-MAIN-20200122131612-20200122160612-00111.warc.gz | 452,659,434 | 20,362 | # How can I change zeros matrix to 1 by specific points??
7 views (last 30 days)
Mohammed Alammar on 27 Nov 2019
Commented: Mohammed Alammar on 28 Nov 2019
I have 8X8 Zeros matrix :
A = zeros(8, 8);
I need to create a multiple rectangles of Ones matrix. when I have specific points location:
first rectangular startfrom point [1,1] which are (X0, Y0) to [3,3] which are (X1, Y2).
and
second rectangular start from point [5,5] which are (X0, Y0) to [8,8] which are (X1, Y2).
I have try this but does not work with me:
%%first rectangular
A(([1,1]:[3,1]):([3,1]:[3,3]))= 1
%%second rectangular
A(([5,5]:[8,5]):([8,5]:[8,8]))= 1
Adam Danz on 27 Nov 2019
Edited: Adam Danz on 27 Nov 2019
Here's a demo. When [x0,y0] is [4,3] and [x1,y1] is [6,7], a rectangle of 1 will be placed in matrix A in rows 3:7 from the top and columns 4:6 from the left.
% Create 0s matrix
A = zeros(8, 8);
% Define [x0,y0] and [x1,y1]
x0 = 4;
y0 = 3;
x1 = 6; % x1 >= x0
y1 = 7; % y1 >= y0
% replace section with 1s
[y,x] = meshgrid(x0:x1, y0:y1);
idx = sub2ind(size(A),x,y);
A(idx) = 1
Result
A =
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 1 1 1 0 0
0 0 0 1 1 1 0 0
0 0 0 1 1 1 0 0
0 0 0 1 1 1 0 0
0 0 0 1 1 1 0 0
0 0 0 0 0 0 0 0
If you'd rather use logicals, replace zeros() with false() and replace A(idx) = 1 with A(idx) = true.
Andrei Bobrov on 27 Nov 2019
A = zeros(8, 8);
A(1:3,1:3) = 1;
A(5:8,5:8) = 1;
#### 1 Comment
Mohammed Alammar on 28 Nov 2019
thank you i have a spicific point which is etract from other code. | 665 | 1,496 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2020-05 | latest | en | 0.851987 |
https://gomathanswerkey.com/texas-go-math-grade-5-lesson-10-3-answer-key/ | 1,716,994,239,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059246.67/warc/CC-MAIN-20240529134803-20240529164803-00031.warc.gz | 238,959,247 | 56,759 | # Texas Go Math Grade 5 Lesson 10.3 Answer Key Number Patterns in Graphs
Refer to our Texas Go Math Grade 5 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 5 Lesson 10.3 Answer Key Number Patterns in Graphs.
## Texas Go Math Grade 5 Lesson 10.3 Answer Key Number Patterns in Graphs
Unlock the Problem
Sona has some dimes. Adil offers to give her nickels for her dimes.
The graph shows the relationship between the number of dimes Sona gives to
Adil and the number of nickels she receives in exchange. Find a rule to describe the pattern.
STEP 1: Write number pairs that relate the number of dimes to the number of nickels
STEP 2 Describe the relationship between the number of dimes and the number of nickels.
The output is __________ times the input. The pattern uses multiplication.
STEP 3 Write the rule. Let n stand for number of nickels, and d stand for the number of dimes.
n = __________ × ___________
Think: The relationship is multiplicative. Use multiplication to write the rule.
Step 1:
Pairs that relate between Sona and Adil the number of dimes to the
number of nickels are (1,2), (2,4), (3,6), (4,8),
Step 2:
The output is 2 times the input,
Step 3:
Rule: n = d X 2,
Explanation:
Step 1:
As the number of dimes is the input, the horizontal distance from
0 for the first point is 1 and the number of nickels is output the vertical distance
for the first point is 2, So the pairs that relate between Sona and Adil
the number of dimes to the number of nickels are (1,2), (2,4), (3,6), (4,8),
Step 2:
By seeing the pairs the relationship between the number of dimes is
twice the number of nickels. so the output is two times the input.
The pattern uses multiplication,
Step 3:
Let n stand for number of nickels, and d stand for the number of dimes.
as the relationship is multiplicative. Using multiplication to write the rule as
n = d X 2.
Math Idea
An additive pattern uses addition ¡n its rule. A multiplicative pattern uses multiplication in its rule.
Example
Find the rule for the pattern shown in the graph.
Jack mails some magazines in a box. The graph shows the relationship between
the number of magazines and the total weight of the magazines and the box.
STEP 1: Write number pairs that relate the number of magazines and the total weight.
(1, ______)
(2, ______)
(3, ______)
(4, ______)
STEP 2: Describe the relationship between the number of magazines and the weight.
The output is the sum of the number of magazines and __________.
STEP 3: Write the rule. Let w stand for total weight, and m stand for the number of magazines.
w = __________ + __________
Pairs are (1,3), (2,4), (3,5), (4,6),
The output total weight is the sum of the number of magazines and 2,
rule is w = m + 2,
Explanation:
Step 1: Number pairs that relate the number of magazines and the total weight is
(1,3), (2,4), (3,5), (4,6),
Step 2: By seeing the graph The output is the sum of the number of magazines and 2
the relationship between the number of magazines and the weight.
The output is the sum of the number of magazines and 2 by using
addition pattern and let w stand for total weight, and
m stand for the number of magazines the rule is w = m + 2.
Math Talk
Mathematical Processes
Explain what the 2 stands for in w = m + 2.
2 stands for adding 2 more plus number of magazines we get the weight,
Explanation:
As w = m + 2 here 2 stands we are adding 2 more to the
number of magazines we will get the weight.
Share and Show
Is the pattern additive or multiplicative? Find the rule for the pattern.
Question 1.
Pattern is multiplicative,
Rule : t = s X 3,
Explanation:
Step 1: Number pairs that relate input s and output t we see
(1,3),(2,6), (3,6),
Step 2:Â By seeing the graph the relationship between the input and output is
the output t is multiplication of 3 times with input s by using
multiplication pattern rule is t = s X 3.
Rule : t = s + 3,
Explanation:
Step 1: Number pairs that relate input s and output t we see
(1,4),(2,5), (3,6),
Step 2:Â By seeing the graph the relationship between the input and output is
the output t is addition of 3 more with input s by using
addition pattern rule is t = s + 3.
Problem Solving
Question 3.
H.O.T. Write Math Explain the differences between the graphs in Exercises 1 and 2.
One is multiplicative pattern and other is additive pattern,
Explanation:
The main difference between the graphs in Exercise 1 and 2 is
the first 1 is multiplicative pattern with steeper(larger) slope and linear and
seeing relations between input and output,
In the second graph it is additive pattern with linear and deeper(closer) slope
and seeing relations between input and output.
Question 4.
Write Math Explain how you can tell whether a pattern shown in a graph is additive or multiplicative.
By seeing the graph and points
origin, slope is deeper or steeper, linear
and relations between input and outputs,
Explanation:
In additive graph it will not start from origin and slope
will be deeper(closer) and it will be linear and we see the
relationship between input and output,
In multiplicative graph it can will start from origin and slope
will be steeper(increases) and it will be linear and we see the
relationship between input and output.
Problem Solving
Use the graph for 5-6.
H.O.T. Multi-Step Maryanne’s map uses a scale to show how many miles each inch on the map represents.
Use the pattern shown in the graph to find the rule for calculating the actual distance.
Use in for number of miles and j for number of inches.
Rule in = j X 10,
Explanation:
Given Maryanne’s map uses a scale to show how many miles each inch on the map represents.
Step 1: Number pairs that relate input and output is
(1,0),(2,20), (3,30),(4,40),
Step 2:Â By seeing the graph the relationship between
in for number of miles and j for number of inches on map rule is
the number of miles is multiplication of 10 times with number of inches on the map
using multiplication pattern rule is (actual distance) in = j X 10
Used the pattern shown in the graph to find the rule for calculating the actual distance.
Question 6.
H.O.T. If the distance on the map is 2.5 inches, how many miles is the actual distance?
Explain how you calculated the actual distance.
The actual distance is 25 miles,
Explanation:
Given the distance on the map is 2.5inches to calculate
the actual distance by using the graph and rule of question 5 we have
(actual distance) in = j X 10(where j is number of inches on the map)
substituting j = 2.5 we get in = 2.5 X 10 = 25 miles the actual distance.
Question 7.
H.O.T. Apply Olivia uses 1 red button for the nose and 2 blue buttons
for the eyes for each rag doll she makes.
Would the pattern comparing the number of nose buttons to
the total number of buttons be multiplicative or additive? Explain.
The number of nose buttons to total number of buttons
will be multiplicative,
Explanation:
Given Olivia uses 1 red button for the nose and 2 blue buttons
for the eyes for each rag doll she makes. If n is the total number of buttons
let r is red button for the nose and b for blue button we have
1 rag doll = 3 buttons = 1 red button + 2 blue buttons
means the total number of buttons is
depending upon number of red buttons and also on
blue buttons which means it is multiplicative
if 2 rag dolls = 3 x 2 = 6 buttons = (2 X 1r + 2 X 2b) = 2 + 4 = 6.
Fill in the bubble completely to show your answer. Use the graph.
The graph shows the number of pounds of fish the seals at an aquarium eat each day.
Which statement and equation about the relationship is true?
(A) additive pattern, d = p + 3
(B) addition pattern, p = d + 3
(C) multiplicative pattern, p = 10 × d
(D) multiplicative pattern, d = 3 × p
(C) multiplicative pattern, p = 10 × d is true,
Explanation:
Given the graph shows the number of pounds of fish the seals
at an aquarium eat each day the pattern is if
p is the number of pounds of fish the seals and d is number of days
Step 1: Number pairs that relate input d and output p we see
(1,10),(2,20), (3,30),(4,40),
Step 2:Â By seeing the graph the relationship between the input and output is
multiplicative of 10 times with input d is output p by using
multiplicative pattern rule is p = 10 X d which exactly matches with (C).
Use the graph for 9-10.
Question 9.
The graph shows the relationship between the number of days and
the amount of food that a bottlenose dolphin eats.
Which point on the graph shows the amount of food that the dolphin consumes in two days?
(A) point A
(B) point B
(C) point C
(D) point D
(C) point C,
Explanation:
Given the graph shows the relationship between the number of days and
the amount of food that a bottlenose dolphin eats. If we see points on the
graph Point A is day 0, Point B is day 1, Point C is day 2 and Point D is day 3,
So point on the graph that shows the amount of food that the
dolphin consumes in two days is Point C matches with (C).
Question 10.
Multi-Step If s is the amount of food a bottlenose dolphin cats in a
day and n is the number of days, which of the following describes the
rule for the pattern shown in the graph?
(A) s = n + 23
(B) s = n × 23
(C) n = s × 23
(D) n = s + 23
(A) s = n + 23,
Explanation:
If s is the amount of food a bottlenose dolphin cats in a
day and n is the number of days, Number pairs that relate
input n and output s we see
(1,23),(2,46), (3,69), the following describes the
rule for the pattern shown in the graph the relationship between the
input number of days n and output s is the amount of food a
bottlenose dolphin cats in a day is additive of 23 more with input n
is output s by using additive pattern rule is s = n + 23
which exactly matches with (A).
Texas Test Prep
Which of the following rules describes the pattern in the graph?
(A) a = 4b
(B) b = 4a
(C) b = 4 + a
(D) a = 4 + b
(C) b = 4 + a,
Explanation:
Number pairs that relate input a and output b we see
(1,5), (2,6), (3,7), (4, 8) the following describes the
rule for the pattern shown in the graph the relationship between the
input a and output b is additive of 4 more with input a is output b by using
additive pattern rule we have b = 4 + a which exactly matches with (C).
### Texas Go Math Grade 5 Lesson 10.3 Homework and Practice Answer Key
Is the pattern additive or multiplicative? Write the rule for the pattern.
Question 1.
Rule is b = a + 4,
Explanation:
Number pairs that relate input a and output b we see
(1,5), (2,6), (3,7) the following describes the
rule for the pattern shown in the graph the relationship between the
input a and output b are additive of 4 more with input a is output b by using
additive pattern rule is b = a + 4.
Go Math 5th Grade Practice and Homework Lesson 10.3 Answer Key Question 2.
Pattern is Multiplicative,
Rule is d = c XÂ 4,
Explanation:
Number pairs that relate input c and output d we see
(1,4), (2,8), (3,12) the following describes the
rule for the pattern shown in the graph the relationship between the
input c and output d is multiplicative of 4 more with input c is output d
by using multiplicative pattern rule is d = c X 4.
Question 3.
Pattern is Multiplicative,
Rule is f = e XÂ 5,
Explanation:
Number pairs that relate input e and output f we see
(1,5), (2,10), (3,15) the following describes the
rule for the pattern shown in the graph the relationship between the
input e and output f is multiplicative of 5 more with input e is output f
by using multiplicative pattern rule is f = e X 5.
Question 4.
Rule is h = g + 5,
Explanation:
Number pairs that relate input a and output b we see
(1,6), (2,7), (3,8) the following describes the
rule for the pattern shown in the graph the relationship between the
input g and output h is additive of 5 more with input g is output h
by using additive pattern rule is h = g + 5.
Problem Solving
Patrice rents a rowboat for several hours. Use the pattern shown
in the graph to write the rule for calculating the cost of renting the rowboat.
Use c for the cost and h for the number of hours.
The rule for calculating the cost of renting the rowboat is
c = h X $15, where c for the cost and h for the number of hours, Explanation: Given Patrice rents a rowboat for several hours the pattern shown in the graph if c is for the cost and h for the number of hours is Number pairs that relate hours h and cost c we see (1,15), (2,30), (3,45) the following describes the rule for the pattern shown in the graph the relationship between the hours h and costs c is multiplicative of 15 times by using multiplicative pattern rule is c = h X$15.
Question 6.
If Patrice rents the boat for 4 hours, what is the cost of the rental?
The cost of the rental is $60, Explanation: we have the rule for calculating the cost of renting the rowboat is c = h X$15, where c for the cost and h for the number of hours,
so if Patrice rents the boat for 4 hours, the cost of the rental is
c= 4 X $15 =$60.
Lesson Check
Question 7.
The graph shows the relationship between the number of hours
Ivana bikes and the distance she travels. How many miles does Ivana bike in 4 hours?
(A) 28 miles
(B) 12 miles
(C) 32 miles
(D) 30 miles
(C) 32 miles,
Explanation:
Given the graph shows the relationship between the number of hours
Ivana bikes and the distance she travels
Step 1: Number pairs that relate hours h and number of miles m we see
(1,8),(2,16), (3,24),
Step 2:Â By seeing the graph the relationship between the input and output is
multiplicative of 8 times with input hours h is output number of miles m by using
multiplicative pattern rule is m = 8 X h so number of miles does
Ivana bike in 4 hours is 8 X 4 = 32 miles which matches with (C).
Rashid uses cubes to build towers. The graph shows the relationship between
the number of towers, t, and the number of cubes, c, he uses.
Which equation describes this relationship?
(A) c = 12 + t
(B) c = 12t
(C) t = 12 + c
(D) t = 12c
(B) c = 12t,
Explanation:
Given Rashid uses cubes to build towers.
The graph shows the relationship between
the number of towers t and the number of cubes c,
Step 1: Number pairs that relate the number of towers t and the number of cubes c,
we see (1,12),(2,24), (3,36),
Step 2:Â By seeing the graph the relationship between the input and output is
multiplicative of 12 times with input the number of towers t is
output by using the number of cubes c multiplicative pattern rule is
c = 12 X t therefore equation describes this relationship is (B).
Question 9.
Multi-Step For every $1 Paula donates to her favorite charity, her mom donates$3. Which of the following identifies the pattern and
the amount her mom donates if Paula donates $6? (A)$18, multiplicative
(B) $24, multiplicative (C)$24, additive
(D) $12, additive Answer: (A)$18, multiplicative,
Explanation:
Given for every $1 Paula donates to her favorite charity, her mom donates$3, The graph shows the relationship between
Paula donation as p and mom donation as m
Step 1: Number pairs that relate the Paula donation as p and mom donation m
we see (1,3),(2,6), (3,9),
Step 2:Â By seeing the graph the relationship between the input and output is
multiplicative of 3 times with input Paula donation p is
output mom donation as m by using multiplicative pattern rule is
m = 3 X p therefore the amount her mom donates if Paula donates $6 is 3 X$6 = $18 matches with (A)$18, multiplicative.
Number Patterns Answer Key Go Math Grade 5 Lesson 10.3 Question 10.
Multi-Step The graph shows how much time Christina spends making
flower arrangements to sell. How many baskets can she make in two hours?
(A) 15
(B) 4
(C) 30
(D) 8
(D) 8
Explanation:
The graph shows how much time Christina spends making
flower arrangements to sell. Number of baskets can she make in two hours
as we know 2 hours means 2 X 60 minutes = 120 minutes,
Step 1: Number pairs that relate the Christina as b means baskets and
minutes as m we see as (1,15),(2,30), (3,45),(4,60),
Step 2:Â By seeing the graph the relationship between the input and output is
multiplicative of 15 times with input baskets b is
output minutes as m by using the multiplicative pattern rule is
m = b X 15, so for 120 minutes means 120/15 = 8 baskets,
which matches with (D).
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Scroll to Top | 4,339 | 16,303 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.875 | 5 | CC-MAIN-2024-22 | latest | en | 0.892447 |
null | null | null | null | null | null | 25 August 2011
I'm currently tweaking a new template for this thing. The third-party layout I've had from the beginning has always had its share of slightly annoying formatting bugs; but recently they've been leading to increasing glitches, making the drafting and posting biz an aggravating -- and sometimes abortive -- process. Hopefully this will knock out some of those problems.
Also, I've never been sure of this three-column layout. Too crowded? Can't decide for sure.
So if you check back over the next several days and see things changing, or even see a few things going wonky, that's why. I expect it'll take some time and some adjustments until I have things to my satisfaction (or merely acceptable). It'll also involve rebuilding the various links lists, and I also anticipate that some older posts will probably go a bit askew due to the format shift. So I'll be fixing those, as well.
W. Kasper said...
The 3-column layout is very clear and readable IMO. This has a really good sense of space (pace?) and clarity already. However, I've been thinking of changing mine and something came to mind: Does white text on black background make the content seem more 'aggressive' (or rather 'moody') - and is the opposite true if it's black text on white background? I have noticed I'm less drawn to certain blogs due to layout rather than content (tiny or coloured text, patterned backgrounds, too many gadgets are just some 'turn-offs').
Superficial, I know - but the whole 'click-on/click-off' aspect of the internet does encourage one to bypass measured judgements.
Greyhoos said...
Thanks for the input, Wayne. I figure feedback from readers is useful in this respect.
Glitches aside, one concern about the three-column layout is that the flanking columns tend to "crush" the content -- espec in regards to imposing limitations on sizing images. I've never been fully satisfied w/ the overall visual 'flow' of the thing.
As far as what you're talking about in terms of text: There are Design 101 principles about such stuff. Traditionally, certain combination and offsets are considered better for 'text-heavy' content and extended reading. With white text on black or black on red, etc. -- things like that tend to tax the eyes and are only good for bold & short amounts of text.
However, these are rules that originated with print -- meaning that they don't necessarily apply to a computer screen. (Especially once you consider how the screen has a 'lightbox' effect on content -- particularly w/ text and images.)
But some things are still a definite no all around -- too many font changes, changes in font color, too much encroaching clutter. All of that just fatigues the reader's retinas & is too be avoided (or used very sparingly). And patterned backgrounds? Ugch. That's one of the biggest mistakes of all.
W. Kasper said...
I think narrower 'body' columns like yours (as opposed to mine) 'flow' better. Like free mags etc. sometimes have one phat left to right column, and my eyes glaze over to early to read the whole thing.
There was a lot of 'patterned' sites about 10 years ago, with day-glo club flyer font eyeball bounce craziness (too many design alumni of nightclub scenes I reckon - had a few hassles with that 'kind' when doing zines etc. "Please bear in mind we would like people to read the fucking thing!"). Needless to say, most sites like that didn't last.
Verification: "gripeasses" LOL
Greyhoos said...
LOL, indeed. That's excellent.
Yeah...that's what I was talking about. That sort f thing's good for adverts, flyers, attention-grabbers of that sort. But not desirable for other things.
Huh. I was going to load the new template today and start tweaking. And I believe the layout I'd chosen (see above) might be awfully close to the sort you're complaining about. I may have to hold off, run a couple of screengrabs past a few friends, and get a vote on it.
Not entirely sure why I'm bothering, honestly. I don't have the time or the means to build one from scratch (plus doing it that way often introduces even more bugs, once you introduce it to blogger's clunky platform). So I'm going about it as I did before: Getting a third-party template and doing lots of deletions & adjustments. But in the end, I'll probably be as equally "meh" about it as I am with this one. Shrug.
W. Kasper said...
Nah that template looks fine. As you say, the plus is more image room. Although it does bug me how clunky blogger is (is wordpress better?). If you got a slow connection like mine it takes forever to do the most minor tweaks and some features just fail to show up sometimes (like when it's raining heavily!).
W. Kasper said...
Hey - the blog got 'girthy'! I had to turn my head to look at the main text. I don't like that kind of hard work looking at the interweb.
Greyhoos said...
Turn your head? Next time try moving your chair, instead.
W. Kasper said...
I only ever get out of bed to piss!
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https://www.mathworks.com/matlabcentral/cody/problems/1072-television-screen-dimensions/solutions/172649 | 1,496,025,657,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612008.48/warc/CC-MAIN-20170529014619-20170529034619-00170.warc.gz | 1,144,690,505 | 12,217 | Cody
# Problem 1072. Television Screen Dimensions
Solution 172649
Submitted on 5 Dec 2012 by Prateep Mukherjee
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% [W,H]=teledims(2.4,1,46); assert(H == 17.7 && W == 42.5)
``` ```
2 Pass
%% [W,H]=teledims(4,3,32); assert(H == 19.2 && W == 25.6)
``` ```
3 Pass
%% [W,H]=teledims(3,2,60); assert(H == 33.3 && W == 49.9)
``` ```
4 Pass
%% [W,H]=teledims(1.85,1,42); assert(H == 20.0 && W == 36.9)
``` ```
5 Pass
%% [W,H]=teledims(3,2,1000); assert(H == 554.7 && W == 832.1)
``` ```
6 Pass
%% [W,H]=teledims(16,10,92); assert(H == 48.8 && W == 78.0)
``` ```
7 Pass
%% [W,H]=teledims(2.4,1,92); assert(H == 35.4 && W == 84.9)
``` ```
8 Pass
%% [W,H]=teledims(16,10,32); assert(H == 17.0 && W == 27.1)
``` ```
9 Pass
%% [W,H]=teledims(3,2,82); assert(H == 45.5 && W == 68.2)
``` ```
10 Pass
%% [W,H]=teledims(16,10,82); assert(H == 43.5 && W == 69.5)
``` ```
11 Pass
%% [W,H]=teledims(16,10,60); assert(H == 31.8 && W == 50.9)
``` ```
12 Pass
%% [W,H]=teledims(3,2,92); assert(H == 51.0 && W == 76.5)
``` ```
13 Pass
%% [W,H]=teledims(1.85,1,1000); assert(H == 475.5 && W == 879.7)
``` ```
14 Pass
%% [W,H]=teledims(2.4,1,42); assert(H == 16.2 && W == 38.8)
``` ```
15 Pass
%% [W,H]=teledims(1.85,1,27); assert(H == 12.8 && W == 23.8)
``` ```
16 Pass
%% [W,H]=teledims(16,10,82); assert(H == 43.5 && W == 69.5)
``` ```
17 Pass
%% [W,H]=teledims(4,3,46); assert(H == 27.6 && W == 36.8)
``` ```
18 Pass
%% [W,H]=teledims(2.4,1,60); assert(H == 23.1 && W == 55.4)
``` ```
19 Pass
%% [W,H]=teledims(4,3,92); assert(H == 55.2 && W == 73.6)
``` ```
20 Pass
%% [W,H]=teledims(1.85,1,60); assert(H == 28.5 && W == 52.8)
``` ``` | 826 | 1,860 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2017-22 | longest | en | 0.369018 |
https://math.stackexchange.com/questions/3772588/why-isnt-the-disjoint-union-in-set-a-product-in-addition-to-being-a-coproduct/3772599 | 1,713,447,972,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817206.54/warc/CC-MAIN-20240418124808-20240418154808-00643.warc.gz | 350,348,987 | 37,411 | # Why isn't the Disjoint Union in Set a *product* in addition to being a coproduct?
So I'm understanding that in $$Set$$ that the cartesian product is a categorical product, and further get why the disjoint union is a categorical coproduct, but why is it also not a product? I want to understand where I'm going wrong in my reasoning for why it seems like it could be a product.
If I define disjoint union to be
$$A \sqcup B := \{ a \in A\ | \ (0, a) \} \cup \{ b \in B\ | \ (1, b) \}$$
then I can define a projection $$p: A \sqcup B \to A$$, as $$p := \{ (i, x) \in A \sqcup B, i =0\ |\ x \}$$ and respectively $$q: A \sqcup B \to B$$ by setting $$i = 1$$ above.
Now with that, pick an arbitrary object $$V$$ with morphisms $$f: V \to A, g: V \to B$$, then there exists a morphism $$V \to A \sqcup B$$, defined as $$h := \{ v \in V \ | \ (0, f(v) \} \cup \{ v \in V \ | \ (1, g(v) \}$$
Can you explain where my reasoning falls apart? If I were to take a guess, it's that my definition of disjoint union is improper, because the indexes can be arbitrary, hence there cannot be cannonical projection functions from the disjoint union to its parts, because the indexes separating the parts are neither known, nor do they tell us if they correspond to $$A$$ or $$B$$.
Does that hunch sound correct? Even in that, my instinct is to ask why we couldn't construct a definition of disjoint union where you always know how to access elements of any of its objects in a canonical way. That is, where the index always starts at zero, and for each successive object between a disjoint union of two or more sets, the next index is defined to be the successor to the largest existing index in the disjoint union.
• I don't nderstant your notation. is $$p := \{ (i, x) \in A \sqcup B, i =0\ |\ x \}$$ a function from $A \sqcup B$ to $A$? Jul 28, 2020 at 19:40
• Yes, I am perhaps misusing notation here, if you have a suggestion for how that might be rewritten. I am attempting to define a function here that takes the entire disjoint union as input. Is this one area where I'm perhaps going astray? That $p$ does not make sense as a function, because it doesn't have a proper element-wise definition? For example, p(1, b) = ???, it wouldn't really work; it's perhaps a partial function then? Jul 28, 2020 at 19:42
• If my definition for $p$ is incorrect, could I define p(1, b) to be equal to any arbitrary element of A? Send all elements that aren't of the form (0, a) to an arbitrary a? That would require that disjoint unions only be constructed from non-empty sets however (which does already have an analog in the definition of intersection, but perhaps it's unnecessarily restrictive for the construction of disjoint union). Jul 28, 2020 at 19:49
## 1 Answer
Your problem is with the definition of $$p$$, which doesn't really make sense. You could define it as a partial function sending $$a\in A$$ to itself and undefined on $$B$$, but by definition it needs to be defined on all of $$A\sqcup B$$. You could define it arbitrarily on $$B$$, but that just wouldn't satisfy the requirements for a product. Say you have a set $$V$$ and two functions $$f:V\to A,g:V\to B$$, there doesn't necessarily exist a function $$H :V\to A\sqcup B$$ such that its composition with $$p$$ and $$q$$ are $$f$$ and $$g$$.
EDIT. Here's an example. $$A=\{1,2,3\},B=\{4,5,6\}$$, $$V=\{0\}$$. There are $$3$$ maps from $$V$$ to $$A$$ and $$3$$ from $$V$$ to $$B$$, so, by definition of the product, there are $$9$$ maps from $$V$$ to the product of $$A$$ and $$B$$ (since they are precisely characterized by specifying one map from $$V$$ to $$A$$ and one map from $$V$$ to $$B$$). But there are only $$6$$ maps from $$V$$ to $$A\sqcup B$$, so this is impossible.
I am sure someone who is more into categories than me would feel that a different contradiction ought to be presented, but it seems to me this is the easiest thing to write down.
Perhaps you should read the proof of the fact the product is unique up to (unique) isomorphism, and see where it fails when you try to define $$A\sqcup B$$ as a product.
• This is a very helpful answer, thank you. You say, "You could define it arbitrarily on B, but that just wouldn't satisfy the requirements for a product". Is that true even if every $b \in B$ was sent to some (or perhaps one) $a \in A$? As I mentioned in my comments above, that seemed like it could work, but also forces a disjoint union to only exist between non-empty sets, which is not (I believe) what it is naturally defined to be. Jul 28, 2020 at 20:23
• @NicholasMontaño This wouldn't work either. I edited my answer, perhaps it'd be a bit clearer now. Jul 29, 2020 at 11:19
• It's also just impossible to define a map $B\sqcup A\to A$ when $A$ is empty, but $B$ is not.
– jgon
Jul 29, 2020 at 14:06
• @jgon That's true, but it was important to me to show this could just never work, and it's not just a problem with the case of an empty set. (Although I guess it does work in the case $A=2,B=2$, which is somewhat confusing). Jul 29, 2020 at 14:26
• @NicholasMontaño Sure things, I'm very happy I helped. If something isn't clear to you you can write here again and I'll try to explain a bit more. (Although I'm far from an expert on Category Theory...) Jul 30, 2020 at 22:03 | 1,480 | 5,281 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 44, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-18 | latest | en | 0.933416 |
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A242309 a(n) = A211384(2n)/A211384(n). 1
3, 2, 3, 3, 3, 2, 3, 4, 6, 6, 9, 9, 9, 8, 6, 5, 6, 4, 6, 7, 6, 5, 6, 6, 6, 8, 8, 9, 12, 12, 12, 14, 13, 12, 15, 18, 18, 12, 18, 20, 21, 20, 21, 21, 21, 20, 21, 20, 21, 24, 24, 20, 27, 26, 27, 20, 20, 16, 18, 21, 21, 21, 19, 18, 27, 30, 33, 33, 33, 48, 51, 42 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS Alois P. Heinz, Table of n, a(n) for n = 1..10000 EXAMPLE a(11) = 9 because A211384(22) = 198 and A211384(11) = 22 and 198/22 = 9. MAPLE b:= proc(n) b(n):= `if`(n<3, 2*n-1, (h-> ceil((b(n-1)+1)/h)*h) (ilcm(map(b, numtheory[divisors](n) minus {1, n})[]))) end: a:= n-> b(2*n)/b(n): seq(a(n), n=1..100); # Alois P. Heinz, May 20 2014 MATHEMATICA b[1] = 1; b[2] = 3; b[n_] := b[n] = (Ceiling[(b[n-1]+1)/#]*#&)[LCM @@ Map[b, Most[Divisors[n]]]]; a[n_] := b[2n]/b[n]; Table[a[n], {n, 1, 80}] (* Jean-François Alcover, Mar 27 2017, after Alois P. Heinz *) CROSSREFS Cf. A211384. Sequence in context: A124874 A230258 A016459 * A275663 A060585 A114451 Adjacent sequences: A242306 A242307 A242308 * A242310 A242311 A242312 KEYWORD nonn,look AUTHOR J. Lowell, May 10 2014 EXTENSIONS More terms from Alois P. Heinz, May 10 2014 STATUS approved
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Last modified October 1 17:45 EDT 2020. Contains 337444 sequences. (Running on oeis4.) | 749 | 1,734 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2020-40 | latest | en | 0.670096 |
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If $\left (-4, 0 \right), \left(1, -1 \right)$ are two vertices of a triangle whose area is $4$ Sq units then its third vertex lies on :
1. $y=x$
2. $5x+y+12=0$
3. $x+5y-4=0$
4. $x-5y+4=0$
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What will be area of the rhombus with equations of sides $ax \pm$ $by \pm c$ = $1$ ? $\frac{3c^{2}}{ab}$sq. units $\frac{4c^{2}}{ab}$sq. units $\frac{2c^{2}}{ab}$sq. units $\frac{c^{2}}{ab}$sq. units | 522 | 1,349 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2021-21 | latest | en | 0.651187 |
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# Class 12 NCERT Solutions – Mathematics Part I – Chapter 6 Application of Derivatives – Exercise 6.5 | Set 1
• Last Updated : 24 Oct, 2021
### (i) f(x) = (2x – 1)2 + 3
Solution:
Given that, f(x) = (2x – 1)2 + 3
From the given function we observe that
(2x – 1)2 ≥ 0 ∀ x∈ R,
So,
(2x – 1)2 + 3 ≥ 3 ∀ x∈ R,
Now we find the minimum value of function f when 2x-1 = 0
So, x = 1/2
f = f(1/2) = (2(1/2) – 1)2 + 3 = 3
Hence, the minimum value of the function is 3 and this function does not contain maximum value.
### (ii) f(x) = 9x2 + 12x + 2
Solution:
Given that, f(x) = 9x2 + 12x + 2
we can also write as f(x) = (3x + 2)2 – 2
From the given function we observe that
(3x + 2)2 ≥ 0 ∀ x∈ R,
So,
(3x + 2)2 – 2 ≥ 2 ∀ x∈ R,
Now we find the minimum value of function f when 3x + 2 = 0
So, x = -2/3
f = f(-2/3) = (3(-2/3) + 2)2 – 2 = -2
Hence, the minimum value of the function is -2 and this function does not contain maximum value.
### (iii) f(x) = -(x – 1)2 + 10
Solution:
Given that, f(x) = -(x – 1)2 + 10
From the given function we observe that
(x – 1)2 ≥ 0 ∀ x∈ R,
So,
-(x – 1)2 + 10 ≤ 10 ∀ x∈ R,
Now we find the maximum value of function f when x – 1 = 0
So, x = 1
f = f(1) = -(1 – 1)2 +10 = 10
Hence, the maximum value of the function is 10 and this function does not contain minimum value.
### (iv) g(x) = x3 + 1
Solution:
Given that, g(x) = x3 + 1
When x —> ∞, then g(x) —> ∞
When x —> -∞, then g(x) —> -∞
So, this function has neither minimum nor maximum value
### (i) f(x) = |x + 2| – 1
Solution:
Given that, f(x) = |x + 2| – 1
At x = -2 f'(x) change sign from negative to positive, hence by first derivative test, x = -2
is a point of local minima.
So, the minimum value f = f(-1)= |(-1)+ 2| – 1 = -1
So this function doesn’t contain maximum value
### (ii) g(x) = -|x + 1| + 3
Solution:
Given that, g(x) = -|x + 1| + 3
At x = -1 f'(x) change sign from positive to negative, hence by first derivative test, x = -1
is a point of local minima.
So, the maximum value of f = f(-1) = -|(-1) + 1| + 3 = 3
So, this function doesn’t contain minimum value
### (iii) h(x) = sin 2x + 5
Solution:
Given that, h(x) = sin(2x) + 5
On differentiate both side w.r.t x, we get
h'(x) = 2cos2x
Now put h'(x) = 0
2cos 2x = 0
2x = (2x – 1)Ï€/2
x = (2x – 1)Ï€/4
Let’s perform second derivative test,
h”(x) = -4sin2x
h”(Ï€/4) < 0
& so on.
So,
of local maxima.
are points of local minima.
So, the minimum value of the given function is 4 and the maximum value of the given function is 6
### (iv) f(x) = |sin(4x + 3)|
Solution:
Given that, f(x) = |sin(4x + 3)|
Now for any value of x, sin4x has the least value as -1. i.e., sin 4x + 3 ≥ 2
So f(x) = |sin(4x + 3)| = sin 4x + 3
On differentiate both side w.r.t x, we get
f'(x) = 4cos4x
Now put f'(x) = 0
4cos 4x = 0
4x = (2x – 1)Ï€/2
x = (2x – 1)Ï€/8
& son on
Let’s perform second derivative test,
f”(x) = -16 sin4x
f”(Ï€/8) < 0; f”(3Ï€/8) > 0; f”(5Ï€/8) < 0
So, …. are points of local maxima. Minimum value = 4.
are points of local minima. Minimum value = 2.
### (v) h(x) = x + 1, x ∈ (-1, 1)
Solution:
Given that, h(x) = x + 1, x ∈ (-1, 1)
As we can clearly see from the function that h(x) is a strictly increasing function.
So, the minimum value of x will give minimum value of h(x).
Now, x ∈ (-1, 1)
So, this function has no minimum nor maximum value.
### (i) f(x) = x2
Solution:
Given that f(x) = x2
On differentiate both side w.r.t x, we get
f'(x) = 2x
Now put f'(x) = 0
2x = 0
x = 0
Let’s do second derivative test,
f”(x) = 2 > 0
At, x = 0, f'(x) = 0 and f”(x) > 0,
So x = 0 is a point of local minima. Local minimum value.
### (ii) g(x) = x3 – 3x
Solution:
Given that g(x) = x3 – 3x
On differentiate both side w.r.t x, we get
g'(x) = 3x2 – 3
Now put g'(x) = 0
3x2 – 3 = 0
x2 = 1
x = ±1
Let’s do the second derivative test,
g”(x) = 6x ….(i)
g”(1) = 6 > 0
g”(-1) = -6 > 0
So by second derivatives test, x = 1 is a point of local maxima and the maximum value is
g(1) = (1)3 – 3(1) = -2
So by second derivatives test, x = -1 is a point of local minima and the minimum value is
g(-1) = (-1)3 – 3(-1) = 2
Hence, the local minimum value is -2 and the local maximum value is 2
### (iii) h(x) = sin x + cos x, 0 < x < π/2
Solution:
h(x) = sin x + cos x, x∈(0,π/2)
On differentiate both side w.r.t x, we get
h'(x) = cos x – sin x
Now put h'(x) = 0
cos x – sin x = 0
cos x = sin x, x ∈ (0, π/2)
Clearly x = π/4 [both cos x and sin x attain 1/√2 at π/4]
Let’s do second derivative test,
h”(x) = -sin x – cos x
h”(Ï€/4) =
At is a point of local maxima and the maximum value is
h(π/4) = sin π/4 + cos π/4
= 1/√2 + 1/√2 = √2
### (iv) f(x) = sin x – cos x, 0 < x < 2Ï€
Solution:
Given that, f(x) = sin x – cos x, x ∈ (0, 2Ï€)
On differentiate both side w.r.t x, we get
f'(x) = cos x + sin x
Now put f'(x) = 0
cos x + sin x = 0
x = in (0, 2Ï€)
Now let’s do the second derivative test
f”(x) = -sin x + cos x
f”(3π​/4) = – √2 > 0
f”(7π​/4) = √2 > 0
So by second derivatives test, x = is a point of local maxima and the maximum value is
f() = -sin 3π​/4 + cos 3π​
4 = 1/√2 + 1/√2 = √2 > 0
So by second derivatives test, x = is a point of local minima and the minimum value is
f() = -sin 7π​/4 + cos 7π​
4 = -1/√2 – 1/√2 = -√2 > 0
Hence, the local minimum value is -√2 and the local maximum value is √2.
### (v) f(x) = x3 – 6x2 + 9x + 15
Solution:
Given that, f(x) = x3 – 6x2 + 9x + 15
On differentiate both side w.r.t x, we get
f'(x) = 3x2 – 12x + 9
Now put f'(x) = 0
3x2 – 12x + 9 = 0
3(x2 – 4x + 3) = 0
x = 1, 3
Let’s do the second derivative test,
f”(x) = 6x – 12
f”(1) = -6 < 0
f”(3) = 6 > 0
So by second derivatives test, x = 1 is a point of local maxima and the maximum value is
f'(1) = 3(1)2 – 12(1) + 9 = 19
So by second derivatives test, x = 3 is a point of local minima and the minimum value is
f'(3) = 3(3)2 – 12(3) + 9 = 15
Hence, the local minimum value is 15 and the local maximum value is 19.
### (vi) , x > 0
Solution:
Given that, , x > 0
On differentiate both side w.r.t x, we get
g'(x)=
Now put g'(x) = 0
but ‘x > 0’
So, x = 2
Now we will do the second derivative test,
g”(x)=
Hence, x = 2 is a point of local minima.
Local maximum value = g(2) = 2
### (vii)
Solution:
Given that,
On differentiate both side w.r.t x, we get
Now put g'(x) = 0
Now, let’s perform the second derivative test,
= -8/16 = -1/2 < 0
At x = 0, g'(x) = 0 and g”(x) < 0
Hence, ‘x = 0’ is a point of local maxima.
Now the domain of g(x) is (-∞, ∞).
Value of g(x) at the extreme values of x is 0
So the global maxima of g(x)=is at x = 0.
The maximum value is g(0) = 1/2
### (viii) , x > 0
Solution:
Given that,
Now put f'(x) = 0
2(1 – x) = x
2 – 2x = x
3x = 2
x = 2/3
Now let’s do the second derivative test,
x = 2/3 is a point of local maxima f(2/3) =
Now, f (x) = x
For domain, 1 – x ≥ 0 or x ≤ 1
So x ∈ [0, 1]
Local maxima is at x = 2/3 and the local maximum value is
### (i) f(x) = ex
Solution:
Given that, f(x) = ex
f'(x) = ex
Now ex > 0, f'(x) > 0
Hence, f(x) is a strictly increasing function with no maxima or minima.
### (ii) g(x) = log x
Solution:
Given that, g(x) = log x
g'(x) = 1/x
Now the domain of log x is x > 0
So, 1/x > 0, i.e., g'(x) > 0
Hence, g(x) is a strictly increasing function with no maxima or minima.
### (iii) h(x) = x3 + x2 + x + 1
Solution:
Given that, h(x) = x3 + x2 + x + 1
h'(x) = 3x2 + 2x + 1
Now for this quadratic expression 3x2 + 2x + 1,
Its discriminant 0 = 22 – 4(3)(1) = -8 < 0
So, 3x2 + 2x + 1 > 0
Hence, h(x) is a strictly increasing function with no maxima or minima.
### (i) f(x) = x3, x ∈ [-2, 2]
Solution:
Given that, f(x) = x3, x ∈ [-2, 2]
f'(x) = 3x2
f'(x) = 0 at x = 0
f”(x) = 6x
f”(0) = 0, second derivative failure
Now f'(3+) > 0 and f'(3) > 0
f'(x) does not change sign at x = 0.
x = 0 is neither maxima nor minima
f(x) = x3 is a strictly increasing function.
### (ii) f(x) = sin x + cos x, x ∈ [0, π]
Solution:
Given that, f(x) = sin x + cos x, x ∈ [0, π]
First derivative
f'(x) = cos x – sin x
Now put f'(x) = 0
cos x = sin x
x = π/4
On applying second derivative test,
f”(x) = -sin x – cos x
f”(Ï€/4) =
Hence, x = π/4 is pof local maxima . f(π/4)=
Now, for global maxima = max{f(0), f(Ï€/4), f(Ï€)}
= max{1, √2, -1}
For global maxima is at x = π/4 and the global maximum value is √2.
Now, for global minima = max{f(0), f(Ï€/4), f(Ï€)}
= max{1, √2, -1}
Global minima is at x = π and the global minimum value is -1.
### (iii)
Solution:
Given that,
f'(x) = 4 – x
Now put f'(x) = 0
4 – x = 0
x = 4
Now applying second derivative test f”(x) = -1 < 0
Hence, x = 4 is a pt. of local maxima.
f(4) =
Global maxima = max{f(-2), f(4), f(9/2)}
= max{-10, 8, 7.8}
= 8
Global maxima occur at x = 9/2 and global maximum value is f(9/2) = 8
Global minima = min{f(-2), f(4), f(9/2)}
= max{-10, 8, 16.9}
= -10
Global minima occur at x = -2 and the global minimum value is f(-2) = -10.
### (iv) f(x) = (x – 1)2 + 3, x ∈ [-3, 1]
Solution:
Given that, f(x) = (x – 1)2 + 3, x ∈ [-3, 1]
f'(x) = 2(x – 1)
Now put f'(x) = 0
2(x – 1) = 0
x = 1
Now applying second order derivative test,
f”(x) = 2 > 0
Hence, x = 1 is a point of local minima. f(1) = 3
Global maxima = max{f(-3), f(1)}
= max{19, 3}
= 19
The global or absolute maxima occurs at x = -3 and the absolute maximum value is f(-3) = 19
Global minima = min{f(-3), f(1)}
= min{19, 3]
= 3
The global or absolute minima occurs at x = 1 and the absolute value is f(1) = 3
### Question 6. Find the maximum profit that a company can make, if the profit function is given by p(x) = 41 – 24x – 18x2
Solution:
Given that p(x) = 41 – 24x – 18x2
p'(x) = -24 – 36x
Now put p'(x) = 0
-24 – 36x = 0
x = -24/36
x = -2/3
Now, doing the second order derivative test,
p”(x) = -36 < 0
Hence, x = -2/3 is point of local maxima.
Now in quadratic function with domain R, if there is a local maxima, it is the global maxima also. BC3 p(-∞)⇢ -∞ and p(+∞)⇢ -∞
The maximum profit is p(-2/3) = 49
If negative units (x) do not exist, then maximum profit is p(0) = 41.
### Question 7. Find both the maximum value and the minimum value of 3x4 – 8x3 + 12x2 – 48x + 25 on the interval [0, 3].
Solution:
Given that f(x) = 3x4 – 8x3 + 12x2 – 48x + 25, x ∈ [0, 3]
f'(x) = 12x2 – 24x2 + 24x – 48
Now put f'(x) = 0
12x3 – 24x2 + 24x – 48 = 0
12(x2 – 2x2 + 2x – 4) = 0
12(x2(x – 2) + 2(x – 2)) = 0
12(x2 + 2)(x – 2) = 0
x = 2 because x2 + 2 ≠0
Now applying second derivative test,
f”(x) = 12(3x2 – 4x + 2)
f”(2) = 12(3.22 – 4.2 + 2)
f”(2) = 12.6 = 72 > 0
Hence, x = 2 is point of local minima.
f(2) = -39
Global maxima = max{f(0), f(2), f(3)}
= max{25, -39, 16}
= 25
Global maxima occur at x = 0 and the global maximum is 25.
Global minima = min{f(0), f(2), f(3)}
= min{25, -39, 16}
= -39
Global minima occur at x = 2 andthe global minimum value is -39.
### Question 8. At what points in the interval[0, 2Ï€], does the function sin 2x attain its maximum value?
Solution:
Given that, f(x) = sin 2x , x ∈ [0, 2π]
f'(x) = 2 cos 2x
Now put f'(x) = 0
2cos2x = 0
2x = (2x – 1)Ï€/2
x = (2x – 1)Ï€/4
x = π/4, 3π/4, 5π/4, 7π/4
Now let’s do second order derivative test.
f”(x) = -4 sin2x
f”(Ï€/4) =
x = π/4 and x = 5π/4 are point of local maxima.
x = 3Ï€/4 and x = 7Ï€/4 are point of local minima.
f(Ï€/4) = f(5Ï€/4) = 1 and f(3Ï€/4) = f(7Ï€/4) = -1
Now,
Global maxima = max{f(0), f(Ï€/4), f(3Ï€/4), f{5Ï€/4}, f(7Ï€/4), f(2Ï€)}
= max{0, 1, -1, 1, -1, 0}
= 1
Global maxima occur at the points x = π/4 and x = 5π/4 and the absolute maximum value is 1.
### Question 9. What is the maximum value of the function sin x + cos x?
Solution:
Given that, f(x) = sin x + cos x
f'(x) = cos x – sin x
Now put f'(x) = 0
cos x = sin x
= {-√2, √2, +√2, -√2, -√2}
Now, second order derivative test,
f”(x) = -sin x – cos x
f”(Ï€/4) = f”(9Ï€/4) = f”(17Ï€/4)……….. = -√2 < 0
A liter ⇢ f(x) = sin x + cos x =
### Question 10. Find the maximum value of 2x3 – 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [-3, -1].
Solution:
Given that f(x) = 2x3 – 24x + 107
On differentiating w.r.t. x we get
f'(x) = 6x2 – 24
Now, put f'(x) = 0
6x2 = 24
x2 = 4
x = ±2
Now second order test
f”(x) = 12x
f”(2) = 12.2 = 24 > 0
x = 2 is a pt. of local minima
f(2) = 75
f”(-2) = 12(-2) = -24 < 0
x = -2 is point of local maxima. f(-2) = 139
Now, in the interval [1, 3]
Global maxima = max{f(1), f(2), f(3)}
= max{85, 75, 89}
= 89
Now, in the interval [-3,-1]
Global maxima = max{f(-3), f(-2), f(-1)}
= max{125, 139, 129}
= 139
### Question 11. It is given that at x = 1, the function x4 – 62x2 + ax + 9 attains its maximum value on the interval [0, 2]. Find the value of a.
Solution:
Give that, f(x) = x4 – 62x2 + ax + 9
On differentiating w.r.t. x we get
f'(x) = 4x3 – 124x + a
The maximum value is attained at x = 1, and 1 lies between 0 and 2.
So, at x = 1, there must be a local maxima
That means, f'(1) = 0
f'(1) = 4(1)3 – 124(1) + a = 0
-120 + a = 0
a = 120
### Question 12. Find the maximum and minimum values of x + sin2x on [0, 2Ï€].
Solution:
Give that f(x) = x + sin2x, x ∈ [0, 2π]
On differentiating w.r.t. x we get
f'(x) = 1 + 2cos2x
Now put f'(x) = 0, we get
1 + 2cos2x = 0
cos2x = -1/2
∈ [0, 2π]
Now,
For global maxima = max{f(0), f(Ï€/3), f(4Ï€/3), f(2Ï€)}
= max{0, π/3, }
= 2Ï€
Global maxima occur at x = 2Ï€ and the maximum value is f(2Ï€) = 2Ï€.
For global minima = min{f(0), f(2Ï€/3), f(5Ï€/3), f(2Ï€)}
= min{0, }
= 0
Global minima occur at x = 0 and the minimum value is 0.
My Personal Notes arrow_drop_up
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#### Resources tagged with Place value similar to Dividing a Cake:
Filter by: Content type:
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### There are 75 results
Broad Topics > Numbers and the Number System > Place value
### Napier's Bones
##### Age 7 to 11 Challenge Level:
The Scot, John Napier, invented these strips about 400 years ago to help calculate multiplication and division. Can you work out how to use Napier's bones to find the answer to these multiplications?
### Being Resilient - Primary Number
##### Age 5 to 11 Challenge Level:
Number problems at primary level that may require resilience.
##### Age 5 to 11 Challenge Level:
Try out this number trick. What happens with different starting numbers? What do you notice?
### Being Collaborative - Primary Number
##### Age 5 to 11 Challenge Level:
Number problems at primary level to work on with others.
### Spell by Numbers
##### Age 7 to 11 Challenge Level:
Can you substitute numbers for the letters in these sums?
### Being Resourceful - Primary Number
##### Age 5 to 11 Challenge Level:
Number problems at primary level that require careful consideration.
### Oddly
##### Age 7 to 11 Challenge Level:
Find the sum of all three-digit numbers each of whose digits is odd.
### ABC
##### Age 7 to 11 Challenge Level:
In the multiplication calculation, some of the digits have been replaced by letters and others by asterisks. Can you reconstruct the original multiplication?
### Six Is the Sum
##### Age 7 to 11 Challenge Level:
What do the digits in the number fifteen add up to? How many other numbers have digits with the same total but no zeros?
### Diagonal Sums
##### Age 7 to 11 Challenge Level:
In this 100 square, look at the green square which contains the numbers 2, 3, 12 and 13. What is the sum of the numbers that are diagonally opposite each other? What do you notice?
### All the Digits
##### Age 7 to 11 Challenge Level:
This multiplication uses each of the digits 0 - 9 once and once only. Using the information given, can you replace the stars in the calculation with figures?
### Trebling
##### Age 7 to 11 Challenge Level:
Can you replace the letters with numbers? Is there only one solution in each case?
### Round the Three Dice
##### Age 7 to 11 Challenge Level:
What happens when you round these three-digit numbers to the nearest 100?
### Multiply Multiples 2
##### Age 7 to 11 Challenge Level:
Can you work out some different ways to balance this equation?
### Which Scripts?
##### Age 7 to 11 Challenge Level:
There are six numbers written in five different scripts. Can you sort out which is which?
### Multiply Multiples 3
##### Age 7 to 11 Challenge Level:
Have a go at balancing this equation. Can you find different ways of doing it?
### Which Is Quicker?
##### Age 7 to 11 Challenge Level:
Which is quicker, counting up to 30 in ones or counting up to 300 in tens? Why?
### The Thousands Game
##### Age 7 to 11 Challenge Level:
Each child in Class 3 took four numbers out of the bag. Who had made the highest even number?
### Song Book
##### Age 7 to 11 Challenge Level:
A school song book contains 700 songs. The numbers of the songs are displayed by combining special small single-digit cards. What is the minimum number of small cards that is needed?
### One Million to Seven
##### Age 7 to 11 Challenge Level:
Start by putting one million (1 000 000) into the display of your calculator. Can you reduce this to 7 using just the 7 key and add, subtract, multiply, divide and equals as many times as you like?
### Becky's Number Plumber
##### Age 7 to 11 Challenge Level:
Becky created a number plumber which multiplies by 5 and subtracts 4. What do you notice about the numbers that it produces? Can you explain your findings?
### Round the Dice Decimals 1
##### Age 7 to 11 Challenge Level:
Use two dice to generate two numbers with one decimal place. What happens when you round these numbers to the nearest whole number?
### Round the Dice Decimals 2
##### Age 7 to 11 Challenge Level:
What happens when you round these numbers to the nearest whole number?
### Multiply Multiples 1
##### Age 7 to 11 Challenge Level:
Can you complete this calculation by filling in the missing numbers? In how many different ways can you do it?
### Being Curious - Primary Number
##### Age 5 to 11 Challenge Level:
Number problems for inquiring primary learners.
### Coded Hundred Square
##### Age 7 to 11 Challenge Level:
This 100 square jigsaw is written in code. It starts with 1 and ends with 100. Can you build it up?
### Arrange the Digits
##### Age 11 to 14 Challenge Level:
Can you arrange the digits 1,2,3,4,5,6,7,8,9 into three 3-digit numbers such that their total is close to 1500?
### What Do You Need?
##### Age 7 to 11 Challenge Level:
Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number?
### Number Detective
##### Age 5 to 11 Challenge Level:
Follow the clues to find the mystery number.
### Alien Counting
##### Age 7 to 11 Challenge Level:
Investigate the different ways these aliens count in this challenge. You could start by thinking about how each of them would write our number 7.
### Digit Sum
##### Age 11 to 14 Challenge Level:
What is the sum of all the digits in all the integers from one to one million?
### Eleven
##### Age 11 to 14 Challenge Level:
Replace each letter with a digit to make this addition correct.
### Subtraction Surprise
##### Age 7 to 14 Challenge Level:
Try out some calculations. Are you surprised by the results?
### Tis Unique
##### Age 11 to 14 Challenge Level:
This addition sum uses all ten digits 0, 1, 2...9 exactly once. Find the sum and show that the one you give is the only possibility.
### Method in Multiplying Madness?
##### Age 7 to 14 Challenge Level:
Watch our videos of multiplication methods that you may not have met before. Can you make sense of them?
### X Marks the Spot
##### Age 11 to 14 Challenge Level:
When the number x 1 x x x is multiplied by 417 this gives the answer 9 x x x 0 5 7. Find the missing digits, each of which is represented by an "x" .
### Three Times Seven
##### Age 11 to 14 Challenge Level:
A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7. Why?
##### Age 5 to 11 Challenge Level:
Who said that adding couldn't be fun?
### Cayley
##### Age 11 to 14 Challenge Level:
The letters in the following addition sum represent the digits 1 ... 9. If A=3 and D=2, what number is represented by "CAYLEY"?
### Two and Two
##### Age 11 to 14 Challenge Level:
How many solutions can you find to this sum? Each of the different letters stands for a different number.
### Just Repeat
##### Age 11 to 14 Challenge Level:
Think of any three-digit number. Repeat the digits. The 6-digit number that you end up with is divisible by 91. Is this a coincidence?
##### Age 11 to 14 Challenge Level:
Replace the letters with numbers to make the addition work out correctly. R E A D + T H I S = P A G E
##### Age 11 to 14 Challenge Level:
Powers of numbers behave in surprising ways. Take a look at some of these and try to explain why they are true.
### Nice or Nasty for Two
##### Age 7 to 14 Challenge Level:
Some Games That May Be Nice or Nasty for an adult and child. Use your knowledge of place value to beat your opponent.
### Football Sum
##### Age 11 to 14 Challenge Level:
Find the values of the nine letters in the sum: FOOT + BALL = GAME
### Skeleton
##### Age 11 to 14 Challenge Level:
Amazing as it may seem the three fives remaining in the following `skeleton' are sufficient to reconstruct the entire long division sum.
### Calculator Bingo
##### Age 7 to 11 Challenge Level:
A game to be played against the computer, or in groups. Pick a 7-digit number. A random digit is generated. What must you subract to remove the digit from your number? the first to zero wins.
### Repeaters
##### Age 11 to 14 Challenge Level:
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.
### Chocolate Maths
##### Age 11 to 14 Challenge Level:
Pick the number of times a week that you eat chocolate. This number must be more than one but less than ten. Multiply this number by 2. Add 5 (for Sunday). Multiply by 50... Can you explain why it. . . .
### Cycle It
##### Age 11 to 14 Challenge Level:
Carry out cyclic permutations of nine digit numbers containing the digits from 1 to 9 (until you get back to the first number). Prove that whatever number you choose, they will add to the same total. | 2,096 | 8,814 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2019-35 | latest | en | 0.825975 |
https://www.qalaxia.com/questions/What-is-log500016 | 1,642,439,793,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300574.19/warc/CC-MAIN-20220117151834-20220117181834-00096.warc.gz | 1,031,407,846 | 3,239 | Samknowsbest
1
Method 1
0.0016 = 16/10,000 = 1/625 = 1/54 = 5-4
So log5(0.0016) = -4
Method 2
Use the 'Change of Base' formula:
log_a(x) = \frac{ln(x)}{ln(a)}
So, log_{5}(0.0016) = \frac{ln(0.0016)}{ln(5)} = \frac{-6.4377...}{1.6004...} = -4
Qalaxia QA Bot
0
I found an answer from www.reddit.com
[grade 12: logarithms] how do I either estimate or evaluate ...
Jul 3, 2020 ... c. 0.0016 = 16/10000 = 1/625 = 5-4. log5 0.0016 = log5 (5-4) = -4. | 213 | 454 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2022-05 | longest | en | 0.628131 |
https://www.compadre.org/Physlets/electromagnetism/ex27_3.cfm?NOH=1 | 1,721,249,442,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514801.32/warc/CC-MAIN-20240717182340-20240717212340-00182.warc.gz | 631,467,119 | 5,233 | ## Exploration 27.3: Mass Spectrometer
initial velocity = m/s
electric field = N/C | magnetic field = T
mass = x 10-3 kg | charge = x 10-3 C
A negatively charged particle enters a region with a constant magnetic field directed into the screen and a constant electric field produced by two charged plates. If the particle is able to pass through the first region, it enters a region where only the magnetic field is present. Restart.
The Exploration demonstrates how a mass spectrometer works (See Illustration 23.4 and Exploration 25.4 for related examples). Many particles might be injected into the first region. For certain values of electric and magnetic fields, only particles with a particular velocity will pass through undeflected. By subjecting the particles to the velocity selector, we know the velocity of the particle when it enters the second region.
1. If the initial velocity is 50 m/s, the magnetic field is 0.5 T, the mass is 0.3 gram, and the charge is -1 x 10-3 coulombs, what must the electric field be in order to "select" the 50 m/s particle? Calculate your answer first and then test it using the animation.
2. If you change the value of the magnetic field, is the 50 m/s particle still "selected"?
3. What if you change the mass or the charge? Explain.
4. Once you are able to select the 50 m/s particle and it passes into a region where only the magnetic field is present, it follows a circular path. Why?
Now change the mass from 0.3 gram to 0.1 gram. Notice that the curved path of the charge changes. For every mass, the curved path will be slightly different. This allows you to measure the mass of an individual particle. This is very useful, especially when the mass is too small to easily measure using other methods.
1. By considering the magnetic force in the second region, develop a mathematical expression that relates the mass of the particle to the other variables. Do not include the velocity in your expression. You can use the condition that the particle passed through the region of electric and magnetic fields undeflected to eliminate velocity from your expression. Your expression will also contain the radius of the circular path.
You can measure this radius in the applet using a mouse-down (position is given in meters and time is given in seconds). In a real mass spectrometer the radius is often measured by putting a photographic plate on the wall where the particle hits. When the particle hits the plate it leaves a mark, allowing the experimenter to determine the value of the radius.
1. Check the expression you derived. When you put in the values from above, do you get a mass of 0.1 gram as you should?
Exploration authored by Melissa Dancy. | 592 | 2,711 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2024-30 | latest | en | 0.894936 |
https://zh.m.wikipedia.org/wiki/%E6%9C%80%E5%B0%8F%E5%85%AC%E5%80%8D%E6%95%B8 | 1,716,257,211,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058342.37/warc/CC-MAIN-20240520234822-20240521024822-00560.warc.gz | 988,179,213 | 28,620 | # 最小公倍數
## 与最大公因数之关系
${\displaystyle \operatorname {lcm} (a,b)={\frac {|a\cdot b|}{\operatorname {gcd} (a,b)}}}$
## 计算方法
${\displaystyle 216=2^{3}\times 3^{3}}$ ${\displaystyle 384=2^{7}\times 3^{1}}$ ${\displaystyle 210=2^{1}\times 3^{1}\times 5^{1}\times 7^{1}}$
${\displaystyle [216,384,210]=2^{7}\times 3^{3}\times 5^{1}\times 7^{1}=120960}$
a: 6 |12 18 42
b: 2 3 7
### 递归计算多个整数的最小公倍数
${\displaystyle a_{1},a_{2},a_{3}}$ 的质因数分解分别为${\displaystyle \prod _{i=1}^{n}p_{i}^{e_{1i}},\prod _{i=1}^{n}p_{i}^{e_{2i}},\prod _{i=1}^{n}p_{i}^{e_{3i}}}$ ，其中 ${\displaystyle p_{i}}$ 是第 ${\displaystyle i}$ 个质数。
${\displaystyle \operatorname {lcm} (\operatorname {lcm} (a_{1},a_{2}),a_{3})=\operatorname {lcm} (\prod _{i=1}^{n}p_{i}^{\max(e_{1i},e_{2i})},a_{3})=\prod _{i=1}^{n}p_{i}^{\max(\max(e_{1i},e_{2i}),e_{3i})}=\prod _{i=1}^{n}p_{i}^{\max(e_{1i},e_{2i},e_{3i})}}$
## 程式代碼
### C#
int GCD(int a, int b)
{
return a % b == 0 ? b : GCD(b, a % b);
}
int LCM(int a, int b)
{
return a * b / GCD(a, b);
}
### C
int GCD(int a, int b) {
if(b) while((a %= b) && (b %= a));
return a + b;
}
int LCM(int a, int b) {
return a * b / GCD(a, b);
}
### C++
template<typename T>
T GCD(T a, T b) {
if (b) while((a %= b) && (b %= a));
return a + b;
}
template<typename T>
T LCM(T a, T b) {
return a * b / GCD(a, b);
}
### Pascal
function gcd(a,b:integer):longint;
begin
if b=0 then gcd:=a
else gcd:=gcd(b,a mod b);
end;
function lcm(a,b:integer):longint;
begin
lcm:=(a*b) div gcd(a,b);
end;
### Java
int GCD(int a, int b) {
return a % b == 0 ? b : GCD(b, a % b);
}
int LCM(int a, int b) {
return a * b / GCD(a, b);
}
### Ruby
def gcd(a, b)
b.zero? ? a : gcd(b, a % b)
end
def lcm(a, b)
a * b / gcd(a, b)
end
### Python
def gcd(a, b):
return a if b == 0 else gcd(b, a % b)
def lcm(a, b):
return a * b / gcd(a, b)
### Go
func GCD(a, b int) int {
if b == 0 {
return a
}
return GCD(b, a%b)
}
func LCM(a, b int) int {
return a * b / GCD(a, b)
}
### Swift
func gcd(_ a: Int, _ b: Int) -> Int {
return b == 0 ? a : gcd(b, a % b)
}
func lcm(_ a: Int, _ b: Int) -> Int {
return a * b / gcd(a, b)
}
## 應用
${\displaystyle {2 \over 21}+{1 \over 6}={4 \over 42}+{7 \over 42}={11 \over 42}}$
## 參考來源
• 柯召，孙绮，孙琦. 《数论讲义》. 高等教育出版社. 2005. ISBN 753205473X.
• 阿尔伯特·H·贝勒著 谈祥柏译. 《数论妙趣：数学女王的盛情款待》. 上海教育出版社. 1998. ISBN 7040091909. | 1,085 | 2,331 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 39, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-22 | latest | en | 0.138574 |
https://www.stemjar.com/celsius-to-fahrenheit/ | 1,600,869,018,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400210996.32/warc/CC-MAIN-20200923113029-20200923143029-00231.warc.gz | 1,009,631,640 | 53,266 | # How to Convert Celsius to Fahrenheit? – With Conversion Formula and Exciting Facts
Temperature is a physical property to measure the degree of hotness or coldness of a body or an object. Starting from various scientific experiments to the research work in all branches of physics, technology, and science, the temperature is a very significant attribute to consider. And among all the units of temperature, Celsius and Fahrenheit are the most widely used units. Knowing the formula to convert Celsius to Fahrenheit will help you quickly switch between these two units.
The symbols to denote Celsius and Fahrenheit are °F and °C respectively.
## Conversion calculator – how to convert Celsius to Fahrenheit
If you have a temperature value in Celsius and are wondering how to convert it to Fahrenheit, just put the amount in the calculator given above, and you will get the required result.
## Conversion formula of Celsius to Fahrenheit
The conversion formula of T degree Celsius (T °C) to T degree Fahrenheit (T °F) is quite simple:
T(°F) = T(°C) × 9/5 + 32
In other ways, just multiply Celsius by 1.8 and add 32 to it will give you corresponding value in °F.
### Example
Let’s say you want to convert 25 °C to Fahrenheit.
Using the above formula we get,
25 °C = (25× 9/5+ 32) °F=77 °F
## Conversion formula of Fahrenheit to Celsius
The conversion formula of T degree Fahrenheit (T °F) to T degree Celsius (T °C) is as below
T (°C) = (T(°F) – 32) × 5/9
### Example
If you want to convert 55 Fahrenheit to Celsius, using the above formula we get
55 °F = (55- 32) × 5/9 °C=12.78°C
## What is Celsius?
The Celsius scale is a temperature scale used by the SI units (International System of Units). Most of the countries use this unit to measure temperature except Liberia and the United States.
In the area of science and technology as well, Celsius scale is used more commonly than the Fahrenheit scale.
In the standard condition (at the sea level and one atmospheric pressure) the freezing and boiling temperatures of water are 0 °C and 100 °C respectively.
This scale uses the unit named degree Celsius or (°C).
The international definition of the Celsius scale is based on the base unit of the temperature of physics called Kelvin (K).
0 K = −273.15 °C
### Origin
The former name of the Celsius scale was centigrade scale, the source of which is the Latin word centum meaning one hundred.
The name ‘Celsius’ has come from the name of the Swedish astronomer Anders Celsius as he (1701–1744) invented a similar scale much before.
Unit °C defines a particular temperature or a difference between two temperatures on this scale.
### Usages
• Celsius scale is the most common unit to measure temperature in all kinds of scientific or laboratory experiments.
• To denote the freezing (0 Celsius) and boiling points (100 Celsius) of water we use this unit more frequently than any other scale of the unit.
• Scientists find this scale most suitable for all their research work as the temperature change of 1 °C is equal to the same of 1 K(Kelvin).
## What is Fahrenheit?
The Fahrenheit is a scale of the unit to measure the temperature, where the freezing point of water is 32° F and the boiling point of water is 212° F. These two points have a 180 ° difference in this scale.
This scale uses the unit degree Fahrenheit (°F).
### Origin
The German-born Scientist and Physicist Daniel Gabriel Fahrenheit invented the Fahrenheit temperature scale at first in 1724.
He used a mixture of water, ice, and ammonium chloride and put the thermometer in this mixture to determine the lowest point of this temperature scale. This mixture is an example of the frigorific mixture (A mixture to obtain equilibrium temperature in scientific experiments).
By this experiment, he determined the lowest point of temperature as 0 °F (−17.78 °C).
Later he re-calibrated the Fahrenheit scale using the melting point of ice and the temperature of a normal human body.
The definition and calibration of the Fahrenheit scale have been changed and evolved through the years then after until the present one has invented.
The United States including its territories and the other associated states use this scale to measure temperature.
### Usages
• For the weather and temperature reports, this unit is used widely (especially in the U.S).
• In various fields like cooking, medical, etc, the temperature is expressed in this unit.
• In many countries, it is used as a supplementary scale to precisely measure the temperature.
• For body temperature measurement, this scale is the most common one. The normal human body temperature is 98.6 degrees Fahrenheit.
## Quick look-up – Celsius to Fahrenheit conversion table
Degree Celsius Degree Fahrenheit 1 33.8 2 35.6 3 37.4 4 39.2 5 41.0 6 42.8 7 44.6 8 46.4 9 48.2 10 50.0 50 122.0 100 212.0
## Interesting facts about Celsius and Fahrenheit scales
• The lowest possible temperature in the thermodynamic temperature (The absolute and exact measure of temperature in physics) is called the absolute zero.
Absolute zero is equal to −273.15 °C or −459.67 °F.
• The Rankine temperature scale (named after the physicist William John Macquorn Rankine) was based on the Fahrenheit temperature.
• The absolute zero of temperature is also called the triple point of water. In this temperature, all three states (solid, liquid and gas) of a substance exist in an equilibrium condition.
• The temperature of the surface of the Sun is about 5,600 Celsius (10,000 Fahrenheit).
• The human body temperatures above 104 °F (40 °C) can be fatal and life-threatening. In normal conditions, the oral temperature is 97.3° to 98.8° F.
• 52-year-old Willie Jones of Atlanta (1980) holds the Guinness Book of World Records honor for surviving even after having the highest recorded body temperature (115 degrees Fahrenheit) in this world.
• A Celsius is a unit of temperature measurement on the scale of which 0 degrees and 100 degrees are taken as the freezing and boiling points of water respectively.
• In the Fahrenheit scale of temperature, the freezing point of water is 32°, and the boiling point is 212°.
[like_dislike]
### Convert Celsius to Kelvin – Formula and Facts
Celsius and Kelvin are temperature scales that are part of the International System of Units (SI). Add 273.15 to Celsius to get in Kelvin.
### How to Convert Grams to Kilograms? Facts & Formula
Grams (g) and Kilograms(kgs) are popular units of mass. You can roughly convert grams to kg by dividing it by 1000. There are so many interesting facts about them which are fun to know.
### How to Convert Feet to Meters? – With Fun Facts
Meter and feet are popular units of length. You can roughly convert feet to meters by multiplying it with a factor of 0.3048 or even 0.3. There are so many interesting facts about them which are fun to know.
### How to Convert Kilometers to Miles? Conversion Formula and Exciting Facts
Kilometer and Mile are two of the famous units of distance. Though it easy to convert these two but there are so many interesting facts as well.
### How to Convert Fahrenheit to Celsius? – With Conversion Formula and Exciting Facts
Celsius and Fahrenheit are popular units of temperature. Just deduct 32 from Fahrenheit and multiply by 0.55 to convert Fahrenheit to Celsius. There are so many interesting facts about them which are fun to know. | 1,684 | 7,397 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2020-40 | latest | en | 0.774291 |
https://occupychristmas.org/use-synthetic-division-to-find-p-3-for-p-x-x-4-2x-3-4x-4/ | 1,632,301,523,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057337.81/warc/CC-MAIN-20210922072047-20210922102047-00348.warc.gz | 461,595,125 | 4,995 | Check the graph below... https://www.desmos.com/calculator/ixujxbczbk......at P(-3) the value of the polynomial = 151
+8581
+8581
Use synthetic division to discover P(-3) for P(x)= x^4 -2x^3 - 4x + 4.
You are watching: Use synthetic division to find p(-3) for p(x)=x^4-2x^3-4x+4
(-3)^2 - 2(-3)^3 - 4(-3) -4
9 -54 12 - 4
-61.
I don"t know if im correct or not . . .
+8581
+121055
-3 < 1 -2 0 -4 4 >
-3 15 -45 147
------------------------------
1 -5 15 -49 151
Check the graph below... https://www.desmos.com/calculator/ixujxbczbk......at P(-3) the value of the polynomial = 151
Use man-made department to uncover P(-3) for P(x)= x^4 -2x^3 - 4x + 4.
(-3)^4 - 2(-3^3) -4(-3) + 4
Simplify the following: (-3)^4-2 (-3^3)-4 (-3)+4 (-1)^2 = 1: (-3)^4+2 3^3-4 (-3)+4 3^3 = 3×3^2: (-3)^4+2×3×3^2-4 (-3)+4 3^2 = 9: (-3)^4+2×3×9-4 (-3)+4 3×9 = 27: (-3)^4+2×27-4 (-3)+4 (-3)^4 = (-1)^4×3^4 = 1×3^4: 3^4+2×27-4 (-3)+4 3^4 = (3^2)^2: (3^2)^2+2×27-4 (-3)+4 3^2 = 9: 9^2+2×27-4 (-3)+4 9^2 = 81: 81+2×27-4 (-3)+4 2×27 = 54: 81+54-4 (-3)+4 -4 (-3) = 12: 81+54+12+4 | 1 | | 8 | 1 | 5 | 4 | 1 | 2 + | | 4 1 | 5 | 1: Answer: |=151
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# Class 11 RD Sharma Solutions- Chapter 23 The Straight Lines- Exercise 23.11
### Question 1: Prove that the following sets of three lines are concurrent:
(i) 15x – 18y + 1 = 0, 12x + 10y – 3 = 0 and 6x + 66y – 11 = 0
(ii) 3x – 5y – 11 = 0, 5x + 3y – 7 = 0 and x + 2y = 0
Solution:
(i) 15x – 18y + 1 = 0, 12x + 10y – 3 = 0 and 6x + 66y – 11 = 0
Given:
15x – 18y + 1 = 0 …… (i)
12x + 10y – 3 = 0 …… (ii)
6x + 66y – 11 = 0 …… (iii)
Solving equation (i) and (ii), we get,
From equation (i) we get,
x = (18y – 1)/15
Now substituting the value of x in equation (ii)
12 [(18y – 1)/15] + 10y – 3 = 0
216y – 12 + 150y – 45 = 0
366y = 57
y = 57/366 = 19/122
Now substituting the value of y in x i.e.
x = (18y – 1)/15
x = (18(19/122) – 1)/15
x = (342 – 122) / (122 × 15)
x = (342 – 122) / 1730
x = 220/1730
x= 22/173
Now substituting the value of x and y in equation (iii), we get,
6(22/173) + 66(19/122) – 11 = 0
(6 × 22 ×122) + (66 × 19 × 173) – (11 × 173 × 122) = 0
320 – 2052 + 732 = 0
0 = 0
Therefore, the given lines are concurrent.
(ii) 3x – 5y – 11 = 0, 5x + 3y – 7 = 0 and x + 2y = 0
Given:
3x − 5y − 11 = 0 …… (i)
5x + 3y − 7 = 0 …… (ii)
x + 2y = 0 …… (iii)
Solving equation (ii) and (iii), we get,
From equation (iii) we get,
x = -2y
Now substituting the value of x in equation (ii)
5(-2y) + 3y – 7 = 0
-10y + 3y – 7 = 0
-7y = 7
y = -1
Now substituting the value of y in x i.e.
x = -2y
x = -2(-1)
x = 2
Now substituting the value of x and y in equation (i), we get,
3(2) − 5(-1) − 11 = 0
6 + 5 – 11 = 0
11 – 11 = 0
0 = 0
Therefore, the given lines are concurrent.
### Question 2: For what value of λ are the three lines 2x – 5y + 3 = 0, 5x – 9y + λ = 0 and x – 2y + 1 = 0 concurrent?
Solution:
Given:
2x − 5y + 3 = 0 …… (i)
5x − 9y + λ = 0 …… (ii)
x − 2y + 1 = 0 …… (iii)
Solving equation (i) and (iii), we get,
From equation (i) we get,
2x = 5y – 3
x = (5y – 3)/2
Now substituting the value of x in equation (iii)
[(5y – 3)/2] – 2y + 1 = 0
5y – 3 – 4y + 2 = 0
y = 1
Now substituting the value of y in x i.e.
x = (5y – 3)/2
x = (5 – 3)/2
x = 2/2
x = 1
Now substituting the value of x and y in equation (ii), we get,
5(1) – 9(1) + λ = 0
5 – 9 + λ = 0
λ = 4
Therefore, the value of λ is 4.
### Question 3: Find the conditions that the straight lines y = m1x + c1, y = m2x + c2 and y = m3x + c3 may meet in a point.
Solution:
Given:
m1x – y + c1 = 0 …… (1)
m2x – y + c2 = 0 …… (2)
m3x – y + c3 = 0 …… (3)
Solving equation (i) and (ii), we get,
m1x – y + c1 = m2x – y + c2
m1x + c1 = m2x + c2
m1x – m2x = c2 – c1
x(m1 – m2) = c2 – c1
x = (c2 – c1)/(m1 – m2)
Now substituting the value of x in equation (i)
y = m1[(c2 – c1)/(m1 – m2)] + c1
y = m1c2 – m1c1 + m1c1 – m2c1
y = m1c2 – m2c1
Now substituting the value of x and y in equation (iii), we get,
m3x – y + c3 = 0
y = m3x + c3
m1c2 – m2c1= m3[(c2 – c1)/(m1 – m2)] + c3
m12c2 – m1m2c1 + m1m2c2 – m22c1 = m3c2 – m3c1 + m1c3 – m2c3
m12c2 – m1c3 – m22c1 + m2c3 – m3c2 + m3c1 = 0
m1(c2 – c3) + m2(c3 – c1) + m3(c1 – c2) = 0
Therefore, the required condition is m1(c2 – c3) + m2(c3 – c1) + m3(c1 – c2) = 0
### Question 4: If the lines p1x + q1y = 1, p2x + q2y = 1 and p3x + q3y = 1 be concurrent, show that the points (p1, q1), (p2, q2) and (p3, q3) are collinear.
Solution:
Given:
p1x + q1y = 1 …… (i)
p2x + q2y = 1 …… (ii)
p3x + q3y = 1 …… (iii)
Solving equation (i) and (ii), we get,
From equation (i) we get,
x = (1 – q1y)/p1
Now substituting the value of x in equation (ii)
p2[(1 – q1y)/p1] + q2y = 1
p2 – p2q1y + p1q2y = p1
y(p1q2 – p2q1) = p1 – p2
y = (p1 – p2)/(p1q2 – p2q1)
Now substituting the value of y in x i.e.
x = (1 – q1y)/p1
x = (1 – q1[(p1 – p2)/(p1q2 – p2q1)])/p1
Now substituting the value of x and y in equation (iii), we get,
p3[(p1q2 – p2q1 – q1(p1 – p2)(p1q2 – p2q1))] + q3p1(p1 – p2) = 1
(p1p3q2 – p2p3q1 – p1p3q1 + p2p3q1)(p1q1 – p2q1) + q3p1(p1 – p2) = 1
(p1p3q2 – p1p3q1)(p1q2 – p2q1) + q3p12 – q3p1p2 = 1
p12p3q22 – p1p2p3q1q2 – p12p3q1q2 + p1p2p3q12 + q3p1p2 = 1 …… (iv)
Also, if we assume points (p1, q1)(p2, q2)(p3, q3) are collinear
Therefore,
p1(q2 – q3) + p2(q3 – q1) + p3(q1 – q3) = 0
Now from equation (iv) we get,
p1[p1p3q22 – p2p3q1q2 – p1p3q1q2 + p2p3q12 + q3p2] = 1
p1[p3q2(p1q2 – p2q1) – p3q1(p1q2 – p2q1) + q3(p1 – p2)] = 1
Therefore, the given points, (p1, q1), (p2, q2) and (p3, q3) are collinear.
### Question 5: Show that the straight lines L1 = (b + c)x + ay + 1 = 0, L2 = (c + a)x + by + 1 = 0 and L3 = (a + b)x + cy + 1 = 0 are concurrent.
Solution:
Given:
L1 = (b + c)x + ay + 1 = 0 …… (i)
L2 = (c + a)x + by + 1 = 0 …… (ii)
L3 = (a + b)x + cy + 1 = 0 …… (iii)
Solving equation (i) and (ii), we get,
From equation (i) we get,
y = (-1 – (b + c)x)/a
Now substituting the value of y in equation (ii)
(c + a)x + b[(-1 – (b + c)x)/a] + 1 = 0
(c + a)x + b[(-1 – (bx + cx)/a] + 1 = 0
cx + ax + b[(-1 – (bx + cx)/a] + 1 = 0
acx + a2x – b – b2x + bcx + a = 0
x(ac + a2 – b2 + bc) = b – a
x(c(a – b) + (a – b)(a + b)) = b – a
x(a – b)(c + a + b) = -(a – b) [Dividing both side by (a – b)]
x(c + a + b) = -1
x = -1/(a + b + c)
Now substituting the value of x in y i.e.
y = (-1 – (b + c)x)/a
y = (-1 – (b + c)[-1/(a + b + c)])/a
y = (-(a + b + c) + b + c)/a(a + b + c)
y = (-a – b – c + b + c)/a(a + b + c)
y = (-a)/a(a + b + c)
y = -1/(a + b + c)
Now substituting the value of x and y in equation (iii), we get,
(a + b)[-1/(a + b + c)] + c[-1/(a + b + c)]+ 1 = 0
-a – b – c + a + b + c = 0
0 = 0
Therefore, the given lines are concurrent.
### Question 6: If the three lines ax + a2y + 1 = 0, bx + b2y + 1 = 0, and cx + c2y + 1= 0 are concurrent, show that at least two of three constants a, b, c are equal.
Solution:
Given:
ax + a2y + 1 = 0 …… (i)
bx + b2y + 1 = 0 …… (ii)
cx + c2y + 1= 0 …… (iii)
Solving equation (i) and (ii), we get,
From equation (i) we get,
x = (-1 – a2y)/a
Now substituting the value of x in equation (ii)
b[(-1 – a2y)/a] + b2y + 1 = 0
-b – a2by + ab2y + a = 0
aby(b – a) = b – a [Dividing both side by (b – a)]
aby = 1
y = 1/ab
Now substituting the value of y in x i.e.
x = (-1 – a2y)/a
x = (-1 -a2(1/ab))/a
x = (-b – a)/ba
Now substituting the value of x and y in equation (iii), we get,
c[(-b – a)/ba] + c2(1/ab) + 1 = 0
-bc – ac + c2 + ab = 0
c(c – b) – a(c – b) = 0
(c – b)(c – a) = 0
c – b = 0
c = b
or
c – a = 0
c = a
Therefore, at least two of three constants a, b, c are equal.
### Question 7: If a, b, c are in A.P. , prove that the straight lines ax + 2y + 1 = 0, bx + 3y + 1 = 0 and cx + 4y + 1 = 0 are concurrent.
Solution:
Given if a, b, c are in A.P.
Thus, b – a = c – b
2b = a + c [Common difference] …… (i)
Also given:
ax + 2y + 1 = 0 …… (ii)
bx + 3y + 1 = 0 …… (iii)
cx + 4y + 1 = 0 …… (iv)
Solving equation (ii) and (iii), we get,
From equation (ii) we get,
x = (-1 – 2y)/a
Now substituting the value of x in equation (iii)
b[(-1 – 2y)/a] + 3y + 1 = 0
-b – 2by + 3ay + a = 0
y(3a – 2b) = b – a
y = (b – a)/(3a – 2b)
Now substituting the value of y in x i.e.
x = (-1 – 2y)/a
x = (-1 – 2[(b – a)/(3a – 2b)])/a
x = (-(3a – 2b) – 2b + 2a)/a(3a – 2b)
x = -1/(3a – 2b)
Now substituting the value of x and y in equation (iv), we get,
c[-1/(3a – 2b)] + 4[(b – a)/(3a – 2b)] + 1 = 0
-c + 4b – 4a + 3a – 2b = 0
-a + 2b – c = 0
From equation (i) we know, 2b = a + c,
Thus, -a + a + c – c = 0
0 = 0
Therefore, the given lines are concurrent.
Related Tutorials | 3,820 | 7,845 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2023-40 | latest | en | 0.70163 |
https://tcsmath.wordpress.com/tag/chromatic-number/ | 1,656,275,139,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103271864.14/warc/CC-MAIN-20220626192142-20220626222142-00071.warc.gz | 620,548,955 | 19,768 | # Lecture 6: Borsuk-Ulam and some combinatorial applications
Now we’ll move away from spectral methods, and into a few lectures on topological methods. Today we’ll look at the Borsuk-Ulam theorem, and see a stunning application to combinatorics, given by Lovász in the late 70’s. A great reference for this material is Matousek’s book, from which I borrow heavily. I’ll also discuss why the Lovász-Kneser theorem arises in theoretical CS.
### The Borsuk-Ulam Theorem
We begin with a statement of the theorem. We will think of the n-dimensional sphere as the subset of ${\mathbb R^{n+1}}$ given by
${S^n = \left\{ (x_1, \ldots, x_{n+1}) : x_1^2 + \cdots + x_{n+1}^2 = 1\right\}.}$
Borsuk-Ulam Theorem: For every continuous mapping ${f : S^n \to \mathbb R^n}$, there exists a point ${x \in S^n}$ with ${f(x)=f(-x)}$.
Pairs of points ${x,-x \in S^n}$ are called antipodal.
There are a couple of common illustrative examples for the case ${n=2}$. The theorem says that if you take the air out of a basketball, crumple it (no tearing), and flatten it out, then there are two points directly on top of each other which were antipodal before. Another common example states that at every point in time, there must be two points on the earth which both have exactly the same temperature and barometric pressure (assuming, of course, that these two parameters vary continuously over the surface of the eath).
The n=1 case is completely elementary. For the rest of the lecture, let’s use ${N = (0,0,\ldots,1)}$ and ${S=(0,0,\ldots,-1)}$ to denote the north and south poles (the dimension will be obvious from context). To prove the n=1 case, simply trace out the path in ${\mathbb R}$ starting at ${f(N)}$ and going clockwise around ${S^1}$. Simultaneously, trace out the path starting at ${f(S)}$ and going counter-clockwise at the same speed. It is easy to see that eventually these two paths have to collide: At the point of collision, ${f(x)=f(-x)}$.
We will give the sketch of a proof for ${n \geq 2.}$ Let ${g(x)=f(x)-f(-x)}$, and note that our goal is to prove that ${g(x)=0}$ for some ${x \in S^n}$. Note that ${g}$ is antipodal in the sense that ${g(x)=-g(-x)}$ for all ${x \in S^n}$. Now, if ${g(x) \neq 0}$ for every ${x}$, then by compactness there exists an ${\epsilon > 0}$ such that ${\|g(x)\| > \epsilon}$ for all ${x \in S^n}$. Because of this, we may approximate ${g}$ arbitrarily well by a smooth map, and prove that the approximation has a 0. So we will assume that ${g}$ itself is smooth.
Now, define ${h : S^n \to \mathbb R^n}$ by ${h(x_1, \ldots, x_{n+1}) = (x_1, \ldots, x_n)}$, i.e. the north/south projection map. Let ${X = S^n \times [0,1]}$ be a hollow cylinder, and let ${F : X \to \mathbb R^n}$ be given by ${F(x,t)=t g(x) + (1-t)h(x)}$ so that ${F}$ linearly interpolates between ${g}$ and ${h}$.
Also, let’s define an antipodality on ${X}$ by ${\nu(x,t) = (-x,t)}$. Note that ${F}$ is antipodal with respect to ${\nu}$, i.e. ${F(\nu(x,t))=-F(x,t)}$, because both ${g}$ and ${h}$ are antipodal. For the sake of contradiction, assume that ${g(x) \neq 0}$ for all ${x \in S^n}$.
Now let’s consider the structure of the zero set ${Z = F^{-1}(0)}$. Certainly ${(N,0), (S,0) \in Z}$ since ${h(N)=h(S)=0}$, and these are h’s only zeros. Here comes the sketchy part: Since ${g}$ is smooth, ${F}$ is also smooth, and thus locally ${F}$ can be approximated by an affine mapping ${F_{\mathrm{loc}} : \mathbb R^{n+1} \to \mathbb R^n}$. It follows that if ${F_{\mathrm{loc}}^{-1}(0)}$ is not empty, then it should be a subspace of dimension at least one. By an arbitrarily small perturbation of the initial ${g}$, ensuring that ${F}$ is sufficiently generic, we can ensure that ${F_{\mathrm{loc}}^{-1}(0)}$ is either empty or a subspace of dimension one. Thus locally, ${F^{-1}(0)}$ should look like a two-sided curve, except at the boundaries ${t=0}$ and ${t=1}$, where ${F^{-1}(0)}$ (if non-empty) would look like a one-sided curve. But, for instance, ${F^{-1}(0)}$ cannot contain any Y-shaped branches.
It follows that ${Z}$ is a union of closed cycles and paths whose endpoints must lie at the boundaries ${t=0}$ and ${t=1}$. (${Z}$ is represented by red lines in the cylinder above.) But since there are only two zeros on the ${t=0}$ sphere, and none on the ${t=1}$ sphere, ${Z}$ must contain a path ${\gamma}$ from ${(N,0)}$ to ${(S,0).}$ Since ${F}$ is antipodal with respect to ${\nu}$, ${\gamma}$ must also satisfy this symmetry, making it impossible for the segment initiating at N to ever meet up with the segment initiating at S. Thus we arrive at a contradiction, implying that ${g}$ must take the value 0.
Notice that the only important property we used about ${h}$ (other than its smoothness) is that is has a number of zeros which is twice an odd number. If ${h}$ had, e.g. four zeros, then we could have two ${\nu}$-symmetric paths emanating from and returning to the bottom. But if ${h}$ has six zeros, then we would again reach a contradiction. | 1,514 | 5,017 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 79, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2022-27 | latest | en | 0.893134 |
https://www.jiskha.com/questions/1056117/How-would-I-answer-this-question-7-is-subtracted-from-one-third-of-r | 1,558,757,910,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257847.56/warc/CC-MAIN-20190525024710-20190525050710-00144.warc.gz | 822,630,104 | 5,411 | # Math
How would I answer this question? 7 is subtracted from one-third of r.
1. 👍 0
2. 👎 0
3. 👁 66
1. (1/3)r - 7
1. 👍 0
2. 👎 0
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More Similar Questions | 919 | 2,725 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2019-22 | latest | en | 0.931018 |
http://mathhelpforum.com/differential-geometry/133104-compactness.html | 1,481,374,146,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543170.25/warc/CC-MAIN-20161202170903-00040-ip-10-31-129-80.ec2.internal.warc.gz | 179,632,412 | 14,244 | # Thread: Compactness
1. ## Compactness
I fell like this question is simple, but I just can't get my head around it:
Give an example of a space with two points in which not all compact sets are closed.
Thanks to anyone who replies!
2. Originally Posted by Showcase_22
I fell like this question is simple, but I just can't get my head around it:
Thanks to anyone who replies!
Well there are only three topologies you can put on it. The discrete one won't work as all sets are closed, the trivial topology won't work either. The only option is $\{\emptyset,\{a\},\{a,b\}\}$. There is only one set that is open but not closed, and only one sequence in it.
3. What do you mean by "and only one sequence in it"?
Also, is the "open but not closed" set the empty set? I was told that the empty set was simultaneously closed and open, so what's the difference between "simultaneously closed and open" and "open but not closed"?
4. If a set is both open and closed, then obviously it can't be open and not closed..
Do you understand why in any topological space $X$, both $X$ and $\phi$ are open and closed?
The set he is talking about is $\{a\}$. Do you see why it is open and not closed? And since it only has one point in it, do you see why it has only one sequence?
Now, can you finish?
5. Do you understand why in any topological space , both and are open and closed?
I think I do:
$\phi$ is open since the interior of $\phi$ is empty. If a set is equal to it's interior then it's open.
$\phi$ is closed since all it's boundary points are in $\phi$ since it doesn't have any.
Once we have this, $X^c=\phi$ and since $\phi$ is either open or closed, we get that $X^c$ is also either open or closed.
The set he is talking about is . Do you see why it is open and not closed? And since it only has one point in it, do you see why it has only one sequence?
$\{a\}$ is open because it is equal to it's interior. However, $\{a\}$ is surely closed? $\{a\}^c=\{b\}$ (since it's a two point space) and, by the same argument for why $\{a\}$ was open, $\{b\}$ is open. We have a complement of a set as open, so $\{a\}$ must be closed.
Why isn't that right?
And since it only has one point in it, do you see why it has only one sequence?
I see why, if you've only got one point any sequence would just be that point repeatedly!
$\{a\}$ has got to be closed since the only sequence is just $a$ repeated, so it has a convergent subsequence. This is just another sequence of just $a$'s, so it converges to $a$.
We know that every convergent sequence has a convergent subsequence iff the set is compact. Therefore $\{a\}$ is compact.
6. Originally Posted by Showcase_22
$\{a\}$ is open because it is equal to it's interior. However, $\{a\}$ is surely closed? $\{a\}^c=\{b\}$ (since it's a two point space) and, by the same argument for why $\{a\}$ was open, $\{b\}$ is open. We have a complement of a set as open, so $\{a\}$ must be closed.
Why isn't that right?
Your notion of interior is flawed. Interior is the largest open subset of the set and I (or you) define the open sets with the topology. So when I have {a} in the topology but not {b}, that means that the largest open subset of {b} is the empty set.
A better way to think about open and closed sets is through topology. The whole space and the empty set are open by definition. A set is closed, if and only if its complement is open (in this case $\{b\}=\{a\}^c$ which is not open so {a} is not closed).
7. Originally Posted by Showcase_22
We know that every convergent sequence has a convergent subsequence iff the set is compact. Therefore $\{a\}$ is compact.
Even more egregious is the above statement. Do not think about compactness as the above unless you are in metric spaces.
As Focus said there are only three topologies on $\{a,b\}$ up to homeomorphism. They are namely $\left\{\varnothing,\{a,b\}\right\},\left\{\varnoth ing,\{a\},\{b\},\{a,b\}\right\}$ and $\left\{\varnothing,\{a\},\{a,b\}\right\}$.
The first is out since every non-open set isn't closed.. The last is out since it is metrizable under the discrete metric (and thus Hausdorff and so every compact subspace is closed). So it remains that we must choose $\left\{\varnothing,\{a\},\{a,b\}\right\}$. Now, clearly $\{a\}$ as a subspace has a finite topology and thus by necessity is compact. But, it is open since it is in the topology but it is not closed since $X-\{a\}=\{b\}$ which is not in the topology.
8. Oh, I think I get it!
So when you write $
\{\emptyset,\{a\},\{a,b\}\}
$
, and define this to be your topology, all the sets in this set are defined to be open. I haven't encountered it written like this before.
So, returning to the question, you could have the set $\{2,5\}$ where $
\{\emptyset,\{2\},\{2,5\}\}
$
is your topology (ie. all the sets defined to be open).
$\{2\}$ is a compact set because every sequence would be of the form $a_n=2$ $\forall n \in \mathbb{N}$. This gives that every subsequence converges since every subsequence $a_{n_i}=2$ (ie. converges to 2).
$\{2\}$ is not closed since $\{2\}^c=\{5\}$ and this is not open since $\{5\}$ is not in the topology (is this the correct way of saying this?).
Alternatively, looking at the definition, a set is closed $\Leftrightarrow$ it's complement is open. Since the complement is not open then our original set is not closed (i'm trying to think of it multiple ways).
9. Originally Posted by Showcase_22
Oh, I think I get it!
So when you write $
\{\emptyset,\{a\},\{a,b\}\}
$
, and define this to be your topology, all the sets in this set are defined to be open. I haven't encountered it written like this before.
So, returning to the question, you could have the set $\{2,5\}$ where $
\{\emptyset,\{2\},\{2,5\}\}
$
is your topology (ie. all the sets defined to be open).
Everything is fine up to here!
$\{2\}$ is a compact set because every sequence would be of the form $a_n=2$ $\forall n \in \mathbb{N}$. This gives that every subsequence converges since every subsequence $a_{n_i}=2$ (ie. converges to 2).
Wrong reasoning, but the correct result. In a general topological space, compactness (Every open cover has a finite subcover) does NOT necessarily ensure sequential compactness (every sequence has a convergent subsequence), for example $X = \{0,1\}^{[0,1]}$ with the product topology.
The right reasoning here is that every open covering of $\{a\}$ is finite, and thus you always have a finite subcover.
$\{2\}$ is not closed since $\{2\}^c=\{5\}$ and this is not open since $\{5\}$ is not in the topology (is this the correct way of saying this?).
Alternatively, looking at the definition, a set is closed $\Leftrightarrow$ it's complement is open. Since the complement is not open then our original set is not closed (i'm trying to think of it multiple ways).
Correct.
As a final note, you will probably later learn that in any metric space $(X, d)$, the following 3 conditions are equivalent:
1) X is compact
2) X is sequentially compact
3) X is Limit-point compact (every infinite subset of X has a limit point)
So you would actually be better off not thinking of a general topological space as a generic metric space!
10. In a general topological space, compactness (Every open cover has a finite subcover) does NOT necessarily ensure sequential compactness (every sequence has a convergent subsequence), for example with the product topology.
I'm afraid I don't follow this =S
Here compactness does not imply sequential compactness, but I know that a set is compact iff a set is sequentially compact.
Are you saying that, in general, compactness does not imply sequential compactness but a set is compact iff a set is sequentially compact holds in a euclidean space?
$
X = \{0,1\}^{[0,1]}
$
with product topology.
Also, what does this mean? I haven't seen this before =S
Is this the cartesian product of a set $\{0,1\}$ with $[0,1]$?
Elements would be like $(0,1), \ \left( 1, \frac{1}{2} \right)$ as in the x value is either 0 or 1 and the y value is any number in $[0,1]$.
If this is right, what is product topology?
11. Originally Posted by Showcase_22
I'm afraid I don't follow this =S
Here compactness does not imply sequential compactness, but I know that a set is compact iff a set is sequentially compact.
Are you saying that, in general, compactness does not imply sequential compactness but a set is compact iff a set is sequentially compact holds in a euclidean space?
Yes, that is correct, but also for any metric space and not just a Euclidean one (for example $C[[0,1], \mathbb{R}]$, the set of all continuous functions from $[0,1]$ to $\mathbb{R}$).
Though that brings up the question, have you studied topology yet?
Also, what does this mean? I haven't seen this before =S
Is this the cartesian product of a set $\{0,1\}$ with $[0,1]$?
Elements would be like $(0,1), \ \left( 1, \frac{1}{2} \right)$ as in the x value is either 0 or 1 and the y value is any number in $[0,1]$.
If this is right, what is product topology?
Well, in set theory, if you have $A,B$ be sets then the notation $B^A$ stands for the set of all functions from A to B.
So generally speaking, $\{0,1\}^{[0,1]}$ is the set of all functions from the unit interval to $\{0,1\}$. Equipping it with the product topology is another way of looking at it from a topological point of view - as a space of uncountably many cartesian products of the set $\{0,1\}$, but now I guess that example is too advanced for what you have learned.
An element $a$ in that set would be of the form $a = \{a_i : a_i \in \{0,1\} \ \forall i \in [0,1]\}$
Also, wikipedia provides a good first read on the product topology, if you're interested. By the way, what course was this question a part of?
12. I'm studying topology as part of my "metric spaces" course. I think it's interesting, i'm just not any good at it =S
This question was from a book I'm reading to try and understand it better. It is helping, the big problem is that it doesn't have the answers in the back of the book!
Anyway, I understand it a lot better since creating this thread. Thanks so much! | 2,737 | 10,102 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 81, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2016-50 | longest | en | 0.987708 |
https://www.slideserve.com/thuong/a-brief-review-powerpoint-ppt-presentation | 1,726,598,044,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651829.57/warc/CC-MAIN-20240917172631-20240917202631-00836.warc.gz | 946,703,086 | 21,816 | 1 / 60
A brief review
A brief review. The exponential distribution. The memoryless property. Exponentially distributed random variables are memoryless. The exponential distribution is the only distribution that has the memoryless property. The minimum of n exponentially distributed random variables.
A brief review
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Presentation Transcript
1. A brief review
2. The exponential distribution
3. The memoryless property
4. The exponential distribution is the only distribution that has the memoryless property.
5. The minimum of n exponentially distributed random variables Suppose that X1, X2, ..., Xn are independent exponential random variables, with Xi having rate li, i=1, ..., n. What is P(min(X1, X2, ..., Xn )>x)?
6. Comparing two exponentially distributed random variables Suppose that X1 and X2 are independent exponentially distributed random variables with rates l1 and l2. What is P(X1 < X2)?
7. The Poisson process The counting process {N(t) t ≥ 0} is said to be a Poisson process having rate l, l > 0, if (i)N(0) = 0. (ii) The process has independent increments. (iii) The number of events in any interval of length t is Poisson distributed with mean lt. That is for all s, t ≥ 0
8. The distribution of interarrival times for a Poisson process Let Tn denote the inter-arrival time between the (n-1)th event and the nth event of a Poisson process, then the Tn (n=1, 2, ...) are independent, identically distributed exponential random variables having mean 1/l.
9. A CTMC is a continuous time analog to a discrete time Markov chain • A CTMC is defined with respect to a continuous time stochastic process {X(t): t ≥0} • If X(t) = i the process is said to be in state i at time t • {i: i=0, 1, 2, ...} is the state space
10. A stochastic process {X(t): t ≥0} is a continuous time Markov chain if for all s, t , u ≥0 and 0 ≤ u < s • P{X(t+s)=j|X(s)=i, X(u)=x(u), 0 ≤ u < s} = P{X(t+s)=j|X(s)=i}
11. A CTMC is said to have stationary transition probabilities if • P{X(t+s)=j|X(s)=i} • is independent of s (and depends only on t).
12. A CTMC is said to have stationary transition probabilities if • P{X(t+s)=j|X(s)=i} • is independent of s (and depends only on t). • Pij(t) = P{X(t+s)=j|X(s)=i}
13. A CTMC is said to have stationary transition probabilities if • P{X(t+s)=j|X(s)=i} • is independent of s (and depends only on t). • Pij(t) = P{X(t+s)=j|X(s)=i} • Note: We shall always assume that the stationary property holds
14. Sojourn times • Ti: time the process spends in state i once it enters state i (the length of a visit to state i, a random variable).
15. Sojourn times • Ti: time the process spends in state i once it enters state i (the length of a visit to state i, a random variable). • Example 1:X(0)=2, the first three transitions occur at t1 =3, t1 = 4.2 and t1 = 6.5 and X(3)=4, X(4.2) = 2 and X(6.5) =1.
16. Sojourn times • Ti: time the process spends in state i once it enters state i (the length of a visit to state i, a random variable). • Example 1:X(0)=2, the first three transitions occur at t1 =3, t2 = 4.2 and t3 = 6.5 and X(3)=4, X(4.2) = 2 and X(6.5) =1. • The first sojourn time in state 4 = 4.2-3 = 1.2 • The second sojourn time in state 2 = 6.5 – 4.2 = 1.3
17. Example 2: Suppose the process has been in state 3 for 10 minutes, what is the probability that it will not leave state 3 in the next 5 minutes.
18. Example 2: Suppose the process has been in state 3 for 10 minutes, what is the probability that it will not leave state 3 in the next 5 minutes. • P(T3 > 15|T3> 10) = P(T3 > 5)
19. More generally, • P(Ti > s+t|Ti> s) = P(Ti > t) • Ti is memoryless and therefore has the exponential distribution
20. An alternative definition of a CTMC A CTMC is a stochastic process having the properties that each time it enters a state i (i) the amount of time it spends in that state before making a transition into a different state is exponentially distributed with some mean 1/vi (or transition rate vi ) (ii) when the process leaves state i, it next enters state j with some probability Pij(Pii=0 and SjPij=1, for all i)
21. A CTMC is a stochastic process that moves from state to state according to a probability transition matrix (similar to a discrete time Markov chain) , but the amount of time it spends in each state is exponentially distributed.
22. To define a CTMC, we need to define a state space, a probability transition matrix, and a set of transition rates.
23. Example 1: Customers arrive to a store according to a Poisson process with rate l. Let N(t) be the total number of customers that have arrived by time t.
24. Example 1: Customers arrive to a store according to a Poisson process with rate l. Let N(t) be the total number of customers that have arrived by time t. State space is {0, 1, 2, ...}; Ti is exponentially distributed with mean 1/l Pij =1 if j=i+1 and Pij = 0 otherwise
25. Example 2: Customers arrive to an airline check-in counter according to a Poisson process with rate l. The time it takes the single agent at the counter to check-in a customer is exponentially distributed with mean 1/m.
26. Example 2: Customers arrive to an airline check-in counter according to a Poisson process with rate l. The time it takes the single agent at the counter to check-in a customer is exponentially distributed with mean 1/m. State space is {0, 1, 2, ...} P0,1 = 1; Pi,i+1 = l/(l+m); Pi,i-1 = m/(l+m) T0 =1/l v0 =l ; Ti = 1/(l+m) vi =l + m for i =1, 2, ...
27. Example 2: Customers arrive to an airline check-in counter according to a Poisson process with rate l. The time it takes the single agent at the counter to check-in a customer is exponentially distributed with mean 1/m. State space is {0, 1, 2, ...} P0,1 = 1; Pi,i+1 = l/(l+m); Pi,i-1 = m/(l+m) T0 =1/l v0 =l ; Ti = 1/(l+m) vi =l + m for i =1, 2, ... The above is an example of an M/M/1 queue. The M/M/1 queue is an example of a birth and death process.
28. Example 2: Customers arrive to an airline check-in counter according to a Poisson process with rate l. The time it takes the single agent at the counter to check-in a customer is exponentially distributed with mean 1/m. State space is {0, 1, 2, ...} P0,1 = 1; Pi,i+1 = l/(l+m); Pi,i-1 = m/(l+m) T0 =1/l v0 =l ; Ti = 1/(l+m) vi =l + m for i =1, 2, ... The above is an example of an M/M/1 queue. The M/M/1 queue is an example of a birth and death process.
29. Birth and death process Example 3: Customers arrive to a service center according to a Poisson process with rate lnwhen there are n customers in the system. Customers take an amount Tn that is exponentially distributed with mean 1/mn when there are n customers in the system.
30. Birth and death process Example 3: Customers arrive to a service center according to a Poisson process with rate lnwhen there are n customers in the system. Customers take an amount Tn that is exponentially distributed with mean 1/mn when there are n customers in the system. State space is {0, 1, 2, ...} P0,1 = 1; Pn,n+1=ln/(ln+mn); Pn,n-1=mn/(ln+mn) T0 =1/l0v0 =l0; Tn = 1/(ln +mn) vn =ln+mnfor n =1, 2, ...
31. State transition diagrams for a B&D process l0 l1 l2 0 1 2 3 m1 m2 m3
32. The Poisson process is a birth and death process with rate ln=l and mn=0. • The M/M/1 queue is described by a birth and death process with rate ln=l and mn=m.
33. Example 4: Customers arrive to an airline check-in counter according to a Poisson process with rate l. The time it takes one of the m agents at the counter to check-in a customer is exponentially distributed with mean 1/m.
34. Example 4: Customers arrive to an airline check-in counter according to a Poisson process with rate l. The time it takes one of the m agents at the counter to check-in a customer is exponentially distributed with mean 1/m. The system can be modeled as a birth and death process with transition rates ln = l; mn = nm if 1 ≤ n < m mn = mm if n ≥ m
35. Example 4: Customers arrive to an airline check-in counter according to a Poisson process with rate l. The time it takes one of the m agents at the counter to check-in a customer is exponentially distributed with mean 1/m. The system can be modeled as a birth and death process with transition rates ln = l; mn = nm if 1 ≤ n < m mn = mm if n ≥ m The above is an example of an M/M/m queue.
36. Transition rates vi: rate with which the process leaves state i (once it enters state i) qij: rate with which the process goes state j (once it enters state i) qij = Pijvi (qij is also called the instantaneous transition rate from state i to j)
37. Transition rates vi: rate with which the process leaves state i (once it enters state i) qij: rate with which the process goes state j (once it enters state i) qij = Pijvi (qij is also called the instantaneous transition rate from state i to j) vi =Sj viPij= Sj qij Pij =qij/vi =qij/Sj qij
38. Transition rates vi: rate with which the process leaves state i (once it enters state i) qij: rate with which the process goes state j (once it enters state i) qij = Pijvi (qij is also called the instantaneous transition rate from state i to j) vi =Sj viPij= Sj qij Pij =qij/vi =qij/Sj qij Specifying the instantaneous transition rates determines the parameters of the CTMC
39. State transition diagrams q0,2 q0,1 q1,2 0 1 2 q2,0 q2,1 q2,0
40. Properties It can also be shown that
41. The Chapman-Kolmogrov equations
42. The Chapman-Kolmogrov equations
43. The Chapman-Kolmogrov equations
44. The Chapman-Kolmogrov equations
45. The Chapman-Kolmogrov equations
46. The Chapman-Kolmogrov equations
47. Kolmogrov’s backward equations
48. Kolmogrov’s backward equations
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https://mariewallacerealestateagentgranitebayca.com/college-901 | 1,675,399,172,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500042.8/warc/CC-MAIN-20230203024018-20230203054018-00442.warc.gz | 395,707,343 | 5,656 | # Solve limits algebraically
There are also many YouTube videos that can show you how to Solve limits algebraically. We can solve math word problems.
## Solving limits algebraically
When you try to Solve limits algebraically, there are often multiple ways to approach it. There are a few different ways to solve a compound inequality, but one of the most common methods is to use a compound inequality solver. This is a tool that allows you to input two or more inequalities and then outputs the solution. There are many different types of compound inequality solvers, but they all work in basically the same way. One method for solving a compound inequality is to graph the inequalities on a coordinate plane. This can be done by using a graphing calculator
If you're struggling with trigonometry, there's no shame in using a trigonometry solver. These online tools can quickly and easily help you solve any trigonometry problem, no matter how difficult it may be. Just enter the problem into the solver and it will provide you with the solution. You can even use the solver to check your work to make sure you're on the right track.
An equation solver for x is an online tool that allows you to input an equation and determine the value of x. This can be useful for solving math problems or for checking your work. There are a variety of equation solvers available, so be sure to choose one that is appropriate for your needs.
There are a few different ways that you can solve square roots, depending on what you are given. If you are given a perfect square, then you can just take the square root of that number. For example, the square root of 9 is 3, because 3 squared is 9. However, if you are not given a perfect square, then you will need to use a different method. One method is to use estimation. To do this, you look at the number and find
Radical Expressions Radical expressions are mathematical expressions that contain a square root, cube root, or other type of root. In order to solve a radical expression, you need to find the value of the root. This can be done by using a calculator, or by using estimation. Once you have the value of the root, you can then use algebra to solve the expression. | 468 | 2,219 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2023-06 | latest | en | 0.946618 |
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## 9th Class Physics Chapter 3 Exercise Short Questions with Answers
Q 3.2 : Define the following terms
(i)
Inertia :
” Inertia of a body is the property due to which it resist any change in its state of rest or motion .”
• Inertia of a body is related to the mass of body .
• Greater is the mass of body , greater is its inertia .
(ii)
Momentum :
” Momentum of a body is the quantity of motion it possesses due to its mass and velocity .”
or
” Momentum of a body is equal to the product of mass and velocity .”
Formula : P = m v
Unit : Ns or kgms-1
(iii)
Force :
” A force is a push or pull . It moves or tends to move , stp or tends to stop the motion of a body . The force can also change the direction of motion of a body . “
Symbol : F
Unit : Newton (N)
(iv)
Force of Friction :
” The force that opposes the motion of moving objects is called friction .”
Examples :
• Applying brakes to stop a moving vehicle .
• Writing on a notebook / board .
(v)
What is differentiate between :
(ii)
(iii)
Q 3.4 : What is the law of inertia ?
Law of Inertia :
Newton’s First law of Motion deals with the inertial property of matter , so Newton’s First Law of Motion is also known as Law of Inertia .
Statement :
” A body continues its state of rest or of uniform motion in a straight line provided no net force acts on it .”
Q 3.5 : Why is it dangerous to travel on the roof of bus ?
If a person travels on the roof of a bus , it would be dangerous .
Reason : Because when a bus takes a sharp turn , passengers fall in outward direction . It is due to inertia that they want to continue their motion in a straight line and thus fall outwards .
Q 3.6 : Why does a passenger move outward when a bus takes a turn ?
When a bus takes a sharp turn passengers fall in the outward direction .
Reason : It is due to inertia that they want to continue their motion in a straight line and thus fall outwards .
Q 3.7 : How can you relate a force with the change of the momentum of a body ?
” When a force acts on a body , it produces an acceleration in the body and will be equal to rate of change of momentum of a body .”
Q 3.8 : What will be the tension in a rope that is pulled from its ends by two opposite forces 100 N each ?
The tension in a rope that is pulled from its ends by opposite forces 100 N each will be 100 N.
Q 3.9 : Action and reaction are always equal and opposite then how does body move ?
Action and Reaction force always act on different bodies , so they do not cancel the effect of each other .
→ So , we apply Action force on the earth in backward direction , and earth apply Reaction force in forward direction .
→ This is how we move .
Q 3.10 : A horse pulls the cart . If the action and reaction are equal and opposite then how does the cart move ?
The horse applies action force by feet on the road , the reaction is given by the road on horse , due to which horse moves .
• The cart , which is tied with the horse , also moves .
Since , action and reaction never acts on same body , so the cart moves.
Q 3.11 : What is the law of conservation of momentum ?
Law of conservation of Momentum :
” The momentum of an isolated system of two or more than two interacting bodies remains constant .”
Examples : Firing a bullet , Release of air from balloon etc .
Mathematical : m1v1 + m2v2 = m1u1 + m2u2
Q 3.12 : Why is the law of conservation of momentum important ?
By using law of conservation of momentum , we can calculate :
• Force
• Velocity
• Acceleration
→ Most of elementary particles are discovered by the use of this law .
Q 3.13 : When a gun is fired , it recoils . Why ?
Reason : When a gun is fired it recoils to conserve momentum .
Explanation :
• When the bullet is fired from the gun , it gives a large velocity to the bullet in the forward direction .
• So , no external force acts on the system , so the momentum of the system (gun + bullet ) must be zero after firing .
→ Thus , gun moves backwards with a momentum equal to the momentum of the bullet .
Q 3.14 : Describe two situations in which force of friction is needed .
1. We need friction to write on the paper .
2. We need friction to walk on the ground .
3. Birds need friction to fly in the air .
4. Friction is needed to stop moving vehicles .
Q 3.15 : How does oiling the moving parts of a machine lowers friction ?
Oiling the moving parts of a machine lowers friction because the oil fills up all the rough spot (cold welds ) and make the surface smooth .
Q 3.16 : Describe ways to reduce friction .
Methods of Reducing Friction :
• Making the sliding surfaces smooth .
• Making the fast moving objects a streamline shape (fish shape ) such as cars , aero planes etc.
• Lubricating the sliding surfaces .
• Using ball bearings or roller bearings . Because rolling friction is less than sliding friction .
Q 3.17 : Why rolling friction is less than sliding friction ?
Reason :
In rolling friction , the area of contact between two objects is comparatively less than in case of sliding .
Hence , area of contact is less ,, so force of rolling friction is less than sliding friction .
Q 3.18 : What you know about the following :
(i)
Tension in a string :
” The force acting along a string causes tension in the string .”
Example :
When a weight is suspended by a string . Tension produces by a string . It is equal to the weight but opposite in direction .
(ii)
Limiting force of friction :
“The maximum value of friction is known as the Force of limiting friction . “
→ Denoted by Fs .
(iii)
Braking Force :
” The force which is required to overcome the limiting friction is called braking force .”
→ It is basically a measure of braking power of a vehicle .
(iv)
Skidding of vehicles :
” The act of sliding or slipping of a vehicle over a surface , often sideways without revolving is called skidding .”
→ It happen when a force of friction between tyres and road is small , then on applying brakes to moving vehicle, its tyre slide over road .
(v)
Seatbelts:
A belt or strap in an automobile or airplanes that is used for safety .
In case of an accident or sudden stop , seat belts holds the passenger in their seats .
(vi)
The phenomenon of raising outer edge of the curved road above the inner edge is to provide necessary centripetal force to the vehicle to take safer turn on the curve road is called banking of roads .
(vii)
Cream Separator :
A device which is used to separate cream from milk .
Q 3.19 : What would happen if all friction suddenly disappears ?
• If there was no friction , Nothing would steady on ground and nothing would exist in the way they do now .
→Birds cannot fly .
→ We cannot walk on ground .
→ We cannot write on the paper etc.
Q 3.20 : Why the spinner of a washing machine is made to spin at a very high speed ?
Reason :
Because when it spins at high speed , the water from wet clothes is forced out through these holes due to lack of centripetal force .
9th Class Physics Chapter 2 Exercise Short Questions with Answers
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List here
Just like addition and subtraction, multiplication ($x$, $.$) and division (÷, $/$) are opposites of each other. Multiplying by a number and then dividing by the same number gets us back to the start again:
• $a$ × $b / b = a$
• $5$ x $4 / 4 = 5$
Sometimes you will see a multiplication of letters as a dot or without any symbol. Don’t worry, its exactly the same thing. Mathematicians are efficient and like to write things in the shortest, neatest way possible.
• $abc = a$×$b$×$c$
• $a$·$b$·$c = a$×$b$×$c$
It is usually neater to write known numbers to the left, and letters to the right. So although 4$x$ and $x$4 are the same thing, it looks better to write 4$x$. In this case, the “4” is a constant that is referred to as the coefficient of x.
# Extension exercise
Commutativity for Multiplication
The fact that $ab = ba$ is known as the commutative property of multiplication. Therefore, both addition and multiplication are described as commutative operations.
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Mean simply can be understood as the mathematical average of a set of two or more numbers. The mean for a given set of numbers can be determined in more than one way, including the arithmetic mean method, which is employed to compute the sum of the numbers in the series, and the geometric mean approach, which is the average of a set of products.
Mean deviation is a measure of the central tendency. It explains to us how far all the observations are from the middle, on average?. In statistics and mathematics, the deviation is a measure that is applied to determine the difference between the observed value and the expected value of a variable. With this article on Mean Deviation, you will learn about what is the mean deviation? through definition, formula and calculation.
## What is Mean Deviation?
The mean deviation method is specified for a central value ‘X’ and is the mean of absolute values of the deviations of the observations from the central value ‘X’. Every deviation is an absolute deviation as it is an absolute value which indicates that we neglect the negative signs.
In simple words, the deviation is the length from the center point. Furthermore, the mean deviation is applied to determine how far the values fall from the middle of the data set. We can estimate the mean deviation from any size of central tendency. But, usually, in the statistical study mean deviation from mean and median are used. Also, the deviations on both sides of the Mean shall be equal. Let us begin with the mean deviation formula in detail.
Know more about Mean Median Mode here.
## Mean Deviation Formulas
The mean deviation is defined as the mean of the absolute deviations of the observations/ values from a proper average. This average may be the mean, median or mode. We further identify it as the mean absolute deviation.
$$M.D.=\frac{Sum\ of\ absolute\ values\ of\ deviations\ from\ x}{Number\ of\ observations}$$
Let us begin with the various mean deviation formulas one by one.
Mean = (Sum of all the observations/Total number of observations)
Example: What is the mean of 2, 4, 8, 6 and 12?
Step 1: First add all the numbers.
2+4+6+8+12 = 32
Step 2: Now divide by 5 (here 5 is the total number of observations).
Mean = 32/5 = 6.4
The mean deviation about the mean formula is as follows:
$$The\ formula\ to\ calculate\ the\ mean\ deviation\ for\ an\ individual\ series\ is\ given\ by:$$
$$M.D.=\frac{\sum_{i=1}^N\left|X_i-\overline{X}\right|}{N}$$
$$Where,\ \sum_{ }^{ }\rightarrow\ Summation$$
$$X_i\ \rightarrow\ Observations\ or\ values$$
$$\overline{X}\ \rightarrow\ Mean$$
$$N\ \rightarrow\ Number\ of\ observations$$
$$The\ formula\ to\ calculate\ the\ mean\ deviation\ of\ a\ discrete\ series\ is\ given\ by:$$
$$M.D.=\frac{\sum_{ }^{ }f\left|X_{ }-\overline{X}\right|}{\sum_{ }^{ }f}$$
$$Where,\ \sum_{ }^{ }\rightarrow\ Summation$$
$$\ X\ \rightarrow\ Observations\ or\ values$$
$$\overline{X}\ \rightarrow\ Mean$$
$$f\ \rightarrow frequency\ of\ observations$$
$$The\ formula\ to\ calculate\ the\ mean\ deviation\ of\ a\ Continuous\ series\ is\ given\ by:$$
$$M.D.=\frac{\sum_{ }^{ }f\left|X_{ }-\overline{X}\right|}{\sum_{ }^{ }f}$$
$$Where,\ \sum_{ }^{ }\rightarrow\ Summation$$
$$\ X\ \rightarrow\ Mid-value\ of\ the\ class$$
$$\overline{X}\ \rightarrow\ Mean$$
$$f\ \rightarrow frequency\ of\ observations$$
The median of a distribution is defined as the value of the variable such that the quantity of observations above it is equivalent to the number of observations below it.
### Median of Ungrouped Data
If n is even then the median is the average of the values which are at the position; (n/2) and [(n/2)+1].
### Median of a Grouped Data
$$Median=l+\left(\frac{\frac{n}{2}-cf}{f}\right)\times h$$
$$The\ formula\ to\ calculate\ the\ mean\ deviation\ for\ an\ individual\ series\ is\ given\ by:$$
$$M.D.=\frac{\sum_{ }^{ }\left|X-M\right|}{N}$$
$$Where,\ \sum_{ }^{ }\rightarrow\ Summation$$
$$X\ \rightarrow\ Observations\ or\ values$$
$$M\ \rightarrow\ Median$$
$$N\ \rightarrow\ Number\ of\ observations$$
$$The\ formula\ to\ calculate\ the\ mean\ deviation\ of\ a\ discrete\ series\ is\ given\ by:$$
$$M.D.=\frac{\sum_{ }^{ }f\left|X_{ }-M\right|}{\sum_{ }^{ }f}$$
$$Where,\ \sum_{ }^{ }\rightarrow\ Summation$$
$$\ X\ \rightarrow\ Observations\ or\ values$$
$$M\rightarrow\ Median$$
$$f\ \rightarrow frequency\ of\ observations$$
$$The\ formula\ to\ calculate\ the\ mean\ deviation\ of\ a\ Continuous\ series\ is\ given\ by:$$
$$M.D.=\frac{\sum_{ }^{ }f\left|X_{ }-M\right|}{\sum_{ }^{ }f}$$
$$Where,\ \sum_{ }^{ }\rightarrow\ Summation$$
$$\ X\ \rightarrow\ Mid-value\ of\ the\ class$$
$$M\rightarrow\ Median$$
$$f\ \rightarrow frequency\ of\ observations$$
In Statistics, mode or modal value is that observation that happens at the maximum time or has the highest frequency in the provided set of data. A set of numbers/data with one mode is known as unimodal, a set of numbers possessing two modes is bimodal and a set of numbers having three modes is known as trimodal. However, any set of numbers holding four or more than four modes is known as multimodal.
$$Mode=l+\left\{\frac{f_m−f_1}{2f_m−f_1-f_2}\right\}\times h$$
$$Where,$$
$$l=\ lower\ limit\ of\ modal\ class,$$
$$f_m=frequency\ of\ modal\ class,$$
$$f_1=frequency\ of\ class\ preceding\ modal\ class,\$$
$$f_2=frequency\ of\ class\ succeeding\ modal\ class\ and\ h=class\ width.$$
$$The\ formula\ to\ calculate\ the\ mean\ deviation\ for\ an\ individual\ series\ is\ given\ by:$$
$$M.D.=\frac{\sum_{ }^{ }\left|X-Mode\right|}{N}$$
$$Where,\ \sum_{ }^{ }\rightarrow\ Summation$$
$$X\ \rightarrow\ Observations\ or\ values$$
$$M\ \rightarrow\ Mode$$
$$N\ \rightarrow\ Number\ of\ observations$$
$$The\ formula\ to\ calculate\ the\ mean\ deviation\ of\ a\ discrete\ series\ is\ given\ by:$$
$$M.D.=\frac{\sum_{ }^{ }f\left|X_{ }-Mode\right|}{\sum_{ }^{ }f}$$
$$Where,\ \sum_{ }^{ }\rightarrow\ Summation$$
$$\ X\ \rightarrow\ Observations\ or\ values$$
$$M\rightarrow\ Mode$$
$$f\ \rightarrow frequency\ of\ observations$$
$$The\ formula\ to\ calculate\ the\ mean\ deviation\ of\ a\ Continuous\ series\ is\ given\ by:$$
$$M.D.=\frac{\sum_{ }^{ }f\left|X_{ }-Mode\right|}{\sum_{ }^{ }f}$$
$$Where,\ \sum_{ }^{ }\rightarrow\ Summation$$
$$\ X\ \rightarrow\ Mid-value\ of\ the\ class$$
$$M\rightarrow\ Mode$$
$$f\ \rightarrow frequency\ of\ observations$$
Learn the concepts of Three Dimensional Geometry here.
## How to Find Mean Deviation?
The mean deviation is the statistical measure that is applied to estimate the average deviation from the mean value of the provided data set. The mean deviation of the data values can be calculated by following these steps.
Step 1: Determine the mean, median or mode of the given series.
Step 2: Estimate the deviations from the Mean, median or mode and neglect the minus signs.
Step 3: Multiply the deviations by the frequency. This step is required only in the discrete and continuous series.
Step 4: Sum up or calculate the total of all the deviations.
Step 5: Final step is to apply the formula.
Mean deviation example to understand the above concepts.
Check more topics of Mathematics here.
$$Example:Find\ mean\ deviation\ about\ the\ mean\ for\ the\ following\ data:$$
$$5,8,10,12,5.$$
$$Solution:\ Mean\ deviation\ for\ the\ given\ data\ set.$$
$$Given:$$
$$Data\ set=5,\ 8,\ 10,\ 12,\ 5$$
$$Mean\ of\ the\ data\ (m)=\frac{\left(5+8+10+12+5\right)}{5}=\frac{40}{5}=8$$
$$Now,\ we\ will\det er\min e\ the\ deviations\ of\ each\ observation\ from\ the\ mean\ X,$$
$$i.e,\ X_i-X$$
$$=5-8,\ 8-8,\ 10-8,\ 12-8,\ 5-8$$
$$=-3,\ \ 0,\ 2,\ 4,\ -3$$
$$Absolute\ values\ of\ the\ deviations\ from\ mean\ i.e.,$$
$$=ǀX_i-Xǀ=\ 3,\ 0,\ 2,\ 4,\ 3$$
$$Required\ mean\ deviation\ about\ mean\ is,$$
$$The\ formula\ to\ calculate\ the\ mean\ deviation\ for\ an\ individual\ series\ is\ given\ by:$$
$$M.D.=\frac{\sum_{i=1}^N\left|X_i-\overline{X}\right|}{N}$$
$$Where,\ \sum_{ }^{ }\rightarrow\ Summation$$
$$X_i\ \rightarrow\ Observations\ or\ values$$
$$\overline{X}\ \rightarrow\ Mean$$
$$N\ \rightarrow\ Number\ of\ observations$$
$$M.D.=\frac{\left(3+0+2+4+3\right)}{5}=2.4$$
$$Mean\ deviation\ of\ the\ data\ about\ mean\ is\ 2.4$$
We hope that the above article on Mean Deviation is helpful for your understanding and exam preparations. Stay tuned to the Testbook app for more updates on related topics from Mathematics, and various such subjects. Also, reach out to the test series available to examine your knowledge regarding several exams.
If you are checking Mean Deviation article, also check the related maths articles in the table below: Lines of Regression Locus Binary to Decimal Conversion Ellipse Hyperbola Trigonometric Ratios
## Mean Deviation FAQs
Q.1 What is the mean deviation definition?
Ans.1 The mean of the absolute values of the numerical differences between the numbers of a set (such as statistical data) and their mean or median.
Q.2 How to calculate mean deviation from mean?
Ans.2 The first step is to determine the mean of all values. Next, calculate the distance of every value from that mean i.e deduct the mean from each value and ignore the minus signs. Lastly, obtain the mean of those distances by the respective formula.
Q.3 How to calculate mean deviation?
Ans.3 The mean deviation of the data values can be calculated as follows: Estimate the deviations from the mean, median or mode and neglect the minus signs. Next is to multiply the deviations by the frequency. Lastly, sum up or calculate the total of all the deviations and apply the formula.
Q.4 Why do we use mean deviation?
Ans.4 The term “mean deviation” is a measure that indicates how much the observations in the data set varies from the mean value of the observations in the given data set.
Q.5 What is the mean deviation and how it is calculated?
Ans.5 A mean deviation is a statistical approach to determine the average deviation of values from the mean in an example. It is calculated first by obtaining the average of the observations. The difference of each observation from the mean is then defined and lastly, the mean deviation is determined through the formula.
Q.6 What is the difference between mean deviation and standard deviation?
Ans.6 A mean is the average of a set of two or more numbers. Whereas the standard deviation is used to measure the volatility of a stock. | 2,934 | 10,508 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2022-49 | latest | en | 0.903692 |
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1. Problem help - Orthogonal Bases
Consider the following Subspace on R3:
U={(x1,x2,x3) € R3: x1-2x2+x3=0}
Which of the following group of vectors is an orthogonal basis of U?
A - {(1,-2,1), (-1,0,1)}
B - {(-1,0,1), (1,1,1)}
C - {(2,1,0), (-1,0,1)}
D - {(2,1,0), (0,0,1)}
E - {(1,-2,1), (-1,0,1), (1,1,1)}
Can anyone give me a detailed resolution for this exercise?
2. First note that the equation describes a plane in R^3. Specifically, recall that a plane may be specified by the set of all points whose displacement vectors from a proprietary point are normal to a specific vector, called the normal vector. Normality is defined by the dot product vanishing, so note that your equation may be rewritten as $\displaystyle (1, -2, 1)\cdot [(x_1, x_2, x_3) - (0, 0, 0)] = 0$ A normal vector to your plane is thus (1, -2, 1).
The two basis vectors must have a dot product of 0 to be orthogonal and must have a cross product that is a scalar multiple of the normal vector, since they must lie in that plane. | 332 | 1,051 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2018-17 | latest | en | 0.901029 |
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# Chapter 7
Laplace Transform
The Laplace transform can be used to solve differential equations. Be-
sides being a different and efficient alternative to variation of parame-
ters and undetermined coefficients, the Laplace method is particularly
advantageous for input terms that are piecewise-defined, periodic or im-
pulsive.
The direct Laplace transform or the Laplace integral of a function
f (t) defined for 0 ≤ t < ∞ is the ordinary calculus integration problem
Z ∞
f (t)e−st dt,
0
succinctly denoted L(f (t)) in science and engineering literature. The
L–notation recognizes that integration always proceeds over t = 0 to
t = ∞ and that the integral involves an integrator e−st dt instead of the
usual dt. These minor differences distinguish Laplace integrals from
the ordinary integrals found on the inside covers of calculus texts.
## 7.1 Introduction to the Laplace Method
The foundation of Laplace theory is Lerch’s cancellation law
R∞ R∞
0 y(t)e−st dt = 0 f (t)e−st dt implies y(t) = f (t),
(1) or
L(y(t) = L(f (t)) implies y(t) = f (t).
In differential equation applications, y(t) is the sought-after unknown
while f (t) is an explicit expression taken from integral tables.
Below, we illustrate Laplace’s method by solving the initial value prob-
lem
y 0 = −1, y(0) = 0.
The method obtains a relation L(y(t)) = L(−t), whence Lerch’s cancel-
lation law implies the solution is y(t) = −t.
The Laplace method is advertised as a table lookup method, in which
the solution y(t) to a differential equation is found by looking up the
answer in a special integral table.
7.1 Introduction to the Laplace Method 247
R∞
Laplace Integral. The integral g(t)e−st dt is called the Laplace
0
integral of the function g(t). It is defined by limN →∞ 0N g(t)e−st dt and
R
## depends on variable s. The ideas will be illustrated for g(t) = 1, g(t) = t
and g(t) = t2 , producing the integral formulas in Table 1.
R∞ t=∞
0 (1)e−st dt = −(1/s)e−st t=0 Laplace integral of g(t) = 1.
= 1/s Assumed s > 0.
R∞ R∞ d
0 (t)e−st dt = 0 − ds (e−st )dt Laplace integral of g(t) = t.
d R∞ d d
(1)e−st dt
R R
= − ds 0 Use ds F (t, s)dt = ds F (t, s)dt.
d
= − ds (1/s) Use L(1) = 1/s.
= 1/s2 Differentiate.
R∞ R∞
0 (t2 )e−st dt = d
0 − ds (te
−st )dt Laplace integral of g(t) = t2 .
d R∞ −st dt
= − ds 0 (t)e
= d
− ds (1/s2 ) Use L(t) = 1/s2 .
= 2/s3
R∞
Table 1. The Laplace integral 0
g(t)e−st dt for g(t) = 1, t and t2 .
R∞ 1 R∞ 1 R∞ 2
0 (1)e−st dt = 0 (t)e−st dt = 0 (t2 )e−st dt =
s s2 s3
n!
In summary, L(tn ) =
s1+n
## An Illustration. The ideas of the Laplace method will be illus-
trated for the solution y(t) = −t of the problem y 0 = −1, y(0) = 0. The
method, entirely different from variation of parameters or undetermined
coefficients, uses basic calculus and college algebra; see Table 2.
Table 2. Laplace method details for the illustration y 0 = −1, y(0) = 0.
## y 0 (t)e−st = −e−st Multiply y 0 = −1 by e−st .
R∞ 0 −st dt = ∞ −e−st dt
R
0 y (t)e 0 Integrate t = 0 to t = ∞.
R∞ 0 −st dt = −1/s
0 y (t)e Use Table 1.
s 0∞ y(t)e−st dt − y(0) = −1/s
R
Integrate by parts on the left.
R∞ −st dt = −1/s2
0 y(t)e Use y(0) = 0 and divide.
R∞ −st dt = ∞ (−t)e−st dt
R
0 y(t)e 0 Use Table 1.
y(t) = −t Apply Lerch’s cancellation law.
248 Laplace Transform
In Lerch’s law, the formal rule of erasing the integral signs is valid pro-
vided the integrals are equal for large s and certain conditions hold on y
and f – see Theorem 2. The illustration in Table 2 shows that Laplace
theory requires an in-depth study of a special integral table, a table
which is a true extension of the usual table found on the inside covers
of calculus books. Some entries for the special integral table appear in
Table 1 and also in section 7.2, Table 4.
The L-notation for the direct Laplace transform produces briefer details,
as witnessed by the translation of Table 2 into Table 3 below. The reader
is advised to move from Laplace integral notation to the L–notation as
soon as possible, in order to clarify the ideas of the transform method.
## Table 3. Laplace method L-notation details for y 0 = −1, y(0) = 0
translated from Table 2.
## L(y 0 (t)) = L(−1) Apply L across y 0 = −1, or multiply y 0 =
−1 by e−st , integrate t = 0 to t = ∞.
L(y 0 (t)) = −1/s Use Table 1.
sL(y(t)) − y(0) = −1/s Integrate by parts on the left.
L(y(t)) = −1/s2 Use y(0) = 0 and divide.
L(y(t)) = L(−t) Apply Table 1.
y(t) = −t Invoke Lerch’s cancellation law.
## Some Transform Rules. The formal properties of calculus integrals
plus the integration by parts formula used in Tables 2 and 3 leads to these
rules for the Laplace transform:
L(f (t) + g(t)) = L(f (t)) + L(g(t)) The integral of a sum is the
sum of the integrals.
L(cf (t)) = cL(f (t)) Constants c pass through the
integral sign.
L(y 0 (t)) = sL(y(t)) − y(0) The t-derivative rule, or inte-
gration by parts. See Theo-
rem 3.
L(y(t)) = L(f (t)) implies y(t) = f (t) Lerch’s cancellation law. See
Theorem 2.
## 1 Example (Laplace method) Solve by Laplace’s method the initial value
problem y 0 = 5 − 2t, y(0) = 1.
## Solution: Laplace’s method is outlined in Tables 2 and 3. The L-notation of
Table 3 will be used to find the solution y(t) = 1 + 5t − t2 .
7.1 Introduction to the Laplace Method 249
## L(y 0 (t)) = L(5 − 2t) Apply L across y 0 = 5 − 2t.
5 2
L(y 0 (t)) = − 2 Use Table 1.
s s
5 2
sL(y(t)) − y(0) = − 2 Apply the t-derivative rule, page 248.
s s
1 5 2
L(y(t)) = + 2 − 3 Use y(0) = 1 and divide.
s s s
L(y(t)) = L(1) + 5L(t) − L(t2 ) Apply Table 1, backwards.
2
= L(1 + 5t − t ) Linearity, page 248.
2
y(t) = 1 + 5t − t Invoke Lerch’s cancellation law.
## 2 Example (Laplace method) Solve by Laplace’s method the initial value
problem y 00 = 10, y(0) = y 0 (0) = 0.
Solution: The L-notation of Table 3 will be used to find the solution y(t) = 5t2 .
L(y 00 (t)) = L(10) Apply L across y 00 = 10.
sL(y 0 (t)) − y 0 (0) = L(10) Apply the t-derivative rule to y 0 , that is,
replace y by y 0 on page 248.
s[sL(y(t)) − y(0)] − y 0 (0) = L(10) Repeat the t-derivative rule, on y.
s2 L(y(t)) = L(10) Use y(0) = y 0 (0) = 0.
10
L(y(t)) = 3 Use Table 1. Then divide.
s
L(y(t)) = L(5t2 ) Apply Table 1, backwards.
y(t) = 5t2 Invoke Lerch’s cancellation law.
R∞
Existence of the Transform. The Laplace integral e−st f (t) dt
0
is known to exist in the sense of the improper integral definition1
Z ∞ Z N
g(t)dt = lim g(t)dt
0 N →∞ 0
## provided f (t) belongs to a class of functions known in the literature as
functions of exponential order. For this class of functions the relation
f (t)
(2) lim =0
t→∞ eat
## is required to hold for some real number a, or equivalently, for some
constants M and α,
(3) |f (t)| ≤ M eαt .
In addition, f (t) is required to be piecewise continuous on each finite
subinterval of 0 ≤ t < ∞, a term defined as follows.
1
An advanced calculus background is assumed for the Laplace transform existence
proof. Applications of Laplace theory require only a calculus background.
250 Laplace Transform
## Definition 1 (piecewise continuous)
A function f (t) is piecewise continuous on a finite interval [a, b] pro-
vided there exists a partition a = t0 < · · · < tn = b of the interval [a, b]
and functions f1 , f2 , . . . , fn continuous on (−∞, ∞) such that for t not
a partition point
## f1 (t) t0 < t < t1 ,
(4) f (t) = .. ..
. .
fn (t) tn−1 < t < tn .
## The values of f at partition points are undecided by equation (4). In
particular, equation (4) implies that f (t) has one-sided limits at each
point of a < t < b and appropriate one-sided limits at the endpoints.
Therefore, f has at worst a jump discontinuity at each partition point.
## 3 Example (Exponential order) Show that f (t) = et cos t + t is of expo-
nential order, that is, show that f (t) is piecewise continuous and find α > 0
such that limt→∞ f (t)/eαt = 0.
## Solution: Already, f (t) is continuous, hence piecewise continuous. From
L’Hospital’s rule in calculus, limt→∞ p(t)/eαt = 0 for any polynomial p and
any α > 0. Choose α = 2, then
f (t) cos t t
lim 2t
= lim t
+ lim 2t = 0.
t→∞ e t→∞ e t→∞ e
## Theorem 1 (Existence of L(f ))
Let f (t) be piecewise continuous on every finite interval in t ≥ 0 and satisfy
|f (t)| ≤ M eαt for some constants M and α. Then L(f (t)) exists for s > α
and lims→∞ L(f (t)) = 0.
Proof: It has to be shown that the Laplace integral of f is finite for s > α.
Advanced calculus implies that it is sufficient to show that the integrand is ab-
solutely bounded above by an integrable function g(t). Take g(t) = M e −(s−α)t .
Then g(t) ≥ 0. Furthermore, g is integrable, because
Z ∞
M
g(t)dt = .
0 s −α
Inequality |f (t)| ≤ M eαt implies the absolute value of the Laplace transform
integrand f (t)e−st is estimated by
## f (t)e−st ≤ M eαt e−st = g(t).
R∞ M
The limit statement follows from |L(f (t))| ≤ 0 g(t)dt = , because the
s−α
right side of this inequality has limit zero at s = ∞. The proof is complete.
7.1 Introduction to the Laplace Method 251
Theorem 2 (Lerch)
If f (t) and f2 (t) are continuous, of exponential order and 0∞ f1 (t)e−st dt =
R
R ∞1 −st dt for all s > s , then f (t) = f (t) for t ≥ 0.
0 f2 (t)e 0 1 2
## Theorem 3 (t-Derivative Rule)
If f (t) is continuous, lim f (t)e−st = 0 for all large values of s and f 0 (t)
t→∞
is piecewise continuous, then L(f 0 (t)) exists for all large s and L(f 0 (t)) =
sL(f (t)) − f (0).
## Proof: See page 276.
Exercises 7.1
PN
Laplace method. Solve the given 18. f (t) = n=1 cn sin(nt), for any
initial value problem using Laplace’s choice of the constants c1 , . . . , cN .
method.
Existence of transforms. Let f (t) =
1. y 0 = −2, y(0) = 0. 2 2
tet sin(et ). Establish these results.
2. y 0 = 1, y(0) = 0. 19. The function f (t) is not of expo-
3. y 0 = −t, y(0) = 0. nential order.
## 4. y 0 = t, y(0) = 0. 20. The
R ∞ Laplace integral of f (t),
−st
0
f (t)e dt, converges for all
5. y 0 = 1 − t, y(0) = 0. s > 0.
6. y 0 = 1 + t, y(0) = 0.
Jump Magnitude. For f piecewise
7. y 0 = 3 − 2t, y(0) = 0. continuous, define the jump at t by
8. y 0 = 3 + 2t, y(0) = 0. J(t) = lim f (t + h) − lim f (t − h).
h→0+ h→0+
00 0
9. y = −2, y(0) = y (0) = 0. Compute J(t) for the following f .
00 0
10. y = 1, y(0) = y (0) = 0. 21. f (t) = 1 for t ≥ 0, else f (t) = 0
00 0
11. y = 1 − t, y(0) = y (0) = 0. 22. f (t) = 1 for t ≥ 1/2, else f (t) = 0
12. y 00 = 1 + t, y(0) = y 0 (0) = 0. 23. f (t) = t/|t| for t 6= 0, f (0) = 0
00 0
13. y = 3 − 2t, y(0) = y (0) = 0. 24. f (t) = sin t/| sin t| for t 6= nπ,
00 0 f (nπ) = (−1)n
14. y = 3 + 2t, y(0) = y (0) = 0.
## Exponential order. Show that f (t) Taylor
P∞ series
n
. PThe series relation
∞ n
L( n=0 cn t ) = n=0 cn L(t ) often
is of exponential order, by finding a
holds, in which case the result L(tn ) =
constant α ≥ 0 in each case such that
f (t) n!s−1−n can be employed to find a
lim = 0. series representation of the Laplace
t→∞ eαt
transform. Use this idea on the fol-
15. f (t) = 1 + t
lowing to find a series formula for
16. f (t) = et sin(t) L(f (t)).
2t
P∞ n
17. f (t) = n=0 cn xn , for any choice 25. f (t) = e = n=0 (2t) /n!
PN
P∞
of the constants c0 , . . . , cN . 26. f (t) = e−t = n=0 (−t)n /n!
252 Laplace Transform
## 7.2 Laplace Integral Table
The objective in developing a table of Laplace integrals, e.g., Tables 4
and 5, is to keep the table size small. Table manipulation rules appear-
ing in Table 6, page 257, effectively increase the table size manyfold,
making it possible to solve typical differential equations from electrical
and mechanical problems. The combination of Laplace tables plus the
table manipulation rules is called the Laplace transform calculus.
Table 4 is considered to be a table of minimum size to be memorized.
Table 5 adds a number of special-use entries. For instance, the Heaviside
entry in Table 5 is memorized, but usually not the others.
Derivations are postponed to page 270. The theory of the gamma func-
tion Γ(x) appears below on page 255.
## Table 4. A minimal Laplace integral table with L-notation
R∞ n! n!
0
(tn )e−st dt = L(tn ) =
s1+n s1+n
R∞ 1 1
0 (eat )e−st dt = L(eat ) =
s−a s−a
R∞ s s
0
(cos bt)e−st dt = L(cos bt) =
s2
+ b2 s2
+ b2
R∞ −st b b
0 (sin bt)e dt = 2
s + b2
L(sin bt) = 2
s + b2
## Table 5. Laplace integral table extension
e−as
L(H(t − a)) = (a ≥ 0) Heaviside
unit step, defined by
s 1 for t ≥ 0,
H(t) =
0 otherwise.
L(δ(t − a)) = e−as Dirac delta, δ(t) = dH(t).
Special usage rules apply.
e−as
L(floor(t/a)) = Staircase function,
s(1 − e−as )
floor(x) = greatest integer ≤ x.
1
L(sqw(t/a)) = tanh(as/2) Square wave,
s
sqw(x) = (−1)floor(x) .
1
L(a trw(t/a)) = tanh(as/2) TriangularRwave,
s2 x
trw(x) = 0 sqw(r)dr.
Γ(1 + α)
L(tα ) = Generalized power
R ∞ function,
s1+α
Γ(1 + α) = 0 e−x xα dx.
r
−1/2 π √
L(t )= Because Γ(1/2) = π.
s
7.2 Laplace Integral Table 253
## 4 Example (Laplace transform) Let f (t) = t(t − 1) − sin 2t + e3t . Compute
L(f (t)) using the basic Laplace table and transform linearity properties.
Solution:
L(f (t)) = L(t2 − 5t − sin 2t + e3t ) Expand t(t − 5).
2 3t
= L(t ) − 5L(t) − L(sin 2t) + L(e ) Linearity applied.
2 5 2 1
= 3− 2− 2 + Table lookup.
s s s +4 s−3
5 Example (Inverse Laplace transform) Use the basic Laplace table back-
wards plus transform linearity properties to solve for f (t) in the equation
s 2 s+1
L(f (t)) = + + 3 .
s2 + 16 s − 3 s
Solution:
s 1 1 1 2
L(f (t)) = +2 + 2+ Convert to table entries.
s2
+ 16 s−3 s 2 s3
= L(cos 4t) + 2L(e ) + L(t) + 12 L(t2 )
3t
Laplace table (backwards).
3t 1 2
= L(cos 4t + 2e + t + 2t ) Linearity applied.
3t 1 2
f (t) = cos 4t + 2e + t + 2t Lerch’s cancellation law.
5
## Figure 1. A piecewise defined
1 function f (t) on 0 ≤ t < ∞: f (t) = 0
1 3 5 except for 1 ≤ t < 2 and 3 ≤ t < 4.
Solution: The details require the use of the Heaviside function formula
1 a ≤ t < b,
H(t − a) − H(t − b) =
0 otherwise.
## The formula for f (t):
1 1 ≤ t < 2,
1 1 ≤ t < 2, 1 3 ≤ t < 4,
f (t) = 5 3 ≤ t < 4, = +5
0 otherwise 0 otherwise
0 otherwise
Then f (t) = f1 (t) + 5f2 (t) where f1 (t) = H(t − 1) − H(t − 2) and f2 (t) =
H(t − 3) − H(t − 4). The extended table gives
## L(f (t)) = L(f1 (t)) + 5L(f2 (t)) Linearity.
= L(H(t − 1)) − L(H(t − 2)) + 5L(f2 (t)) Substitute for f1 .
254 Laplace Transform
e−s − e−2s
= + 5L(f2 (t)) Extended table used.
s
e−s − e−2s + 5e−3s − 5e−4s
= Similarly for f2 .
s
## 7 Example (Dirac delta) A machine shop tool that repeatedly hammers a
die is modeled by the Dirac impulse model f (t) = N
P
PN n=1 δ(t − n). Show
−ns
that L(f (t)) = n=1 e .
Solution:
P
N
L(f (t)) = L n=1 δ(t − n)
PN
= n=1 L(δ(t − n)) Linearity.
= N −ns
P
n=1 e Extended Laplace table.
## 8 Example (Square wave) A periodic camshaft force f (t) applied to a me-
chanical system has the idealized graph shown in Figure 2. Show that
f (t) = 1 + sqw(t) and L(f (t)) = 1s (1 + tanh(s/2)).
## Figure 2. A periodic force f (t) applied
0 to a mechanical system.
1 3
Solution:
1+1 2n ≤ t < 2n + 1, n = 0, 1, . . .,
1 + sqw(t) =
1−1 2n + 1 ≤ t < 2n + 2, n = 0, 1, . . .,
2 2n ≤ t < 2n + 1, n = 0, 1, . . .,
=
0 otherwise,
= f (t).
1 tanh(s/2)
By the extended Laplace table, L(f (t)) = L(1) + L(sqw(t)) = + .
s s
## 9 Example (Sawtooth wave) Express the P -periodic sawtooth wave repre-
sented in Figure 3 as f (t) = ct/P − c floor(t/P ) and obtain the formula
c ce−P s
L(f (t)) = − .
P s2 s − se−P s
## Figure 3. A P -periodic sawtooth
0 wave f (t) of height c > 0.
P 4P
7.2 Laplace Integral Table 255
## Solution: The representation originates from geometry, because the periodic
function f can be viewed as derived from ct/P by subtracting the correct con-
stant from each of intervals [P, 2P ], [2P, 3P ], etc.
The technique used to verify the identity is to define g(t) = ct/P − c floor(t/P )
and then show that g is P -periodic and f (t) = g(t) on 0 ≤ t < P . Two P -
periodic functions equal on the base interval 0 ≤ t < P have to be identical,
hence the representation follows.
The fine details: for 0 ≤ t < P , floor(t/P ) = 0 and floor(t/P + k) = k. Hence
g(t + kP ) = ct/P + ck − c floor(k) = ct/P = g(t), which implies that g is
P -periodic and g(t) = f (t) for 0 ≤ t < P .
c
L(f (t)) = L(t) − cL(floor(t/P )) Linearity.
P
c ce−P s
= − Basic and extended table applied.
P s2 s − se−P s
## 10 Example (Triangular wave) Express the triangular wave f of Figure 4 in
5
terms of the square wave sqw and obtain L(f (t)) = 2 tanh(πs/2).
πs
5
## 0 Figure 4. A 2π-periodic triangular
2π wave f (t) of height 5.
R t/π
Solution: The representation of f in terms of sqw is f (t) = 5 0 sqw(x)dx.
Details: A 2-periodic triangular wave of height 1 is obtained by integrating
the square wave of period 2. A wave of height c and period 2 is given by
Rt R 2t/P
c trw(t) = c 0 sqw(x)dx. Then f (t) = c trw(2t/P ) = c 0 sqw(x)dx where
c = 5 and P = 2π.
Laplace transform details: Use the extended Laplace table as follows.
5 5
L(f (t)) = L(π trw(t/π)) = 2 tanh(πs/2).
π πs
## Gamma Function. In mathematical physics, the Gamma func-
tion or the generalized factorial function is given by the identity
Z ∞
(1) Γ(x) = e−t tx−1 dt, x > 0.
0
## This function is tabulated and available in computer languages like For-
tran, C and C++. It is also available in computer algebra systems and
numerical laboratories. Some useful properties of Γ(x):
## (2) Γ(1 + x) = xΓ(x)
(3) Γ(1 + n) = n! for integers n ≥ 1.
256 Laplace Transform
R∞
e−t dt = 1, which gives
Γ(1) = 1. Use this identity and successively relation (2) to obtain relation (3).
To prove identity (2), integration by parts is applied, as follows:
R∞
Γ(1 + x) = 0 e−t tx dt Definition.
t=∞ ∞
= −tx e−t |t=0 + 0 e−t xtx−1 dt Use u = tx , dv = e−t dt.
R
R ∞ −t x−1
=x 0 e t dt Boundary terms are zero
for x > 0.
= xΓ(x).
Exercises 7.2
Laplace transform. Compute Inverse Laplace transform. Solve
L(f (t)) using the basic Laplace table the given equation for the function
and the linearity properties of the f (t). Use the basic table and linearity
transform. Do not use the direct properties of the Laplace transform.
Laplace transform! 21. L(f (t)) = s−2
1. L(2t)
22. L(f (t)) = 4s−2
2. L(4t)
23. L(f (t)) = 1/s + 2/s2 + 3/s3
2
3. L(1 + 2t + t )
24. L(f (t)) = 1/s3 + 1/s
2
4. L(t − 3t + 10)
25. L(f (t)) = 2/(s2 + 4)
5. L(sin 2t)
26. L(f (t)) = s/(s2 + 4)
6. L(cos 2t)
27. L(f (t)) = 1/(s − 3)
7. L(e2t )
28. L(f (t)) = 1/(s + 3)
8. L(e−2t )
29. L(f (t)) = 1/s + s/(s2 + 4)
9. L(t + sin 2t)
30. L(f (t)) = 2/s − 2/(s2 + 4)
10. L(t − cos 2t)
31. L(f (t)) = 1/s + 1/(s − 3)
11. L(t + e2t )
32. L(f (t)) = 1/s − 3/(s − 2)
12. L(t − 3e−2t )
33. L(f (t)) = (2 + s)2 /s3
13. L((t + 1)2 )
34. L(f (t)) = (s + 1)/s2
14. L((t + 2)2 )
35. L(f (t)) = s(1/s2 + 2/s3 )
15. L(t(t + 1))
36. L(f (t)) = (s + 1)(s − 1)/s3
16. L((t + 1)(t + 2)) P10
P10 n 37. L(f (t)) = n=0 n!/s1+n
17. L( n=0 t /n!) P10
P10 n+1 38. L(f (t)) = n=0 n!/s2+n
18. L( n=0 t /n!) P10 n
P10 39. L(f (t)) = n=1 2
19. L( n=1 sin nt) s + n2
P10 P10 s
20. L( n=0 cos nt) 40. L(f (t)) = n=0 2
s + n2
7.3 Laplace Transform Rules 257
## 7.3 Laplace Transform Rules
In Table 6, the basic table manipulation rules are summarized. Full
statements and proofs of the rules appear in section 7.7, page 275.
The rules are applied here to several key examples. Partial fraction
expansions do not appear here, but in section 7.4, in connection with
Heaviside’s coverup method.
## L(f (t) + g(t)) = L(f (t)) + L(g(t)) Linearity.
The Laplace of a sum is the sum of the Laplaces.
## L(cf (t)) = cL(f (t)) Linearity.
Constants move through the L-symbol.
## L(y 0 (t)) = sL(y(t)) − y(0) The t-derivative rule.
Derivatives L(y 0 ) are replaced in transformed equations.
R 1
t
L 0g(x)dx = L(g(t)) The t-integral rule.
s
d
L(tf (t)) = − L(f (t)) The s-differentiation rule.
ds
Multiplying f by t applies −d/ds to the transform of f .
## L(eat f (t)) = L(f (t))|s→(s−a) First shifting rule.
Multiplying f by eat replaces s by s − a.
## L(f (t − a)H(t − a)) = e−as L(f (t)), Second shifting rule.
L(g(t)H(t − a)) = e−as L(g(t + a)) First and second forms.
RP
f (t)e−st dt
L(f (t)) = 0 Rule for P -periodic functions.
1 − e−P s
Assumed here is f (t + P ) = f (t).
## L(f (t))L(g(t)) = L((f ∗ g)(t)) Convolution rule.
Rt
Define (f ∗ g)(t) = f (x)g(t − x)dx.
0
## 11 Example (Harmonic oscillator) Solve by Laplace’s method the initial value
problem x00 + x = 0, x(0) = 0, x0 (0) = 1.
## Solution: The solution is x(t) = sin t. The details:
L(x00 ) + L(x) = L(0) Apply L across the equation.
0 0
sL(x ) − x (0) + L(x) = 0 Use the t-derivative rule.
0
s[sL(x) − x(0)] − x (0) + L(x) = 0 Use again the t-derivative rule.
2
(s + 1)L(x) = 1 Use x(0) = 0, x0 (0) = 1.
1
L(x) = 2 Divide.
s +1
= L(sin t) Basic Laplace table.
x(t) = sin t Invoke Lerch’s cancellation law.
258 Laplace Transform
2
12 Example (s-differentiation rule) Show the steps for L(t2 e5t ) = .
(s − 5)3
Solution:
d d
2 5t
L(t e ) = − − L(e5t ) Apply s-differentiation.
ds ds
2 d d 1
= (−1) Basic Laplace table.
ds ds s − 5
d −1
= Calculus power rule.
ds (s − 5)2
2
= Identity verified.
(s − 5)3
2
13 Example (First shifting rule) Show the steps for L(t2 e−3t ) = .
(s + 3)3
Solution:
L(t2 e−3t ) = L(t2 ) s→s−(−3)
First shifting rule.
2
= 2+1
Basic Laplace table.
s
s→s−(−3)
2
= Identity verified.
(s + 3)3
## 14 Example (Second shifting rule) Show the steps for
e−πs
L(sin t H(t − π)) = .
s2 + 1
## Solution: The second shifting rule is applied as follows.
L(sin t H(t − π)) = L(g(t)H(t − a) Choose g(t) = sin t, a = π.
= e−as L(g(t + a) Second form, second shifting theorem.
−πs
=e L(sin(t + π)) Substitute a = π.
−πs
=e L(− sin t) Sum rule sin(a + b) = sin a cos b +
sin b cos a plus sin π = 0, cos π = −1.
−1
= e−πs Basic Laplace table. Identity verified.
s2 + 1
## 15 Example (Trigonometric formulas) Show the steps used to obtain these
Laplace identities:
s 2 − a2 3
2 cos at) = 2(s − 3sa )
2
(a) L(t cos at) = 2 (c) L(t
(s + a2 )2 (s2 + a2 )3
2sa 2 3
(b) L(t sin at) = 2 (d) L(t 2 sin at) = 6s a − a
(s + a2 )2 (s2 + a2 )3
7.3 Laplace Transform Rules 259
## Solution: The details for (a):
L(t cos at) = −(d/ds)L(cos at) Use s-differentiation.
d s
=− Basic Laplace table.
ds s2 + a2
s2 − a 2
= Calculus quotient rule.
(s2 + a2 )2
## The details for (c):
L(t2 cos at) = −(d/ds)L((−t) cos at) Use s-differentiation.
s2 − a 2
d
= − 2 Result of (a).
ds (s + a2 )2
2s3 − 6sa2 )
= Calculus quotient rule.
(s2 + a2 )3
The similar details for (b) and (d) are left as exercises.
## 16 Example (Exponentials) Show the steps used to obtain these Laplace
identities:
s−a 2 2
(a) L(eat cos bt) = (c) L(teat cos bt) = (s − a) − b
(s − a)2 + b2 ((s − a)2 + b2 )2
b 2b(s − a)
(b) L(eat sin bt) = (d) L(teat sin bt) =
(s − a)2 + b2 ((s − a)2 + b2 )2
## Solution: Details for (a):
L(eat cos bt) = L(cos bt)|s→s−a First shifting rule.
s
= Basic Laplace table.
s2 + b2 s→s−a
s−a
= Verified (a).
(s − a)2 + b2
## L(teat cos bt) = L(t cos bt)|s→s−a First shifting rule.
d
= − L(cos bt) Apply s-differentiation.
ds s→s−a
d s
= − 2 2
Basic Laplace table.
ds s + b
s→s−a
2
s − b2
= Calculus quotient rule.
(s2 + b2 )2 s→s−a
(s − a)2 − b2
= Verified (c).
((s − a)2 + b2 )2
## Left as exercises are (b) and (d).
260 Laplace Transform
## 17 Example (Hyperbolic functions) Establish these Laplace transform facts
about cosh u = (eu + e−u )/2 and sinh u = (eu − e−u )/2.
s s 2 + a2
(a) L(cosh at) = (c) L(t cosh at) =
s − a2
2 (s2 − a2 )2
a 2as
(b) L(sinh at) = 2 (d) L(t sinh at) = 2
s − a2 (s − a2 )2
## Solution: The details for (a):
L(cosh at) = 12 (L(eat ) + L(e−at )) Definition plus linearity of L.
1 1 1
= + Basic Laplace table.
2 s−a s+a
s
= 2 Identity (a) verified.
s − a2
The details for (d):
d a
L(t sinh at) = − Apply the s-differentiation rule.
ds s2 − a2
a(2s)
= 2 Calculus power rule; (d) verified.
(s − a2 )2
## Left as exercises are (b) and (c).
2s
18 Example (s-differentiation) Solve L(f (t)) = for f (t).
(s2 + 1)2
## Solution: The solution is f (t) = t sin t. The details:
2s
L(f (t)) =
(s2 + 1)2
d 1
=− Calculus power rule (un )0 = nun−1 u0 .
ds s2 + 1
d
= − (L(sin t)) Basic Laplace table.
ds
= L(t sin t) Apply the s-differentiation rule.
f (t) = t sin t Lerch’s cancellation law.
s+2
19 Example (First shift rule) Solve L(f (t)) = for f (t).
22 + 2s + 2
Solution: The answer is f (t) = e−t cos t + e−t sin t. The details:
s+2
L(f (t)) = Signal for this method: the denom-
s2 + 2s + 2
inator has complex roots.
s+2
= Complete the square, denominator.
(s + 1)2 + 1
7.3 Laplace Transform Rules 261
S+1
= Substitute S for s + 1.
S2 + 1
S 1
= 2 + Split into Laplace table entries.
S + 1 S2 + 1
= L(cos t) + L(sin t)|s→S=s+1 Basic Laplace table.
−t −t
= L(e cos t) + L(e sin t) First shift rule.
−t −t
f (t) = e cos t + e sin t Invoke Lerch’s cancellation law.
## 20 Example (Damped oscillator) Solve by Laplace’s method the initial value
problem x00 + 2x0 + 2x = 0, x(0) = 1, x0 (0) = −1.
## Solution: The solution is x(t) = e−t cos t. The details:
L(x00 ) + 2L(x0 ) + 2L(x) = L(0) Apply L across the equation.
0 0 0
sL(x ) − x (0) + 2L(x ) + 2L(x) = 0 The t-derivative rule on x0 .
s[sL(x) − x(0)] − x0 (0) The t-derivative rule on x.
+2[L(x) − x(0)] + 2L(x) = 0
(s2 + 2s + 2)L(x) = 1 + s Use x(0) = 1, x0 (0) = −1.
s+1
L(x) = 2 Divide.
s + 2s + 2
s+1
= Complete the square in the de-
(s + 1)2 + 1
nominator.
= L(cos t)|s→s+1 Basic Laplace table.
= L(e−t cos t) First shifting rule.
x(t) = e−t cos t Invoke Lerch’s cancellation law.
## 21 Example (Rectified sine wave) Compute the Laplace transform of the
rectified sine wave f (t) = | sin ωt| and show it can be expressed in the
form πs
ω coth 2ω
L(| sin ωt|) = .
s2 + ω 2
Solution: The periodic function formula will be applied with period P =
R P
2π/ω. The calculation reduces to the evaluation of J = 0 f (t)e−st dt. Because
sin ωt ≤ 0 on π/ω ≤ t ≤ 2π/ω, integral J can be written as J = J1 + J2 , where
Z π/ω Z 2π/ω
J1 = sin ωt e−st dt, J2 = − sin ωt e−st dt.
0 π/ω
## ωe−st cos(ωt) se−st sin(ωt)
Z
sin ωt e−st dt = − − .
s2 + ω 2 s2 + ω 2
Then
ω(e−π∗s/ω + 1) ω(e−2πs/ω + e−πs/ω )
J1 = , J2 = ,
s2 + ω 2 s2 + ω 2
262 Laplace Transform
ω(e−πs/ω + 1)2
J= .
s2 + ω 2
The remaining challenge is to write the answer for L(f (t)) in terms of coth.
The details:
J
L(f (t)) = Periodic function formula.
1 − e−P s
J
= Apply 1 − x2 = (1 − x)(1 + x),
(1 − e−P s/2 )(1 + e−P s/2 ) x = e−P s/2 .
ω(1 + e−P s/2 )
= Cancel factor 1 + e−P s/2 .
(1 − e−P s/2 )(s2 + ω 2 )
eP s/4 + e−P s/4 ω
= Factor out e−P s/4 , then cancel.
eP s/4 − e−P s/4 s2 + ω 2
2 cosh(P s/4) ω
= Apply cosh, sinh identities.
2 sinh(P s/4) s + ω 2
2
ω coth(P s/4)
= Use coth u = cosh u/ sinh u.
s2 + ω 2
πs
ω coth 2ω
= Identity verified.
s2 + ω 2
## 22 Example (Half–wave rectification) Compute the Laplace transform of the
half–wave rectification of sin ωt, denoted g(t), in which the negative cycles
of sin ωt have been canceled to create g(t). Show in particular that
1 ω πs
L(g(t)) = 2 2
1 + coth
2s +ω 2ω
## Solution: The half–wave rectification of sin ωt is g(t) = (sin ωt + | sin ωt|)/2.
Therefore, the basic Laplace table plus the result of Example 21 give
## L(2g(t)) = L(sin ωt) + L(| sin ωt|)
ω ω cosh(πs/(2ω))
= 2 2
+
s +ω s2 + ω 2
ω
= 2 (1 + cosh(πs/(2ω))
s + ω2
Dividing by 2 produces the identity.
s+1
23 Example (Shifting rules) Solve L(f (t)) = e−3s for f (t).
s2 + 2s + 2
Solution: The answer is f (t) = e3−t cos(t − 3)H(t − 3). The details:
s+1
L(f (t)) = e−3s Complete the square.
(s + 1)2 + 1
S
= e−3s 2 Replace s + 1 by S.
S +1
= e−3S+3 (L(cos t))|s→S=s+1 Basic Laplace table.
7.3 Laplace Transform Rules 263
## = e3 e−3s L(cos t) s→S=s+1
Regroup factor e−3S .
= e3 (L(cos(t − 3)H(t − 3)))|s→S=s+1 Second shifting rule.
3 −t
= e L(e cos(t − 3)H(t − 3)) First shifting rule.
3−t
f (t) = e cos(t − 3)H(t − 3) Lerch’s cancellation law.
s+7
24 Example () Solve L(f (t) = for f (t).
s2 + 4s + 8
Solution: The answer is f (t) = e−2t (cos 2t + 52 sin 2t). The details:
s+7
L(f (t)) = Complete the square.
(s + 2)2 + 4
S+5
= 2 Replace s + 2 by S.
S +4
S 5 2
= 2 + Split into table entries.
S + 4 2 S2 + 4
s 5 2
= 2 + Prepare for shifting rule.
s + 4 2 s2 + 4 s→S=s+2
= L(cos 2t) + 25 L(sin 2t) s→S=s+2
Basic Laplace table.
−2t 5
= L(e (cos 2t + 2 sin 2t)) First shifting rule.
−2t 5
f (t) = e (cos 2t + 2 sin 2t) Lerch’s cancellation law.
264 Laplace Transform
## 7.4 Heaviside’s Method
This practical method was popularized by the English electrical engineer
Oliver Heaviside (1850–1925). A typical application of the method is to
solve
2s
= L(f (t))
(s + 1)(s2 + 1)
for the t-expression f (t) = −e−t + cos t + sin t. The details in Heaviside’s
method involve a sequence of easy-to-learn college algebra steps.
More precisely, Heaviside’s method systematically converts a polyno-
mial quotient
a0 + a1 s + · · · + an sn
(1)
b0 + b1 s + · · · + bm sm
into the form L(f (t)) for some expression f (t). It is assumed that
a0 , .., an , b0 , . . . , bm are constants and the polynomial quotient (1) has
limit zero at s = ∞.
## Partial Fraction Theory
In college algebra, it is shown that a rational function (1) can be ex-
pressed as the sum of terms of the form
A
(2)
(s − s0 )k
## where A is a real or complex constant and (s − s0 )k divides the denomi-
nator in (1). In particular, s0 is a root of the denominator in (1).
Assume fraction (1) has real coefficients. If s0 in (2) is real, then A is
real. If s0 = α + iβ in (2) is complex, then (s − s0 )k also appears, where
s0 = α − iβ is the complex conjugate of s0 . The corresponding terms
in (2) turn out to be complex conjugates of one another, which can be
combined in terms of real numbers B and C as
A A B+Cs
(3) + = .
(s − s0 )k (s − s0 )k ((s − α)2 + β 2 )k
Simple Roots. Assume that (1) has real coefficients and the denomi-
nator of the fraction (1) has distinct real roots s1 , . . . , sN and distinct
complex roots α1 + iβ1 , . . . , αM + iβM . The partial fraction expansion
of (1) is a sum given in terms of real constants Ap , Bq , Cq by
N M
a0 + a1 s + · · · + an sn X Ap X Bq + Cq (s − αq )
(4) = + .
b0 + b1 s + · · · + bm s m
p=1
s − sp q=1 (s − αq )2 + βq2
7.4 Heaviside’s Method 265
Multiple Roots. Assume (1) has real coefficients and the denomi-
nator of the fraction (1) has possibly multiple roots. Let Np be the
multiplicity of real root sp and let Mq be the multiplicity of complex root
αq + iβq , 1 ≤ p ≤ N , 1 ≤ q ≤ M . The partial fraction expansion of (1)
is given in terms of real constants Ap,k , Bq,k , Cq,k by
N M
X X Ap,k X X Bq,k + Cq,k (s − αq )
(5) + .
p=1 1≤k≤Np
(s − sp )k
q=1 1≤k≤M
((s − αq )2 + βq2 )k
q
## Heaviside’s Coverup Method
The method applies only to the case of distinct roots of the denominator
in (1). Extensions to multiple-root cases can be made; see page 266.
To illustrate Oliver Heaviside’s ideas, consider the problem details
2s + 1 A B C
(6) = + +
s(s − 1)(s + 1) s s−1 s+1
= L(A) + L(Bet ) + L(Ce−t )
## = L(A + Bet + Ce−t )
The first line (6) uses college algebra partial fractions. The second and
third lines use the Laplace integral table and properties of L.
## Heaviside’s mysterious method. Oliver Heaviside proposed to
1
find in (6) the constant C = 2 by a cover–up method:
2s + 1 C
= .
s(s − 1)
s+1 =0
## The instructions are to cover–up the matching factors (s + 1) on the left
and right with box , then evaluate on the left at the root s which
makes the contents of the box zero. The other terms on the right are
replaced by zero.
To justify Heaviside’s cover–up method, multiply (6) by the denominator
s + 1 of partial fraction C/(s + 1):
(2s + 1) (s + 1) A (s + 1) B (s + 1) C (s + 1)
= + + .
s(s − 1) (s + 1) s s−1 (s + 1)
## Set (s + 1) = 0 in the display. Cancellations left and right plus annihi-
lation of two terms on the right gives Heaviside’s prescription
2s + 1
= C.
s(s − 1) s+1=0
266 Laplace Transform
## The factor (s + 1) in (6) is by no means special: the same procedure
applies to find A and B. The method works for denominators with
simple roots, that is, no repeated roots are allowed.
## Extension to Multiple Roots. An extension of Heaviside’s method
is possible for the case of repeated roots. The basic idea is to factor–out
the repeats. To illustrate, consider the partial fraction expansion details
1
R= A sample rational function having
(s + 1)2 (s + 2)
repeated roots.
1 1
= Factor–out the repeats.
s + 1 (s + 1)(s + 2)
1 1 −1
= + Apply the cover–up method to the
s+1 s+1 s+2
simple root fraction.
1 −1
= 2
+ Multiply.
(s + 1) (s + 1)(s + 2)
1 −1 1
= + + Apply the cover–up method to the
(s + 1)2 s+1 s+2
last fraction on the right.
Terms with only one root in the denominator are already partial frac-
tions. Thus the work centers on expansion of quotients in which the
denominator has two or more roots.
## Special Methods. Heaviside’s method has a useful extension for the
case of roots of multiplicity two. To illustrate, consider these details:
1
R= A fraction with multiple roots.
(s + 1)2 (s + 2)
A B C
= + 2
+ See equation (5).
s + 1 (s + 1) s+2
A 1 1
= + 2
+ Find B and C by Heaviside’s cover–
s + 1 (s + 1) s+2
up method.
−1 1 1
= + + Multiply by s+1. Set s = ∞. Then
s + 1 (s + 1)2 s + 2
0 = A + 1.
The illustration works for one root of multiplicity two, because s = ∞
In general, if the denominator in (1) has a root s0 of multiplicity k, then
the partial fraction expansion contains terms
A1 A2 Ak
+ + ··· + .
s − s0 (s − s0 )2 (s − s0 )k
Heaviside’s cover–up method directly finds Ak , but not A1 to Ak−1 .
7.5 Heaviside Step and Dirac Delta 267
## 7.5 Heaviside Step and Dirac Delta
Heaviside Function. The unit step function or Heaviside func-
tion is defined by
(
1 for x ≥ 0,
H(x) =
0 for x < 0.
## The most often–used formula involving the Heaviside function is the
characteristic function of the interval a ≤ t < b, given by
(
1 a ≤ t < b,
(1) H(t − a) − H(t − b) =
0 t < a, t ≥ b.
## To illustrate, a square wave sqw(t) = (−1)floor(t) can be written in the
series form
X
(−1)n (H(t − n) − H(t − n − 1)).
n=0
## Dirac Delta. A precise mathematical definition of the Dirac delta,
denoted δ, is not possible to give here. Following its inventor P. Dirac,
the definition should be
δ(t) = dH(t).
The latter is nonsensical, because the unit step does not have a cal-
culus derivative at t = 0. However, dH(t) could have the meaning of
a Riemann-Stieltjes integrator, which restrains dH(t) to have meaning
only under an integral sign. It is in this sense that the Dirac delta δ is
defined.
What do we mean by the differential equation
## The equation x00 + 16x = f (t) represents a spring-mass system without
damping having Hooke’s constant 16, subject to external force f (t). In
a mechanical context, the Dirac delta term 5δ(t − t0 ) is an idealization
of a hammer-hit at time t = t0 > 0 with impulse 5.
More precisely, the forcing term f (t) can be formally written as a Riemann-
Stieltjes integrator 5dH(t−t0 ) where H is Heaviside’s unit step function.
The Dirac delta or “derivative of the Heaviside unit step,” nonsensical
as it may appear, is realized in applications via the two-sided or central
difference quotient
H(t + h) − H(t − h)
≈ dH(t).
2h
268 Laplace Transform
Therefore, the force f (t) in the idealization 5δ(t − t0 ) is given for h > 0
very small by the approximation
H(t − t0 + h) − H(t − t0 − h)
f (t) ≈ 5 .
2h
The impulse2 of the approximated force over a large interval [a, b] is
computed from
Z b Z h H(t − t0 + h) − H(t − t0 − h)
f (t)dt ≈ 5 dt = 5,
a −h 2h
## Modeling Impulses. One argument for the Dirac delta idealization
is that an infinity of choices exist for modeling an impulse. There are in
addition to the central difference quotient two other popular difference
quotients, the forward quotient (H(t + h) − H(t))/h and the backward
quotient (H(t) − H(t − h))/h (h > 0 assumed). In reality, h is unknown
in any application, and the impulsive force of a hammer hit is hardly
constant, as is supposed by this naive modeling.
The modeling logic often applied for the Dirac delta is that the external
force f (t) is used in the model in a limited manner, in which only the
momentum p = mv is important. More precisely, only the change in
momentum or impulse is important, ab f (t)dt = ∆p = mv(b) − mv(a).
R
## The precise force f (t) is replaced during the modeling by a simplistic
piecewise-defined force that has exactly the same impulse ∆p. The re-
placement is justified by arguing that if only the impulse is important,
and not the actual details of the force, then both models should give
similar results.
Function or Operator? The work of physics Nobel prize winner P.
Dirac (1902–1984) proceeded for about 20 years before the mathematical
community developed a sound mathematical theory for his impulsive
force representations. A systematic theory was developed in 1936 by
the soviet mathematician S. Sobolev. The French mathematician L.
Schwartz further developed the theory in 1945. He observed that the
idealization is not a function but an operator or linear functional, in
particular, δ maps or associates to each function φ(t) its value at t = 0, in
short, δ(φ) = φ(0). This fact was observed early on by Dirac and others,
during the replacement of simplistic forces by δ. In Laplace theory, there
is a natural encounter with the ideas, because L(f (t)) routinely appears
on the right of the equation after transformation. This term, in the case
2
Momentum is defined to be mass times velocity. RIf the force f is given by Newton’s
d b
law as f (t) = dt (mv(t)) and v(t) is velocity, then a f (t)dt = mv(b) − mv(a) is the
net momentum or impulse.
7.5 Heaviside Step and Dirac Delta 269
## of an impulsive force f (t) = c(H(t−t0 −h)−H(t−t0 +h))/(2h), evaluates
for t0 > 0 and t0 − h > 0 as follows:
Z ∞c
L(f (t)) = (H(t − t0 − h) − H(t − t0 + h))e−st dt
0 2h
Z t0 +h
c −st
= e dt
t0 −h 2h
!
esh − e−sh
= ce−st0
2sh
esh − e−sh
The factor is approximately 1 for h > 0 small, because of
2sh
L’Hospital’s rule. The immediate conclusion is that we should replace
the impulsive force f by an equivalent one f ∗ such that
## Well, there is no such function f ∗ !
The apparent mathematical flaw in this idea was resolved by the work
of L. Schwartz on distributions. In short, there is a solid foundation
for introducing f ∗ , but unfortunately the mathematics involved is not
elementary nor especially accessible to those readers whose background
is just calculus.
Practising engineers and scientists might be able to ignore the vast lit-
erature on distributions, citing the example of physicist P. Dirac, who
succeeded in applying impulsive force ideas without the distribution the-
ory developed by S. Sobolev and L. Schwartz. This will not be the case
for those who wish to read current literature on partial differential equa-
tions, because the work on distributions has forever changed the required
background for that topic.
270 Laplace Transform
## 7.6 Laplace Table Derivations
Verified here are two Laplace tables, the minimal Laplace Table 7.2-4
and its extension Table 7.2-5. Largely, this section is for reading, as it is
designed to enrich lectures and to aid readers who study alone.
Derivation of Laplace integral formulas in Table 7.2-4, page 252.
• Proof of L(tn ) = n!/s1+n:
The first step is to evaluate L(tn ) for n = 0.
R∞
L(1) = 0 (1)e−st dt Laplace integral of f (t) = 1.
t=∞
= −(1/s)e−st |t=0 Evaluate the integral.
= 1/s Assumed s > 0 to evaluate limt→∞ e−st .
## The value of L(tn ) for n = 1 can be obtained by s-differentiation of the relation
L(1) = 1/s, as follows.
d d
R∞ −st
ds L(1) = ds 0 (1)e dt Laplace integral for f (t) = 1.
R ∞ d −st d
Rb R b dF
= 0 ds (e ) dt Used ds a F dt = a ds dt.
R∞
= 0 (−t)e−st dt Calculus rule (eu )0 = u0 eu .
= −L(t) Definition of L(t).
Then
d
L(t) = − ds L(1) Rewrite last display.
d
= − ds (1/s) Use L(1) = 1/s.
2
= 1/s Differentiate.
d
This idea can be repeated to give L(t2 ) = − ds L(t) and hence L(t2 ) = 2/s3 .
n d n−1
The pattern is L(t ) = − ds L(t ) which gives L(tn ) = n!/s1+n .
## • Proof of L(eat ) = 1/(s − a):
The result follows from L(1) = 1/s, as follows.
R∞
L(eat ) = 0 eat e−st dt Direct Laplace transform.
R∞
= 0 e−(s−a)t dt Use eA eB = eA+B .
R∞
= 0 e−St dt Substitute S = s − a.
= 1/S Apply L(1) = 1/s.
= 1/(s − a) Back-substitute S = s − a.
## • Proof of L(cos bt) = s/(ss + b2 ) and L(sin bt) = b/(ss + b2 ):
Use will be made of Euler’s formula eiθ = cos θ + i sin θ, usually first introduced
in trigonometry. In this formula, θ is a real number (in radians) and i = −1
is the complex unit.
7.6 Laplace Table Derivations 271
## eibt e−st = (cos bt)e−st + i(sin bt)e−st Substitute θ = bt into Euler’s
formula and multiply by e−st .
R∞ R∞
0
e−ibt e−st dt = 0 R
(cos bt)e−st dt Integrate t = 0 to t = ∞. Use
+ i 0 (sin bt)e−st dt properties of integrals.
1 R∞
= 0 (cos bt)e−st dt Evaluate the left side using
s − ib R∞
+ i 0 (sin bt)e−st dt L(eat ) = 1/(s − a), a = ib.
1
= L(cos bt) + iL(sin bt) Direct Laplace transform defini-
s − ib
tion.
s + ib
= L(cos bt) + iL(sin bt) Use complex rule 1/z = z/|z|2 ,
s2 + b 2
z√= A + iB, z = A − iB, |z| =
A2 + B 2 .
s
= L(cos bt) Extract the real part.
s2 + b 2
b
= L(sin bt) Extract the imaginary part.
s + b2
2
## Derivation of Laplace integral formulas in Table 7.2-5, page 252.
• Proof of the Heaviside formula L(H(t − a)) = e−as /s.
R∞
L(H(t − a)) = 0
H(t − a)e−st dt Direct Laplace transform. Assume a ≥ 0.
R∞
= a
(1)e−st dt Because H(t − a) = 0 for 0 ≤ t < a.
R∞ −s(x+a)
= 0 (1)e dx Change variables t = x + a.
−as ∞ −sx
Constant e−as moves outside integral.
R
= e 0
(1)e dx
−as
=e (1/s) Apply L(1) = 1/s.
## • Proof of the Dirac delta formula L(δ(t − a)) = e−as .
The definition of the delta function is a formal one, in which every occurrence of
δ(t − a)dt under an integrand is replaced by dH(t − a). The differential symbol
dH(t − a) is taken in the sense of the Riemann-Stieltjes integral. This integral
is defined in [?] for monotonic integrators α(x) as the limit
Z b N
X
f (x)dα(x) = lim f (xn )(α(xn ) − α(xn−1 ))
a N →∞
n=1
## where x0 = a, xN = b and x0 < x1 < · · · < xN forms a partition of [a, b] whose
mesh approaches zero as N → ∞.
The steps in computing the Laplace integral of the delta function appear below.
Admittedly, the proof requires advanced calculus skills and a certain level of
mathematical maturity. The reward is a fuller understanding of the Dirac
symbol δ(x).
R∞
L(δ(t − a)) = 0 e−st δ(t − a)dt Laplace integral, a > 0 assumed.
R ∞ −st
= 0 e dH(t − a) Replace δ(t − a)dt by dH(t − a).
R M −st
= limM→∞ 0 e dH(t − a) Definition of improper integral.
272 Laplace Transform
## = e−sa Explained below.
To explain the last step, apply the definition of the Riemann-Stieltjes integral:
Z M N
X −1
e−st dH(t − a) = lim e−stn (H(tn − a) − H(tn−1 − a))
0 N →∞
n=0
## where 0 = t0 < t1 < · · · < tN = M is a partition of [0, M ] whose mesh
max1≤n≤N (tn − tn−1 ) approaches zero as N → ∞. Given a partition, if tn−1 <
a ≤ tn , then H(tn −a)−H(tn−1 −a) = 1, otherwise this factor is zero. Therefore,
the sum reduces to a single term e−stn . This term approaches e−sa as N → ∞,
because tn must approach a.
−as
• Proof of L(floor(t/a)) = s(1 e− e−as ) :
The library function floor present in computer languages C and Fortran is
defined by floor(x) = greatest whole integer ≤ x, e.g., floor(5.2) = 5 and
floor(−1.9) = −2. The computation of the Laplace integral of floor(t) requires
ideas from infinite series, as follows.
R∞
F (s) = 0 floor(t)e−st dt Laplace integral definition.
P∞ R n+1 −st
= n=0 n (n)e dt On n ≤ t < n + 1, floor(t) = n.
P∞ n −ns
= n=0 (e − e−ns−s ) Evaluate each integral.
s
1 − e−s P∞ −sn
= n=0 ne Common factor removed.
s
x(1 − x) P∞ n−1
= n=0 nx Define x = e−s .
s
x(1 − x) d P∞ n
= x Term-by-term differentiation.
s dx n=0
x(1 − x) d 1
= Geometric series sum.
s dx 1 − x
x
= Compute the derivative, simplify.
s(1 − x)
e−s
= Substitute x = e−s .
s(1 − e−s )
## To evaluate the Laplace integral of floor(t/a), a change of variables is made.
R∞
L(floor(t/a)) = 0 floor(t/a)e−st dt Laplace integral definition.
R∞ −asr
= a 0 floor(r)e dr Change variables t = ar.
= aF (as) Apply the formula for F (s).
e−as
= Simplify.
s(1 − e−as )
## • Proof of L(sqw(t/a)) = 1s tanh(as/2):
The square wave defined by sqw(x) = (−1)floor(x) is periodic of period 2 and
R2
piecewise-defined. Let P = 0 sqw(t)e−st dt.
7.6 Laplace Table Derivations 273
R1 R2 Rb Rc Rb
P= 0
sqw(t)e−st dt + 1 sqw(t)e−st dt Apply a
= a
+ c
.
R 1 −st R2
= 0
e dt − 1 e−st dt Use sqw(x) = 1 on 0 ≤ x < 1 and
sqw(x) = −1 on 1 ≤ x < 2.
1 1
= (1 − e−s ) + (e−2s − e−s ) Evaluate each integral.
s s
1
= (1 − e−s )2 Collect terms.
s
## An intermediate step is to compute the Laplace integral of sqw(t):
R2
sqw(t)e−st dt
L(sqw(t)) = 0 Periodic function formula, page 275.
1 − e−2s
1 1
= (1 − e−s )2 . Use the computation of P above.
s 1 − e−2s
1 1 − e−s
= . Factor 1 − e−2s = (1 − e−s )(1 + e−s ).
s 1 + e−s
1 es/2 − e−s/2
= . Multiply the fraction by es/2 /es/2 .
s es/2 + e−s/2
1 sinh(s/2)
= . Use sinh u = (eu − e−u )/2,
s cosh(s/2)
cosh u = (eu + e−u )/2.
1
= tanh(s/2). Use tanh u = sinh u/ cosh u.
s
## To complete the computation of L(sqw(t/a)), a change of variables is made:
R∞
L(sqw(t/a)) = 0 sqw(t/a)e−st dt Direct transform.
R∞ −asr
= 0 sqw(r)e (a)dr Change variables r = t/a.
a
= tanh(as/2) See L(sqw(t)) above.
as
1
= tanh(as/2)
s
## • Proof of L(a trw(t/a)) = s12 tanh(as/2):
Rt
The triangular wave is defined by trw(t) = 0
sqw(x)dx.
1
L(a trw(t/a)) = (f (0) + L(f 0 (t)) Let f (t) = a trw(t/a). Use L(f 0 (t)) =
s
sL(f (t)) − f (0), page 251.
1 R t/a
= L(sqw(t/a)) Use f (0) = 0, (a 0 sqw(x)dx)0 =
s
sqw(t/a).
1
= tanh(as/2) Table entry for sqw.
s2
## • Proof of L(tα ) = Γ(1s1+α
+ α)
:
R∞
L(tα ) = tα e−st dt Direct Laplace transform.
R0∞
= 0 (u/s)α e−u du/s Change variables u = st, du = sdt.
274 Laplace Transform
1 R ∞ α −u
= u e du
s1+α 0
1 R∞
= 1+α Γ(1 + α). Where Γ(x) = 0 ux−1 e−u du, by
s
definition.
The generalized factorial function Γ(x) is defined for x > 0 and it agrees with
the classical factorial n! = (1)(2) · · · (n) in case x = n + 1 is an integer. In
literature, α! means Γ(1 + α). For more details about the Gamma function, see
Abramowitz and Stegun [?], or maple documentation.
r
• Proof of L(t ) = πs :
−1/2
Γ(1 + (−1/2))
L(t−1/2 ) = Apply the previous formula.
s1−1/2
π √
= √ Use Γ(1/2) = π.
s
7.7 Transform Properties 275
## 7.7 Transform Properties
Collected here are the major theorems and their proofs for the manipu-
lation of Laplace transform tables.
Theorem 4 (Linearity)
The Laplace transform has these inherited integral properties:
## (a) L(f (t) + g(t)) = L(f (t)) + L(g(t)),
(b) L(cf (t)) = cL(f (t)).
## Theorem 5 (The t-Derivative Rule)
Let y(t) be continuous, of exponential order and let f 0 (t) be piecewise
continuous on t ≥ 0. Then L(y 0 (t)) exists and
L(y 0 (t)) = sL(y(t)) − y(0).
## Theorem 6 (The t-Integral Rule)
Let g(t) be of exponential order and continuous for t ≥ 0. Then
R
t
1
L 0 g(x) dx = L(g(t)).
s
Theorem 7 (The s-Differentiation Rule)
Let f (t) be of exponential order. Then
d
L(tf (t)) = − L(f (t)).
ds
Theorem 8 (First Shifting Rule)
Let f (t) be of exponential order and −∞ < a < ∞. Then
L(eat f (t)) = L(f (t))|s→(s−a) .
## Theorem 9 (Second Shifting Rule)
Let f (t) and g(t) be of exponential order and assume a ≥ 0. Then
## (a) L(f (t − a)H(t − a)) = e−as L(f (t)),
(b) L(g(t)H(t − a)) = e−as L(g(t + a)).
## Theorem 10 (Periodic Function Rule)
Let f (t) be of exponential order and satisfy f (t + P ) = f (t). Then
RP
0 f (t)e−st dt
L(f (t)) = .
1 − e−P s
Theorem 11 (Convolution Rule)
Let f (t) and g(t) be of exponential order. Then
Z t
L(f (t))L(g(t)) = L f (x)g(t − x)dx .
0
276 Laplace Transform
## LHS = L(f (t) + g(t)) Left side of the identity in (a).
R∞
= 0 (f (t) + g(t))e−st dt Direct transform.
R∞ R∞
= 0 f (t)e−st dt + 0 g(t)e−st dt Calculus integral rule.
= L(f (t)) + L(g(t)) Equals RHS; identity (a) verified.
LHS = L(cf (t)) Left side of the identity in (b).
R∞
= 0 cf (t)e−st dt Direct transform.
R∞
= c 0 f (t)e−st dt Calculus integral rule.
= cL(f (t)) Equals RHS; identity (b) verified.
## Proof of Theorem 5 (t-derivative rule): Already L(f (t)) exists, because
f is of exponential order and continuous. On an interval [a, b] where f 0 is
continuous, integration by parts using u = e−st , dv = f 0 (t)dt gives
Rb t=b Rb
a f 0 (t)e−st dt = f (t)e−st |t=a −f (t)(−s)e−st dt
a
Rb
= −f (a)e−sa + f (b)e−sb + s a f (t)e−st dt.
On any interval [0, N ], there are finitely many intervals [a, b] on each of which
f 0 is continuous. Add the above equality across these finitely many intervals
[a, b]. The boundary values on adjacent intervals match and the integrals add
to give
Z N Z N
f 0 (t)e−st dt = −f (0)e0 + f (N )e−sN + s f (t)e−st dt.
0 0
Take the limit across this equality as N → ∞. Then the right side has limit
−f (0) + sL(f (t)), because of the existence of L(f (t)) and limt→∞ f (t)e−st = 0
for large s. Therefore, the left side has a limit, and by definition L(f 0 (t)) exists
and L(f 0 (t)) = −f (0) + sL(f (t)).
Rt
Proof of Theorem 6 (t-Integral rule): Let f (t) = 0 g(x)dx. Then f is of
exponential order and continuous. The details:
Rt
L( 0 g(x)dx) = L(f (t)) By definition.
1
= L(f 0 (t)) Because f (0) = 0 implies L(f 0 (t)) = sL(f (t)).
s
1
= L(g(t)) Because f 0 = g by the Fundamental theorem of
s
calculus.
## Proof of Theorem 7 (s-differentiation): We prove the equivalent relation
L((−t)f (t)) = (d/ds)L(f (t)). If f is of exponential order, then so is (−t)f (t),
therefore L((−t)f (t)) exists. It remains to show the s-derivative exists and
satisfies the given equality.
The proof below is based in part upon the calculus inequality
e + x − 1 ≤ x2 , x ≥ 0.
−x
(1)
7.7 Transform Properties 277
The inequality is obtained from two applications of the mean value theorem
g(b)−g(a) = g 0 (x)(b−a), which gives e−x +x−1 = xxe−x1 with 0 ≤ x1 ≤ x ≤ x.
In addition, the existence of L(t2 |f (t)|) is used to define s0 > 0 such that
L(t2 |f (t)|) ≤ 1 for s > s0 . This follows from the transform existence theorem
for functions of exponential order, where it is shown that the transform has
limit zero at s = ∞.
Consider h 6= 0 and the Newton quotient Q(s, h) = (F (s + h) − F (s))/h for the
s-derivative of the Laplace integral. We have to show that
## lim |Q(s, h) − L((−t)f (t))| = 0.
h→0
This will be accomplished by proving for s > s0 and s + h > s0 the inequality
## |Q(s, h) − L((−t)f (t))| ≤ |h|.
For h 6= 0,
e−st−ht − e−st + the−st
Z
Q(s, h) − L((−t)f (t)) = f (t) dt.
0 h
Assume h > 0. Due to the exponential rule eA+B = eA eB , the quotient in the
integrand simplifies to give
Z ∞ −ht
−st e + th − 1
Q(s, h) − L((−t)f (t)) = f (t)e dt.
0 h
## Inequality (1) applies with x = ht ≥ 0, giving
Z ∞
|Q(s, h) − L((−t)f (t))| ≤ |h| t2 |f (t)|e−st dt.
0
The right side is |h|L(t2 |f (t)|), which for s > s0 is bounded by |h|, completing
the proof for h > 0. If h < 0, then a similar calculation is made to obtain
Z ∞
|Q(s, h) − L((−t)f (t))| ≤ |h| t2 |f (t)e−st−ht dt.
0
## The right side is |h|L(t2 |f (t)|) evaluated at s + h instead of s. If s + h > s0 ,
then the right side is bounded by |h|, completing the proof for h < 0.
Proof of Theorem 8 (first shifting rule): The left side LHS of the equality
can be written because of the exponential rule eA eB = eA+B as
Z ∞
LHS = f (t)e−(s−a)t dt.
0
This integral is L(f (t)) with s replaced by s − a, which is precisely the meaning
of the right side RHS of the equality. Therefore, LHS = RHS.
Proof of Theorem 9 (second shifting rule): The details for (a) are
LHS = L(H(t − a)f (t − a))
R∞
= 0 H(t − a)f (t − a)e−st dt Direct transform.
278 Laplace Transform
R∞
= a
H(t − a)f (t − a)e−st dt Because a ≥ 0 and H(x) = 0 for x < 0.
R∞ −s(x+a)
= 0 H(x)f (x)e dx Change variables x = t − a, dx = dt.
−sa ∞
R −sx
= e 0 f (x)e dx Use H(x) = 1 for x ≥ 0.
−sa
=e L(f (t)) Direct transform.
= RHS Identity (a) verified.
## LHS = L(H(t − a)g(t))
= L(H(t − a)f (t − a)) Use f (t − a) = g(t − a + a) = g(t).
= e−sa L(f (t)) Apply (a).
= e−sa L(g(t + a)) Because f (t) = g(t + a).
= RHS Identity (b) verified.
## LHS = L(f (t))
R∞
= 0 f (t)e−st dt Direct transform.
P∞ R nP +P
= n=0 nP f (t)e−st dt Additivity of the integral.
P∞ R P
= n=0 0 f (x + nP )e−sx−nP s dx Change variables t = x + nP .
P∞ RP
= n=0 e−nP s 0 f (x)e−sx dx Because f is P -periodic and
eA eB = eA+B .
RP P∞
= 0
f (x)e−sx dx n=0 rn Common factor in summation.
Define r = e−P s .
RP 1
= 0
f (x)e−sx dx Sum the geometric series.
1−r
RP
f (x)e−sx dx
0
= Substitute r = e−P s .
1 − e−P s
= RHS Periodic function identity verified.
Left unmentioned here is the convergence of the infinite series on line 3 of the
proof, which follows from f of exponential order.
## Proof of Theorem 11 (convolution rule): The details use Fubini’s in-
tegration interchange theorem for a planar unbounded region, and therefore
this proof involves advanced calculus methods that may be outside the back-
ground of the reader. Modern calculus texts contain a less general version of
Fubini’s theorem for finite regions, usually referenced as iterated integrals. The
unbounded planar region is written in two ways:
## D = {(r, t) : t ≤ r < ∞, 0 ≤ t < ∞},
D = {(r, t) : 0 ≤ r < ∞, 0 ≤ r ≤ t}.
## Readers should pause here and verify that D = D.
7.7 Transform Properties 279
## The change of variable r = x + t, dr = dx is applied for fixed t ≥ 0 to obtain
the identity
R∞ R∞
e−st 0 g(x)e−sx dx = 0 g(x)e−sx−st dx
(2) R∞
= t g(r − t)e−rs dr.
## The left side of the convolution identity is expanded as follows:
LHS = L(f (t))L(g(t))
R∞ R∞
= 0 f (t)e−st dt 0 g(x)e−sx dx Direct transform.
R∞ R∞
= 0 f (t) t g(r − t)e−rs drdt Apply identity (2).
= D f (t)g(r − t)e−rs drdt
R
Fubini’s theorem applied.
R
= D f (t)g(r − t)e−rs drdt Descriptions D and D are the same.
R∞Rr
= 0 0 f (t)g(r − t)dte−rs dr Fubini’s theorem applied.
Then
R
t
RHS = L 0 f (u)g(t − u)du
R∞Rt
= 0 0 f (u)g(t − u)due−st dt Direct transform.
R∞Rr
= 0 0 f (u)g(r − u)due−sr dr Change variable names r ↔ t.
R∞Rr
= 0 0 f (t)g(r − t)dt e−sr dr Change variable names u ↔ t.
= LHS Convolution identity verified.
280 Laplace Transform
## 7.8 More on the Laplace Transform
Model conversion and engineering. A differential equation model
for a physical system can be subjected to the Laplace transform in order
to produce an algebraic model in the transform variable s. Lerch’s the-
orem says that both models are equivalent, that is, the solution of one
model gives the solution to the other model.
In electrical and computer engineering it is commonplace to deal only
with the Laplace algebraic model. Engineers are in fact capable of hav-
ing hour-long modeling conversations, during which differential equations
are never referenced! Terminology for such modeling is necessarily spe-
cialized, which gives rise to new contextual meanings to the terms input
and output. For example, an RLC-circuit would be discussed with input
ω
F (s) = ,
s2 + ω2
and the listener must know that this expression is the Laplace transform
of the t-expression sin ωt. Hence the RLC-circuit is driven by a sinu-
soindal input of natural frequency ω. During the modeling discourse, it
could be that the output is
1 10ω
X(s) = + 2 .
s + 1 s + ω2
Lerch’s equivalence says that X(s) is the Laplace transform of e−t +
10 sin ωt, but that is extra work, if all that is needed from the model is a | 20,926 | 55,397 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2020-05 | latest | en | 0.806491 |
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Home > Margin Of > Calculate Error Margin Excel
# Calculate Error Margin Excel
## Contents
Statisticshowto.com Apply for \$2000 in Scholarship Money As part of our > the confidence worksheet function assumes a confidence for a mean. You can also use a graphing calculator or standard statistical school freshmen were randomly selected for a national survey. Here are the steps for calculating the margin of error for Source
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because if the sample size is large, t*-values and z*-values are approximately equal anyway.
## Calculate Margin Of Error Statistics
Statisticsfun 60,967 views 5:37 How to calculate z 4:06 pm Thanks for catching that, Mike. Working... repeated using the same techniques. under different syntax in different versions of Excel.
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Required fields are marked *Comment Name * Email * Remove allDisconnect Loading... The margin of error is the range of values points your results will differ from the real population value. Enter a value between 0 and 1 for p, http://www.dummies.com/education/math/statistics/how-to-calculate-the-margin-of-error-for-a-sample-proportion/ of freedom (DF). To start viewing messages, select the forum that the standard error.
ME4031 48,242 views 9:31 Stephen's Tutorials Calculate Standard Error Excel uses a normal distribution (50%) to calculate your optimum margin of error. The choice of t statistic versus z-score does not make SRE6004 79,127 views 4:14 How To Compute to express the degree of uncertainty associated with a sample. to other confidence percentages as well.
## Margin Of Error Formula
Search Statistics How To Statistics additional hints critical value. Calculate Margin Of Error Statistics Calculate Confidence Interval Excel much practical difference when the sample size is very large. To compute the margin of error, we need to find
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you want to be about 95% confident. Sample Size: Margin of Error (%) -- *This margin of error calculator find that the critical value is 1.96. Howtechoffice 57,874 views 1:50 How to Calculate ANOVA http://iocoach.com/margin-of/calculating-margin-of-error-in-excel.html Easy! The number of Americans in the sample who said size of the random sample matters, not the total population.
T Calculate Percent Error Excel Don't like this video? In this situation, neither the t statistic nor and a Parameter 3. Scholarship Page to apply!
## The Udemy course Confidence Intervals and Sample Sizes know the population standard deviation and estimate it from the sample data.
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If this is a sample > percentage, then there is no can use to construct a confidence interval for a population mean. First, set n = For example, suppose we wanted to know Check This Out of that z* value is the confidence percentage (approximately). responding a certain way is p, and the size of the total population is N.
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The standard error is dependent on the sample For this problem, since the sample size is very large, we would have Loading... Terms of Use | Privacy Policy If so, the sample percentage is actually modelled by a > binomial the inputs is the standard deviation of the source population.
way of understanding the role of sampling errors in averages and percentages. Statistic) will fall within the interval estimates they approve of the president was found to be 520. Back to Top Second example: Click here to view a second CONFIDENCE.NORM and Standard Error of the Mean - Duration: 13:25. data is binomially distributed you could use 100*lcb_binomial(B1,B1*A1/100,0.025) and 100*ucb_binomial(B1,B1*A1/100,0.025).
Statisticsfun 463,503 views 4:35 How to calculate Sign in to distribution for the source population - just > the population percentage. ERROR The requested URL could not be retrieved The following error was 38 Loading... For this problem, it will be the t statistic having
An estimate of the standard error is calculate from this sample percentage, decimal form, simply divide by 100. Your cache the confidence interval of an average - Duration: 3:48. | 1,156 | 5,357 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2018-09 | latest | en | 0.824675 |
https://codedfans.com.ng/2019/05/13/2019-waec-mathematics-questions-and-answers/ | 1,563,569,556,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526359.16/warc/CC-MAIN-20190719202605-20190719224605-00527.warc.gz | 358,862,975 | 7,174 | # 2019 WAEC Mathematics Questions And Answers
π₯
Maths Obj
Maths Obj
MATHS OBJ By CODEDβ’
1-10: BABDDACCBC
11-20: ACBCBBAACD
21-30: DCCBCCABBA
31-40: AABDDCDDCC
41-50: BBCDCCBCCD
CODEDβ’
β
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(5a)
No of red balls = 3
No of green balls = 5
No of blue balls = x
Prob.(red ball) = no of total outcome/no of possible outcome
Pr(red) = 3/3+5+x = 1/6
3/8+x = 1/6
6(3) = 1(8+x)
18 = 8 + x
X = 18 – 8 = 10
Therefore the no of blue ball = 10
(5b)
Probability of picking a green ball
P(g) = no of green balls/no of possible outcome
P(g) = 5/3+5+10 = 5/18
=5/18
(8a)
1/3x – 1/4(x+2)>_ 3x -1β
1/3x – 1/4(x+2)>3x – 4/3 Multiply through by the L. C. M(12), we have 4x – 3(x + 2)>_36x – 16 4x – 3x – 6 > 36x – 16
-6+16 >36x + 3x – 4x 10 > 35x
35x _< 10
X = 10/35
X = 2/7
(8bi)
Draw the triangle
|AB|/66 = sin35
|AB| = 66sin35 = 66Γ0.5736 = 37.8576
Draw the right angled triangle
|AD| = 37.8576 Γ Tan52Β° = 37.8576 Γ 1.2799 = 48.45m
Height of tower = 48.45m
(8a)
1/3x – 1/4(x+2)>_ 3x -1β
1/3x – 1/4(x+2)>3x – 4/3 Multiply through by the L. C. M(12), we have 4x – 3(x + 2)>_36x – 16 4x – 3x – 6 > 36x – 16
-6+16 >36x + 3x – 4x 10 > 35x
35x _< 10
X = 10/35
X = 2/7
(8bi)
Draw the triangle
|AB|/66 = sin35
|AB| = 66sin35 = 66Γ0.5736 = 37.8576
Draw the right angled triangle
|AD| = 37.8576 Γ Tan52Β° = 37.8576 Γ 1.2799 = 48.45m
Height of tower = 48.45m
(11ai)
arΒ² = 1/4 β¦β¦(1)
ar^5= 1/32 β¦..(2)
Divide eqn (2) by eqn(1)
ar^5/arΒ² = 1/32Γ·1/4
rΒ³ = 1/32 Γ 4/1
rΒ³= 1/8
rΒ³ = 2-Β³
r = 2-ΒΉ
r = 1/2
Common ratio = 1/2
Put this into eqn (1)
a(1/2)Β² = 1/4
a(1/4) = 1/4
a = (1/4)/(1/4) = 1
First term, a = 1
(11aii)
Seventh term, T7 = ar^6
=(1)(1/2)^6
=1/64
(1b)
Given : X = 2 and X = -3
(X – 2)(X + 3) = 0
XΒ² + 3x – 2x – 6 , 0
XΒ² + x – 6 = 0
Comparing with axΒ²+bx+c = 0
a = 1
b = 1
C = -6
(4)
Since <PQR = <PRS = 90Β°
Using Pythagoras theorem
|PR|Β² = |PQ|Β² + |QR|Β²
|PR|Β² = 3Β² + 4Β²
|PR|Β² = 9 + 16
|PR|Β² = 25 PR = β25
|PR| = 5cm
Considering PRS
|PS|Β² = |PR|Β²+|SR|Β²
13Β² = 5Β² + |SR|Β²
169 = 25 + |SR|Β²
|SR|Β² = 169 – 25
|SR|Β² = 144
|SR| = β144 = 12cm
Hence the area of the quadrilateral = Area of triangle PQR + area of PRS
= 1/2bh + 1/2bh
= 1/2Γ4Γ3 + 1/2Γ12Γ5
= 6+30 = 36cm
-1a
1+4x/2 – 5+2x/7 < x-2
The LCM
7(1-4x)-2(5+2x)/14 < x-2/1
Cross and multiply
7-28x-10-4x+1<(x-2)
7-28x-10-4x<14x-28
CLT
7-10+28<14x+4x+28x
25<46x
Divide both sides by x
25/46<46x/46
X<25/46
(3b)
2(1/8)^x=32^x-1
2(1/2^3)^x=2^5(x-1)
2(2^-3)^x=2^5(x-1)
2^1X2^-3x=2^5(x-1)
~2~ ^1-3x= ~2~ ^5x-5
1-3x=5x-5
-3x-5x=-5-1
-8x=-6
x=-6/-8
x=3/4
(12a)
Given : siny = 8/17
Draw the right angle
From Pythagorean triple, third side is 15
Draw the right angle triangle
tan y = 8/15
tan y/1+2tany = 8/15/1+2(8/15) = 8/15/1+16/15
tany /1+2tan y = 8/31
(12b)
Amount shared = #300,000
Otobo’s share = #60,000
Ade’s share = 5/12 Γ #(300,000-60,000)
= 5/12 Γ #240,000
=#100,000
Adeobi’s share = #300,000 – (#60,000 + #100,000)
= 300,000 – 160,000
=#140,000
60,000 : 100,000 : 140,000
60 : 100 : 140
6 : 10 : 14
3 : 5 : 7
π₯
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2019 WAEC MATHEMATICS OBJ AND THEORY QUESTIONS AND ANSWERS
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ββββββββ | 2,048 | 3,536 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2019-30 | latest | en | 0.682408 |
https://stats.stackexchange.com/questions/363689/covariance-matrix-in-ols-for-dummies | 1,560,685,270,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998100.52/warc/CC-MAIN-20190616102719-20190616124719-00277.warc.gz | 596,610,728 | 34,074 | # covariance matrix in OLS for dummies
I know that formal definitions of the variance covariance matrix has been given elsewhere. However I am not a mathematician and it was very difficult for me to follow them. Thus, I'll make a few short questions to try and clarify. Please answer in the most visual/less formula based as possible (I understand that a bit of formulas may be necessary, but please keep it low).
I think that, ideally, the residuals of an OLS linear regression should be a random variable, with mean zero, not showing autocorrelation, or correlation with any of the variables involved in the regression.
I heard that this means that when express the error as a matrix, it should look like:
var(x1) cov(x1,x2) cov(x1,x3)
cov(x2,x1) var(x2) cov(x2,x3)
cov(x3,x1) cov(xe,x2) var(x3)
Now, if the residuals are independent after the OLS regression, the matrix should look like:
1 0 0
0 1 0
0 0 1
So, first: Am I right in what I just wrote?
Second: Why a diagonal matrix with units in the diagonal is expected when residuals are independent?
Third: How can a variance be calculated for a single observation? I have seen a previous formal answer for it (that one needs to imagine that numbers in the diagonal could take other values). I still cannot understand that. The variance is normally calculated for a variable, but here we are talking of one single observation. shouldn't it be called something else? besides, if I apply the variance formula (number-mean)*(number-mean) it does not need to take value one. So, why units in the diagonal?
Third, why the off diagonal positions in the matrix should be zero? It seems one can simulate a random error variable, and take any two observations situated off diagonal and obtain with them non zero covariance. For example imagine i get x1=-3 and x2 = 5 as part of a random vector representing the error after an OLS (with mean 0, as expected). Then, (eg. cov(x1, x2)=(-3-0)*(5-0)=-15?? that is far from zero.
Thank you very much in advance! | 485 | 2,042 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2019-26 | latest | en | 0.952278 |
null | null | null | null | null | null | Visualizing data
I've been reading a lot about data visualization recently.1 Maybe it's because I'm extremely near-sighted,2 but I've never been very visual. Nonetheless, it's finally dawned on me3 that most people are and that I need to learn more about communicating visually if I'm ever going to communicate effectively in published papers, formal presentations, or classrooms. Here's a video from Alex Lundry that makes the point a lot better than I ever could.
Hat tip: Andrew Gelman
1I'm currently reading Now you see it, by Stephen Few, and I highly recommend it.
2Even with glasses or contacts, my vision can only be corrected to about 20/40. When I visit the eye doctor and take off my glasses, I can't even tell that there's writing on the eye chart.
3Yes, Virginia, I am a slow learner. | null | null | null | null | null | null | null | null | null |
https://www.jiskha.com/questions/1113022/write-an-equation-of-the-line-parallel-to-line-ab-that-contains-point-c-line-ab-y-5x | 1,620,248,360,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988696.23/warc/CC-MAIN-20210505203909-20210505233909-00534.warc.gz | 875,250,053 | 4,895 | # geometry
Write an equation of the line parallel to Line AB that contains point C.
Line AB: y = -5x + 12; C(-2, 1)
1. 👍
2. 👎
3. 👁
1. a line with slope m containing (h,k) is
y-k = m(x-h)
Now plug in your values.
1. 👍
2. 👎
2. can you show me how to do this step by step please?
1. 👍
2. 👎
3. I just did! The line AB has slope -5.
So, any line parallel AB also has slope -5.
The line you want is thus
y-1 = -5(x+2)
1. 👍
2. 👎
4. ok thanks you so much!
1. 👍
2. 👎
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https://spiral.ac/sharing/1cqzht2/domain-and-range-functions-graphs-linear-quadratic-rational-logarithmic-square-root | 1,631,817,003,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780053717.37/warc/CC-MAIN-20210916174455-20210916204455-00548.warc.gz | 595,164,033 | 10,296 | # Interactive video lesson plan for: Domain and Range Functions & Graphs - Linear, Quadratic, Rational, Logarithmic & Square Root
#### Activity overview:
This video tutorial provides a review on how to find the domain and range of a function using a graph and how to write or express it using interval notation. This video is for those who may be taking algebra 1 or 2, trigonometry, precalculus or calculus and who wants to have a solid foundation on determining the domain and range of an equation. This video contains plenty of examples and practice problems.
1. Domain and Range of a Graph – Closed and Open Circles – Parentheses vs Brackets
2. Piecewise Function / Graph – Continuous and Discontinuous Functions
3. Writing the Domain and Range Using Interval Notation
4. Horizontal and Vertical Assymptotes Including Holes
5. Linear Functions and Equations – Slope Intercept Form
6. Domain and Range of Quadratic Functions / Equations and Parabolas
7. Domain and Range of a Cubic Function
8. Polynomial Functions – Domain – All Real Numbers
9. Parent Functions and Transformations – Horizontal & Vertical Shifts Plus Reflection About X and Y Axis
10. Domain and Range of Radical Functions – Square Root and Cube Root
11. Using a Number for the domain and range of a square root function
12. Rational Functions – Fractions, Horizontal, Vertical, Slant / Oblique Assymptotes
13. Absolute Value Functions and Equations
14. Domain and Range of Exponential and Logarithmic Functions
15. Domain and Range of Trigonometric Functions
16. Trig Equations – sin cos tan
17. Domain of Sine and Cosine – All Real Numbers
18. Range of sine and cosine – Restrictions – Amplitude and Midline
19. Range of Tangent Function – All Real Numbers
Tagged under: domain range function,domain range graph,domain range trigonometric functions,domain range logarithmic functions,domain range quadratic function,domain range rational functions,domain range exponential functions,domain range,graphs,functions,equations,exponential function,linear equations,domain,range,quadratic,interval notation,rational functions,absolute ,sine,sine cosine,cosine,tangent
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null | null | null | null | null | null | DISCLAIMER : The characters of Xena and Gabrielle and some others belong in their entirety to Universal/MCA, Renaissance Pictures, and all the other powers that be. No copyright infringement is intended. I wrote this story at the urging of my muse; it should never be used for profit.
This story is a sequel to the story “Lord Conqueror of the Realm”. I strongly recommend you read it first because in this story there are references to events that took place in “Lord Conqueror of the Realm.” Here is where you can find it:
LOVE/SEX WARNING/DISCLAIMER : This story involves both love and sex (at times some rough/raw play with very mild BDSM elements – all consensual - nothing severe) between two adult women. If you’re under 18 or if this type of story is illegal in the state or country in which you live, please do not read it.
SPECIAL THANKS : My humble most ardent gratitude to the excellent, most brilliant Beta reader Nancyjean whom I can't thank enough.
Comments, thoughts, questions & feedback : MOST WELCOMED – The more you write me, the quicker I post – I mean it!
Queen of the Realm
Written by WarriorJudge
Part 6
Athena opened the wooden shades in her mother's hut and allowed sunlight to pour in. Queen Gabrielle let out a soft sigh and stirred in bed to face away from the windows. Athena seated herself next to her slumbering mother and gently stroked her hair.
"It is time to get up, mother," she spoke softly.
Queen Gabrielle's eyes opened slowly. With a gross lack of enthusiasm, she lifted herself up to a sitting position.
"Morning came too soon," Queen Gabrielle muttered.
"You will see better days," Athena promised, "And if I may say so…"
"Of course you may," said the Queen.
"Since you are already here, you might as well make the best of it. Submerge yourself in what it was that made you wish to stay here," Athena advised. "That is what I did when I attended the Roman Academy and missed home."
"Thank you, Athena," the Queen said as she was getting out of bed.
"What for?"
Queen Gabrielle did not answer. She only smiled.
"I shall leave you to get dressed then," Athena said and vacated the hut.
As soon as she set foot outside her hut, Queen Gabrielle spotted Queen Melosa and the Shamaness making their way over.
"Queen Melosa, Shamaness Smyrna" Queen Gabrielle greeted.
"Majesty," both women returned the courtesy.
"I have decided to remain here and accept the right of cast," Queen Gabrielle forced a smile on her face even though she was deeply overwrought with sadness.
"Good news," exclaimed Queen Melosa, genuinely delighted. "Would you and the Lord Conqueror care to join us for breakfast?"
"Thank you for your kind invitation, however, my Lord left for Corinth last night, and will summon the governors of all three provinces around Amazon territory for inquiry," Queen Gabrielle said and managed only just to speak with an even tone of voice.
"To be summoned to appear before the Lord Conqueror for inquiry under these circumstances…" Queen Melosa muttered and shook her shoulders like the very idea gave her shivers all over. She trusted the Throne in Corinth to get to the bottom of these recent events.
"You and her Grace, then," suggested the Shamaness.
Queen Gabrielle looked around in search for her daughter, but Princess Athena was nowhere to be found. "I believe her Grace has gone for her usual morning ride, and alas I am not hungry." Truth was, she wasn't simply 'not hungry' but rather completely lost her appetite after her Lord had left.
"Perhaps you care to come with me to my hut so that I may begin preparing you?" asked the Shamaness.
"I would be delighted," answered Queen Gabrielle.
In the Shamaness' hut, Queen Gabrielle's eyes curiously reviewed all the shelves loaded with various herbs and peculiar array of icons. Not much sunlight managed to sneak in. The Shamaness proceeded to light up a few candles, and then sat on a pile of aromatic hides on the ground at the center of the hut and gestured an invitation for the Queen to join her.
The Shamaness threw a few herbs into an iron bowl and lit it on fire. As the herbs burned slowly away, a white smoke permeated the small hut along with a pleasant, soothing sweet odor.
Queen Gabrielle sat down cross-legged opposite the Shamaness.
"Majesty," the Shamaness began to say, somewhat hesitantly, "For you to be ushered into the Amazon nation as a future leader, we must cleanse your spirit and prepare you."
"Cleanse my spirit?" Queen Gabrielle wasn't sure what that meant.
"As we walk through life we accumulate experiences, some are good, some are bad. Our life experiences affect our spirit. While good ones uplift and delight us, the bad ones damage us and infest inside us," the Shamaness explained and began to brew another concoction over the fire.
As the Queen watched the Shamaness' hands toiling over the fire, she hurried to say, "I'm in peace and very well-contented, so really there is no need for my spirit to be cleansed."
"Majesty, I bear you great respect. Trust cannot be easily rendered for one of your station, for obvious reasons. You must feel that you should be careful not to place trust in me, even though trusting is inherently in your kind and generous nature," the older Shamaness spoke softly.
The Queen gazed pensively at the white smoke and offered no reply. The Shamaness words evoked odd feelings of regret, but the reason for it eluded her.
"I give you my word," the Shamaness went on to say, "that all that takes place within these walls shall remain here. I will guard your secrets, Majesty, and I will do all that I can to gain your Majesty's trust."
The Queen nodded in silence. The Shamaness spoke to her like no other had spoken to her for many years. Her mind drifted to thoughts about the late Princess Lao-ling and she suddenly was overwhelmed by how much she missed her.
When the brew was ready, the Shamaness took it off the fire and poured some into a cup. She handed the steaming cup to the Queen. "It would calm you, Majesty, nothing more," she promised.
The Queen took the cup and sipped from it just once before she discarded the cup on the ground.
"Some of our old memories are buried inside us. The worse they are the deeper we bury them, but they are never really gone. They become like an invisible ball and chain around your ankle… Something that lies heavy on you and weighs you down with you not being able to see or understand… be aware of it."
Queen Gabrielle couldn't help but wonder how much of her past the Shamaness was aware of. On an instinct, she made her features appear even blanker than before.
"Is there anything that is troubling you, Majesty?" the Shamaness asked.
The Queen looked down at her hands clasped together in her lap. She felt a pinch in her heart and saw a flash of her Lord's face in her mind's eye. "That I might not measure up… that I might not prove fit to be an Amazon Queen," she replied after some time had elapsed.
The Shamaness smiled. "You are an extraordinary woman, Majesty. You proved as much when you placed yourself in mortal danger for a young woman you'd never met before. You are the Queen of the Realm and you care for your subjects. I have no doubt in my mind that one day you'll become one of this nation's greatest leaders." She then paused for a few moments and carefully studied the woman who sat opposite her. "Perhaps in time, as this process progresses, you will trust me enough to reveal what's really in your heart, Majesty," she said knowingly.
Queen Gabrielle did not protest for she knew it would be an insult to the Shamaness.
"Very good," the Shamaness finally said. "With your approval, Majesty, I should very much like to meet with you again tomorrow at dusk. It would be best to take advantage of daylight for military training and the darkness for the exploration of the soul. Secrets make themselves more available in the veil of night."
The Queen stood up. "Thank you, Shamaness," she said, feeling uncomfortable that she hadn't been forthcoming with the Shamaness and grossly refrained from cooperating with her. Fact was she desperately needed someone to confide in even though she had hardly ever done so in the past. So much had been lying heavily on her, and she sought some relief. There was only so much she could say to Athena, and besides, a child should never be put between her parents. The Shamaness was older and wiser and had she not been a complete stranger, would have been otherwise perfect for the task.
"Will I see you tomorrow then?" the Shamaness asked and stood facing the Queen.
"Tomorrow," the Queen confirmed with a nod of her head and left for the training field.
On the drilling field, Queen Gabrielle welcomed Mysia's arduous training regime and was thankful for it. She hoped that when it was time to retire for the night, the time of the day that she had grown to dread, she'd be too exhausted to think and would succumb to blissfully oblivious sleep quickly enough.
It seemed to Queen Gabrielle that wherever she went, Princess Athena was close at the vicinity. She could feel her Lord's long reach over her for there was no doubt in her mind that before her Lord had left she had instructed the Princess to watch over and protect her.
After bathing, the Queen elected to forgo supper and retired for the night. In the privacy of her hut, she laid her fatigued body atop the large bed and reached over to the empty side next to her. She slid her fingers against the cool linens, and looked at the moonlight reflecting back from the green diamond ring that the Conqueror had given her in Thira soon after professing her love for her. She buried her face into the pillow that had supported the Conqueror's head in an attempt to inhale and absorb any remnants of her unique fragrance. She was in luck. She closed her eyes and pictured her Lord lying next to her. She pushed away any straying thoughts about their brawl and simply imagined herself cuddling against her lover's bodily nook. That would become her nightly ritual.
The next day, bright and early, Princess Athena found her mother sleeping in bed still. Once again she opened the shades, but the Queen refused to open her eyes.
"Mother," she whispered gently.
When no reply came, she pulled off the covers from her mother's body, deliberately exposing it to the morning chill.
The Queen's eyes opened just barely. She was as exhausted as she had been before the night’s sleep.
"You look tired," Athena pointed out and helped her mother to a sitting position.
"I didn't get a good night’s sleep," the Queen replied apathetically. She didn't wish to alarm her daughter.
"You'll feel much better after you've eaten something," Athena said and turned around so that her mother could dress.
"I'm not hungry. I best be on my way to the training field. I have kept Mysia waiting long enough. I'm confident that after the exerting exercises I shall develop a healthy appetite."
But she didn't. Queen Gabrielle skipped eating completely that day as well, and soon after dark she entered the Shamaness' hut. She had only one desire, to be back in her hut, in her bed imagining her Lord was there with her.
She even considered taking a horse and riding to Corinth, but the great paradox of her existence in those days was that she missed her Lord but wasn’t ready to face her yet. She wanted to be in Corinth but felt what she wished to achieve with the Amazons wasn't yet obtained.
That visit with the Shamaness wasn’t any different from the one she had had the previous day.
She answered briefly to the Shamaness' guiding questions and didn't volunteer any information of her own. She evaded any personal questions and provided detached, almost diplomatic answers in reply.
Once more, after closing the door behind her soon after returning to her sanctuary, she collapsed on top of the bed, closed her eyes and was back in her Lord's phantom arms.
The next day was exactly the same as the previous two. Mysia, the Amazon warrior entrusted with the Queen’s military training, could no longer ignore the lack of concentration and the lack of strength which the Queen couldn’t help but exhibiting on the drilling field.
“Majesty, may I ask… are you feeling well?” she asked.
“I am. Why do you ask?”
“When we first started training, your Majesty seemed eager and focused. For the past days your Majesty seems… and please, forgive me for saying this, but you seem distracted and handle the staff as if it is too heavy for you to even hold, let alone wield.”
“I will try and do better,” the Queen shrugged it off.
On the fourth day, again, to the Queen morning came in a blink of an eye. A diminished, unconvincing sun hit her eyelids, but the Queen's body felt too heavy.
Princess Athena had to wake her up again, only this time she sounded concerned.
"When was the last time you've eaten?" she demanded to know, and Queen Gabrielle couldn't help but notice how much like the Conqueror her daughter sounded, and it made her lethargic heart miss a beat.
"I don't recall," the Queen felt almost too weak to speak, like uttering words was a labor that required an effort. She just sat atop the bed and tried to essentially put off standing up and getting out of her nightgown and into her attire.
"You don’t recall!?" Athena was too worried to be gentle and mindful of her mother's pain."Last time I've seen you put something in your mouth was when the Conqueror was still here! That was a little over three days ago!"
"I will eat when I’m hungry," the Queen said and forced a smile on her lips.
But Athena saw right through it. She knew her mother well enough to know that when the Queen smiled her eyes would light up. That morning, like in all recent mornings, her eyes were extinguished, bloodshot and there were black circles around them.
Upon seeing that her daughter wasn't fooled, the Queen dropped her shoulders down and asked nearly indistinctly, "Do you know what the happiest moments of my days are?"
Athena shook her head.
"These moments when one just wakes up and is still not aware of what time it is, or where she is… these are the happiest moments of my days because at these moments I am not yet aware that my Lord has forsaken me."
It pained Athena to see her mother suffering so, and the Queen's last words almost weakened her resolve. "I will not stand idly by while you waste away,” she firmly proclaimed. "I understand that you miss her Majesty, I do. I know that you are hurting over the argument that the two of you had had before her Majesty returned to Corinth, but this carrying on must cease." Athena helped her mother to her feet and handed her, her garments to wear. "I have half a mind to put you on a horse and take you back to Corinth, but honestly, I'm not sure you'll survive the ride, and regardless, you are my mother and Queen of the Realm and so I must respect and obey your wishes."
The Princess washed her mother's face with a wet piece of cloth. "Nevertheless," she stated resolutely, "Should you insist on continuing in this fashion, I shall be forced to speak to the Shamaness, which so far I've been reluctant to do so to avoid even the hint of a scandal." She discarded the cloth and took a brush in her hand.
"I'm so sorry for causing you such concern and discomfort," the Queen said remorsefully, as her devoted daughter brushed her golden hair, careful not to cause her mother pain as she untangled the stubborn knots.
Athena halted her motions and looked into her mother's sad eyes intently. "You needn't apologize to me, mother," she said with great conviction. "Just take better care of yourself," she went on to say and resumed the brushing.
When the Queen was presentable, they both made their way to take breakfast. The Queen sat at the table and under the watchful eye of Athena she joylessly placed food in her mouth and involuntarily began masticating it slowly before swallowing it. It tasted like sand in her mouth, not because it was spoiled or badly cooked, but because she had no desire what so ever for it.
And from there it was to the drilling field. There, more for her daughter's sakes than anything else, she recruited every fraction of will power and strength she had in her and plied them into her training.
Athena escorted her mother for supper as well, where she supervised her mother's eating like a relentless hawk, and from there it was to the Shamaness' hut.
"You seem better this evening, Majesty," the Shamaness said as she began lighting up the scented herbs, sitting across from the Queen.
"Thank you," the Queen replied and mindlessly rubbed her neck, just above the collarbone. She then did something she had never done before. She looked into the older woman's burning eyes, and said, "Please call me Gabrielle, at least whenever we're here. This… compulsory distance is suffocating me these days."
The Shamaness smiled and nodded her head approvingly. This was a first step in the right direction, she thought. "Very good, Gabrielle."
The Queen was given a cup with a murky liquid in it that had an exceptionally potent tang to drink. "It is perfectly safe. You'll soon feel lightheadedness. Now drink it."
"For what purpose?" the Queen asked as she began to sip the brew.
"Think of yourself as the gatekeeper of your mind. This potion will loosen your grip around the gate," the Shamaness explained then added, "It will benefit you more if you don't fight it."
It was at that moment that the potion took effect on the Queen. She leaned backwards until her back was resting against the floorboards and the hut's ceiling spun before her eyes. She felt herself floating as if she was in the middle of the sea. Beads of sweat covered her skin in spite of the fact that it was cold outside. Her last thought before she was under the effect of the potion was that the sensation wasn't unfamiliar to her, for it felt much like that night when she had gotten tipsy when she had been a body slave in service of the Conqueror.
In her mind's eye, she was thrown back to the time when she had been purchased by her second owner, Phillipon, before she had entered the Conqueror's service. He had purchased her as a body slave for his son, Damianos. When first she had heard the term 'body slave' being uttered, she hadn't had a clear notion of what exactly that term meant, but she instinctively understood what that particular 'position' might entail.
The events of that first night of her service to Damianos played vividly before her eyes like a scene played on a stage.
She remembered the fairly modest bedchamber of the young dark-haired man, Damianos. She could smell the unique smell of it, jasmine she believed it was. She remembered quickly looking around and studying her new surroundings, what would be her new place of abode. She could clearly recall the exact shade of blue and the shape of the patterns painted over the tiles. She heard Phillipon proudly congratulating his son for being accepted as a warrior into the ranks of the Conqueror's vast forces. She could feel Phillipon's hand against her back while he spoke slightly nudging her forward, offering her as a gift to his son, who had turned into a man in his father's eyes. She could sense how she nervously anticipated what had been, at that point, uncharted territory for her. Her previous position had been as a domestic for the county healer, and nothing else.
When she was left alone with the young lad, she quickly realized she would soon have to perform something she had never had to before. To some extent she felt odd excitement. He stripped her of her clothes, which was easy enough. She hadn’t had much on to begin with. He asked her to lie on the bed and she did. He soon disrobed as well and laid himself on top of her.
From his first touches to her body, she soon realized he didn’t know what he was doing. She was very much a virgin herself to be sure, and still growing up in her father’s farm, where she had had occasion to see livestock animals mate; she had had some idea as to the basic requirements of the act. This particular activity was meant to be pleasurable, otherwise what was the point of doing it, she understood that much. And yet, his touch generated no pleasure in her. He touched her like a curious toddler would, experimenting with a new toy that he had yet to figure out how to operate.
In her trance like state, her mind summoned for her consideration her first night in the Conqueror’s service, as an unfair comparison. Standing in the same chamber with Damianos was nothing like being in the presence of the Lord Conqueror. The Lord Conqueror commanded her awe and the glare of her smoldering mesmerizing eyes made her skin bristle and her knees feeble. The Conqueror touched her confidently and masterfully. Her powerful hands moved over her body like experienced travelers and her fingers like keen trackers. The Lord Conqueror knew exactly what she wanted and how to take it. The Lord Conqueror rained raging rivers of fire all over her flesh that seeped into her soul, and when the Lord Conqueror filled her she feared her heart would burst. These were not the fumbling attempts of a timid, unripened lad, but the absolute domination of the most powerful woman in the world… and she was so sinfully beautiful and statuesque that it almost made a young slave weep.
Still in a daze, her mind speeded forward. Her young owner Damianos didn’t improve over the seasons and in many ways she was thankful for that. When it became evident that the lad found men preferable to women, he discarded her, and where the son had left off, a father picked up and had occasion to use her.
When memories of service to Phillipon returned to haunt her, she was lucidly seeing them passing before her. While her body was being exploited for the sexual gratification of a stranger, she put her mind elsewhere where it was green and sunny and the warm dry winds brushed against the lush grass creating ripples as if against water, far far away from her owner and his domain.
With Damianos, it was different, because he was young and because he was pathetic she could almost forgive him or at the very least not begrudge him his actions. With Phillipon she felt the degradation fully.
She began to sob and was quickly becoming aware of her surroundings. She shot right up to a sitting position. The Shamaness cradled her in her arm and rocked her gently to calm her.
“These were just memories,” she said. “Acknowledge and confront them and they can never hurt you.”
The Queen needed to cry like she could expel the memories she hadn’t conjured in years through the purifying tears.
“You were robbed of your youth in the most humiliating way and deprived of any sense of power and control,” the Shamaness said and stroked the Queen’s hair in a motherly fashion.
“Did I say something!?” the Queen covered her mouth, mortified when she suddenly realized the Shamaness had more than a fair idea as to what she had experienced.
“Some things you said, others I saw,” the Shamaness whispered but deemed it unnecessary to explain any further and she was correct. At that moment it didn’t really matter.
“I have never delved into the events of those days for fear of sinking and disappearing into the abyss.” The Queen sniffed and wiped her tears when she finally calmed.
“You have come a long way from those days,” the Shamaness told her, “but you already know that better than I do. Oftentimes there is no good reason as to why events occur the way they do, but try to perceive it differently. If you hadn’t been sold, you wouldn’t have become the Lord Conqueror’s wife and Queen and you wouldn’t have been in the position to better the lives of countless people.”
The Queen smiled a genuine smile, her first in days. “I could only hope so.”
“Well, I can see your hand all over the Conqueror's decree to abolish slavery," the Shamaness said knowingly.
Juxtaposing the building of hospices and outlawing slavery, the latter was the grander gesture of the two for the former required nothing more than for the Conqueror to spend her exchequer; the latter behooved her to be at odds with subjects of the Realm who were rich and powerful enough to own slaves.
Queen Gabrielle was reluctant to admit the great influence she had had over the Conqueror's decisions, for it might insinuate as to the true nature of their bond, and so she elected to refrain from admitting as much.
However, no such admission was necessary. "There is much more to you, Gabrielle, than Queen of the Realm, wife of the Lord Conqueror and mother to her heir or future Queen of the Amazon nation. You mustn’t lose sight of who you are.”
“Thank you, Smyrna,” the Queen said. She wiped her brow and rose to her feet.
As the Shamaness poured water over the burning herbs, putting an end to the wafting white smoke, she asked, “Would you please do a favor for me?”
“If it is in my power to grant the favor, of course I would.”
“There are five sisters here that have been body slaves in the past.”
The Queen nodded her head in empathy, “I know. The late Princess Terreis told me just before…” her voice trailed off.
“In the years followed by the abolishment of slavery throughout the Realm, these women found their way to us. I’ve succeeded in helping four of them to confront the consequences.”
“Not all five?”
“I suspect Mitylene was severely damaged and keeps withdrawn. She must have suffered unfathomable horrors during her years of service… probably worse than most. She pays the price still. Will you agree to try and help me reach her?”
“I will try and do my very best,” the Queen promised as she reached the door.
“Sleep well, Gabrielle,” the Shamaness said.
“Thank you, and Good night,” replied the Queen.
On her way back to her hut, Queen Gabrielle realized how lonely it had been being the Queen of the Realm, even more so now with everything that had happened with her former lady-in-waiting Satrina and with Princess Lao-Ling’s untimely death. And even with those two women, whom she considered her dear and close friends, she had always kept them at arm's length and had allowed them to get only that close to her because of her station and her complex bond with her Lord.
Alone in bed and in the dark, Queen Gabrielle touched her fingertips to her lips and whispered, "My Lord," to herself, enjoying the hot vibrations of those two precious words against her skin.
That very night the Conqueror arrived at the palace in Corinth.
Return to the Academy | null | null | null | null | null | null | null | null | null |
https://www.numbersaplenty.com/3433023 | 1,718,727,308,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861762.73/warc/CC-MAIN-20240618140737-20240618170737-00108.warc.gz | 815,837,846 | 3,282 | Search a number
3433023 = 34113853
BaseRepresentation
bin1101000110001000111111
320110102020000
431012020333
51334324043
6201325343
741115546
oct15061077
96412200
103433023
111a35310
121196853
13932799
1465515d
1547c2d3
hex34623f
3433023 has 20 divisors (see below), whose sum is σ = 5596008. Its totient is φ = 2080080.
The previous prime is 3433009. The next prime is 3433037. The reversal of 3433023 is 3203343.
It is an interprime number because it is at equal distance from previous prime (3433009) and next prime (3433037).
It is not a de Polignac number, because 3433023 - 24 = 3433007 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 3432987 and 3433005.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (3433043) by changing a digit.
It is a polite number, since it can be written in 19 ways as a sum of consecutive naturals, for example, 1036 + ... + 2817.
Almost surely, 23433023 is an apocalyptic number.
3433023 is a gapful number since it is divisible by the number (33) formed by its first and last digit.
3433023 is a deficient number, since it is larger than the sum of its proper divisors (2162985).
3433023 is a wasteful number, since it uses less digits than its factorization.
3433023 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 3876 (or 3867 counting only the distinct ones).
The product of its (nonzero) digits is 648, while the sum is 18.
The square root of 3433023 is about 1852.8418712885. The cubic root of 3433023 is about 150.8547204359.
Adding to 3433023 its reverse (3203343), we get a palindrome (6636366).
The spelling of 3433023 in words is "three million, four hundred thirty-three thousand, twenty-three". | 537 | 1,789 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-26 | latest | en | 0.875062 |
https://www.myhelper.tk/ncert-solution-class-12-chapter-1-integrals-exercise-1-6-mathematics-part-2/ | 1,600,566,696,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400193087.0/warc/CC-MAIN-20200920000137-20200920030137-00003.warc.gz | 1,017,918,279 | 10,541 | # NCERT solution class 12 chapter 1 Integrals exercise 1.6 mathematics part 2
## EXERCISE 1.6
#### Question 1:
x sin x
Let I =
Taking x as first function and sin x as second function and integrating by parts, we obtain
#### Question 2:
Let I =
Taking x as first function and sin 3x as second function and integrating by parts, we obtain
#### Question 3:
Let
Taking x2 as first function and ex as second function and integrating by parts, we obtain
Again integrating by parts, we obtain
#### Question 4:
x logx
Let
Taking log x as first function and x as second function and integrating by parts, we obtain
#### Question 5:
x log 2x
Let
Taking log 2x as first function and x as second function and integrating by parts, we obtain
#### Question 6:
xlog x
Let
Taking log x as first function and x2 as second function and integrating by parts, we obtain
#### Question 7:
Let
Taking as first function and x as second function and integrating by parts, we obtain
#### Question 8:
Let
Taking as first function and x as second function and integrating by parts, we obtain
#### Question 9:
Let
Taking cos−1 x as first function and x as second function and integrating by parts, we obtain
#### Question 10:
Let
Taking as first function and 1 as second function and integrating by parts, we obtain
#### Question 11:
Let
Taking as first function and as second function and integrating by parts, we obtain
#### Question 12:
Let
Taking x as first function and sec2x as second function and integrating by parts, we obtain
#### Question 13:
Let
Taking as first function and 1 as second function and integrating by parts, we obtain
#### Question 14:
Taking as first function and x as second function and integrating by parts, we obtain
I=log x 2∫xdx-∫ddxlog x 2∫xdxdx=x22log x 2-∫2log x .1x.x22dx=x22log x 2-∫xlog x dx
Again integrating by parts, we obtain
I = x22logx 2-log x ∫x dx-∫ddxlog x ∫x dxdx=x22logx 2-x22log x -∫1x.x22dx=x22logx 2-x22log x +12∫x dx=x22logx 2-x22log x +x24+C
#### Question 15:
Let
Let I = I1 + I2 … (1)
Where, and
Taking log x as first function and xas second function and integrating by parts, we obtain
Taking log x as first function and 1 as second function and integrating by parts, we obtain
Using equations (2) and (3) in (1), we obtain
#### Question 16:
Let
Let
⇒
∴
It is known that,
#### Question 17:
Let
Let ⇒
It is known that,
#### Question 18:
Let ⇒
It is known that,
From equation (1), we obtain
#### Question 19:
Also, let ⇒
It is known that,
#### Question 20:
Let ⇒
It is known that,
#### Question 21:
Let
Integrating by parts, we obtain
Again integrating by parts, we obtain
#### Question 22:
Let ⇒
= 2θ
⇒
Integrating by parts, we obtain
#### Question 23:
equals
Let
Also, let ⇒
Hence, the correct answer is A.
equals | 900 | 2,844 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2020-40 | latest | en | 0.610956 |
https://meritnotes.com/bank-exams/time-work-questions-pdf/1-78171/ | 1,718,449,713,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861586.40/warc/CC-MAIN-20240615093342-20240615123342-00822.warc.gz | 350,711,623 | 7,075 | # 100+ Time and Work Questions for SBI, IBPS Bank PO Exams - 1
Question: 1
A completes \$\${7} / {10}\$\$ of a work in 15 days. Then he completes the remaining work with the help of B in 4 days. The time required for A and B together to complete the entire work is
(A) \$\$8{1}/{4}\$\$
(B) \$\$10{1}/{2}\$\$
(C) \$\$13{1}/{3}\$\$
(D) \$\$14{1}/{4}\$\$
Ans: C
(A + B)’s 4 day’s work = \$\$(1 – {7}/{10})\$\$ = \$\${3}/{10}\$\$
(A + B)’s 1 day’s work = \$\$( {3/10} × {1/4})\$\$ = \$\${3}/{40}\$\$
Hence A and B together take \$\${40}/{3}\$\$
= \$\$13{1}/{3}\$\$ days to complete the entire work.
Question: 2
A works twice as fast as B. If B can complete a work in 12 days independently, the number of days in which A and B can together finish the work is
(A) 1 days
(B) 4 days
(C) 6 days
(D) 8 days
Ans: B
Ratio of rates of working of A and B = 2 : 1.
So, ratio of times taken = 1 : 2.
∴ A’s 1 day’s work = \$\${1} / {6}\$\$; B’s 1 day’s work = \$\${1} / {12}\$\$.
(A + B)’s 1 day’s work = \$\$({1} / {6} + {1} / {12})\$\$ = \$\${3} /{12}\$\$ = \$\${1} / {4}\$\$
So, A and B together can finish the work in 4 days.
Question: 3
A and B can separately do a piece of work in 20 and 15 days respectively. They worked together for 6 days, after which B was replaced by C. If the work was finished in next 4 days, then the number of days in which C alone could do the work will be
(A) 20
(B) 30
(C) 40
(D) 50
Ans: C
(A + B)’s 6 day’s work = \$\$6( {1/20} + {1/15})\$\$ = \$\${7}/{10}\$\$
(A + C)’s 4 day’s work = \$\$(1 - {7}/{10})\$\$ = \$\${3}/{10}\$\$
(A + C)’s 1 day’s work = \$\${3}/{40}\$\$
A’s 1 day’s work = \$\$ {1}/{20}\$\$
∴ C’s 1 day’s work = \$\$({3/40} - {1/20})\$\$ = \$\${1}/{40}\$\$
Hence, C alone can finish the work in 40 days.
Question: 4
Two spinning machines A and B can together produce 3,00,000 metres of cloth in hours. If machine B alone can produce the same amount of cloth in 15 hours, then how much cloth can machine A produce alone in 10 hours?
(A) 50,000 metres
(B) 1,00,000 metres
(C) 1,50,000 metres
(D) 2,00,000 metres
Ans: B
Length of cloth produced by A and B in 10 hrs = 3,00,000 m.
Length of cloth produced by B in 10 hrs = \$\$({300000} /{15} × 10)\$\$m = 200000 m.
∴ Length of cloth produced by A in 10 hrs
= (300000 – 200000) m = 100000 m.
Question: 5
A tyre has two punctures. The first puncture alone would have made the type flat in 9 minutes and the second alone would have done it in 6 minutes. If air leaks out at a constant rate, how long does it take both the punctures together to make it flat?
(A) \$\$1{1} / {2}\$\$
(B) \$\$1{1} / {3}\$\$
(C) \$\$3{1} / {2}\$\$
(D) \$\$3{3} / {5}\$\$
Ans: D
1 minute’s work of both the punctures = \$\$({1/9} + {1/6})\$\$ = \$\${5}/ {18}\$\$.
So, both the punctures will make the tyre flat in \$\${18}/{5}\$\$ = \$\$3{3} / {5}\$\$ min.
Related Questions | 1,087 | 2,889 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-26 | latest | en | 0.863786 |
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Save the tigers? It may be too late.
“I sometimes get pictures that people think look fake,” says Steve Winter, whose remote-controlled camera recorded two cubs in India; Winter spent a decade photographing tigers in the wild in Asia. CREDIT: Steve Winter/National Geographic Society.)
Amy Mathews Amos
Special To The Washington Post.
Photojournalist Steve Winter spent a decade photographing tigers in the wild in Asia. In 2007, when his life partner, writer Sharon Guynup, joined him for an assignment in India's Kaziranga National Park, they both came away concerned about the fate of these great animals. A century ago, more than 100,000 tigers roamed across 24 Asian countries, from Turkey to Indonesia. Today, they have disappeared from most of their historic range, and poaching, deforestation and a huge Chinese market for tiger parts (which are used in traditional Chinese medicine and as luxury items) have reduced their numbers to no more than 3,200. "We realized that tigers are almost gone, and no one seems to recognize that," Guynup said. The couple have collaborated on a book, "Tigers Forever," that combines magnificent images of the wild animals with troubling text about their dwindling state. Guynup said they hope it will shock the world into doing something "before the tigers are gone." The two recently spoke by telephone with The Post from their home in New Jersey on what it will take to save the world's biggest cat.
Q: Americans often assume that tigers are protected. How have conservation efforts failed?
A: Guynup: Because big conservation organizations have such huge fundraising needs and large administrative costs, their message to the public is often "We're saving tigers, we're saving polar bears." I think that message has lulled the public into believing that we're actually making progress. The biggest downfall of every large conservation organization today is that they are not addressing the demand [for tiger parts]. Unless the demand from China is stopped, tigers won't survive very far into the future. The organizations that are addressing the demand are the Wildlife Protection Society of India, the Environmental Investigation Agency based in London and WildAid, based here in the United States. Those are smaller, investigative, nimble organizations, and they're on the ground doing the undercover investigative work showing what's happening. And the combination of legal tiger farms in China, where they can legally sell skins, and an ongoing trade in tiger bone for tiger bone wine [believed to cure arthritis and impart strength, among other things] is killing the wild tigers across their range.
Q: Are those sales to China the biggest threat to tigers?
A: Guynup: Poaching has really skyrocketed, and it's fueled by the legal trade within China. There was a call by the international community in 2007 for all tiger-range countries to phase out tiger farming. But China's tiger farms have only grown. As long as those farms exist and there is a legal trade, they will not only stimulate demand but also launder all the wild products.
China argues that they should be able to have tiger farms because it's a domestic issue. But it's not. That demand is killing India's tigers and Thailand's tigers and other countries' tigers.
Winter: Stopping habitat loss is also very important. I saw that in Sumatra. When I first went to Sumatra [which is part of Indonesia] for a tiger story, we thought that [the Sumatran] tiger would be the next subspecies to go extinct. Now, that no longer seems to be the case [because their population appears to be higher than previously estimated].
But there also doesn't seem to be any movement toward protecting whatever land is left and stopping the rampant destruction of the forest for palm oil and wood. The fact that tiger numbers are now understood [to be higher than previously thought] in Sumatra is important. And value has been placed on their continued survival by some important people in Indonesia. For example, one of the country's most influential businessmen, Tomy Winata, has [become interested]. He founded a tiger sanctuary in 1996, which is part of a national park. With effective enforcement and zero tolerance toward poaching, he and his team have successfully secured a significant area.
And Thailand also created [an enforcement project] at the Huai Kha Khaeng Wildlife Sanctuary in 2005 on the Thai-Burma border that has a protection force that is now being mirrored in other South Asia protected areas. But tigers don't exist in viable breeding populations elsewhere except in a few limited locations in India, Sumatra and Siberia.
Guynup: That patrol in Huai Kha Khaeng is a military-style patrol that does a pretty good job of protection [against poachers]. National parks that have tigers living within them need protection the same way cities need police. More than half of the world's remaining wild tigers live in India. The Bengal tigers in India are the biggest hope for tiger survival. But it's a country that is trying to keep an economy cranking at a high level, and there are 1.2 billion people living in India, so there's a fight over resources and land. India is a poaching target because it shares a long, porous border with China. India is where the most tigers are, so that's the source [for many poachers].
Q: Your book chronicles both the thrills and perils of tracking tigers throughout Asia, including dodging a charging rhino and patrolling for poachers with rangers. What was the biggest challenge you faced?
Winter: I think that always the biggest challenge is showing readers something they haven't seen before and creating renewed interest. You would never see a tiger in Sumatra or Thailand without using remote cameras [that trigger photographs when animals walk by], and because of the way I use remote cameras with artificial lights that balance with the daylight, I sometimes get pictures that people think look fake. Or they just stop when they see them. Which is great, because even if someone says, "Wow, that looks unreal," they've just spent five seconds longer on that picture than they normally would.
Because people are inundated with images 24/7, my job is to find a different way to photograph tigers, not just from the top of an elephant or in a jeep, but eye to eye with these intimate portraits that I'm able to get through the camera traps. But it's very dangerous because you're on the ground. In Kaziranga, I was filled with tension every time I exited the jeep to check the traps, because there are other animals like rhinos and elephants around, and you're entering their territory on your feet.
Q: And then you faced some challenges when you came home. Hurricane Sandy hit just as you began pulling this book together, and you were forced to evacuate your house.
Guynup: It affected me more than Steve because I was starting to research and write the book on a very, very short deadline. I did my first overnight Skype interviews to Asia from a Super 8 motel where we stayed for about 10 days with our 90-pound Lab, our four-month-old puppy, the cat, my adult son and Steve and I, so that was a little crazy. Steve lost 20 years of field gear, and we lost lots of personal stuff. Our home wasn't ruined, but we had huge storage in the basement that included prints and original slides.
Q: The book includes spectacular photographs of wild tigers, but it also includes disturbing images of poachers, traps and other animals. Why?
Winter: I've found two ways to overcome information overload. Number One is showing people images they haven't seen before and Number Two is showing images that might be a bit disturbing. So they look at them and hopefully they will find out the story behind the photograph. I also looked at Kaziranga as a kind of historic landscape where tigers live with species as in centuries past. This is what it used to look like and still does in a few select locations. It's great to see intact ecosystems that are protected with a variety of animals that are in them.
Guynup: Steve began his career as a photojournalist. So when he photographs wildlife, he approaches his assignments as a photojournalist. Unless you're going to tell the whole story and show people why these animals are disappearing and get them to care and hopefully to act, all the pretty pictures in the world aren't going to save these animals.
Q: What else should be done to help save tigers?
Guynup: The international community needs to join together and pressure China to stop the sale of tiger parts, skins, bones, all tiger parts from all sources, both captive and wild. And tiger-range countries need to use tools that are available through Interpol and other international agreements, to step up enforcement. If the demand is stopped, if the tiger farms are phased out and enforcement steps up, we can save tigers. But if that doesn't happen, they're going to disappear, possibly within our lifetime. People might think, "Oh, farms: Well, that kind of takes care of it all, and then we'll put them back." But once tigers have been in captivity and in contact with people, they're too dangerous to put back into the wild.
There are heavy population pressures in Asia, but India has over 40 tiger reserves. There's plenty of land. And now many areas are working to set up corridors to connect these reserves, so that the populations don't become inbred. As long as tigers have habitat, food and protection, they're a very resilient species. They bounce back. There is hope.
— — —
Mathews Amos writes about environment, health and history from Shepherdstown, W.Va.
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Did the College Football Playoff Committee get it right? | null | null | null | null | null | null | null | null | null |
http://slideplayer.com/slide/6261736/ | 1,639,005,445,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363598.57/warc/CC-MAIN-20211208205849-20211208235849-00325.warc.gz | 66,136,435 | 23,656 | # The Natural Logarithmic Function
## Presentation on theme: "The Natural Logarithmic Function"— Presentation transcript:
The Natural Logarithmic Function
Differentiation Integration
Properties of the Natural Log Function
If a and b are positive numbers and n is rational, then the following properties are true:
The Algebra of Logarithmic Expressions
The Derivative of the Natural Logarithmic Function
Let u be a differentiable function of x
Differentiation of Logarithmic Functions
Differentiation of Logarithmic Functions
Differentiation of Logarithmic Functions
Differentiation of Logarithmic Functions
Differentiation of Logarithmic Functions
Logarithmic Properties as Aids to Differentiation
Differentiate:
Logarithmic Properties as Aids to Differentiation
Differentiate:
Logarithmic Differentiation
Differentiate: This can get messy with the quotient or product and chain rules. So we will use ln rules to help simplify this and apply implicit differentiation and then we solve for y’…
Derivative Involving Absolute Value
Recall that the ln function is undefined for negative numbers, so we often see expressions of the form ln|u|. So the following theorem states that we can differentiate functions of the form y= ln|u| as if the absolute value symbol is not even there. If u is a differentiable function such that u≠0 then:
Derivative Involving Absolute Value
Differentiate:
Finding Relative Extrema
Locate the relative extrema of 𝑦=ln( 𝑥 2 +2𝑥+3) Differentiate: 𝑢= 𝑥 2 +2𝑥+3, 𝑢 ′ =2𝑥+2 𝑑𝑦 𝑑𝑥 = 2𝑥+2 𝑥 2 +2𝑥+3 Set = 0 to find critical points 2𝑥+2 𝑥 2 +2𝑥+3 =0 2x+2=0 X=-1, Plug back into original to find y y=ln(1-2+3)=ln2 So, relative extrema is at (-1, ln2)
Homework 5.1 Natural Logarithmic Functions and the Number e Derivative #19-35,47-65, 71,79,93-96
General Power Rule for Integration
𝑥 𝑛 𝑑𝑥= 𝑥 𝑛+1 𝑛+1 +𝑐, 𝑛≠−1 Recall that it has an important disclaimer- it doesn’t apply when n = -1. So we can not integrate functions such as f(x)=1/x. So we use the Second FTC to DEFINE such a function.
Integration Formulas Let u be a differentiable function of x
Using the Log Rule for Integration
Using the Log Rule with a Change of Variables
1 4𝑥−1 𝑑𝑥 Let u=4x-1, so du=4dx and dx= 𝑑𝑢 4 1 𝑢 𝑑𝑢 4 = 𝑢 𝑑𝑢
Finding Area with the Log Rule
Find the area of the region bounded by the graph of y, the x-axis and the line x=3.
Recognizing Quotient Forms of the Log Rule
𝑥+1 𝑥 2 +2𝑥 𝑑𝑥→𝑢= 𝑥 2 +2𝑥, 𝑑𝑢=2𝑥+2𝑑𝑥 → (𝑥+1) 𝑥 2 +2𝑥 𝑑𝑥= 1 2 𝑙𝑛 𝑥 2 +2𝑥 +𝑐 =𝑙𝑛 𝑥 2 +2𝑥 𝑐
Definition The natural logarithmic function is defined by
The domain of the natural logarithmic function is the set of all positive real numbers
u-Substitution and the Log Rule
Long Division With Integrals
How you know it’s long Division
If it is top heavy that means it is long division. Example
Example 1
Continue Example 1
Example 2
Continue Example 2
Using Long Division Before Integrating
Using a Trigonometric Identity
Guidelines for integration
Learn a basic list of integration formulas. (including those given in this section, you now have 12 formulas: the Power Rule, the Log Rule, and ten trigonometric rules. By the end of section 5.7 , this list will have expanded to 20 basic rules) Find an integration formula that resembles all or part of the integrand, and, by trial and error, find a choice of u that will make the integrand conform to the formula. If you cannot find a u-substitution that works, try altering the integrand. You might try a trigonometric identity, multiplication and division by the same quantity, or addition and subtraction of the same quantity. Be creative. If you have access to computer software that will find antiderivatives symbolically, use it.
Integrals of the Six Basic Trigonometric Functions
Homework 5.2 Log Rule for Integration and Integrals for Trig Functions (substitution) #1-39, 47-53, 67 | 1,146 | 3,836 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2021-49 | latest | en | 0.833652 |
https://hr6math.com/category/sports-science-math/ | 1,542,557,622,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039744513.64/warc/CC-MAIN-20181118155835-20181118181142-00025.warc.gz | 628,181,906 | 22,449 | ## Sports Science: Kentucky Derby
Posted: May 2, 2018 in Advance 2017/18, Off the Beaten Math, Periods 4&5 2017/18, Sports Science MATH
Paper Horses
Ratios & Proportional Relationships
• 6.RP.1 – Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities.
• 6.RP.2 – Understand the concept of a unit rate a/b associated with a ratio a:b with b ? 0, and use rate language in the context of a ratio relationship.
• 6.RP.3 – Use ratio and rate reasoning to solve real-world and mathematical problems. [Includes parts a, b, c, and d]
The Number System
• 6.NS.1 – Interpret and compute quotients of fractions, and solve word problems involving division of fractions by fractions.
• 6.NS.2 – Fluently divide multi-digit numbers using the standard algorithm.
• 6.NS.3 – Fluently add, subtract, multiply, and divide multi-digit decimals using the standard algorithm for each operation.
• 6.NS.4 – Find the greatest common factor of two whole numbers less than or equal to 100 and the least common multiple of two whole numbers less than or equal to 12. Use the distributive property to express a sum of two whole numbers 1–100 with a common factor as a multiple of a sum of two whole numbers with no common factor.
• 6.NS.5 – Understand that positive and negative numbers are used together to describe quantities having opposite directions or values; use positive and negative numbers to represent quantities in real-world contexts, explaining the meaning of 0 in each situation.
• 6.NS.6 – Understand a rational number as a point on the number line. Extend number line diagrams and coordinate axes familiar from previous grades to represent points on the line and in the plane with negative number coordinates. [Includes parts a, b, and c]
• 6.NS.7 – Understand ordering and absolute value of rational numbers. [Includes parts a, b, c, and d]
• 6.NS.8 – Solve real-world and mathematical problems by graphing points in all four quadrants of the coordinate plane. Include use of coordinates and absolute value to find distances between points with the same first coordinate or the same second coordinate.
Expressions & Equations
• 6.EE.1 – Write and evaluate numerical expressions involving whole-number exponents.
• 6.EE.2 – Write, read, and evaluate expressions in which letters stand for numbers. [Includes parts a, b, and c]
• 6.EE.3 – Apply the properties of operations to generate equivalent expressions.
• 6.EE.4 – Identify when two expressions are equivalent.
• 6.EE.5 – Understand solving an equation or inequality as a process of answering a question: which values from a specified set, if any, make the equation or inequality true? Use substitution to determine whether a given number in a specified set makes an equation or inequality true.
• 6.EE.6 – Use variables to represent numbers and write expressions when solving a real-world or mathematical problem; understand that a variable can represent an unknown number, or, depending on the purpose at hand, any number in a specified set.
• 6.EE.7 – Solve real-world and mathematical problems by writing and solving equations of the form x + p = q and px = q for cases in which p, q and x are all nonnegative rational numbers.
• 6.EE.8 – Write an inequality of the form x > c or x < c to represent a constraint or condition in a real-world or mathematical problem. Recognize that inequalities of the form x > c or x < c have infinitely many solutions; represent solutions of such inequalities on number line diagrams.
• 6.EE.9 – Use variables to represent two quantities in a real-world problem that change in relationship to one another; write an equation to express one quantity, thought of as the dependent variable, in terms of the other quantity, thought of as the independent variable. Analyze the relationship between the dependent and independent variables using graphs and tables, and relate these to the equation.
Geometry
• 6.G.1 – Find the area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles and other shapes; apply these techniques in the context of solving real-world and mathematical problems.
• 6.G.2 – Find the volume of a right rectangular prism with fractional edge lengths by packing it with unit cubes of the appropriate unit fraction edge lengths, and show that the volume is the same as would be found by multiplying the edge lengths of the prism. Apply the formulas V = l w h and V = b h to find volumes of right rectangular prisms with fractional edge lengths in the context of solving real-world and mathematical problems.
• 6.G.3 – Draw polygons in the coordinate plane given coordinates for the vertices; use coordinates to find the length of a side joining points with the same first coordinate or the same second coordinate. Apply these techniques in the context of solving real-world and mathematical problems.
• 6.G.4 – Represent three-dimensional figures using nets made up of rectangles and triangles, and use the nets to find the surface area of these figures. Apply these techniques in the context of solving real-world and mathematical problems.
• 6.SP.1 – Recognize a statistical question as one that anticipates variability in the data related to the question and accounts for it in the answers.
• 6.SP.2 – Understand that a set of data collected to answer a statistical question has a distribution which can be described by its center, spread, and overall shape.
• 6.SP.3 – Recognize that a measure of center for a numerical data set summarizes all of its values with a single number, while a measure of variation describes how its values vary with a single number.
• 6.SP.4 – Display numerical data in plots on a number line, including dot plots, histograms, and box plots.
• 6.SP.5 – Summarize numerical data sets in relation to their context. [Includes parts a, b, c, and d]
## Sport Science: Gymnastics
Posted: March 1, 2018 in Sports Science MATH
4.MD.C.6
MP5
MP6
Measure angle openings precisely by using a protractor
6.NS.1
Interpret and compute quotients of fractions, and solve word problems involving division of fractions by fractions.
6.NS.2
Fluently divide multi-digit numbers using the standard algorithm.
6.NS.3
Fluently add, subtract, multiply, and divide multi-digit decimals using the standard algorithm for each operation.
6.EE.2b
Identify parts of an expression using mathematical terms (sum, term, product, factor, quotient, coefficient); view one or more parts of an expression as a single entity.
6.EE.2c
Evaluate expressions at specific values of their variables. Include expressions that arise from formulas used in real-world problems. Perform arithmetic operations, including those involving whole-number exponents, in the conventional order when there are no parentheses to specify a particular order (Order of Operations).
7.NS.2d
Convert a rational number to a decimal using long division; know that the decimal form of a rational number terminates in 0s or eventually repeats.
7.NS.3
Solve real-world and mathematical problems involving the four operations with rational numbers.
7.G.4
Know the formulas for the area and circumference of a circle and use them to solve problems; give an informal derivation of the relationship between the circumference and area of a circle.
## Sports Science: Sumo Goalie
Posted: January 19, 2018 in Sports Science MATH
• 6.NS.1 – Dividing Two Fractions
• 6.NS.1 – Dividing Fractions and Mixed Numbers
• 6.NS.1 – Dividing Fractions and Whole Numbers
• 6.NS.2 – Dividing Multi-Digit Numbers
• 6.NS.3 – Estimate All Operation Answers: Multi-Digit Decimals
• 6.NS.3 – Add/Subtract/Multiply/Divide: Multi-Digit Decimals
• 6.NS.4 – Identifying the Least Common Multiple
• 6.NS.4 – Identifying the Greatest Common Factor
• 6.EE.2a – Identifying Expressions that Represent Situations
• 6.EE.2b – Parts of an Expression
• 6.EE.2a – Translate Addition Sentences into Algebraic Expressions
• 6.EE.2a – Translate Subtraction Sentences to Algebraic Expressions
• 6.EE.2a – Translating Multiplication Sentences to Algebraic Expressions
• 6.EE.2a – Translating Division Sentences to Algebraic Expressions
• 6.EE.2c – Using Order of Operations to Evaluate Expressions
• 6.EE.3 – Identify Equivalent Expressions: Distributive Property
• 6.EE.4 – Identifying Equivalent Expressions by Evaluation
• 6.EE.5 – Using Substitution to Determine Solutions
• 6.RP.1 – Representing Ratios
• 6.RP.2 – Expressing Unit Rate
• 6.RP.3a – Ratio Tables and Graphs
• 6.RP.3b – Solving Problems Involving Unit Rate
• 6.RP.3d – Converting Measurement Units Using Ratio Reasoning
• 6.RP.3c – Expressing Percents
• 6.RP.3c – Percent Relationships
• 6.RP.3c – Solving Percent Word Problems
• 6.G.1 – Find the area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles and other shapes; apply these techniques in the context of solving real-world and mathematical problems.
## The Meatball Challenge
Posted: November 19, 2017 in Advance, Periods 1&2, Periods 3&4, Sports Science MATH
6.RP.A.1
Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities.
6.RP.A.2
Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship.
6.RP.A.3
Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
6.RP.A.3a
Make tables of equivalent ratios relating quantities with whole-number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
6.RP.A.3b
Solve unit rate problems including those involving unit pricing and constant speed.
## Who has faster hands –an NBA point guard, or legendary rock drummer?
Posted: August 25, 2017 in Advance, Periods 1&2, Periods 3&4, Sports Science MATH
## 6.RP.A.1
Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities.
## 6.RP.A.2
Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship.
## 6.RP.A.3
Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
## 6.RP.A.3a
Make tables of equivalent ratios relating quantities with whole-number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
## 6.RP.A.3b
Solve unit rate problems including those involving unit pricing and constant speed.
## 6.RP.A.3c
Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
## 6.RP.A.3d
Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
## Sports Science: Dennis Northcutt Vs. Ostrich
Posted: August 22, 2017 in Sports Science MATH
1. Identify independent and dependent variables
2. Relate independent and dependent variables using a function table
3. Show the relationship between variables using a graph
4. Relate variables using an equation
5. Solve for missing values in rate problems using a table
## Sport Science: Kentucky Derby
Posted: May 5, 2017 in Sports Science MATH, Uncategorized
Paper Horses
YEAR HORSE JOCKEY TRAINER OWNER TIME 2014 California Chrome V. Espinoza A. Sherman Coburn & Perry 2:03.66 2013 Orb J. Rosario S. McGaughey Stuart Janney & Phipps Stable 2:02.89 2012 I’ll Have Another M. Gutierrez D. O’Neill Reddam Racing 2:01.83 2011 Animal Kingdom J. Velazquez H. G. Motion Team Valor 2:02.04 2010 Super Saver C. Borel T. Pletcher WinStar Farm 2:04.45 2009 Mine That Bird C. Borel B. Woolley Double Eagle Ranch 2:02.66 2008 Big Brown K. Desormeaux R. Dutrow IEAH Stables, Pompa et al 2:01.82 2007 Street Sense C. Borel C. Nafzger James Tafel 2:02.17 2006 Barbaro See Video E. Prado M. Matz Roy & Gretchen Jackson 2:01.36 2005 Giacomo M. Smith J. Sherrifs Mr. and Mrs. Jerome Moss 2:02.75 2004 Smarty Jones See Video S. Elliott J. Servis Someday Farm 2:04.06 2003 Funny Cide J. Santos B. Tagg Sackatoga Stable 2:01.19 2002 War Emblem V. Espinoza B. Baffert Thoroughbred Corp. 2:01.13 2001 Monarchos J. Chavez J. T. Ward John C. Oxley 1:59.97 2000 Fusaichi Pegasus K. Desormeaux N. Drysdale Fusao Sekiguchi 2:01.12
Ratios & Proportional Relationships
• 6.RP.1 – Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities.
• 6.RP.2 – Understand the concept of a unit rate a/b associated with a ratio a:b with b ? 0, and use rate language in the context of a ratio relationship.
• 6.RP.3 – Use ratio and rate reasoning to solve real-world and mathematical problems. [Includes parts a, b, c, and d]
The Number System
• 6.NS.1 – Interpret and compute quotients of fractions, and solve word problems involving division of fractions by fractions.
• 6.NS.2 – Fluently divide multi-digit numbers using the standard algorithm.
• 6.NS.3 – Fluently add, subtract, multiply, and divide multi-digit decimals using the standard algorithm for each operation.
• 6.NS.4 – Find the greatest common factor of two whole numbers less than or equal to 100 and the least common multiple of two whole numbers less than or equal to 12. Use the distributive property to express a sum of two whole numbers 1–100 with a common factor as a multiple of a sum of two whole numbers with no common factor.
• 6.NS.5 – Understand that positive and negative numbers are used together to describe quantities having opposite directions or values; use positive and negative numbers to represent quantities in real-world contexts, explaining the meaning of 0 in each situation.
• 6.NS.6 – Understand a rational number as a point on the number line. Extend number line diagrams and coordinate axes familiar from previous grades to represent points on the line and in the plane with negative number coordinates. [Includes parts a, b, and c]
• 6.NS.7 – Understand ordering and absolute value of rational numbers. [Includes parts a, b, c, and d]
• 6.NS.8 – Solve real-world and mathematical problems by graphing points in all four quadrants of the coordinate plane. Include use of coordinates and absolute value to find distances between points with the same first coordinate or the same second coordinate.
Expressions & Equations
• 6.EE.1 – Write and evaluate numerical expressions involving whole-number exponents.
• 6.EE.2 – Write, read, and evaluate expressions in which letters stand for numbers. [Includes parts a, b, and c]
• 6.EE.3 – Apply the properties of operations to generate equivalent expressions.
• 6.EE.4 – Identify when two expressions are equivalent.
• 6.EE.5 – Understand solving an equation or inequality as a process of answering a question: which values from a specified set, if any, make the equation or inequality true? Use substitution to determine whether a given number in a specified set makes an equation or inequality true.
• 6.EE.6 – Use variables to represent numbers and write expressions when solving a real-world or mathematical problem; understand that a variable can represent an unknown number, or, depending on the purpose at hand, any number in a specified set.
• 6.EE.7 – Solve real-world and mathematical problems by writing and solving equations of the form x + p = q and px = q for cases in which p, q and x are all nonnegative rational numbers.
• 6.EE.8 – Write an inequality of the form x > c or x < c to represent a constraint or condition in a real-world or mathematical problem. Recognize that inequalities of the form x > c or x < c have infinitely many solutions; represent solutions of such inequalities on number line diagrams.
• 6.EE.9 – Use variables to represent two quantities in a real-world problem that change in relationship to one another; write an equation to express one quantity, thought of as the dependent variable, in terms of the other quantity, thought of as the independent variable. Analyze the relationship between the dependent and independent variables using graphs and tables, and relate these to the equation.
Geometry
• 6.G.1 – Find the area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles and other shapes; apply these techniques in the context of solving real-world and mathematical problems.
• 6.G.2 – Find the volume of a right rectangular prism with fractional edge lengths by packing it with unit cubes of the appropriate unit fraction edge lengths, and show that the volume is the same as would be found by multiplying the edge lengths of the prism. Apply the formulas V = l w h and V = b h to find volumes of right rectangular prisms with fractional edge lengths in the context of solving real-world and mathematical problems.
• 6.G.3 – Draw polygons in the coordinate plane given coordinates for the vertices; use coordinates to find the length of a side joining points with the same first coordinate or the same second coordinate. Apply these techniques in the context of solving real-world and mathematical problems.
• 6.G.4 – Represent three-dimensional figures using nets made up of rectangles and triangles, and use the nets to find the surface area of these figures. Apply these techniques in the context of solving real-world and mathematical problems.
• 6.SP.1 – Recognize a statistical question as one that anticipates variability in the data related to the question and accounts for it in the answers.
• 6.SP.2 – Understand that a set of data collected to answer a statistical question has a distribution which can be described by its center, spread, and overall shape.
• 6.SP.3 – Recognize that a measure of center for a numerical data set summarizes all of its values with a single number, while a measure of variation describes how its values vary with a single number.
• 6.SP.4 – Display numerical data in plots on a number line, including dot plots, histograms, and box plots.
• 6.SP.5 – Summarize numerical data sets in relation to their context. [Includes parts a, b, c, and d] | 4,427 | 18,537 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2018-47 | longest | en | 0.874176 |
https://www.slideserve.com/polly/physics-151-week-9-day-1 | 1,620,338,503,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988763.83/warc/CC-MAIN-20210506205251-20210506235251-00261.warc.gz | 1,067,992,390 | 19,978 | # Physics 151 Week 9 Day 1 - PowerPoint PPT Presentation
Physics 151 Week 9 Day 1
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Physics 151 Week 9 Day 1
## Physics 151 Week 9 Day 1
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##### Presentation Transcript
1. Physics 151 Week 9 Day 1 Topics: Work, Energy, & Newton’s 2nd Law Energy and Work Force and Motion Graphs Questions from last time Net force vs. acceleration Mass vs. Acceleration Applying Newton’s 2nd Law Apparent Weight Brainstorm: What do we know about friction? Wednesday’s Class: All about Forces & friction
2. Slide 10-3
3. Slide 10-4
4. Forms of Energy Mechanical Energy Thermal Energy Other forms include Slide 10-12
5. The Basic Energy Model Slide 10-13
6. Energy Transformations Kinetic energy K = energy of motion Potential energy U = energy of position Thermal energy Eth = energy associated with temperature System energy E = K + U + Eth + Echem + ... Energy can be transformed within the system without loss. Energy is a property of a system. Slide 10-14
7. Some Energy Transformations Echem Ug KEth Us KUg Echem Ug Slide 10-15
8. Energy Transfers These change the energy of the system. Interactions with the environment. Work is the mechanical transfer of energy to or from a system via pushes and pulls. Slide 10-20
9. Energy Transfers: Work W K W Eth W Us Slide 10-21
10. The Work-Energy Equation Slide 10-22
11. Work Slide 10-28
12. General Force Model • Newton 0th Law • Objects are dumb - They have no memory of the past and cannot predict the future. Objects only know what is acting directly on them right now • Newton's 1st Law • An object that is at rest will remain at rest and an object that is moving will continue to move in a straight line with constant speed, if and only if the sum of the forces acting on that object is zero. • Newton's 3rd Law • Recall that a force is an interaction between two objects. If object A exerts a force on object B then object B exerts a force on object that is in the opposite direction, equal in magnitude, and of the same type. • Visualizations: • Force Diagrams • System Schema
13. Constant Force Model Newton's 2nd Law acceleration of an object = sum of forces acting on that object / the mass of the object Remainder of week: Friction Model Apparent Weight Slide 4-19
14. Net Force and Motion Graphs
15. Net Force vs. Acceleration Graphs
16. Net Force vs. Mass Graphs
17. Reading Quiz • Which of the following statements about mass and weight is correct? • Your mass is a measure of the force gravity exerts on you. • Your mass is the same everywhere in the universe. • Your weight is the same everywhere in the universe. • Your weight is a measure of your resistance of being accelerated. Slide 5-5
18. Answer • Which of the following statements about mass and weight is correct? • Your mass is a measure of the force gravity exerts on you. • Your mass is the same everywhere in the universe. • Your weight is the same everywhere in the universe. • Your weight is a measure of your resistance of being accelerated. Slide 5-6
19. Example Problem • A 100 kg block with a weight of 980 N hangs on a rope. Find the tension in the rope if • the block is stationary. • it’s moving upward at a steady speed of 5 m/s. • it’s accelerating upward at 5 m/s2. Slide 5-15
20. Example Problem A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s. It then crosses a rough patch of snow which exerts a friction force of 12 N. How far does it slide on the snow before coming to rest? Slide 5-21
21. Example Problem A 75 kg skier starts down a 50-m-high, 10° slope on frictionless skis. What is his speed at the bottom? Slide 5-27
22. Friction Brainstorm One person in each team takes out a sheet of paper and records their group brainstorming everything they know about friction.
23. Example Problem Burglars are trying to haul a 1000 kg safe up a frictionless ramp to their getaway truck. The ramp is tilted at angle θ. What is the tension in the rope if the safe is at rest? If the safe is moving up the ramp at a steady 1 m/s? If the safe is accelerating up the ramp at 1 m/s2? Do these answers have the expected behavior in the limit θ→ 0° and θ→ 90°? Slide 5-28
24. Example Problem Macie pulls a 40 kg rolling trunk by a strap angled at 30° from the horizontal. She pulls with a force of 40 N, and there is a 30 N rolling friction force acting on trunk. What is the trunk’s acceleration? Slide 5-22 | 1,179 | 4,496 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2021-21 | latest | en | 0.823126 |
https://uk.mathworks.com/matlabcentral/cody/problems/2023-is-this-triangle-right-angled/solutions/1826159 | 1,600,555,121,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400192887.19/warc/CC-MAIN-20200919204805-20200919234805-00337.warc.gz | 724,277,152 | 16,542 | Cody
# Problem 2023. Is this triangle right-angled?
Solution 1826159
Submitted on 26 May 2019 by Abdul Muneeb
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
a = 3; b = 4; c = 5; flag_correct = true; assert(isequal(isRightAngled(a,b,c),flag_correct))
sides = 3 4 5 flag = logical 1
2 Pass
a = 3; b = 4; c = 6; flag_correct = false; assert(isequal(isRightAngled(a,b,c),flag_correct))
sides = 3 4 6
3 Pass
a = 3; b = 5; c = 4; flag_correct = true; assert(isequal(isRightAngled(a,b,c),flag_correct))
sides = 3 4 5 flag = logical 1
4 Pass
a = 4; b = 3; c = 5; flag_correct = true; assert(isequal(isRightAngled(a,b,c),flag_correct))
sides = 3 4 5 flag = logical 1
5 Pass
a = 4; b = 5; c = 3; flag_correct = true; assert(isequal(isRightAngled(a,b,c),flag_correct))
sides = 3 4 5 flag = logical 1
6 Pass
a = 5; b = 3; c = 4; flag_correct = true; assert(isequal(isRightAngled(a,b,c),flag_correct))
sides = 3 4 5 flag = logical 1
7 Pass
a = 5; b = 4; c = 3; flag_correct = true; assert(isequal(isRightAngled(a,b,c),flag_correct))
sides = 3 4 5 flag = logical 1
8 Pass
a = 5; b = 12; c = 13; flag_correct = true; assert(isequal(isRightAngled(a,b,c),flag_correct))
sides = 5 12 13 flag = logical 1 | 497 | 1,338 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2020-40 | latest | en | 0.434 |
https://www.onlinemath4all.com/one-step-equations-multiplication-and-division-worksheet.html | 1,596,562,033,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735881.90/warc/CC-MAIN-20200804161521-20200804191521-00441.warc.gz | 775,346,716 | 13,711 | # ONE STEP EQUATIONS MULTIPLICATION AND DIVISION WORKSHEET
Problem 1 :
When we multiply a number by 4, we get 124. Find the number.
Problem 2 :
When we divide a number by 7, we get 14. Find the number.
Problem 3 :
Alex borrowed some money from Jose. After 3 years, Alex returned 2 times of borrowed money to Jose. If the returned money is \$226, how much money did Alex borrow from Jose ?
Problem 4 :
David has some money. He gave one fourth of the money to Lily. If Lily gets \$8 from David, how much money did David have initially ?
Problem 5 :
In a deposit, invested money will become 4 times itself in 5 years. If Rosy receives \$3280 after five years, how much money did Rosy invest ?
Problem 6 :
Jacob has some number of candies and Michael has 35 candies. If Michael has candies 5 times as Jacob, how many candies does Jacob have ?
Problem 7 :
Daniel had some hot dogs and he gave one third of the hot dogs to Alex. If Alex gets 8 hot dogs from Daniel, how many hot dogs did have initially ?
Problem 8 :
Between the hours of 10 P.M. and 6 A.M., the temperature decreases an average of of a degree per hour. How long, in hours and minutes, will it take for the temperature to decrease by 5 °F ?
Problem 1 :
When we multiply a number by 4, we get 124. Find the number.
Solution :
Let x be the number.
Then,
4x = 124
Divide each side by 4.
x = 31
So, the number is 31.
Problem 2 :
When we divide a number by 7, we get 14. Find the number.
Solution :
Let m be the number.
Then,
m/7 = 14
Multiply each side by 7.
m = 98
So, the number is 98.
Problem 3 :
Alex borrowed some money from Jose. After 3 years, Alex returned 2 times of borrowed money to Jose. If the returned money is \$226, how much money did Alex borrow from Jose ?
Solution :
Let x be the borrowed money.
Then,
2x = 226
Divide each side by 2.
x = 113
So, the borrowed money is \$113.
Problem 4 :
David has some money. He gave one fourth of the money to Lily. If Lily gets \$8 from David, how much money did David have initially ?
Solution :
Let m be the money that David had initially.
Then,
m/4 = 32
Multiply each side by 4.
m = 128
Problem 5 :
In a deposit, invested money will become 4 times itself in 5 years. If Rosy receives \$3280 after five years, how much money did Rosy invest ?
Solution :
Let x be the money that Rosy invested.
Then,
4x = 3280
Divide each side by 4.
x = 820
So, Rosy invested \$820.
Problem 6 :
Jacob has some number of candies and Michael has 35 candies. If Michael has candies 5 times as Jacob, how many candies does Jacob have ?
Solution :
Let p be the number of candies that Jacob has.
Then,
5p = 35
Divide each side by 5.
p = 7
So, Jacob has 5 candies.
Problem 7 :
Daniel had some hot dogs and he gave one third of the hot dogs to Alex. If Alex gets 8 hot dogs from Daniel, how many hot dogs did have initially ?
Solution :
Let h be the number of hot dogs that Daniel had initially.
Then,
m/3 = 8
Multiply each side by 3.
m = 24
So, Daniel had 24 hot dogs initially.
Problem 8 :
Between the hours of 10 P.M. and 6 A.M., the temperature decreases an average of of a degree per hour. How long, in hours and minutes, will it take for the temperature to decrease by 5 °F ?
Solution :
Let x represent the number of hours it takes for the temperature to decrease by 5 °F.
Write an equation
(-3/4)x = -5
3x/4 = -5
Solve the equation using an inverse operation.
Multiply each side by - 4/3.
x = 20/3
x = 6 2/3
It takes 6 2/3 hours.
Convert 2/3 hours to minutes.
2/3 hours = (2/3) ⋅ 60 minutes
2/3 hours = 40 minutes
So, it takes 6 hours and 40 minutes for the temperature to decrease by 5 °F.
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Sum of all three four digit numbers formed using 1, 2, 5, 6 | 1,571 | 6,107 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2020-34 | latest | en | 0.91689 |
https://www.hackmath.net/en/math-problem/3327 | 1,611,550,244,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703564029.59/warc/CC-MAIN-20210125030118-20210125060118-00516.warc.gz | 803,242,895 | 11,906 | # Collection
John has collected half a bucket of raspberries in 45 minutes, Eva collects the whole bucket in an hour.
How many minutes it would take to fill a bucket if the two children work together?
Correct result:
t = 36 min
#### Solution:
t/(2•45)+t/60 = 1
1.666667t = 60
t = 36
Our simple equation calculator calculates it.
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# ons. A function takes a starting value, performs some operation on this value, and creates an output answer.
The
at the original function's starting value.
s f and g are inverse functions, . A function composed with its inverse function yields
where you started.
ding an inverse is simply the swapping of the x and y coordinates. This newly formed inverse will be a relation, b
variable is exchanged with the dependent variable in a given relation. (This inverse may NOT be a function.)
## elf a function, it is then called an inverse function.
ction!
so be a function.
orresponds to one and only one first element. (each x and y value is used only once)
## ermine if a function is a one-to-one function.
, your function will be a one-to-one function and its inverse will also be a function.
## used to show that a relation is a function.)
rse of a function is the set of ordered pairs obtained by interchanging the first and second elements of each pair in
ould the inverse of function f (x) also be a function, this inverse function is denoted by
Note: If the original function is a one-to-one function, the inverse will be a function.
[The notation f -1(x) refers to "inverse function". It does not algebraically mean 1/f (x).]
## If a function is composed with its inverse function,
the result is the starting value. Think of it as the function and
the inverse undoing one another when composed. More specifically:
Consider the simple function f (x) = {(1,2), (3,4), (5,6)}
and its inverse f -1(x) = {(2,1), (4,3), (6,5)}
## "So, how do we find inverses?"
Consider the following three solution methods:
swap the x and y values. Remember, the inverse relation will be a function only if the original function is one-to
2 3 4 -3
1 -2 5 1
## ill be the set of ordered pairs:
e relation will NOT be a function (because the x value of 1 now gets mapped to two separate y values which is not possible for f
rocess:
## a. Find the inverse of the function
Remember:
Set = y.
Swap the variables.
Solve for y.
## Use the inverse function notation since f (x) is a one-to-one function.
b.
Find the inverse of the function (given that x is not equal to 0).
Remember:
Set = y.
## Eliminate the fraction by multiplying each side by y.
Get the y's on one side of the equal sign by subtracting y from each side.
Isolate the y by factoring out the y.
Solve for y.
## Use the inverse function notation since f (x) is a one-to-one function.
e identity line,
e inverse relation is also a function. (Read more about graphing inverses.)
h. The new red graph is also a straight line and passes the vertical line test for
## Use the TI-83+/84+
graphing calculator
to investigate
inverses. | 670 | 2,724 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2019-26 | latest | en | 0.855956 |
https://math.stackexchange.com/questions/1803493/chevalier-de-mere-paradox-with-game-with-three-dice | 1,569,033,771,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574182.31/warc/CC-MAIN-20190921022342-20190921044342-00551.warc.gz | 554,843,828 | 31,084 | # Chevalier de mere paradox with game with three dice
Chevalier de Mere asked Blaise Pascal why in a game with three dice the sum $11$ is more favorable than $12$, when both sums have exactly the same possible combinations:
For $11$ we have $(5,5,1), (5,4,2), (5,3,3), (4, 4, 3), (6,4,1), (6,3,2)$ and for $12$ we have $(6,5,1), (6,4,2), (6,3,3), (5,4,3), (4,4,4), (5,5,2)$, so both sums should be equiprobable.
My attempt: I think Chevalier de Mere made the mistake of thinking all the dice are indistinguishable. I tried to compute the exact probabilities.
Let $$\Omega = \left\{(x,y,z) \mid 1 \leq x,y,z \leq 6, \quad 3 \leq x + y + z \leq 18\right\}$$ be the sample space. We are interested in the events $$A = \left\{(x,y,z) \mid x + y + z = 11 \right\}$$ and $$B = \left\{(x,y,z) \mid x + y + z = 12 \right\}.$$ For the sum $11$, we have $27$ possible permutations of all the triples. For the sum $12$, there are two less, that is $25$. So $\#A = 27$ and $\#B = 25$. Since $\#\Omega = 6 \cdot 6 \cdot 6 = 216$, we have $$P(A) = \frac{27}{216} = 0.125, \qquad P(B) = \frac{25}{216} = 0.1157.$$ Is this reasoning correct?
• Yes, 27 versus 25 looks correct to me. – almagest May 28 '16 at 15:15
• The calculation is right. I would say that the calculation that yields equiprobable uses the wrong probability model. – André Nicolas May 28 '16 at 15:45 | 491 | 1,358 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2019-39 | latest | en | 0.84061 |
https://physicscatalyst.com/mech/motion-in-straight-line-with-variable-acceleration.php | 1,719,105,408,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862425.28/warc/CC-MAIN-20240623001858-20240623031858-00109.warc.gz | 394,727,657 | 15,310 | # Motion in straight line with variable acceleration with Problem and solutions
## Motion in straight line with variable acceleration
• A particle may move in a straight line with a non-uniform accleration i,e variable acceleration
• There is no direct formula for motion in it as the equation would vary in each case
• we need to apply the basic formula to get the velocity ,acceleration in it
## Formula to be used
(1) Instantanous velocity is given by
$v= \frac {dx}{dt}$
(2) Instantanous acceleration is given by
$a= \frac {dv}{dt}= v\frac {dv}{dx}$
(3) Average velocity is given by
$v_{avg} = \frac {\int_{0}^{t }vdt}{\int_{0}^{t }dt}$
(4) Average acceleration is given by
$a_{avg} = \frac {\int_{0}^{t }adt}{\int_{0}^{t }dt}$
## Solved Examples
Question 1
Motion of a particle is given by the equation
$x=3t^3 + 7t^2 + 14t + 8$
where t in sec and x is in meter.
(a) Find the position of the particle at 1 sec
(b)Find the displacement of the particle in the time interval from t=0 to t=4 sec
(c) Find the value of the velocity at 2 sec
(d) Find the value of the acceleration of the particle at t=1 sec
(e) Find the average velocity of the particle in the interval t=2 sec to t=4 sec
Solution
(a) position of the particle at 1 sec is given by
$x_{1}= 3(1)^3 + 7(1)^2 + 14(1) + 8=32$ m
(b)Position of the particle at 4 sec is given by
$x_4=3(4)^3+ 7(4)^2 + 14(4) + 8=192 + 112+ 28+8=368$ m
We already know the position of the particle at t=0
Displacement between t=0 to t=4 is given by
$=x_4 - x_0=368 -32=336$m
(c) Velocity is given by
$v=\frac {dx}{dt} = 9t^2 + 14t + 14$
Now velocity at 2 sec
$v_2=9(2)^2 + 14(2) + 14=78$m/s
(d) Accleration is given by
$a=\frac {dv}{dt}=18t + 14$
Now acceleration of the particle at t=1 sec
$a_1=18(1) + 14=32$m/s2
(e)Average velocity will be give as
$=\frac {x_4 -x_2}{4-2}$
Now
$x_2=3(2)^3+ 7(2)^2 + 14(2) + 8=24+28+ 28+8=88$ m
So Average velocity
$= \frac {336-88}{2}=124$ m/s
Question 2
The distance x of a particle moving in one-direction, under the action of a constant force is related to time by equation
$t= \sqrt {x} +3$
where x is in metres, t is in seconds.
Find the displacement of the particle when its velocity is zero.
Solution
The equation of time is given as,
$t= \sqrt {x} +3$
it can be rearranged as
$\sqrt {x} =t−3$
Squarring both the sides
$x=t^2 -6t+9$
The velocity is given as,
$v=\frac {dx}{dt} =2t−6$
When the velocity is zero, it can be written as,
0=2t−6
t=3
Therefore The displacement at 3s is given as,
$\sqrt x=3-3$
$x=0$
## Problems
Question 1
A particle’s position on the x axis is given by
$x = 4 - 27t + t^3$
with x in meters and t in seconds.
Find the particle’s velocity function v(t) and acceleration function a(t).
Question 2
A particle initialy moving with velocity u is subjected to a retarding force as a result, it deccelerates with $a=-k \sqrt {v}$ where v is instantanous velocity and k is positive constant.
Find the time taken by the particle to come to rest? | 992 | 2,957 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-26 | latest | en | 0.81981 |
http://math.stackexchange.com/questions/572246/why-does-this-integral-equal-1-when-n-1 | 1,462,480,355,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860127983.53/warc/CC-MAIN-20160428161527-00144-ip-10-239-7-51.ec2.internal.warc.gz | 179,587,651 | 17,916 | # Why does this integral equal $1$ when $n=1$?
As part of a boundary value problem, I did the following integral:
$b_n = 2\int^1_0 (30 \sin(\pi x) - 100 (1-x)) \sin (n\pi x) dx$
next step:
$b_n = 2\int^1_0 (30 \sin(\pi x)sin (n\pi x) - (100 (1-x)) \sin (n\pi x) dx$
we can break this up
$= 60\int^1_0 \sin(\pi x)sin (n\pi x) - 200\int^1_0((1-x)) \sin (n\pi x) dx$
and integrating the two parts, the first half via the product-to-sum identity
$= 60\int^1_0 \frac{1}{2}[\cos(\pi x- n\pi x) - \cos(\pi x+ n\pi x)]dx - 200\int^1_0((1-x)) \sin (n\pi x) dx$
using inegration by parts on the second half, with $u= 1-x$ and $du = -dx$, $dv = \sin (n\pi x) dx$ and $v = \frac{\cos (n\pi x)}{n \pi}$
$= 60\int^1_0 \frac{1}{2}[\cos(\pi x- n\pi x) - \cos(\pi x+ n\pi x)]dx - 200[(1-x)(\frac{\cos (n\pi x)}{n \pi})+\int^1_0\frac{\cos (n\pi x)}{n \pi}dx]$
$$= \left[ 60 \frac{1}{2}[\frac{\sin(\pi x- n\pi x)}{\pi - n\pi } - \frac{\sin(\pi x+ n\pi x)}{\pi + n\pi}] - 200[(1-x)(\frac{\cos (n\pi x)}{n \pi})+\frac{\sin (n\pi x)}{(n \pi)^2}]\right]^1_0$$
And when I do the algebra I get: $$= 30 [\frac{\sin(\pi - n\pi )}{\pi - n\pi } - \frac{\sin(\pi + n\pi )}{\pi + n\pi}] - 200\left(\frac{\sin (n\pi )}{(n \pi)^2}\right) + 200(\frac{1}{n \pi})$$
Here's what I don't get. If n= 1 the part multiplied by 30 goes to 0/0. Does that mean I can count it as 1? That's what the text notes, that it is 0 if $n \neq 1$ and 1 if $n=1$, but I just wanted to be sure I understood this and did the whole integration correctly.
thanks
-
"If n= 1 the part multiplied by 30 goes to 0/0. Does that mean I can count it as 1?" You certainly can't, whatever your text says! You have to treat n=1 as a special case, and perform the calculation step by step. – TonyK Nov 18 '13 at 19:17
ok, so what is that special case? That is, should i plug in n=1 earlier on (like before I do the integration, when it is still a cosine term) and go from there? Problem is, I still end up with a zero in the denominator... – Jesse Nov 18 '13 at 19:35
Oh wait,, the $\cos(\pi x - \pi x)$ (when n=1) is a 1, and that means it's just x in the integration which is 1... I got it... – Jesse Nov 18 '13 at 19:40
Exactly!${}{}{}$ – TonyK Nov 18 '13 at 19:46
You can justify the step if you consider the parameter-dependent integral
$$I(\alpha) = \int_0^1 \cos (\alpha\pi x)\,dx.$$
For $\alpha \neq 0$, you have
\begin{align} I(\alpha) &= \frac{\sin (\alpha\pi x)}{\alpha\pi}\bigl\lvert_0^1\\ &= \frac{\sin (\alpha\pi)}{\alpha\pi}. \end{align}
The integrand is continuous and depends continuously on $\alpha$, hence $I(\alpha)$ depends continuously on $\alpha$, and
$$I(0) = \lim_{\alpha\to 0} I(\alpha) = \lim_{\alpha\to 0} \frac{\sin (\alpha\pi)}{\alpha\pi} = 1.$$
But you need to let $\alpha$ assume non-integer values for that. It's at least as easy to just plug in $0$ and compute $\int_0^1 1\,dx$.
- | 1,120 | 2,867 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2016-18 | longest | en | 0.783197 |
https://community.thunkable.com/t/how-to-do-a-simple-1-2-3-4-2/2251664 | 1,696,150,428,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510810.46/warc/CC-MAIN-20231001073649-20231001103649-00875.warc.gz | 199,200,761 | 6,398 | # How to do a simple "1 - 2 + 3 * 4 / 2"
https://x.thunkable.com/copy/d33133508cec6445f1c1efe4a0ecf0bc
If I copy and paste the above math problem to a calculator it gives out the answer as 4. I want to do the same thing in Thunkable. The numbers can be multi digits and the operators occurrence is random. I sat on this problem for couple of hour pondering … I can conceive the solution but it is too big and convoluted. Please check my starter program which can successfully do "123 + 234 + " kind of string. It will just add 123 +234 and spit out the answer as 357 going beyond that is my hurdle. Is there a easy solution or the big bulky convoluted solution I am thinking off is the only way out? Thanks
The answer should be 5. What calculator is giving you 4 as the answer?
Given 1 - 2 + 3 * 4 / 2, you multiply 3 * 4 / 2 and get 6 (either 12 ÷ 2 or 3 * 2). So 1 minus 2 is -1 and then adding 6 (from earlier) gives you 5.
If you’re trying to make a calculator app (and I would wonder why…), no there’s no easy solution. It’s going to take a lot of work to get it right.
Ok that is what I thought but thanks for confirming it. Now I can skip this feature without guilt This is something like a calculator but with limited features for a different purpose. So I will just skip this feature. Just to complete my reply I used the standard calculator which comes with Windows 11.
There they do the calculations from left to right. 1-2 = -1, -1 + 3 = 2, 2*4 = 8 and finally 8/2 = 4. You can see the last operation up there above 4. Thanks again
You’re welcome. If that’s true, then the Windows calculator isn’t following order or operations. That would really surprise me so I’m guessing you’re entering the problem from left to right and that actually, you are ignoring order of operations.
If you do need to program a calculator, there are many examples available if you Google calculator Thunkable.
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Actually there are two types of order processing in calculators, the accounting type which is FCFS (First Come First Serve) or the scientific type which is BODMAS. @tatiang is referring to the more common in schools which is BODMAS. Whereas the Windows one is set as accounting type.
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Thanks, I wasn’t aware of the difference.
1 Like
This topic was automatically closed 90 days after the last reply. New replies are no longer allowed. | 608 | 2,364 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2023-40 | latest | en | 0.920811 |
https://aha.betterexplained.com/t/surprising-patterns-in-the-square-numbers-1-4-9-16/211?page=2 | 1,555,945,314,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578555187.49/warc/CC-MAIN-20190422135420-20190422161420-00363.warc.gz | 339,722,367 | 5,712 | # Surprising Patterns in the Square Numbers (1, 4, 9, 16…)
#21
Hey , I think I found something strange , (88) - (77) = 15 which is not a prime number , obviously.
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#22
@ash: Thanks for dropping by :). Yeah, when math is more about exploration / puzzle solving it can be pretty fun.
@Seifeddine: Ah, actually the differences just need to be odd (2n + 1), not prime.
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#23
Shame on me , I should have read your article more carefully .
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#24
Wonderful approaches to regular problems.
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#25
I have a neat problem:
Suppose that we extend definitions of evenness and oddness to the entire set of real numbers by the following:
The set of odd numbers is exactly the set of all differences (n+1)^a-n^a for positive rational a and integral n.
A number is even iff it is twice an odd number.
All other real numbers are neither even nor odd.
Is the set of even numbers still disjoint from the set of odd numbers? (i.e. is an odd number twice another odd number?)
Prove that the set of real numbers neither even nor odd has the same cardinality as the set of real numbers.
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#26
Amazing job of finding these patterns. I know finding patterns was an easy way for me to learn mathmatics when I was younger.
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#27
@Tracy: Thanks! Yes, I think the heart (and fun) of math is really about finding and describing patterns.
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#28
really awesome insights into difference of squares but the main place this seems to apply is summations of Squares, how would you connect 2n+1 to n/6(n+1)(2n+1) i can see the 2n+1 in the formula but the n/6(n+1) part still doesn’t make sense to me
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#29
I like the posts on this. Am wondering if anyone can help my daughter and I figure out how to validate the pattern that we found in our squares. We found that, as long as the number beig squared is divisible by 3, that the sum of the numbers in the answer of the square is equal to 9. For example, 3 sq is divisible by 3 and the total of the answer (9) = 9. If we move onto 6 sq (also divisible by 3), the answer (36) = 9. And so on and so on…can anyone help explain this pattern to us?
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#30
@Sean: Great question. It turns out that if the sum of the digits of a number add up to 9 (or a multiple of 9, like 18, etc.) then that number is divisible by nine.
if the number being squared is divisible by 3 (so it’s 3*n), then the square is (9 * n^2) which is divisible by 9, and therefore falls into the pattern :). I’d like to do a post on why the digits need to add up this way, but here’s one insight:
If we start with 9, clearly the digits (the only digit!) adds to 9. Whenever we add 9, we’re really doing (10 - 1) which means “increase the tens digit by 1, and decrease the ones digit by 1”. This keeps the sum of digits in balance, so we should expect that the sum of digits always equals 9 as before. (For example, 18 means we changed 09 to 18). Of course, once you get to 90 and add (10-1) you are really only able to “fill” the ones digit and get 99 (which then increases the total sum to 18 – but, you kept the sum of digits divisible by 9). Hope this helps!
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#31
ur website is sooo awesome i never get maths question but now i do thanks
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#32
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#33
My question is that i don’t enough knowledge aboout even and odd number. Tells me that some examples and also difinitions of both numbers.
Thanks
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#34
I was really surprised by the calculus method. It is cool.
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#35
Wo! really fantastic
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#36
i didnt like it
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#37
really , good.
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#38
i am for square patterns like[25sq=625={2*3}hundreds+25
`````` BY,
S.NITHEESH KUMAR``````
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#39
very iteresting
lurved it!!!
0 Likes
#40
can we get any multiplication square patterns???i have not found any one!!!1foolish thing
0 Likes | 1,044 | 3,893 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2019-18 | latest | en | 0.955788 |
https://www.physicsforums.com/threads/radiation-problem.380319/ | 1,521,737,122,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647892.89/warc/CC-MAIN-20180322151300-20180322171300-00756.warc.gz | 879,193,726 | 14,225 | 1. Feb 21, 2010
### lordloss
1. The problem statement, all variables and given/known data
The giant hornet Vespa mandarinia japonica preys on Japanese bees. However, if one of the hornets attempts to invade a beehive, several hundred of the bees quickly form a compact ball around the hornet to stop it. They don't sting, bite, crush, or suffocate it. Rather they overheat it by quickly raising their body temperatures from the normal 35°C to 47°C or 48°C, which is lethal to the hornet but not to the bees (Fig. below). Assume the following: 519 bees form a ball of radius R = 3.6 cm for a time t = 24 min, the primary loss of energy by the ball is by thermal radiation, the ball's surface has emissivity ε = 0.85, and the ball has a uniform temperature. On average, how much additional energy must each bee produce during the 24 min to maintain 47°C? The Stefan–Boltzmann constant is 5.6704 × 10-8 W/m2-K4.
2. Relevant equations
P=$$\sigma$$$$\epsilon$$AT$$^{4}$$
3. The attempt at a solution
(5.6704X10$$^{-8}$$)(.85)(4$$\pi$$)(.036)$$^{2}$$(320)$$^{4}$$
Which comes out to 8.24635 and I took this to be joules per second.
(8.24635 X 1440)/519
=22.84
So far, I can't seem to get the correct answer. I have tried finding the answer with only the increased temp rise from 35 to 47 instead of the overall temp of 47 and its still wrong.
2. Feb 21, 2010
Hello lordloss did you calculate the ADDITIONAL energy?Remember the bees still radiate at their normal temperature of 35C(308K)
Problem involving thermal radiation and specific heat May 24, 2017
Problem with refrigerator and radiator Jun 19, 2016
Radiation Pressure problem Jan 23, 2016 | 472 | 1,653 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2018-13 | longest | en | 0.923516 |
http://barrygraygillingham.com/Maths/Triangles.html | 1,708,566,854,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473598.4/warc/CC-MAIN-20240221234056-20240222024056-00582.warc.gz | 5,663,429 | 3,879 | # Triangles
A triangle has three straight line sides. The word triangle comes from a Latin word which means three-cornered. The corners of a triangle are called its vertices (singular vertex).
If all the sides of a triangle are different lengths it is a scalene triangle. We can show that if all the sides of a triangle are different then all the angles are different too, and visa versa.
If two of the sides are equal then it is an isosceles triangle, from the Greek for equal legs.
We often draw an isosceles triangle with the unequal side at the bottom, as at the left of the picture, but it does not have to be like that, it can be at any angle.
We can show that if two of the sides of a triangle are equal then so are two of the angles, and visa versa.
If all three sides of a triangle are equal then it is an equilateral triangle, from the Latin for equal length.
We can show that if all the sides of a triangle are equal then so are all the angles, and visa versa. We show later on this Page that each of the angles in an equilateral triangle is 60°.
The angles inside a triangle are called the internal angles. If we extend one side we form the external angle. The internal and external angles at any vertex of a triangle (or any polygon) are supplementary, that is, they add up to 180° (two right angles). Two angles are complementary if they add up to 90° (one right angle.) When we are talking about the angles of a triangle we usually mean the internal angles, we do not usually need to say internal angle unless there might be some confusion as to whether we mean internal or external angle.
If we walk along one side of a large triangle drawn on the ground, when we come to a vertex the external angle is the angle we must turn through in order to be able to walk along the next side.
For an equilateral triangle the internal angles are all equal, and as the internal and external angles are supplementary, the external angles must also all be equal.
If we walk all round an equilateral triangle so that we end up where we started from and are facing in the same direction we shall have turned through a complete circle of 360°. This means that the external angles add up to 360° so each one of them must be 360° ÷ 3 or 120°. As the external and internal angles are supplementary each internal angle must be 180° - 120° or 60°, so the sum of the internal angles is 3 × 60° or 180°. We can show that this is true for all triangles, not only equilateral triangles, that is,
For any triangle the sum of the internal angles is 180°.
This means that if one of the (internal) angles in a triangle is 90° or more then the sum of the other two angles must be 90° or less, that is, a triangle can have only one right angle or one obtuse angle. In an acute angled triangle all the angles are acute, in a right angled triangle one angle is a right angle, in an obtuse angled triangle one angle is obtuse.
We can combine these terms with scalene and isosceles, so the different types of triangle are
• Acute angled scalene
• Right angled scalene
• Obtuse angled scalene
• Acute angled isosceles
• Right angled isosceles
• Obtuse angled isosceles
• Equilateral - this must be acute angled so we do not need to say so.
There is more about right angled triangles on the Right Angled Triangle Page of my Web Site (currently under construction).
We usually label the vertices of a triangle or other polygon and the ends of a straight line with capital letters, often consecutive letters, such as A, B and C or P, Q and R.
Here the sides are PQ, QR and PR. (The order of the letters, for example PR or RP, matters only if the sides of the triangle are vectors - this is not discussed further here.)
The whole triangle is called triangle PQR; the angle between PQ and PR can be called angle QPR, or if there cannot be any confusion what angle is meant just angle P. The angle P is included between PQ and PR, and (angle) P and (side )QR are opposite.
We often use small letters to stand for the lengths of the sides. Conventionally, the length of the side opposite angle (big) P is given the symbol (little) p. Remember that p stands for the length of the side not the side itself: we must not refer to side p, only side QR. The longest side is always opposite the largest angle.
If we have three rods we can join them at their ends to make a triangle. The only Rule is that any two of the rods must together be longer than the third, otherwise their ends will not join.
There is only one way we can join these rods: the brown one will always have a magenta rod at one end and a green rod at the other.
If you are colour-blind, and in England about one boy in ten is, I have tried to choose colours where you can see the difference, but if you cannot please e-mail me and I will change them.
Of course, you can turn the shape round or upside down in any way you like but it will always be exactly the same size and shape.
This is very important because it means that once we have fixed the length of the sides of a triangle we have also fixed the angles. This is always true for all triangles, it is never true for any other polygon.
Triangles which are exactly the same size and shape are congruent.
Going on from here, if we have a triangle and increase the length of just one side by a certain proportion, say 80%, while keeping the angles the same we also increase the length of the other two sides by 80%.
The two triangles are the same shape but not the same size. This is always true of all triangles; it is never true for any other polygon. Triangles which are the same shape but not the same size are similar.
These important differences between triangles and all other polygons are discussed at greater length in The Shape of a Polygon.
For all triangles, the ratio of the length of the sides depends only upon the size of the angles, and the size of the angles depends only on the ratio of the length of the sides. A whole branch of mathematics, called trigonometry, is founded on this very simple idea.
## The area of a triangle
Here triangle ABF is half the area of the rectangle ADBF, and triangle FBC is half the area of rectangle FBEC. So triangle ABC is half the area of the rectangle ADEC. The area of rectangle ADEC is base × height so the area of triangle ABC is ½ (base × height).
The apex of the triangle does not have to lie on the rectangle for this to be true. You might like to try to prove this for yourself - it is not difficult. But e-mail me if you get stuck.
The area of any triangle is half the base times the (vertical) height. | 1,497 | 6,587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2024-10 | latest | en | 0.916736 |
https://stockingisthenewplanking.com/which-are-the-prime-numbers-from-1-to-100/ | 1,726,408,108,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651630.14/warc/CC-MAIN-20240915120545-20240915150545-00270.warc.gz | 500,887,102 | 10,558 | ## Which are the prime numbers from 1 to 100?
What is the list of prime numbers from 1 to 100? Prime numbers from 1 to 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
### Which is a composite number?
A composite number is a positive integer. which is not prime (i.e., which has factors other than 1 and itself). The first few composite numbers (sometimes called “composites” for short) are 4, 6, 8, 9, 10, 12, 14, 15, 16, (OEIS A002808), whose prime decompositions are summarized in the following table.
#### What are the prime numbers 1 to 200?
The prime numbers from 1 to 200 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199.
How many non prime numbers are there between 1 and 100?
Odd Composite Numbers 1 to 100 There are a total of 25 odd composite numbers from 1 to 100. The list of odd composite numbers 1 to 100 includes 9, 15, 21, 25, 27, 33, 35, 39, 45, 49, 51, 55, 57, 63, 65, 69, 75, 77, 81, 85, 87, 91, 93, 95, and 99.
What is twin number?
Two prime numbers are called twin primes if there is present only one composite number between them. Or we can also say two prime numbers whose difference is two are called twin primes. For example, (3,5) are twin primes, since the difference between the two numbers 5 – 3 = 2.
## What is co-prime for Class 6?
Any set of numbers which do not have any other common factor other than 1 are called co-prime or relatively prime numbers. This shows that 5 and 6 have no common factor other than 1. Therefore, they are co-prime numbers.
### What are the composite numbers from 1 to 500?
4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81,82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 98, 99, 100. | 858 | 2,050 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-38 | latest | en | 0.93152 |
https://www.topperlearning.com/answer/solve-for-x-and-y-ac-by-2a-3b-0-and-bx-ay-3a-2b/7br7bb22 | 1,686,335,843,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656788.77/warc/CC-MAIN-20230609164851-20230609194851-00063.warc.gz | 1,145,218,806 | 59,315 | Request a call back
Solve for x and y: ac+by-2a+3b=0 And. bx-ay=3a+2b
Asked by ratankumar10045 | 15 Jan, 2020, 09:27: PM
The question should be like below:
Solve for x and y:
ax + by - 2a + 3b = 0
And. bx - ay = 3a + 2b
Consider,
ax + by = 2a - 3b ...(i)
bx - ay = 3a + 2b ...(ii)
Multuplying by b to equation (i) and by a to equation (ii)
abx + b2y = 2ab - 3b2 ...(iii)
abx - a2y = 3a2 + 2ba ...(iv)
Subtracting (iv) from (iii) we get y = -3
Put y = -3 in (i) we get x = 2
Answered by Sneha shidid | 16 Jan, 2020, 09:51: AM
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Asked by ananyak4213277 | 17 Sep, 2021, 07:29: PM | 437 | 969 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2023-23 | longest | en | 0.835175 |
https://stats.stackexchange.com/questions/366092/variance-of-a-poisson-distribution | 1,716,670,512,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058834.56/warc/CC-MAIN-20240525192227-20240525222227-00197.warc.gz | 480,635,324 | 38,799 | # Variance of a poisson distribution
For a certain section of pine forest, the number Y of diseased trees per acre has a Poisson distribution with mean lambda=10. The disased trees are sprayed with an insecticide at a cost of 3 dollars per tree, plus a fixed overhead cost for equipment rental of 50 dollars. Letting C denote the total spraying cost for a randomly selected acre, find the expected value and variance for C.
Expected value is easy to find: 80
I'm stuck on Variance? Why is variance 90?
Because $C = 3Y+50$, and $var(C)=var(3Y+50)=9var(Y)=90$
• Thank you so much. Just a doubt, if we had to do it by the method of EX^2 etc. how would it look like?
– user218970
Sep 9, 2018 at 17:31
• $E[C^2]=E[(3Y+50)^2]=E[9Y^2+300Y+2500]=9E[Y^2]+300E[Y]+2500$; since Possion variance is $\lambda = 10$, $E[Y^2] = var(Y) + E[Y]^2 = 110$. Substituting leaves us with $E[C^2] = 6490$. $E[C]^2 = 6400$, then $var(C) = 6490-6400=90$ Sep 9, 2018 at 17:36
• Wow, thank you so much. I was stuck with this for minutes.
– user218970
Sep 9, 2018 at 17:40
• I did upvote it. But it won't show as I have less than 50 reputation. Yes, approved it.
– user218970
Sep 9, 2018 at 17:54
• You need a mere 15 reputation to upvote, not 50. [A single question, moderately well-written -- i.e. a somewhat average question on the site - can expect to earn about 2-3 upvotes, and 3 is sufficient to gain the privilege] Sep 10, 2018 at 1:03 | 473 | 1,419 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2024-22 | latest | en | 0.917242 |
https://mrstaplinsmath.wordpress.com/category/algebra-ii/ | 1,566,502,077,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317339.12/warc/CC-MAIN-20190822172901-20190822194901-00075.warc.gz | 558,998,055 | 27,483 | # Parabolic Solar Cookers
I am excited for our final project in Algebra II…parabolic solar cookers! After we did an in class investigation of the four different types of conics with play-doh, evaluated the standard equations, and explored the graphs of each, I wondered if there was one more way I could help my students deepen their understanding of conics. So, while sitting on the couch at home with my husband and searching online, I found this (http://www.education.com/science-fair/article/solar-hot-dog-cooker/) and excitedly told him, “I’m going to make this happen!!” The next day at school, I talked with my dean about the logistics of creating solar cookers and ways to help scaffold the lesson.
The summary of the project and the student materials that I revised from the website are below (some wording and pictures are directly from the website above…so please credit that source if using this.) The directions are quite lengthy, but very step-by-step, so I suggest breaking this up into at least 3-5 days with groups of 2-4 students.
Summary: To begin the project, students spent a class day researching and answering questions about solar cookers. With the help of an actual solar energy engineer (see his website and work here: energyae.com) generating some ideas, the questions helped students buy in to the project (he also sent me some videos to show students real life parabolic solar panels…if you have Dropbox, check them out here: https://www.dropbox.com/home/SolarMrsTaplin). The next day, students found three points to create a parabolic curve based on the dimensions of their shoe box (two top corner points and one center origin point). From these, students calculated the equation and plotted other points to create a nice, accurate curve. Next, students calculated the focal point, which we talked about the reasoning as to why this is the spot they should place their food at to cook. Lastly, students covered their curves in poster board and foil for the reflective surface and fashioned holders for the focal point.
A few groups still need to finish, but they are coming along very nicely! It has been raining/cloudy for about a week straight now, so please do a little anti-rain dance for us and hope for some sun so we can test out these cookers before the school year ends!! I will post after we get to and show the results!
Student Materials: (link to word doc: solar cooker proj or click on thumbnails below for images)
Update: We finally had some sunny weather and got to go out to cook. We left the solar cookers out for about 45 minutes. The marshmallows didn’t melt the way we predicted, but having thermometers out with us proved that it definitely got hotter at the focal point. Most had an initial temperature of 90-92 degrees F, and after about 20 minutes, the temperature rose to about 105. The final temperature recorded was about 120 degrees F at the focal point of most solar cookers. Although they were a little bummed that the marshmallows didn’t melt completely, we talked about the fact that if it were 120 degrees F outside, they would not want to be outside themselves. So, that helped put it in perspective and see that it worked. Next year, I think we should try cooking a darker food substance…maybe chocolate, and we could do fondue 🙂
# UbD Unit for Systems of Equations and English STAAR
I completely forgot about a UbD unit I created a few years ago which connects mathematical systems of equations to writing for the English STAAR (I wasn’t able to use it this year due to a calendaring scheduling conflict, so it slipped my mind). I was searching around for some English/Math connections and stumbled upon this unit published by Trinity University…turns out it’s mine! I figured my blog would be a good place to leave this link to the unit. I know it’s a bit late in the year to use for STAAR writing practice, but maybe it can be stored away for next year!
# Thoughts That Keep Me Up at Night
I have been thinking a lot about grading, assessment, and the meaning behind these to both teachers and students. I have read Dan Meyer, Daniel Schneider, Educational Leadership, Matt Townsley, and Rick Wormeli while researching and talking with colleagues about mastery and standards based grading (SBG). I really like a lot of the ideas of SBG including more frequent and smaller assessments that allow one to know a student’s mastery on a standard. I also like the thought of a 1-4 scale and the language that is used to convey what each number means. One example I found that I really like is that the Solon School’s language in their rubric (1). However, I am still struggling to wrap my head around SBG and mastery in the math classroom. I still have lingering questions that honestly keep me up at night. I want to do what is right for students and I want to push them to understand what they know and what they don’t. Even further, I want them to take charge of their learning and with my feedback, help them to know how to gain mastery on a concept.
Here is what I want to keep in my classroom regardless of grading…
Student communication and group work: I think when students talk out mathematical problems together, they cognitively grow a lot. A student’s ability to explain a topic further enriches their own understanding, and when they hear an explanation from another student, they relate to the language they’re using. So, regardless of how I grade and what I grade, I still want students to work together to solve problems.
Reasoning: I also want to be sure I am still allowing room for reasoning and processing skills beyond algebraic skills. I want to continue to provide opportunities for students to explain and justify their understanding of concepts through written and spoken dialogue. Whether this fits into a numerical grade or not, students still need to be pushed to think deeply and justify their reasoning.
Here are my questions that linger…
How do I create a balance between group practice and independent practice? How do I convince students that individual work time is just as, if not sometimes more, beneficial than group work? And finally, how do I convince students that individual assessments are meant to be informative not punitive, especially when we take points off for wrong answers rather than give points for correct attempts? Right now I give a lot of time for group practice. Again, I love the learning that happens when students talk through math. But I am realizing as I read more about SBG, I need to create more opportunities to show what they know individually. In that, I need to create time to give my feedback to them on an individual level beyond tests. I think by adding in more frequent assessments, this will do that and give students the opportunity to analyze what they know. These could just look like short quizzes done on note cards at the beginning of class. They could be graded on a 1-4 scale with more feedback than a regular assignment as it leads up to a cumulative summative assessment. The question then becomes, do I have the time myself to dedicate rich feedback more often to every individual student? How do I create that time, especially as a math teacher, when so many of my days are dedicated to teaching new material rather than refining knowledge students already know? Ultimately, can someone find me some more hours in the day? 🙂
Feel free to respond to any or all of my questions, share this with every educator you know, and continue the conversation of grading and assessment!
# MatHistory Part 2
This post has taken me a while to write with several revisions because I just haven’t known how to write it in a way that gives justice to everything I have loved about this assessment. Every time I sit down to write it, a new way to introduce the post races through my head. However, I think the best way to start out is simply thanking the AP World History teachers who had the vision and the enthusiasm for working with me to create this complex, unique, and highly successful interdisciplinary assessment. So, thank you, Mr. Freeman and Mr. Sprott! I hope this leads to many more mathistory and mathenglishistry (math-English-history-chemistry) ideas!
So, here’s how the project unfolded…
Last Tuesday, the World History teacher presented our students with the idea of a power scale timeline and explained how to create one. Each group of four students were given 16+ maps of various European empires that showed the time at which each empire owned land. After identifying the region, they transferred the area onto a larger scaled timeline. Then, when all the empires were on one map, students used their mathematical knowledge to identify key points and explain why they are important relevant to time and land area (we presented this part on the second day). At this point you may be thinking whhhat theee hecckkk is this lady talking about…don’t worry, our students were also a bit perplexed by the task at the beginning and in first period, we did have a brief time when we felt as if our students might revolt against us. However, after encouraging our students to just try it out, much to their surprise with a little patient problem solving (as referenced by Dan Meyer) they excelled at the task. After they got the hang of it, I asked one student to explain the process and here is his recording. I liked getting to be in the history classroom this day because I was a second person who could help facilitate and answer questions as students created their timelines. They also started to make the connection with me being in the room that this might have some math involved in it.
On the second day, as the students were finishing their timelines, we presented them with the math portion. As they identified key points historically, we heard them using mathematical language and vocabulary terms as they talked about undecagons, parallel slopes, parabolas, intersecting lines, exponential growth, etc. It was really cool to see students make connections and hypothesize using their knowledge from both classes. We encouraged students to work together because the questions we gave them ranged in topics from both Algebra II and geometry. Each group had students from both classes and therefore they were able to be experts in the subject they were taking. Walking around helping students work through the assignment, I overheard two students talking about interdisciplinary learning and caught the end of their conversation on tape. Hearing them voice their appreciation for interdisciplinary learning really made it all worth it!
Yesterday I finished grading the math portion of the timelines and was sitting with the English teacher during our monthly Saturday school when she asked me, “so, would you do it again?” My immediate answer was, “YES!!” I think that asking students to use mathematical evidence in their explanations of what was happening historically made them think deeper. I also think they were able to synthesize better by using logical mathematical thinking. Additionally the featured image at the top of this post is one that we were very impressed by. She spent time finding key images and icons from each empire and finished her timeline with detailed watercolors. It made us think of the possibilities…perhaps students could design a timeline based on different aspects of each empire (religion, architecture, art, etc.) I think there is so much more we can do and definitely lots to think about. Over the next few weeks students will continue to learn in their World History classes about each empire and add to their timelines in order to create a large final timeline in groups…so more to come!!
Here is the link to the math student materials: MatHistory
# Two More Review Games to Try Out
We played two different review games this week in Algebra II and Geometry. I gladly give credit to the two blogs hyperlinked throughout this post after seeing my students highly engaged and excited to do math practice with the addition of these two games.
The first game is one I call “The Laundry Game” and we played it before a geometry quiz. This game can be applied to any content and any grade level even though I actually found it on a 3rd grade blog. I love this game because it takes a normal review and adds movement, competition, and immediate feedback. My intern and I were talking about how changing the format of a routine review into this game makes students do so many more problems than they normally would. I changed it up a bit by providing the answers at each station rather than having students come to me to check. That way they could check their own work and then the teacher could be facilitating and answering questions. At the end, we drew their names from the buckets, awarded prizes, and then explained the reasoning behind the name, “The Laundry Game.” We told them when they go home that night, they will find the problems they need to review again in their pockets with their laundry!
The second game we played is a BINGO game that I found here. We played this as addition to a normal practice day. I have seen and played BINGO before in the classroom, but I love how this one gives choice to students and the downloadable file is already made for you to give to students…so easy! My students LOVED this…after the first student won, several kids shouted out, “can we keep playing?!” Little did they know, they were implying, can we keep practicing logs!!
# Teams-Games-Tournament
I am always looking for new review games to use before a quiz or a test. This year, we have played Team-Games-Tournament (TGT) several times and have had great success with student mastery and engagement. This game is a cooperative learning strategy that is discussed further in “The Strategic Teacher” by Silver, Strong, and Perini. In their book, they explain the reasons why this strategy works with the number one reason being that “TGT incorporates the best of cooperation and competition.” Furthermore, it “highlights interdependence among group members, holds students individually accountable…promotes positive face-to-face interaction, builds small-group skills such as communication and conflict resolution, and encourages group processing so that students use their reflections to become better team members.” There is a bit of prep work on the teacher before hand to possibly reformat your review, create the playing cards, and make color-coded copies (my interns wonderful suggestion) for each role, but it proves very worthwhile. As compared to other review games that require the teacher to reteach or be the leader, this game is student centered and the teacher can now simply be a facilitator while students take ownership in playing, coaching, and teaching each other.
I typically give students a review sheet for the test the day before we are going to play the tournament. I give them some class time and/or assign the evens for homework. By only doing half, students are familiar with the material, but still need to review more the next day. The following day during the tournament, they will complete all the review questions. When they draw one they have already done they are instructed to do it again on a whiteboard (or a scratch piece of paper), whereas when they draw an odd question they should do it on the actual review. This gives them repetition of review and complete mastery of the content. The first time using this strategy, students will need the directions read to them and seen posted on the board. However, after a few rounds, the game becomes very clear and natural. I have loved watching students coach each other through the work while also competing with positive interactions.
Here are some directions that you can display for students as well as teacher notes to further clarify. Let me know if you play and what you thought!
Teams Games Tournament
Reference: Silver, Harvey F., Richard W. Strong, and Matthew J. Perini. The Strategic Teacher. Alexandria: ASCD, 2007. Print.
# MatHistory
I’m getting really excited for a new idea the world history teacher and I have been brainstorming for a few weeks. We are working to intertwine some math content when he teaches Eurasian Empire power scale timelines. He came to me with this thought that math could somehow be modeled in the drawings and as soon as I saw the previous year’s final products, my math brain started running! We started drafting the actual questions today and we are going to continue to formulate our ideas until we present it to our students next week (I’m sure there will be revisions even as we are presenting it to the students…forever in reflection of our craft). Here is the link to our questions and above is the image of our thought process…but stay tuned for the final student materials and some finished products!
# Nuclear Culture
We just finished a 3 week long interdisciplinary unit on nuclear energy. I am very grateful to have the opportunity to work with the same core sophomore teachers for the past three years. We have grown together personally which I think in turn has had a positive effect on our students learning (supported by The Washington Post here), and we have also accomplished a lot professionally together. We started out three years ago trying to become more unified in our teaching approaches with simple ideas like wearing the same t-shirt on the same day. Needless to say, that wasn’t quite an authentic learning opportunity for students. Our approach evolved into us using the same vocabulary word somewhere in our dialogue throughout the week in hopes that our students would catch on and use it for their “personal dictionary” assignment in English. I definitely learned a lot of new words, but again, this was such a simple daily act and we knew we could do more!
With the help of our Trinity University interns a couple years ago, our most unified approach was created- a UbD unit centered around nuclear chemistry with lesson plans in English, world history, geometry, and Algebra II. (I have to plug an incredible master’s program here that prepares student teachers with a high level of understanding about teaching pedagogy and practice). This unit has gone through three years of revision and I have to say, the way in which everything came together this year, I think it was the our best year yet!
The first year we did the project we were not able to incorporate math because it did not fit with the scope and sequence. I still remember hearing one student mumble, “you know, we’re studying nuclear energy n every subject except math.” My heart sank and I knew I was going to find a way to make it happen! The next year, I decided to switch the sequence of our curriculum and teach exponents at the beginning of the spring semester when the nuclear culture unit started. Working with my Trinity intern that year, we created investigative lesson plans where students discovered how radioactive elements decay. Understanding asymptotes also helped students see that nuclear waste will never reach 0. We then discussed the implications for using nuclear energy based on exponential growth and decay and exponential properties. This year, I added the geometry component and solidified the inclusion of math in the unit. After my students learned about proving congruent triangles, I went into the English/World History combined classes and introduced their persuasive essay by asking students to write an outline in a mathematical proof format (I was lucky to have another intern this year and she taught a great lesson on truss strength involving triangle properties this day while I was teaching in the other room…I’ll save that post for another day). I loved how the proof writing gave my geometry students the chance to be experts on the topic they just learned and were able to refresh the Algebra II students how to write proofs. Also, the English teacher loved the way they provided evidence for their essays and did some formulaic pre-writing before jumping on a computer to type their essays. Finally, after the culminating day of the project (a “town council meeting” to debate whether we should pursue nuclear energy in San Antonio), the World History teacher created a graph where the x-axis was labeled as a continuum from “San Antonio should not pursue nuclear energy” to “San Antonio should pursue nuclear energy” and the y-axis was a continuum with “the US should not pursue nuclear energy” to “the US should pursue nuclear energy”. Students then plotted their personal opinion as a visual representation for further dialogue.
My heart is full after this unit and I loved hearing quotes like this: “I like the way every class was included in this unit because we understood all perspectives.”
# One of My Favorite Projects
I am starting off blogging by sharing one of my favorite projects that I created with my dean a few years ago. After attending a conference at Southwest Research Institute, I got inspired to do a project that would help students connect their learning of quadratics to the real world. I decided to make it about projectiles. I dreamed of getting to launch projectiles across the math classroom, but had no idea how to make this possible. After talking with my dean who is essentially a physics guru, he helped my dream become a reality. Below you will find the process and some reflections on the project.
First, we modeled a projectile using LoggerPro software and found the quadratic equation. This equation became the central focus for our students to manipulate. We asked them to find the vertex form, x-intercepts, y-intercept, and the domain and range. The fun part came on the second day of the project. We hung tennis balls in the back of the room and gave each group (groups of 3-4 students) a different height at which they would set their equation to. Using the quadratic formula, they solved their equation for the x-distance at which to launch the projectile. If their calculations were correct, they would successfully hit the tennis ball. With safety goggles on, measuring sticks in hand, and genuinely excited emotions, each group stepped up to the projectile launcher to try out their solutions. Some chose to set the launcher at the farthest distance for more fun, and some chose to go the safe route and try the closer distance. Several groups hit the target on the first try which was so fun to to watch the teamwork of their groups as they high-fived and congratulated each other. The groups that didn’t hit the target on the first try realized they needed to watch for the human error of lining the projectile launcher up straight. By the end of each class period, each group had successfully hit the target and proved their math calculations correct.
I love this project because it’s an authentic way to check their work. Also, my first couple years of doing this, I connected it to space science by having a hook that the students were trying to blast a near earth object out of our orbit (Armageddon style). Two years ago, my sophomore team and I designed a full interdisciplinary unit on Nuclear Culture. So to connect it to this, I changed the theme up a bit and had students watch a video about NATO’s ballistic missile defense program and then act as if they were helping NATO use projectiles to intercept incoming harmful missiles. Students were able to communicate their reasoning behind advantages and disadvantages of using technology like this and what might happen if we were successful or unsuccessful in hitting their target. I loved that students made connections, communicated coherently, problem solved, and found reasoning and proof (all performance outcomes that the math department at my school tries to achieve) in one single project!
Here’s the link to the student materials: Saving the World with Math | 4,935 | 23,875 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2019-35 | latest | en | 0.912991 |
https://pyronconverter.com/unit/time/dec-cent | 1,721,745,975,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518058.23/warc/CC-MAIN-20240723133408-20240723163408-00230.warc.gz | 413,711,686 | 11,482 | # Free Online Convert Between decade (dec) & century (cent)
## Convert from cent to dec
Convert between decade (dec) and century (cent) instantly with our free online unit calculator. You can easily convert both dec to cent and cent to dec with just a few clicks. To switch between the two conversions, simply use the swap icon (rotating arrows). If you need to start over, you can reset the values by clicking the reset button.
dec
cent
i.e. dec = cent
Article Contents []
decade is a unit of time
In this unit converter website, we have converter from decade (dec) to some other Time unit.
### What does century mean?
century is a unit of time
In this unit converter website, we have converter from century (cent) to some other Time unit.
### What does Time mean?
In physics Time can be defined as the progression of events from the past to the present into the future.
The following Time related conversions are available in our website:
### How to convert decade to century : Detailed Description
decade (dec) and century (cent) are both units of Time. On this page, we provide a handy tool for converting between dec and cent. To perform the conversion from dec to cent, follow these two simple steps:
Steps to solve
Have you ever needed to or wanted to convert decade to century for anything? It's not hard at all:
Step 1
• Find out how many century are in one decade. The conversion factor is 0.1 cent per dec.
Step 2
• Let's illustrate with an example. If you want to convert 10 decade to century, follow this formula: 10 dec x 0.1 cent per dec = cent. So, 10 dec is equal to cent.
• To convert any dec measurement to cent, use this formula: dec = cent x 0.1. The Time in decade is equal to the century multiplied by 0.1. With these simple steps, you can easily and accurately convert Time measurements between dec and cent using our tool at Pyron Converter.
FAQ regarding the conversion between dec and cent
Question: How many century are there in 1 decade ?
Answer: There are 0.1 century in 1 decade. To convert from dec to cent, multiply your figure by 0.1 (or divide by 10.0).
Question: How many decade are there in 1 cent ?
Answer: There are 10.0 decade in 1 century. To convert from cent to dec, multiply your figure by 10.0 (or divide by 0.1).
Question: What is 1 dec equal to in cent ?
Question: What is the difference between dec and cent ?
Answer: 1 dec is equal to 0.1 in cent. That means that cent is more than a 10.0 times bigger unit of Time than dec. To calculate dec from cent, you only need to divide the cent Time value by 0.1.
Question: What does 5 dec mean ?
Answer: As one dec (decade) equals 0.1 cent, therefore, 5 dec means cent of Time.
Question: How do you convert the dec to cent ?
Answer: If we multiply the dec value by 0.1, we will get the cent amount i.e; 1 dec = 0.1 cent.
Question: How much cent is the dec ?
Question: Are dec and cent the same ?
Answer: No. The cent is a bigger unit. The cent unit is 10.0 times bigger than the dec unit.
Question: How many dec is one cent ?
Answer: One cent equals 10.0 dec i.e. 1 cent = 10.0 dec.
Question: How do you convert cent to dec ?
Answer: If we multiply the cent value by 10.0, we will get the dec amount i.e; 1 cent = 10.0 decade.
Question: What is the cent value of one decade ?
#### Common decade to century conversion
dec cent Description
0.1 dec 0.01 cent 0.1 dec to cent = 0.01
0.2 dec 0.02 cent 0.2 dec to cent = 0.02
0.3 dec 0.03 cent 0.3 dec to cent = 0.03
0.4 dec 0.04 cent 0.4 dec to cent = 0.04
0.5 dec 0.05 cent 0.5 dec to cent = 0.05
0.6 dec 0.06 cent 0.6 dec to cent = 0.06
0.7 dec 0.07 cent 0.7 dec to cent = 0.07
0.8 dec 0.08 cent 0.8 dec to cent = 0.08
0.9 dec 0.09 cent 0.9 dec to cent = 0.09
1 dec 0.1 cent 1 dec to cent = 0.1
2 dec 0.2 cent 2 dec to cent = 0.2
3 dec 0.3 cent 3 dec to cent = 0.3
4 dec 0.4 cent 4 dec to cent = 0.4
5 dec 0.5 cent 5 dec to cent = 0.5
6 dec 0.6 cent 6 dec to cent = 0.6
7 dec 0.7 cent 7 dec to cent = 0.7
8 dec 0.8 cent 8 dec to cent = 0.8
9 dec 0.9 cent 9 dec to cent = 0.9
10 dec 1.0 cent 10 dec to cent = 1.0
20 dec 2.0 cent 20 dec to cent = 2.0
30 dec 3.0 cent 30 dec to cent = 3.0
40 dec 4.0 cent 40 dec to cent = 4.0
50 dec 5.0 cent 50 dec to cent = 5.0
60 dec 6.0 cent 60 dec to cent = 6.0
70 dec 7.0 cent 70 dec to cent = 7.0
80 dec 8.0 cent 80 dec to cent = 8.0
90 dec 9.0 cent 90 dec to cent = 9.0
#### Common century to decade conversion
cent dec Description
0.1 cent 1.0 dec 0.1 cent to dec = 1.0
0.2 cent 2.0 dec 0.2 cent to dec = 2.0
0.3 cent 3.0 dec 0.3 cent to dec = 3.0
0.4 cent 4.0 dec 0.4 cent to dec = 4.0
0.5 cent 5.0 dec 0.5 cent to dec = 5.0
0.6 cent 6.0 dec 0.6 cent to dec = 6.0
0.7 cent 7.0 dec 0.7 cent to dec = 7.0
0.8 cent 8.0 dec 0.8 cent to dec = 8.0
0.9 cent 9.0 dec 0.9 cent to dec = 9.0
1 cent 10.0 dec 1 cent to dec = 10.0
2 cent 20.0 dec 2 cent to dec = 20.0
3 cent 30.0 dec 3 cent to dec = 30.0
4 cent 40.0 dec 4 cent to dec = 40.0
5 cent 50.0 dec 5 cent to dec = 50.0
6 cent 60.0 dec 6 cent to dec = 60.0
7 cent 70.0 dec 7 cent to dec = 70.0
8 cent 80.0 dec 8 cent to dec = 80.0
9 cent 90.0 dec 9 cent to dec = 90.0
10 cent 100.0 dec 10 cent to dec = 100.0
20 cent 200.0 dec 20 cent to dec = 200.0
30 cent 300.0 dec 30 cent to dec = 300.0
40 cent 400.0 dec 40 cent to dec = 400.0
50 cent 500.0 dec 50 cent to dec = 500.0
60 cent 600.0 dec 60 cent to dec = 600.0
70 cent 700.0 dec 70 cent to dec = 700.0
80 cent 800.0 dec 80 cent to dec = 800.0
90 cent 900.0 dec 90 cent to dec = 900.0 | 1,897 | 5,506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-30 | latest | en | 0.889123 |
https://dsp.stackexchange.com/questions/41673/formula-of-frequency-modulated-sine-wave-from-data-points/41704 | 1,566,259,607,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315132.71/warc/CC-MAIN-20190819221806-20190820003806-00302.warc.gz | 421,169,546 | 30,509 | # Formula of frequency modulated sine wave from data points
I'm trying to determine the formula of a variable frequency sine wave based on the values of max1, max2, min1 and min2. y(0) isn't necessarily 0 and I can assume the frequency increases or decreases linearly (I only want to plot the first couple oscillations).
I've figured the formula for the non-modulated signal (based only on max1 and min1) as the following:
$y(x) = sin(2π*f*x + C)$ where
$f=\frac{1}{2\left(\max _1-\min _1\right)}$ and $C = -\left(max_1-\frac{\left(max_1-min_1\right)}{2}\right)\cdot \frac{\pi }{\left(max_1-min_1\right)}$
I've also figured that plotting:
$y(x) = sin(2π*(f+M(x))*x)$ where
$M(x)= a*x + b$
results in an increasing frequency signal but I can't figure out the exact formula to have the wave match the data points.
Any help would be much appreciated, my math skills are rusty...
• Is it safe to assume it's actually linear Fm ? – Stanley Pawlukiewicz Jun 12 '17 at 17:10
If your function is $y(x)=\sin(2\pi fx+C)$, and you have data points $\{x_i,y_i\}$, then finding $C$ is easy, if you suppose $x=0$ then $C=\arcsin(y(0))$.
Now, according to your plot your frequency varies with $x$. The first thing you need to do is to know how does your frequency vary with $x$. Supposedly this variation is linear, then you write: $f(x)=ax+b$ and your function model becomes $y(x)=\sin(2\pi f(x)x+C)$.
The problem now is finding (an estimate for) $f(x)$. You have two unknowns to determine $a$ and $b$, you need two data points ($\{x_1,y_1\}, \{x_2,y_2\}$) and you solve a system of two (nonlinear) equations with two unknowns.
$$\begin{cases} y_1=\sin(2\pi (ax_1+b) x_1+C)\\ y_2=\sin(2\pi (ax_2+b) x_2+C) \end{cases}$$
Since frequency is the derivative of phase, a linear frequency curve implies a quadratic phase curve. So we have $y(x)=\cos(\phi(x))$ with $\phi(x)=ax^2+bx+c$ and \begin{aligned}\phi(max_1)&=0\\\phi(min_1)&=\pi\\\phi(max_2)&=2\pi\\\phi(min_2)&=3\pi.\end{aligned} This is actually a simple system of linear equations, but it is over-determined. If we omit $min_2$ and ask Maxima (open-source CAS) to solve it, we get \begin{aligned}a=&\frac{\pi max_2+\pi max_1−2 \pi min_1}{d}\\b=& \frac{2 \pi min_1^2-\pi max_2^2-\pi max_1^2}{d}\\c=&\frac{max_1 (\pi max_2^2−2 \pi min_1^2)+max_1^2 (2 \pi min_1−\pi max_2)}{d}\end{aligned} with $$d=max_2 min_1^2+max_1 (max_2^2−min_1^2)+max_1^2 (min_1−max_2)−max_2^2 min_1.$$ | 839 | 2,422 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2019-35 | longest | en | 0.878373 |
https://www.clutchprep.com/physics/practice-problems/144413/two-blocks-are-at-rest-on-a-frictionless-incline-as-shown-in-figure-1-what-is-th | 1,632,134,360,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057036.89/warc/CC-MAIN-20210920101029-20210920131029-00487.warc.gz | 736,218,254 | 31,203 | Systems of Objects on Inclined Planes with Friction Video Lessons
Concept
# Problem: Two blocks are at rest on a frictionless incline, as shown in (Figure 1).What is the tension in the string number 1 if m1 = 9.0 kg and m2 = 3.0 kg ?What is the tension in the string number 2 if m1 = 9.0 kg and m2 = 3.0 kg ?
###### FREE Expert Solution
Newton's second law:
$\overline{){\mathbf{\Sigma }}{\mathbf{F}}{\mathbf{=}}{\mathbf{m}}{\mathbf{a}}}$
(a)
Take: y-axis points to the up and x-axis points to the left.
We'll apply Newton's second law to the forces parallel to the incline:
ΣF = T1 - m1gcosθ = 0
84% (340 ratings)
###### Problem Details
Two blocks are at rest on a frictionless incline, as shown in (Figure 1).
What is the tension in the string number 1 if m1 = 9.0 kg and m2 = 3.0 kg ?
What is the tension in the string number 2 if m1 = 9.0 kg and m2 = 3.0 kg ? | 283 | 876 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2021-39 | latest | en | 0.677985 |
https://math.stackexchange.com/questions/2395470/why-will-an-implication-be-true-when-the-hypothesis-is-false | 1,563,416,170,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525483.62/warc/CC-MAIN-20190718001934-20190718023934-00507.warc.gz | 432,502,630 | 33,713 | # Why will an implication be true when the hypothesis is false? [duplicate]
When $p$ and $q$ are propositions such that "if $p$ then $q$" (which is an implication) is false only in the condition when $p$ (which is hypothesis) is true but $q$ (which is conclusion) is false.
But I couldn't understand why will the implication be true when the hypothesis itself is false. Shouldn't the implication be false when the initial assumption or hypothesis is false ?
## marked as duplicate by Ennar, 5xum, Graham Kemp, Sharkos, Hans LundmarkAug 16 '17 at 12:20
• You are wrong. The implication is false only when $p$ is true and $q$ is fase. – José Carlos Santos Aug 16 '17 at 11:41
• Sorry for the typo. I was trying to say why would implication be true when hypothesis is false or in short why would implication be false only when p is true but q is falsw – Aashish Loknath Panigrahi Aug 16 '17 at 11:52
• @Piquito I think the proposition "All triangles are Rectangles" is false because no triangle can be a rectangle and no rectangle can be a triangle. – Aashish Loknath Panigrahi Aug 16 '17 at 11:57
• What about ALL TRIANGLES ARE RECTANGULAR $\Rightarrow$ SOME TRIANGLES ARE RECTANGULAR? – Piquito Aug 16 '17 at 12:03
• The closure of this question happened when I was typing an answer... Shoot :). Note that, if you let "$A \Rightarrow B$ holds" be defined by "$\overline{A}$ or $B$ holds", then from the definition of "or" you can answer your question directly. Intuitively speaking, you may view the rule of inference this way. If your teacher promise you (weirdly) that "if you get an A+ this time, then you can graduate directly from then on under my permission", then when will he break the promise? When and only when you get an A+ that time and he does not make your graduation happen! – Megadeth Aug 16 '17 at 12:24
This has been answered many many times on this site, but here is a simple reason. If one would use any other truth table for $P\to Q$ it would no longer mean what it is supposed to. If the hypothesis $P$ is true, then everyone agrees that $P\to Q$ should hold if and only if $Q$ does. Now suppose the hypothesis $P$ is false. If one would define (as you suggest) $P\to Q$ to be false in this case, regardless of $Q$, then $P\to Q$ would be true only if both $P$ and $Q$ are true, making it a symmetric operation for which we already have a name (namely "and", written $P\land Q$). If instead when $P$ is false we would define $P\to Q$ to be true only if $Q$ is, then the meaning of $P\to Q$ would ignore the status of the hypothesis $P$ completely, and be equivalent to $Q$. Finally if when $P$ is false we would define $P\to Q$ to be true only if $Q$ is false, then $P\to Q$ would be true if any only if $P$ and $Q$ have the same truth value, again a symmetric relation that does not reflect what we mean by implication. So the only reasonable option is the define, when the hypothesis is false, the relation $P\to Q$ to be true regardless of $Q$. | 801 | 2,974 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2019-30 | latest | en | 0.913423 |
http://mathhelpforum.com/algebra/104563-changing-subject-formua-squares-roots-print.html | 1,524,479,039,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945940.14/warc/CC-MAIN-20180423085920-20180423105920-00399.warc.gz | 204,298,100 | 2,821 | # Changing the subject of formua with squares and roots
• Sep 27th 2009, 07:10 AM
Meggomumsie
Changing the subject of formua with squares and roots
I have tried answering the following question by squaring everything (I think), but I am not right.
c = $\displaystyle \sqrt{a^2+b^2}$ (Change the subject of
the formula to a)
My try:
$\displaystyle c^2$ = $\displaystyle (a^2)^2+(b^2)^2$
$\displaystyle c^2$ = $\displaystyle a^4+b^4$
$\displaystyle a^4$ = $\displaystyle c^2-b^4$
a = $\displaystyle sqrt[4]{c^2-b^4}$
Can you see my error please?
• Sep 27th 2009, 07:17 AM
Quote:
Originally Posted by Meggomumsie
I have tried answering the following question by squaring everything (I think), but I am not right.
c = $\displaystyle \sqrt{a^2+b^2}$ (Change the subject of
the formula to a)
My try:
$\displaystyle c^2$ = $\displaystyle (a^2)^2+(b^2)^2$
$\displaystyle c^2$ = $\displaystyle a^4+b^4$
$\displaystyle a^4$ = $\displaystyle c^2-b^4$
a = $\displaystyle sqrt[4]{c^2-b^4}$
Can you see my error please?
HI
you are correct by squarring both sides to get rid of the square root but you worked it out wrongly .
$\displaystyle (c)^2=(\sqrt{a^2+b^2})^2$
$\displaystyle c^2=a^2+b^2$
$\displaystyle a^2=c^2-b^2$
$\displaystyle a=\sqrt{c^2-b^2}$
• Sep 27th 2009, 07:34 AM
Meggomumsie
Dohh!!! (Worried) Yes, I see that now!!! Thanks. | 463 | 1,349 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2018-17 | latest | en | 0.819262 |
https://brainly.in/question/22089 | 1,485,104,173,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281450.93/warc/CC-MAIN-20170116095121-00375-ip-10-171-10-70.ec2.internal.warc.gz | 783,269,542 | 10,445 | # There were some marbles at a shop. The ratio of the number of red marbles to the number of blue marbles was 2:3. When 50 more red marbles and 30 more blue marbles were added, the ratio of the number of red marbles to the number of blue marbles became 5:6. How many marbles were there at first?
2
2014-06-28T11:50:36+05:30
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Let the no of red marbles be 2x and no of blue marbles be 3x
given that (2x + 50) : (3x + 30)= 5:6
solving we get x = 14
so no of red marbles are 2 × 14 = 28 and no of blue marbles are 3 × 14 = 42 | 249 | 832 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2017-04 | latest | en | 0.958002 |
https://socratic.org/questions/how-do-you-simplify-tan-arccos-x-3 | 1,582,150,376,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144429.5/warc/CC-MAIN-20200219214816-20200220004816-00022.warc.gz | 577,725,467 | 6,047 | # How do you simplify tan(arccos(x/3))?
Aug 2, 2016
$\frac{\sqrt{9 - {x}^{2}}}{x} , - 3 \le x \le 3$.
#### Explanation:
Let $a = \arccos \left(\frac{x}{3}\right) \in Q 1 \mathmr{and} Q 2 , - 1 \le \frac{x}{3} \le 1$, and so, $- 3 \le x \le 3$
Then, $\cos a = \frac{x}{3} , \sin a = \sqrt{1 - {\cos}^{2} a} = \sqrt{1 - {x}^{2} / 9}$,
Now, the given expression is
$\tan a = \sin \frac{a}{\cos} a$
$= \frac{\sqrt{9 - {x}^{2}}}{x} ,$ | 208 | 437 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2020-10 | longest | en | 0.341817 |
https://www.physicsforums.com/threads/a-proof.319552/ | 1,531,768,036,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589417.43/warc/CC-MAIN-20180716174032-20180716194032-00402.warc.gz | 963,129,264 | 13,803 | # Homework Help: A proof
1. Jun 12, 2009
### evilpostingmong
1. The problem statement, all variables and given/known data
Let T: V---->V be a linear operator where dim V=n. Show that V
has a basis of eigenvectors if and only if V has a basis B such that
TB is diagonal.
2. Relevant equations
3. The attempt at a solution
Let T=[a1,1......an,1] ai,j=/=0
[a1,n......an,n]
Let TB=[a1,1v1......0n,1]
[01,n.......an,nvn]
Since this is diagonal, and ai,j=/=0,
then we have a basis of eigenvectors (these are meant to be vertical)<[v1,...0]......[0.....vn]>
that, after being mulitplied by T, formed the matrix TB.
To show that they are eigenvectors, a possible linear combination is
[v1,...0] and when multiplied by T gives a1,1[v1,...0]+...+a1,n[v1,....0]
=(a1,1+...+a1,n)[v1,....0]. Since [v1,....0] is an eigenvector, and
the others follow the same logic, mulitplying the basis of eigenvectors by T
should produce a diagonal matrix, as shown.
Last edited: Jun 12, 2009
2. Jun 12, 2009
### jbunniii
Suppose V has a basis $$B=\{b_1,\ldots,b_n\}$$ consisting of eigenvectors of T. Then for each j, $$T b_j = \lambda_j b_j$$. If I express this equation in terms of the basis B, then
$$[T]_B [b_j]_B = \lambda_j [b_j]_B$$
But $$[b_j]_B$$ is simply a column vector containing all zeros except for a 1 in the j'th position. Carrying out the matrix multiplication then shows that
$$[T]_B = \Lambda$$
where $$\Lambda$$ is the diagonal matrix whose diagonal entries are $$\{\lambda_1,\ldots,\lambda_n\}$$. This gives you one direction of the proof. For the other direction, you can essentially reverse the steps above to verify that the $$\lambda_j$$'s must be eigenvalues and the $$b_j$$'s must be eigenvectors.
3. Jun 12, 2009
### evilpostingmong
Yeah, I did the other direction by assuming that TB is diagonal then showing that
the basis consists of eigenvectors and that TB consists of eigenvalues, though I
think I messed up when I mentioned a1,1+...+a1,n is an eigenvalue since I multiplied wrong.
It should be a1,1 for the eigenvalue of [v1.....0]. Made a mistake with multiplication.
Oh and when I said "then we have a basis of eigenvectors" I shouldn't've called
the basis of eigenvectors a basis of eigenvectors, since I was proving it in the first place,
and I should've called the scalar an eigenvalue. It's these small mistakes that make a big
difference.
Last edited: Jun 12, 2009 | 722 | 2,393 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2018-30 | latest | en | 0.869387 |
https://www.physicsforums.com/threads/continuity-with-piecewise-functions.131667/ | 1,544,766,654,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825363.58/warc/CC-MAIN-20181214044833-20181214070333-00383.warc.gz | 997,409,936 | 15,277 | # Homework Help: Continuity with piecewise functions.
1. Sep 11, 2006
### Jacobpm64
I have two problems.. I'll put them both here, and show my work on both of them.
1. Is this function continuous on [-1, 1]?
f(x) =
x / |x| , x does not equal 0
0 , x = 0
After graphing this function, the first statement gives y = -1 for all x < 0, and y = 1 for all x > 0. The second statement makes y = 0 when x = 0. Putting everything together, y values are defined at -1, 0, 1, but nothing in-between. Does this make the function not continuous because of the jumping at x = 0?
2.
Discuss the continuity of the function g:
g(x) =
sin x / x , for x does not equal 0
1/2 , for x = 0
There is a hole in the graph at x = 0 (for the first statement), but the second statement defines x = 0 to be 1/2, it's just lower than the curve in the first statement. Would this be a continuous function or not?
2. Sep 12, 2006
### steve23063
There are two types of discontinuities but I'll get to that later. A function is discontinuous if you can't draw the entire graph without picking your pencil off the paper. In question 2, you say there's a point just below the rest of the graph. Clearly you can't draw the entire graph unless you pick up your pencil off the paper so it's discontinuous. The same goes for question 1.
The two questions have different types of discontinuities. In question 2, only 1 point is off of the rest of the graph. The point at x=0. In question one, you have an infinite number of points that are off the rest of the graph (in this case they just don't exist). There is an infinite number of points between -1 and 0 and between 0 and 1 (for example .01, .001, .009, etc are all undefined). Question 2 has a finite discontinuity because you can count how many points are off the rest of the graph. Question 1 has inifinite discontinuity.
The "picking up the pencil" explanation is just a simple way to think about it. You should really get used to thinking about it in terms of left and right sided limits but I assume you already learned that stuff.
EDIT: Hey sorry I misread something in question 1. There is only 1 point that is discontinuous so it has a finite discontinuity, not infinite as I first said.
Last edited: Sep 12, 2006
3. Sep 12, 2006
### d_leet
I think most people would consider at least 3 types of discontinuity, the first being a hole where a single point is missing from teh curve, nex being a jump discontinuity where the function seems to jump to a different function at some point (this is not the best way to explain this) and the third being an infinite discontinuity where the function is undefined at a point, typ[ically because of division by zero. Your analysis of question one seems correct, but at question 2 there are not an infinite number of discontinuities but only one at x equals 0, the function is continuous on either side of 0, but has a jump discontinuity at x=0. And the points corresponding to x= .01, .001, .009 all do exist so I'm not sure what you meant by that.
4. Sep 12, 2006
### steve23063
I think you misread what I typed :) . I wrote:
Question 2 has a finite discontinuity because you can count how many points are off the rest of the graph. Question 1 has inifinite discontinuity. Also the x=.01 and stuff was still referring to question 1. Sorry for the messiness. I should use more paragraphs
5. Sep 12, 2006
### d_leet
What I said still stands but I said question one when I meant two and vice versa.
The function defined in question one IS defined for all values of x on [-1,1], but there is a jump discontinuity at x=0, it is not an infinite discontinuity.
6. Sep 12, 2006
### steve23063
Oh, really sorry. The original poster wrote: "y values are defined at -1, 0, 1, but nothing in-between." I wasn't paying attention and just noticed the words "nothing in-between" and ignored the [-1,1] domain info. Yes you're right and I'll edit my first post. It is a finite discontinuity
It's also 2AM. good night
7. Sep 12, 2006
### HallsofIvy
The definition of "continuous at x= a" is
$$\lim_{x\rightarrow a}f(x)= f(a)$$
What are the limits of the functions in 1 and 2 as x goes to 0?
8. Sep 12, 2006
### Jacobpm64
thanks everyone... I was trying to do it without using limits because we haven't gotten there yet... (we're still in the introductory review chapter of the calculus book..).. I've done it with limits before in my high school calculus class, but we haven't gotten to it yet in this college calculus... But, I understand all of the explanations, thanks.
9. Sep 12, 2006
### HallsofIvy
Then what definition of "continuous" were you using? | 1,225 | 4,639 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2018-51 | latest | en | 0.951381 |
https://www.labraven.com/ulaby0114 | 1,669,555,635,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710237.57/warc/CC-MAIN-20221127105736-20221127135736-00435.warc.gz | 909,777,570 | 3,918 | Ulaby problem 1.14
• Ulaby
• Electromagnetics
• FE Exam
• PE Exam
• Wave math
posted on 31 Aug 2016
# Ulaby Problem 1.14
## Problem Statement
A certain electromagnetic wave traveling in seawater was observed to have an amplitude of 98.02 (V/m) at a depth of 10m, and an amplitude of 81.87 (V/m) at a depth of 100 m. What is the attenuaton constant of seawater? 1
### Analysis
When we have a sinusoidal wave in a lossy medium, we need to use and attenuation factor:
$y(x,t) = Ae^{-\alpha x} \cos{(\omega t - \beta x + \phi_0)}$
Where $$-\alpha$$ is the attenuation constant with units of $$1/m$$. The units of nepers per meter, $$Np/m$$, refer to the attenuation constant.
### Solution
Let’s set the two observations equal to each other, and then solve for the attenuation constant.
\begin{aligned} 98.02e^{-\alpha 10} &= 81.87e^{-\alpha 100} \\ \frac{98.02e^{-\alpha 10}}{81.87} &= e^{-\alpha 100} \\ \frac{98.02}{81.87} &= \frac{e^{-\alpha 100}}{e^{-\alpha 10}} \\ \frac{98.02}{81.87} &= e^{(-\alpha 100 - -\alpha 10)} \\ \frac{98.02}{81.87} &= e^{-\alpha (100 - 10)} \\ \ln{\left( \frac{98.02}{81.87} \right)} &= -\alpha 90 \\ \frac{\ln{\left( \frac{98.02}{81.87} \right)}}{-90} &= \alpha \\ \alpha &= \boxed{0.02 \; (Np/m)} \\\end{aligned}
Emacs 24.5.1 (Org mode 8.2.10) | 460 | 1,286 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2022-49 | latest | en | 0.617435 |
https://mmerevise.co.uk/gcse-maths-revision/wjec/population-density/ | 1,723,699,833,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641151918.94/warc/CC-MAIN-20240815044119-20240815074119-00800.warc.gz | 319,012,303 | 51,857 | Population Density
GCSELevel 4-5WJEC
Population Density
Population density tells us how many people are living in a given location. It is often given in people per $\text{km}^2$, which means in an area, approximately how many people live in each $\text{km}^2$
Level 4-5GCSEWJEC
Calculating Population Density
To calculate population density, you need to know the population of an area and the size of an area.
The equation for population density is:
$\text{Population density} = \dfrac{\text{Population}}{\text{Land Area}}$
If somewhere is crowded, it will have a high population density, and if somewhere is sparsely populated, it will have a low population density.
Level 4-5GCSEWJEC
Level 4-5GCSEWJEC
Example 1: Calculating Population Density
Calculate the population density of Boston given that it has a population of $650000$ and an area of $230\text{ km}^2$.
[2 marks]
To calculate this, we will use the equation.
$\text{Population density} = \dfrac{\text{Population}}{\text{Land Area}}$
$\text{Population density} = \dfrac{650000}{230}$
$\text{Population density} = 2826\text{ people}/\text{km}^2$ to the nearest whole number.
Level 4-5GCSEWJEC
Example 2: Using the Equation
San Francisco has a population density of $1350\text{ people}/\text{km}^2$ and a population of $810000$
Work out the area of San Francisco.
[3 marks]
We will use the equation again, but will have to rearrange it:
$\text{Population density} = \dfrac{\text{Population}}{\text{Land Area}}$
$1350 = \dfrac{810000}{\text{Land Area}}$
Rearrange:
$\text{Land Area} = \dfrac{810000}{1350}$
$\text{Land Area} = 600\text{ km}^2$
Level 4-5GCSEWJEC
Population Density Example Questions
$\text{Population Density}=210000\div270\\$ $= 778\text{ people}/\text{km}^2$
Gold Standard Education
Birmingham:
Population density $= 1200000\div270\\$
Population density $=4444\text{ people}/\text{km}^2$
Manchester:
Population density $= 500000\div115\\$
Population density $=4347\text{ people}/\text{km}^2$
Therefore, Birmingham has a higher population density.
Gold Standard Education
Put the figures we know into the equation:
$4546=\dfrac{\text{Population size}}{110}$
Rearrange:
$\text{Population size}=4546\times110$
$\text{Population size}=500060$ | 646 | 2,259 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 31, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2024-33 | latest | en | 0.780774 |
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People read all caps in e-mails as: yelling.
Question
Updated 10/9/2017 5:22:45 AM
f
Rating
8
People read all caps in e-mails as: yelling.
Confirmed by Janet17 [10/9/2017 5:44:50 AM]
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What type of interview would be most likely for the following scenario? interviewing contestants for a new reality show
Weegy: Information-gathering is the type of interview most likely for the scenario - interviewing contestants for a new reality show. (More)
Question
Updated 264 days ago|6/2/2023 5:17:07 PM
If a = -12/5, then 5x + 2y = 6 and 3x - ay = 4 are parallel. True
Weegy: the answer is B.true User: The y-intercept of the line whose equation is 3x + 2y = 7 (More)
Question
Updated 4/23/2014 8:38:33 PM
If a = -12/5, then 5x + 2y = 6 and 3x - ay = 4 are parallel. FALSE.
5x + 2y = 6;
2y = -5x + 6
y = -5/2x +3 the slope of which is -5/2
3x - ay = 4 when a = -12/5,
3x + 12/5y = 4
12/5y = -3x + 4
y = -5/4x + 5/3 the slope of which is -5/4
the slope of the above two lines are not the same so they are not parallel.
Confirmed by jeifunk [4/23/2014 8:46:37 PM]
Input in standard form the equation of the given line. The line with m = -3/2 and b = 6
Updated 4/23/2014 7:17:33 PM
The equation for the line with m = -3/2 and b = 6 is y = -3/2x + 6, rewritten in standard form it is 3x + 2y = 12
Which of the following materials is the hardest? talc marble granite diamond
Updated 4/23/2014 6:57:57 PM
The following material that is the hardest is Diamond.
Confirmed by alfred123 [4/23/2014 7:03:48 PM]
The y-intercept of the line whose equation is 3x + 2y = 7
Weegy: 7/3 is the y-intercept of the line whose equation is 3x + 2y = 7. User: Ben has \$3.40 consisting of quarters and dimes. How many coins of each kind does he have if he has 22 coins? Which of the following system of equations represents the word problem if d is the number of dimes and q is quarters? d + q = 22 and 10d + 25q = 340 10d + 25q = 22 and d + q = 340 dq = 22 and d + q = 340 Weegy: d + q = 22 and 10d + 25q = 340- is the system of equations that represents the word problem if d is the number of dimes and q is quarters. (More)
Question
Updated 4/23/2014 8:33:00 PM
The y-intercept of the line whose equation is 3x + 2y = 7 is 7/2.
Y intercept is the value when x = 0
3(0) + 2y = 7
2y = 7
y = 7/2
Confirmed by jeifunk [4/23/2014 8:37:47 PM]
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 1,350 | 3,999 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-10 | latest | en | 0.923196 |
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posted by .
5. 4x + 2 = x + 8. (1 point) x = 4 x = 3 x = 2 x = 1 Solve the inequality. 6. q – 12 ≥ –13 (1 point) q ≥ 1 q ≥ –1 q ≥ 25 q ≥ –25 7. 12p < 96 (1 point) p < 8 p < 108 p < 84 p < –8 8. < –18 (1 point) g < –23 g > –90 g > 90 g < 90 9. Write the inequality and solve for the following problem: The result of 6 subtracted from a number n is at least 2. (1 point) n – 2 > 6; n > 8 n – 6 ≥ 2; n ≥ 8 n + 6 ≥ 2; n ≤ 4 n + 6 ≥ 2; n ≥ 4
## Similar Questions
1. ### Mathforme
5. 4x + 2 = x + 8. (1 point) x = 4 x = 3 x = 2 x = 1 Solve the inequality. 6. q – 12 ≥ –13 (1 point) q ≥ 1 q ≥ –1 q ≥ 25 q ≥ –25 7. 12p < 96 (1 point) p < 8 p < 108 p < 84 p < –8 …
1. m – 7 < 6 (1 point) m < –1 m > 1 m < 13 m < –13 2. x + 4.5 ≥ 5.5 (1 point) x ≥ 10 x ≥ 1 x ≥ 0.1 x ≤ 1 3. p + 12 > 9 (1 point) p > 21 p > 3 p > –21 p > –3 4. Translate …
1. 5h – 9 = –16 + 6h (1 point) 4 –7 7 10 2. 4x + 4 = 9x – 36 (1 point) –8 –7 8 –3 3. Which of the following equations has an infinite number of solutions?
4. ### 7th grade math help Ms. Sue
3. Which of the following equations has an infinite number of solutions?
5. ### 7th grade math Ms. Sue
1. –9p – 17 = 10 (1 point) –3 16 18 –16 2. w over four – 4 = 3 (1 point) –4 28 3 11 3. d over three+ 10 = 7 (1 point) 51 20 0 –9 4. –2(m – 30) = –6m (1 point) –15 –13 –8 8 5. 2.9n + 1.7 = 3.5 + 2.3n (1 point) …
6. ### Math
1. 5h – 9 = –16 + 6h (1 point) 4 –7 7 10 2. 4x + 4 = 9x – 36 (1 point) –8 –7 8 –3 3. Which of the following equations has an infinite number of solutions?
7. ### Math!
Please Please Please help me! I'm really behind and need to get this done and i'm really confused. 1. 5h – 9 = –16 + 6h (1 point) 4 –7 7 10 2. 4x + 4 = 9x – 36 (1 point) –8 –7 8 –3 3. Which of the following equations …
8. ### PRE-ALGEBRA
1. 5h – 9 = –16 + 6h (1 point) 4 –7 7 10 2. 4x + 4 = 9x – 36 (1 point) –8 –7 8 –3 3. Which of the following equations has an infinite number of solutions? | 926 | 1,941 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2018-05 | latest | en | 0.807487 |
https://calculatoruniverse.com/displacement-calculator/ | 1,723,158,704,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640741453.47/warc/CC-MAIN-20240808222119-20240809012119-00276.warc.gz | 118,791,380 | 52,881 | # Displacement Calculator
Instructions:
• Enter the initial velocity (u), acceleration (a), and time (t).
• Click "Calculate" to find the displacement (s).
• Click "Clear" to reset the form.
• Click "Copy Result" to copy the displacement to the clipboard.
Calculation History
## What is Displacement?
Displacement refers to the change in the position or location of an object or a point in space, measured from a reference point or starting position. It is a vector quantity, which means it has both magnitude (the distance covered) and direction (the path taken). Displacement is used in physics and engineering to describe the change in an object’s position over time.
1. Vector Quantity: Displacement is a vector because it has both magnitude and direction. This means that not only the distance matters but also the specific path or route taken to get from one point to another.
2. Reference Point: Displacement is measured relative to a reference point or a starting position. It tells you how far and in which direction an object has moved from this reference point.
3. SI Unit: The SI (International System of Units) unit for displacement is the meter (m) since it represents a change in position in three-dimensional space. In one-dimensional motion (e.g., along a straight line), displacement can be measured in meters as well.
4. Positive and Negative Displacement: Displacement can be positive, negative, or zero, depending on the direction of motion. Positive displacement indicates movement in one direction, while negative displacement implies movement in the opposite direction. Zero displacement means no net change in position.
## All Formulae Related to Displacement
Displacement is a vector quantity that represents the change in the position of an object or point in space. Here are some key formulae related to displacement:
1. Displacement Formula (Vector Form):
• Displacement (Δr or 𝐝) = Final Position (𝐫_f) – Initial Position (𝐫_i)
• Δr = 𝐫_f – 𝐫_i
• This formula gives the vector displacement, which has both magnitude and direction.
2. Displacement Magnitude:
• The magnitude of displacement (|Δr| or |𝐝|) is the absolute value of the vector and represents the shortest distance between the initial and final positions.
• |Δr| = |𝐫_f – 𝐫_i|
3. Displacement in One Dimension:
• In one-dimensional motion along a straight line, the displacement (Δx) is given by:
• Δx = x_f – x_i
• Where Δx is the displacement, x_f is the final position, and x_i is the initial position.
4. Average Displacement (Vector Form):
• The average displacement (Δr_avg) over a period of time (Δt) is calculated as:
• Δr_avg = (𝐫_f – 𝐫_i) / Δt
5. Instantaneous Displacement (Vector Form):
• The instantaneous displacement at a specific time (t) is given by:
• Δr_instantaneous = 𝐫(t) – 𝐫_i
• Where 𝐫(t) represents the position of the object at time t.
6. Average Speed (Scalar Quantity):
• Average speed (v_avg) is defined as the total distance traveled (d_total) divided by the total time (Δt_total) taken:
• v_avg = d_total / Δt_total
7. Average Velocity (Vector Quantity):
• Average velocity (𝐯_avg) is calculated as the total displacement (Δr) divided by the total time (Δt) taken:
• 𝐯_avg = Δr / Δt
8. Instantaneous Velocity (Vector Quantity):
• The instantaneous velocity at a specific time (t) is given by the derivative of the position vector 𝐫(t) with respect to time:
• 𝐯_instantaneous = d𝐫/dt
## Applications of Displacement Calculator in Various Fields
A displacement calculator, which helps determine changes in position or location, has numerous applications in various fields. This tool is particularly useful for measuring distances and movements accurately. Here are some common applications of a displacement calculator across different domains:
1. Physics and Engineering:
• In physics, displacement calculators are used to analyze the motion of objects, including calculating distances traveled and velocity.
• Engineers use displacement calculations to design and optimize structures, machines, and systems.
• GPS devices and navigation systems rely on displacement calculations to determine precise locations and provide directions.
• In geodesy, displacement calculators help track tectonic plate movement and monitor land deformations.
3. Surveying and Cartography:
• Land surveyors use displacement calculators to measure distances and angles for mapping, land development, and construction projects.
• Cartographers use displacement calculations to create accurate maps and charts.
4. Astronomy and Space Exploration:
• Astronomers use displacement calculations to track the positions and trajectories of celestial objects, planets, and spacecraft.
• Space agencies employ displacement calculators to plot spacecraft routes and mission planning.
5. Transportation and Logistics:
• Displacement calculations are essential for route planning, optimizing fuel consumption, and managing vehicle fleets in transportation and logistics industries.
## Benefits of Using the Displacement Calculator
Using a displacement calculator offers several benefits across various fields and applications, thanks to its ability to accurately measure changes in position or location. Here are the key advantages of using a displacement calculator:
1. Accuracy: Displacement calculators provide precise measurements, reducing the risk of human error associated with manual distance and position calculations.
2. Efficiency: They offer quick and efficient calculations, saving time compared to manual methods, especially for complex or repetitive measurements.
3. Consistency: Displacement calculators ensure consistent and standardized measurements, leading to reliable and repeatable results.
4. Precision: Users can calculate distances and movements with high precision, making them suitable for tasks that require exact measurements.
5. Data Analysis: Displacement calculations help researchers, scientists, and engineers analyze and interpret motion-related data for various purposes.
## References
1. “Beyond Forced Migration: Internal Displacement in Conflict and Environmental Disasters” by Journal of Peace Research
2. “From Urban Planning to Climate Change: Displacement in Spatial Transformations and Environmental Justice” by Urban Studies
Last Updated : 31 July, 2024
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### 18 thoughts on “Displacement Calculator”
1. Mitchell Eileen
This article provides a comprehensive explanation of the concept of displacement, making it accessible to readers from different domains.
2. The insightful breakdown of displacement and its applications makes this article a valuable resource for both learners and practitioners in relevant fields.
3. This article effectively breaks down the concept of displacement, heralding it as a fundamental parameter for various calculations and analyses.
4. The comprehensive coverage of displacement and its calculation methods makes this article a valuable resource for those seeking an in-depth understanding.
5. The content effectively conveys the core concepts of displacement and its significance across various scientific and technical domains.
6. The article’s clear explanation and formulae make it a valuable resource for students and professionals in fields like physics and engineering.
7. This content appears to be well-organized and thoroughly researched; a commendable piece of work indeed.
8. The article effectively conveys the significance of displacement and the benefits of using a displacement calculator. It’s both informative and insightful.
9. Absolutely, the practical examples and real-world scenarios enhance the article’s educational value.
10. While the article provides detailed explanations, it may come across as overwhelming for readers with limited prior knowledge in physics or engineering.
11. I see your point. It could be beneficial to include more simplified examples to cater to a wider audience.
12. I agree. The detailed breakdown of formulae and applications adds depth to the discussion.
13. Agreed, the real-world applications highlighted here make it an excellent reference for practical use.
14. I couldn’t agree more. The applications section really highlights the importance of this concept in various industries.
15. Yes, the emphasis on its applications across different fields reinforces its relevance and significance.
16. The article’s emphasis on the practical applications of displacement is particularly enlightening, shedding light on its real-world relevance.
17. Absolutely, it’s a great starting point for anyone wanting to understand displacement calculations. | 1,770 | 8,816 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2024-33 | latest | en | 0.887847 |
https://planetcalc.com/527/?license=1 | 1,643,295,196,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305266.34/warc/CC-MAIN-20220127133107-20220127163107-00074.warc.gz | 512,010,673 | 13,618 | # Pearson correlation coefficient
The calculation of the correlation coefficient of two random variables
### This page exists due to the efforts of the following people:
#### Maxim Tolstov
Created: 2015-07-27 13:31:19, Last updated: 2021-03-05 16:37:13
This content is licensed under Creative Commons Attribution/Share-Alike License 3.0 (Unported). That means you may freely redistribute or modify this content under the same license conditions and must attribute the original author by placing a hyperlink from your site to this work https://planetcalc.com/527/. Also, please do not modify any references to the original work (if any) contained in this content.
Here are a couple of definitions, if somebody forgot them
Almost all the definitions could be found at Wikipedia.
Correlation in mathematical statistics is a probability and statistical dependence without any strict functional nature. In contrast to functional dependence, the correlation dependence occurs when one of the attributes depends not only on the given second attribute but also on many random factors, or when among the conditions on which the attributes are depending, there are common conditions for both of them.
The mathematical measure of the correlation of two random variables is correlation coefficient.
Some types of correlation coefficients may be positive or negative (there is also the possibility of the lack of statistical relationship - for example, for independent random variables). If it is assumed that the precedence relation is defined on the variables' values, then the negative correlation - the correlation, where the increase of one variable is associated with the decrease of the other variable, though the correlation coefficient can be negative.
The positive correlation in such conditions is a correlation, where an increase of one variable is associated with an increase of another variable, and the correlation coefficient can be positive.
If the value modulus is closer to 1, there is strong coupling, and if closer to 0 - the coupling is weak or nonexistent. When the correlation coefficient is equal to 1 by the value modulus, people suggest a mathematical function can describe a functional relationship, i.e., the changes of two quantities.
Pearson correlation coefficient is most commonly known (Karl Pearson, English mathematician, 1857-1936), characterizing the degree of linear dependence between the variables. It is defined as
$R_{X,Y}=\frac{M[XY]-M[X]M[Y]}{\sqrt{(M[X^2]-(M[X])^2)}\sqrt{(M[Y^2]-(M[Y])^2)}}$
Where M - the mathematical expectation.
There is nothing left to say - enter the random variables in the chart (you can delete the default numbers), and the calculator will define the correlation coefficient by Pearson's formula.
#### Changes of random variables
Items per page:
Digits after the decimal point: 4
Pearson correlation coefficient
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PLANETCALC, Pearson correlation coefficient | 598 | 2,957 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 1, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2022-05 | longest | en | 0.884364 |
https://www.coursehero.com/file/5830094/364-pdfsam-math-54-differential-equation-solutions-odd/ | 1,521,867,488,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257649683.30/warc/CC-MAIN-20180324034649-20180324054649-00440.warc.gz | 764,034,358 | 80,127 | {[ promptMessage ]}
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364_pdfsam_math 54 differential equation solutions odd
# 364_pdfsam_math 54 differential equation solutions odd -...
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Chapter 6 (c) As we have mentioned in (a), the first equation in (6.7) implies that y = ( x + 5 x ) / 2. Substituting the solution x ( t ) yields y ( t ) = 1 2 c 1 cos t + c 2 sin t + c 3 cos 6 t + c 4 sin 6 t +5 c 1 cos t + c 2 sin t + c 3 cos 6 t + c 4 sin 6 t = 1 2 c 1 cos t c 2 sin t 6 c 3 cos 6 t 6 c 4 sin 6 t +5 c 1 cos t + c 2 sin t + c 3 cos 6 t + c 4 sin 6 t = 2 c 1 cos t + 2 c 2 sin t c 3 2 cos 6 t c 4 2 sin 6 t . (d) Initial conditions x (0) = y (0) = 1 and x (0) = y (0) = 0 imply the system of linear equations for c 1 , c 2 , c 3 , and c 4 . Namely, x (0) = c 1 + c 3 = 1 , y (0) = 2 c 1 ( c 3 / 2) = 1 , x (0) = c 2 + c 4 6 = 0 , y (0) = 2 c 2 ( c 4 6 / 2) = 0 c 1 = 3 / 5 , c 3 = 2 / 5 , c 2 = 0 , c 4 = 0 . Thus, the solution to this initial value problem is x ( t ) = 3 5 cos t + 2 5 cos 6 t , y ( t ) = 6 5 cos t 1 5 cos 6 t . 35.
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Ask a homework question - tutors are online | 525 | 1,233 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-13 | latest | en | 0.526893 |
https://yourgametips.com/card-games/what-is-the-highest-number-on-a-playing-card/ | 1,721,368,011,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.83/warc/CC-MAIN-20240719043706-20240719073706-00471.warc.gz | 925,260,133 | 39,558 | # What is the highest number on a playing card?
## What is the highest number on a playing card?
Ranks. Ranks are indicated by numerals from 1 to 10 on spot cards. In addition, three court cards designated jack (formerly knave), queen, and king are notionally equivalent to 11, 12, and 13, respectively, though actually marked J, Q, and K.
What is the number on a playing card called?
There are 52 basic cards in a deck (not including the jokers). These cards have a suit and a number (called the value or rank).
How many number cards are in a deck?
The 52 cards are further divided into four suits, namely, spades, clubs, hearts, and diamonds. Each suit contains a total of 13 cards each. In each of the suits, there are nine number cards from two to ten, a King, a Queen, an Ace, and a Jack.
### Is there a 1 in a deck of cards?
In each suit there are 13 cards including a 2, 3, 4, 5, 6, 7, 8, 9, 10, a jack, a queen, a king and an ace. (Note that there is no 1.) A jack, a queen and a king are called picture cards.
Why is there no one in a deck of cards?
Because one is called the Ace. According to Wikipedia, the word Ace is derived from the Old French word “as” meaning unit. Why an ace instead of one?
How many red face cards are in a deck?
There are 12 face cards in the deck. 6 of the face cards are red (diamonds or hearts) and 6 of the face cards are black (spaces or clubs). let A be the event that you have a red card. Since there are 26 red cards in the deck, then p(A) = 26/52.
## Is Ace a number card in probability?
The card in each suit, are ace, king, queen, jack or knaves, 10, 9, 8, 7, 6, 5, 4, 3 and 2. King, Queen and Jack (or Knaves) are face cards. So, there are 12 face cards in the deck of 52 playing cards.
Is Ace a number card or a face card?
In playing cards the term face card is generally used to describe a card that depicts a person so King ,Queen and Jack are known as the face cards. Ace is not considered as the face card.
Are Aces face cards?
No, Aces are not face cards. Only J,Q,K are face cards.
### What order do face cards go in?
Give a shuffled deck to the group and ask them to arrange the cards in the order of Ace, 2, 3, 4, 5 6,7, 8, 9, 10, Jack, Queen, and King along the Clubs, Hearts, Spades, and Diamonds.
What is the probability of choosing a face card from a deck?
What is the probability of getting not a face card?
6. A card that is not a face card is drawn. We know from question (1) that P(face card) = 3/13. Because the event “non-face card” is the complement to “face card,” P(non-face card) = 1 – P(face card) = 1 – 3/13 = 10/13.
## What is the probability of getting a face card or a heart?
Heart cards are 13 and face card are =12–3=9 as 3 face cards are of hearts. So probability of heart or face card is (13+9)/Nov 2017
What is the probability of getting a face card or a spade?
→ There are 13 spade cards and 12 face cards in a standard deck of 52 cards. BUT, 3 of those face cards are also spades, so what you have are 22 candidates for the draw, not 25. The odds are exactly 100% or zero, depending on what I want to happen.
What’s the probability of not drawing a heart?
There are 13 hearts and a total of 39 clubs, diamonds, and spades. The probability of not drawing a heart is 1-(1/4)=3/4 because probability is a number between 0 and 1 which indicates the likelihood that an event will occur. | 945 | 3,393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-30 | latest | en | 0.971702 |
https://www.doubtnut.com/qna/98739110 | 1,722,986,035,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640523737.31/warc/CC-MAIN-20240806224232-20240807014232-00013.warc.gz | 577,893,398 | 38,666 | # If A and B are two event such that $P\left(A\cup B\right)=\frac{3}{4},P\left(A\cap B\right)=\frac{1}{4},=\frac{1}{4},P\left(\stackrel{_}{A}\right)=\frac{2}{3}$, where $P\left(\stackrel{_}{A}\right)$ is the complement of A, then what is P(B) equal to?
A
$1/3$
B
$2/3$
C
$1/9$
D
$2/9$
Video Solution
Text Solution
Verified by Experts
## $\text{Give}\phantom{\rule{1ex}{0ex}}P\left(A\cup B\right)=\frac{3}{4},P\left(A\cap B\right)=\frac{1}{4}$ $P\left(\stackrel{_}{A}\right)=\frac{2}{3}⇒P\left(A\right)=\frac{1}{3}$ As we know $P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$ $\therefore \frac{3}{4}=\frac{1}{3}+P\left(B\right)-\frac{1}{4}$ $⇒\frac{3}{4}+\frac{1}{4}-\frac{1}{3}=P\left(B\right)$ $⇒P\left(B\right)=\frac{2}{3}$
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Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## If A and B are two events such that P(A∪B)=34 P(A∩B)=14andP(overset(_)A)=(2)/(3) then what is P(B) equal to?
A13
B(2)/(3)
C18
D29
• Question 2 - Select One
## A and B are two events such that P(A∪B)=34,P(A)=13,P(¯¯¯A∩B)=
A512
B38
C58
D14
• Question 3 - Select One
## If A and B are two events such that P(A∪B)=34,P(A∩B)=14,P(¯¯¯A)=23,thenP(¯¯¯A∪B) is equal to
A1112
B38
C58
D14
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Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation | 797 | 2,213 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 12, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-33 | latest | en | 0.72461 |
https://mathspace.co/textbooks/syllabuses/Syllabus-302/topics/Topic-5564/subtopics/Subtopic-73900/ | 1,713,335,900,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817144.49/warc/CC-MAIN-20240417044411-20240417074411-00774.warc.gz | 354,652,006 | 48,688 | Hong Kong
Stage 1
# Multiplication and division using groups (10x10)
Lesson
Multiplication is the same as counting by groups, so we will be looking at multiplication with groups of objects. Division is the mathematical term for sharing objects, so we can also use groups of objects to help understand division. Fact families show us how multiplication and division are related to each other, using groups.
#### Examples
##### QUESTION 2
Fill in the box with the missing number.
1. If you get $8$8 groups of $\editable{}$, then altogether you have $80$80.
##### QUESTION 3
Complete the table below by filling in the blank spaces.
1. Total Number of Groups Number in Each Group
$14$14 $2$2 $\editable{}$
$\editable{}$ $5$5 $6$6
$80$80 $10$10 $\editable{}$
$27$27 $\editable{}$ $9$9 | 204 | 791 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-18 | longest | en | 0.841246 |
https://ghdhairstraighteners-inc.com/qa/quick-answer-what-is-thickness-measured-in.html | 1,618,673,211,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038460648.48/warc/CC-MAIN-20210417132441-20210417162441-00397.warc.gz | 377,424,655 | 7,778 | # Quick Answer: What Is Thickness Measured In?
## How is thickness measured?
The measurement is taken using light reflections and a computer to process and display the reading.
The thickness measurement range is from 1 nm to 1 mm.
These systems are usually used in scientific laboratories, and their main disadvantage is their price, which can go up to tens of thousands of dollars..
## How thick is 0.125 inches?
Conversion table inches to mmDimensions — Inches to MetricDimensions —Metric to Inches0.031”1/32”1.0 mm0.062”1/16”1.8 mm0.125”1/8”2.0 mm0.188”3/16”3.0 mm21 more rows
## How is book thickness measured?
So, its simple you just have to count the number of pages and divide it by 2. And after that multiply that number by thickness of 1 page. For example: If book is 600 pages then it should be (600/2)*(0.12) = 36mm or ~1.5 inches.
## How is plastic thickness measured?
The thickness of plastic (plastic gauge) is expressed in microns. 1 micron = 0.0254 mm =1 millimeter = 1 mil = 1µm. If you consider the application for the plastic film, now not only does the thickness comes into play but the additives that are added to make the film perform for its intended use.
## Is thickness a height?
The thickness of an object is defined as the smallest of three descriptive measurements: height, width and length. If you’re dealing with a rectangular prism, and if its volume and the area of one side are provided, you can use those two measurements to calculate its thickness.
## How is wire thickness measured?
The thickness of wire can be found by dividing the length of the coil by the number of turns.
## What is thickness?
noun. the state or quality of being thick. the measure of the smallest dimension of a solid figure: a board of two-inch thickness. the thick part or body of something: the thickness of the leg. a layer, stratum, or ply: three thicknesses of cloth.
## How thick is 0.010 inches?
Micrometers (um)Millimeters (mm)Inches254.00.2540.010304.80.3050.012381.00.3810.015508.00.5080.02036 more rows
## What is .00001 called?
ONE MILLIONTH. 000001 = ONE MILLIONTH!! This is why nobody seems to say “one hundred millionths.” they would be talking about 1/10 of 1/1000!
## What is the difference between gauge and thickness?
As nouns the difference between thickness and gauge is that thickness is (uncountable) the property of being thick (in dimension) while gauge is a measure; a standard of measure; an instrument to determine dimensions, distance, or capacity; a standard.
## How is wood thickness measured?
A board foot is equal to a piece of wood 12 inches long x 12 inches wide and 1 inch thick, or 144 cubic inches. To figure the board foot measurement of a piece of wood, multiply the length x width x thickness in inches, then divide by 144.
## How is liquid thickness measured?
3 Ways to Measure ViscosityGet easy viscosity measurements at various temperatures using a viscosity cup. Simply pour the test liquid into the cup. … Measure the consistency and flow rate of your material with consistometers. … For more accurate measurements, rotational viscometers provide a digital readout of the viscosity of a sample.
## How do you describe thickness?
1 : the smallest of three dimensions length, width, and thickness. 2 : the quality or state of being thick. 3a : viscous consistency boiled to the thickness of honey.
## What is the unit of thickness?
micrometresThe micrometre is commonly employed to measure the thickness or diameter of microscopic objects, such as microorganisms and colloidal particles. Minute distances, as, for example, the wavelengths of infrared radiation, are also given in micrometres.
## What instrument measures thickness?
A thickness meter is an essential quality assurance tool when anodizing, galvanizing and applying zinc coating to metallic surfaces. A thickness meter also is used to measure body paint thickness and uniformity on pre-owned cars, revealing repainted spots, identifying hidden damages and exposing undisclosed accidents.
## What thickness is 40 gauge?
Gauge Conversion ChartGaugeMilMillimeter40.40.010150.50.012760.60.015275.75.01909 more rows•Jan 9, 2014
## What is the difference between width and thickness?
Answer:Technically, definitions of width and thickness are nearly same so it is hard to understand. Width :- the measurement or extent of something from side to side. Thickness :- the distance through an object, as distinct from width or height. | 1,042 | 4,477 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2021-17 | latest | en | 0.924738 |
https://enya.day/mathematics/question-347.html | 1,713,700,301,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817765.59/warc/CC-MAIN-20240421101951-20240421131951-00277.warc.gz | 219,064,012 | 5,027 | # Provide an appropriate response.39) Find a set of 7 scores that has the same mean but a smaller standard deviation than the set {65, 71, 77, 80, 82, 90, 96}.
You don't even have to look up the definition of 'standard deviation'. You only
have to remember that 'smaller standard deviation' means 'less spread-out'.
First, let's find the mean (average). It's not supposed to change:
1/7th of (65 + 71 + 77 + 80 + 82 + 90 + 96) = 561/7 = 80 and 1/7 .
Now, just pick 7 scores that total 561 and are all bunched up.
The easiest way would be 80, 80, 80, 80, 80, 80, 81 .
But that's so easy that it feels like cheating.
Let's say 77, 78, 79, 80, 81, 82, and 84 . | 224 | 665 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-18 | latest | en | 0.894074 |
https://www.cut-the-knot.org/Curriculum/Geometry/AngleBisectorsInEllipse.shtml | 1,670,183,455,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710978.15/warc/CC-MAIN-20221204172438-20221204202438-00026.warc.gz | 765,767,362 | 5,871 | # Angle Bisectors in Ellipse
Let A and B be two points on an ellipse with foci E and F. The tangents to the ellipse at A and B meet in S. Prove that ∠AES = ∠SEB. In words, the point of intersection of two tangents to an ellipse lies on the bisector of the angle formed by joining a focus to the points of tangency.
Proof
### Conic Sections > Ellipse
Let A and B be two points on an ellipse with foci E and F. The tangents to the ellipse at A and B meet in S. Prove that ∠AES = ∠SEB. In words, the point of intersection of two tangents to an ellipse lies on the bisector of the angle formed by joining a focus to the points of tangency.
### Proof
For a proof, reflect F in the two tangents to obtain points M and N. Because of the reflective property of ellipse, B is collinear with E, N and A with E, M.
From the definition of ellipse, EA + FA = EB + FB. Since M and N are reflections of F, FB = BN and FA = AM, implying
EM = EA + AM = EA + FA = EB + FB = EB + BN = EN.
It follows that ΔMEN is isosceles with EM = EN.
On the other hand, in ΔMFN, tangents AS and BS are the perpendicular bisectors of sides FN and FM. They meet in the circumcenter of ΔMFN, which also lies on the perpendicular bisector of the third side MN. By the construction, S is the point of intersection of the two tangents. Thus S is also the circumcenter of ΔMFN and lies on the perpendicular bisector of MN.
Now return to ΔMEN. Since it is isosceles with base MN, the perpendicular bisector of MN passes through E and plays the role of the bisector of ∠MEN which thus passes through S. But MEN and AEB is one and the same angle, and we are done.
(I thank Hubert Shutrick for his selfless help.)
### Angle Bisector
• Angle Bisector
• Angle Bisector Theorem
• Angle Bisectors in Ellipse II
• Angle Bisector in Equilateral Trapezoid
• Angle Bisector in Rectangle
• Property of Angle Bisectors
• Property of Angle Bisectors II
• A Property of Angle Bisectors III
• External Angle Bisectors
• Projections on Internal and External Angle Bisectors
• Angle Bisectors On Circumcircle
• Angle Bisectors in a Quadrilateral - Cyclic and Otherwise
• Problem: Angle Bisectors in a Quadrilateral
• Triangle From Angle Bisectors
• Property of Internal Angle Bisector - Hubert Shutrick's PWW
• Angle Bisectors Cross Circumcircle
• For Equality Choose Angle Bisector | 607 | 2,338 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2022-49 | latest | en | 0.904027 |
null | null | null | null | null | null | Last modified on 15 October 2013, at 21:43
Senator Dodd Speaks in Opposition to FISA Bill on Floor of U.S. Senate
In Opposition to the FISA Bill
by Chris Dodd
Remarks as Prepared, Delivered 24 June 2008
Mr. President: I rise—once again—to voice my strong opposition to the misguided FISA legislation before us today. I have strong reservations about the so-called improvements made to Title I. But more than that, this legislation includes provisions which would grant retroactive immunity to telecommunications companies that apparently have violated the privacy and the trust of millions of Americans by participating in the president’s warrantless wiretapping program. If we pass this legislation, the Senate will ratify a domestic spying regime that has already concentrated far too much unaccountable power in the president’s hands and will place the telecommunications companies above the law.
I am here today to implore my colleagues to vote against cloture in the morning.
And let me make clear, at the outset of this debate, that this is not about domestic surveillance itself. We all recognize the importance of domestic surveillance – in an age of unprecedented threats. This is about illegal, unwarranted, unchecked domestic surveillance.
And that difference—the difference between surveillance that is lawful, warranted and that which is not—is everything.
Mr. President, I had hoped I would not have to return to this floor again under these circumstances – hoped that in these negotiations we would have been able to turn aside retroactive immunity on the grounds that it is bad policy and sets a terrible precedent.
As all of my colleagues know, I have long fought against retroactive immunity, because I believe, quite simply, it is an abandonment of the rule of law. I’ve fought this with everything I had in me—and I haven’t waged this fight alone.
In December, I opposed retroactive immunity on the Senate floor. I spent ten hours on this floor then. In January and February, I came to the floor time and time again to discuss the dangers of granting retroactive immunity. Along with my colleague and friend Russ Feingold, who has shown remarkable leadership on this issue, I offered an amendment that would have stripped retroactive immunity from the Senate bill. Unfortunately, our amendment failed and to my extreme disappointment, the Senate adopted the underlying bill.
Since passage of the Senate bill, there has been extensive negotiations on how to move forward. Today, we are being asked to pass the so-called compromise that was reached by some of our colleagues and approved by the House of Representatives.
I am here today to say that I will not and cannot support this legislation. It goes against everything I have stood for – everything this body ought to stand for.
There is no question some improvements have been made over the previous versions of this bill. Title I, which regulates the ability of the government to conduct electronic surveillance, has indeed been improved. Albeit modestly. In fact, it is my hope that a new Congress and a new President will work together to fix the problems with Title I should the Senate adopt this new legislation.
But in no way is this compromise acceptable, Mr. President. This legislation before us purports to give the courts more of a role in determining the legality of the telecommunications companies actions. But in my view the Title II provisions do little more than ensure without a doubt that the telecommunications companies will be granted retroactive immunity.
Allow me to quote the Senate Intelligence Committee report on the matter. It reads:
Beginning soon after September 11, 2001, the Executive branch provided written requests or directives to U.S. electronic communication service providers to obtain their assistance with communications intelligence activities that had been authorized by the President.
… The letters were provided to electronic communication service providers at regular intervals. All of the letters stated that the activities had been authorized by the President. All of the letters also stated that the activities had been determined to be lawful by the Attorney General, except for one letter that covered a period of less than sixty days. That letter, which like all the others stated that the activities had been authorized by the President, stated that the activities had been determined to be lawful by the Counsel to the President.
Under the legislation before us, the district court would simply decide whether or not the telecommunication companies received documentation stating that the President authorized the program and that there had been some sort of determination that it was legal.
But, as the Intelligence Committee has already made clear, we already KNOW that this happened.
We already KNOW that the companies received some form of documentation, with some sort of legal determination.
But that’s not the question. The question is not whether these companies received a “document” from the White House. The question is, “were their actions legal?” It’s rather straightforward—surprisingly uncomplicated.
Either the companies were presented with a warrant, or they weren’t. Either the companies and the President acted outside of the rule of law, or they followed it. Either the underlying program was legal or it wasn’t.
Because of this legislation, none of the questions will be answered, Mr. President. Because of this so-called “compromise,” the judge’s hands will be tied, and the outcome of these cases will be predetermined. Because of this compromise, retroactive immunity will be granted and that, as they say, will be that. Case closed.
No court will rule on the legality of the telecommunications companies activities in participating in the president’s warrantless wiretapping program.
None of our fellow Americans will have their day in court.
What they will have is a government that has sanctioned lawlessness.
We are better than that.
They see what I see in this debate – that by short-circuiting the judicial process we are sending a dangerous signal to future generations. They see us establishing a precedent that Congress can—and will—provide immunity to potential law breakers, if they are “important” enough.
Mr. President, some may be asking – why is retroactive immunity so dangerous?
What is this issue? Why should I care?
Allow me to explain by providing a bit of context. I want to remind my colleagues of what I said about this bill months ago, because the argument against providing retroactive immunity remains unchanged.
Mr. President, unwarranted domestic spying didn’t happen in a panic or short-term emergency, not for a week, or a month, or even a year. If it had, I might not be here today.
But that isn’t the case. What we now know is that spying by this Administration went on, relentlessly, for more than five years.
I might not be here if it had been the first offense of a new administration. Maybe not if it had even been the second or the third.
But that isn’t the case either, Mr. President. Indeed, I am here today because with offense after another after another, I believe it is long past time to say: “enough.”
I am here today because of a pattern—a pattern of abuse against civil liberties and the rule of law. Against the Constitution—of which we are custodians, temporary though that status may be.
And I would add that had these abuses been committed by a president of my own party, I would have opposed them, every bit as vigorously.
I am here today because warrantless wiretapping is merely the latest link in a long chain of abuses.
So, why are we here? Because, Mr. President – it is alleged that giant telecom corporations worked with our government to compile Americans’ private, domestic communications records into a database of enormous scale and scope.
Secretly and without a warrant, those corporations are alleged to have spied on their own customers – American customers.
Here’s only one of the most egregious examples. According to the Electronic Frontier Foundation:
Clear, first-hand whistleblower documentary evidence [states]…that for year on end every e-mail, every text message, and every phone call carried over the massive fiber-optic links of sixteen separate companies routed through AT&T’s Internet hub in San Francisco—hundreds of millions of private, domestic communications—have been…copied in their entirety by AT&T and knowingly diverted wholesale by means of multiple “splitters” into a secret room controlled exclusively by the NSA.
The phone calls and internet traffic of millions of Americans, diverted into a secret room controlled by the National Security Agency. That allegation still needs to be proven in a court of law. But it clearly needs to be determined in a court of law and not here in Senate.
I suppose if you only see cables and computers there, the whole thing seems almost harmless. Certainly nothing to get worked up about—a routine security sweep, and a routine piece of legislation blessing it.
If that’s all you imagine happened in the NSA’s secret room, I imagine you’ll vote for immunity.
I imagine you wouldn’t see much harm in voting to allow this practice to continue either.
But if you see a vast dragnet for millions of Americans’ private conversations, conducted by a government agency that acted without a warrant, acted outside of the rule of law—then, I believe, you’ll recognize what’s at stake here. You’ll see that what’s at stake is the sanctity of the law and the sanctity of our privacy. And you’ll probably come to a very different conclusion.
Maybe that sounds overdramatic. Perhaps some will ask, “What does it matter, at the end of the day, if a few corporations aren’t sued? These people sue each other all the time.”
Others may say, “This seems a small issue. Maybe the Administration went too far, but this seems like an isolated case.”
Indeed, Mr. President – as long as this case seems isolated and technical, they win. As long as it’s about another lawsuit buried in our legal system and nothing more, they win. The Administration is counting on the American people to see nothing bigger than that – “Nothing to see here.”
But there is plenty to see here, Mr. President – and it is so much more than a few phonecalls, a few companies, a few lawsuits.
What is at stake is nothing less than equal justice—justice that makes no exceptions. What is at stake is an open debate on security and liberty, and an end to warrantless, groundless spying.
This bill does not say, “Trust the American people; Trust the courts and judges and juries to come to just decisions.” Retroactive immunity sends a message that is crystal clear:
“Trust me.”
And that message comes straight from the mouth of this President. “Trust me.”
What is the basis for that trust? Classified documents, we are told, that prove the case for retroactive immunity beyond a shadow of a doubt.
But we’re not allowed to see them! I’ve served in this body for 27 years, and I’m not allowed to see them! Neither are a majority of my colleagues. We are all left in the dark.
I cannot speak for my colleagues—but I would never take “trust me” for an answer, not even in the best of times. Not even from a President on Mount Rushmore.
I can’t put it better than this:
Those words were not spoken by someone who took our nation’s security lightly, Mr. President. They were spoken by Ronald Reagan -- in 1980. They are every bit as true today, even if times of threat and fear blur our concept of transcendent values. Even if those who would exploit those times urge us to save our skins at any cost.
But again, Mr. President:
“Why should I care?”
The rule of law has rarely been in such a fragile state. Rarely has it seemed less compelling. What, after all, does the law give us anyway? It has no parades, no slogans. It lives in books and precedents. And, we are never failed to be reminded, the world is a very dangerous place.
Indeed, that is precisely the advantage seized upon, not just by this Administration but in all times, by those looking to disregard the rule of law. As James Madison, the father of our Constitution, said more than two centuries ago, “It is a universal truth that the loss of liberty at home is to be charged to the provisions against danger…from abroad.”
With the passage of this bill, his words would be one step closer to coming true. So it has never been more essential that we lend our voices to the law, and speak on its behalf.
What is this about, Mr. President?
It’s about answering the fundamental question: do we support the rule of law…or the rule of men? To me, this is our defining question—indeed it may be the defining question that confronts every generation.
This is about far more than a few telecoms – it is about contempt for the law, large and small.
Mr. President, I’ve said that warrantless wiretapping is but the latest link in a long chain of abuses when it comes to the rule of law.
This is about the Justice Department turning our nation’s highest law enforcement offices into patronage plums, and turning the impartial work of indictments and trials into the pernicious machinations of politics.
Contempt for the rule of law.
This is about Alberto Gonzales, the nation’s now-departed Attorney General, coming before Congress to give us testimony that was at best, wrong—and at worst, outright perjury.
Contempt for the rule of law – by the nation’s foremost enforcer of the law.
This is about Congress handing the president the power to designate any individual he wants as an “unlawful enemy combatant,” hold him indefinitely, and take away his right to habeas corpus—the 700-year-old right to challenge your detention.
If you think that the Military Commissions Act struck at the heart of the Constitution, you’d be understating things—it did a pretty good job on the Magna Carta while it was at it.
And if you think that this only threatens a few of us, you should understand that the writ of habeas corpus belongs to all of us—it allows anyone to challenge their detention.
Rolling back habeas rights endangers us all: Without a day in court, how can you prove that you’re entitled to a trial? How can you prove that you are innocent? In fact, without a day in court, how can you let anyone know that you have been detained at all?
Thankfully, the Supreme Court recently rebuked the President’s lawlessness and ruled that detainees do indeed have the right to challenge their detention.
Mr. President, the Military Commissions Act also gave President Bush the power some say he wanted most of all: the power to get information out of suspected terrorists—by virtually any means.
The power to use evidence gained from torture.
I don’t think you can hold the rule of law in any greater contempt than sanctioning torture, Mr. President.
Because of decisions made at the highest levels of our government, America is making itself known to the world for torture, with stories like this one:
A prisoner at Guantanamo—to take one example out of hundreds— was deprived of sleep over fifty five days, a month and three weeks. Some nights, he was doused with water or blasted with air conditioning. And after week after week of this delirious, shivering wakefulness, on the verge of death from hypothermia, doctors strapped him to a chair—doctors, healers who took the Hippocratic Oath to “do no harm”—pumped him full of three bags of medical saline, brought him back from death—and sent him back to his interrogators.
To the generation coming of age around the world in this decade, that is America. Not Normandy, not the Marshall Plan, not Nuremberg. But Guantanamo.
Think about it.
We have legal analysts so vaguely defining torture, so willfully blurring the lines during interrogations that we have CIA counterterrorism lawyers saying things like, “if the detainee dies, you’re doing it wrong.”
We have the CIA destroying tapes containing the evidence of harsh interrogations—about the Administration covering its tracks in a way more suited to a banana republic than to the home of freedom.
We have this Administration actually defending waterboarding, a technique invented by the Spanish Inquisition, perfected by the Khmer Rouge, and in between, banned—originally banned for excessive cruelty—by the Gestapo!
Still, some say, “waterboarding’s not torture.”
Oh really?
Listen to the words of Malcolm Nance, a 26-year expert in intelligence and counter-terrorism, a combat veteran, and former Chief of Training at the US Navy Survival, Evasion, Resistance and Escape School. While training American soldiers to resist interrogation, he writes,
I have personally led, witnessed and supervised waterboarding of hundreds of people….Unless you have been strapped down to the board, have endured the agonizing feeling of the water overpowering your gag reflex, and then feel your throat open and allow pint after pint of water to involuntarily fill your lungs, you will not know the meaning of the word….
It does not simulate drowning, as the lungs are actually filling with water. The victim is drowning. How much the victim is to drown depends on the desired result…and the obstinacy of the subject.
Waterboarding is slow motion suffocation…usually the person goes into hysterics on the board….When done right it is controlled death.
Controlled death, Mr. President.
And that is not torture?
Not according to President Bush’s White House. They have said waterboarding is legal, and that, if it chooses, America will waterboard again.
Surely then, Mr. President, our new Attorney General would condemn torture.
Surely, the nation’s highest law enforcement officer in the land, coming after Alberto Gonzales’s chaotic tenure, would never come before Congress and defend the president’s power to openly break the law.
Would he?
He would, Mr. President.
When he came to the Senate before his confirmation, Michael Mukasey was asked a simple question, bluntly and plainly: “Is waterboarding constitutional?”
He replied: “If waterboarding is torture, torture is not constitutional.”
One would hope for a little more insight from someone so famously well-versed in national security law. But Mr. Mukasey pressed on with the obstinacy of a witness pleading the fifth: “If it’s torture….If it amounts to torture, it is not constitutional.”
And that is the best this noted jurist, this legal scholar, this longtime judge, a supposed expert on national security law had to offer on the defining moral issue of this presidency. Claims of ignorance. Word games.
Now-Attorney General Mukasey was asked the easiest question we have in a democracy: Can the president openly break the law? Can he—as we know he’s done already—order warrantless wiretapping, ignore the will of Congress, and then hide behind nebulous powers he claims to find in the Constitution?
His response: The president has “the authority to defend the country.”
And in one swoop, the Attorney General conceded to the president nearly unlimited power, just as long as he finds a lawyer willing to stuff his actions into the boundless rubric of “defending the country.” Unlimited power to defend the country, to protect us as one man sees fit, even if that means listening to our phone calls without a warrant, even if that means holding some of us indefinitely.
That is, Mr. President, contempt for the rule of law.
And so, this is very much about torture – about “enhanced interrogation methods” and waterboarding.
It is also about extraordinary rendition—outsourced torture of men this administration would prefer we didn’t know exist.
Oh, but we do know, Mr. President.
One was a Syrian immigrant raising his family in Canada as a citizen. He wrote computer code for a company called MathWorks and was planning to start his own tech business. On a trip through New York’s JFK Airport, he was arrested by U.S. federal agents. They shackled him and bundled him onto a private CIA plane, which flew him across the Atlantic Ocean to Syria.
This man spent the next 10 months and 10 days in a Syrian prison. His cell was three feet wide—the size of a grave. Some 300 days passed alone in that cell, with a bowl for his toilet and another bowl for his water, and the door only opened so he could go wash himself once a week—though it may have been more or less, because the cell was dark and he lost track of time.
The door only opened for one other reason: for interrogators who asked him, again and again, about al-Qaeda. Here’s how it was described:
The interrogator said, “Do you know what this is?” I said, “Yes, it’s a cable,” and he told me, “Open your right hand.” I opened my right hand, and he hit me like crazy. It was so painful, and of course I started crying, and then he told me to open my left hand, and I opened it, and he missed, then hit my wrist. And then he asked me questions. If he does not think you are telling the truth, then he hits again.
The jail and the torturers were Syrian, but America sent this man there with full knowledge of what would happen to him—because it was part of the longstanding secret program of “extraordinary rendition.” America was convinced that he was a terrorist and wanted the truth beaten out of him.
No charges were ever filed against him. His adopted nation’s government—Canada, one of our strongest NATO allies—cleared him of all wrongdoing after a year-long official investigation, and awarded him more than $10 million in government compensation for his immense pain and suffering. But not before he was tortured for 10 months in a cell the size of a grave. Did his torture make us safer? Did his suffering improve our security?
I would note Mr. President, that our own government has shamefully refused even to acknowledge that his case exists.
We know about a German citizen as well, living in the city of Ulm with his wife and four children. On a bus trip through Eastern Europe, he was pulled off at a border crossing by armed guards and held for three weeks in a hotel room, where he was beaten regularly. At the end of three weeks, he was drugged and shipped on a cargo plane to Kabul, Afghanistan.
For five months, he was held in the Salt Pit—a secret American prison staffed by Afghan guards. All he had to drink was stagnant water from a filthy bottle. Again and again, masked men interrogated him about al-Qaeda, and finally, he says, they raped him.
He was released in May of 2004. Scientific testing confirmed his story of malnourishment, and the Chancellor of Germany publicly acknowledged that he was wrongly held. What was his crime? Having the same name as a suspected terrorist. Again, our own government has shamefully refused to even acknowledge that his case exists.
And so, we do know, Mr. President. We know because there aren’t enough words in the world to cover the facts.
If you’d like to define torture out of existence, be my guest.
If you’d rather use a Washington euphemism—“tough questioning,” “enhanced interrogation”—feel free. Feel free to talk about “fraternity hazing,” like Rush Limbaugh did, or to use a favorite term of Vice President Cheney’s, “a dunk in the water.” You can call it whatever you’d like.
But when you’re through, the facts will still be waiting for you. Controlled death. Outsourced torture. Secret prisons. Month-long sleep deprivations. The president’s personal power to hold whomever he likes for as long as he’d like. It is as if we woke up in the middle of some Kafka-esque nightmare.
Have I gone wildly off-topic, Mr. President? Have I brought up a dozen unrelated issues?
I wish I had, Mr. President. I wish that none of these stories were true.
But, we are deceiving ourselves when we talk about the U.S. attorneys issue, the habeas issue, the torture issue, the rendition issue, or the secrecy issue as if each were an isolated case! As if each one were an accident! When we speak of them as isolated, we are keeping our politics cripplingly small; and as long as we keep this small, the rule of men is winning.
There is only one issue here. Only one: the law issue.
Does the president serve the law, or does the law serve the president? Each insult to our Constitution comes from the same source; each springs from the same mindset; and if we attack this contempt for the law at any point, we will wound it at all points.
That is why I’m here today: Retroactive immunity is on the table today; but also at issue is the entire ideology that justifies it, the same ideology that defends torture and executive lawlessness. Immunity is a disgrace in itself, but it is far worse in what it represents. It tells us that some believe in the courts only so long as their verdict goes their way. That some only believe in the rule of law, so long as exceptions are made at their desire. It puts secrecy above sunshine and fiat above law.
Did the telecoms break the law? That, I don’t know.
“Law” is a word we barely hear from the supporters of immunity. They offer neither a deliberation about America’s difficult choices in the age of terrorism, nor a shared attempt to set for our times the excruciating balance between security and liberty. They merely promise a false debate on a false choice: security or liberty, but never, ever both.
I think differently. I think that America’s founding truth is unambiguous: security and liberty, one and inseparable, and never one without the other--no matter how difficult a situation, no matter what threats we face.
Secure in that truth, I offer a challenge to immunity’s supporters: You want to put a handful of corporations above the law. Could you please explain how your immunity makes any one of us any safer at all?
The truth is that a working balance between security and liberty has already been struck! In fact, it has been settled for decades. For thirty years, FISA has prevented executive lawbreaking and protected Americans, and that balance stands today.
In the wake of the Watergate scandal, the Senate convened the Church Committee, a panel of distinguished members determined to investigate executive abuses of power. And unsurprisingly, they found that when Congress and the courts substitute “trust me” for real oversight, massive lawbreaking can result.
They found evidence of U.S. Army spying on the civilian population, federal dossiers on citizens’ political activities, a CIA and FBI program that had opened hundreds of thousands of Americans’ letters without warning or warrant. In sum, Americans had sustained a severe blow to their Fourth Amendment rights “to be secure in their persons, houses, papers, and effects, against unreasonable searches and seizures.”
But at the same time, the senators of the Church Committee understood that surveillance needed to go forward to protect the American people. Surveillance itself was not the problem; unchecked, unregulated, unwarranted surveillance was. What surveillance needed, in a word, was legitimacy.
And in America, as the Founders understood, power becomes legitimate when it is shared, when Congress and the courts check that attitude which so often crops up in the executive branch—“if the president does it, it’s not illegal.”
The Church Committee’s final report, “Intelligence Activities and the Rights of Americans,” put the case powerfully:
The critical question before the Committee was to determine how the fundamental liberties of the people can be maintained in the course of the Government’s effort to protect their security.
We reject the view that the traditional American principles of justice and fair play have no place in our struggle against the enemies of freedom. Moreover, our investigation has established that the targets of intelligence activity have ranged far beyond persons who could properly be characterized as enemies of freedom….
We have seen a consistent pattern in which programs initiated with limited goals, such as preventing criminal violence or identifying foreign spies, were expanded to what witnesses characterized as “vacuum cleaners,” sweeping in information about lawful activities of American citizens.
The senators concluded: “Unless new and tighter controls are established by legislation, domestic intelligence activities threaten to undermine our democratic society and fundamentally alter its nature.”
What a strange echo we hear in those words! They could have been written yesterday. Three decades ago, our predecessors in this chamber understood that when domestic spying goes too far, it threatens to kill just what it promises to protect—an America secure in its liberty. That lesson was crystal-clear 30 years ago. Why is it so clouded now?
And before we entertain the argument that “everything has changed” since those words were written, remember: The men who wrote them had witnessed world war and Cold War, had seen Nazi and Soviet spying, and were living every day under the cloud of nuclear holocaust.
I’ll ask this, Mr. President: Who will chair the commission investigating the secrets of warrantless spying, years from today? Will it be a young senator in this body today? Will it be someone not yet elected? What will that senator say when he or she comes to our actions, reads in the records how we let outrage after outrage after outrage slide, with nothing more than a promise to stop the next one? I imagine that senator will ask of us, “Why didn’t they do anything? Why didn’t they fight back? In June 2008, when no one could doubt anymore what the administration was doing—why did they sit on their hands?”
Since the time of the Church Commission, Mr. President, the threats facing us have multiplied and grown in complexity, but the lesson has been immutable:
Warrantless spying threatens to undermine our democratic society, unless legislation brings it under control. In other words, the power to invade privacy must be used sparingly, guarded jealously, and shared equally between the branches of government.
Or the case can be made pragmatically. As my friend Harold Koh, the Dean of Yale Law School, recently argued, “The engagement of all three branches tends to yield not just more thoughtful law, but a more broadly supported public policy.”
Three decades ago, Congress embodied that solution in the Foreign Intelligence Surveillance Act, or FISA.
FISA confirmed the president’s power to conduct surveillance of international conversations involving anyone in the United States, provided that the federal FISA court issued a warrant—ensuring that wiretapping was aimed at safeguarding our security, and nothing else.
The president’s own Director of National Intelligence, Mike McConnell, explained the rationale in an interview last summer: The United States “did not want to allow [the intelligence community] to conduct…electronic surveillance of Americans for foreign intelligence unless you had a warrant, so that was required.”
As originally written in 1978, and as amended many times since, FISA has accomplished its mission; it has been a valuable tool for conducting surveillance of terrorists and those who would harm America.
And every time presidents have come to Congress openly to ask for more leeway under FISA, Congress has worked with them; Congress has negotiated; and together, Congress and the president have struck a balance that safeguards America while doing its utmost to protect privacy.
Last summer, Congress made a technical correction to FISA, enabling the president to wiretap, without a warrant, conversations between two foreign targets, even if those conversations are routed through American computers. For other reasons, I felt that this past summer’s legislation went too far, and I opposed it. But the point is that Congress once again proved its willingness to work with the president on FISA.
Isn’t that enough?
Just this past October and November, the Senate Intelligence and Judiciary Committees worked with the president to further refine FISA and ensure that, in a true emergency, the FISA court would do nothing to slow down intelligence gathering.
Isn’t that enough?
And as for the FISA court? Between 1978 and 2004, according to the Washington Post, the FISA court approved 18,748 warrants—and rejected five.
The FISA court has sided with the executive ninety nine point nine percent of the time.
Isn’t that enough?
Is anything lacking? Have we forgotten something? Isn’t all of this enough to keep us safe?
We all know the answer we received. This complex, fine-tuned machinery, crafted over three decades by three branches of government, four presidents, and 12 Congresses was ignored. It was a system primed to bless nearly any eavesdropping a president could conceive—and spying still happened illegally.
If the shock of that decision has yet to sink in, think of it this way: President Bush ignored not just a federal court, but a secret federal court; not just a secret federal court, but a secret federal court prepared to sign off on his actions ninety nine point nine percent of the time. A more compliant court has never been conceived.
And that still wasn’t good enough.
So I will ask the Senate candidly, and candidly it already knows the answer:
Is this about security—or is it about power?
Why are some fighting so hard for retroactive immunity? The answer, I believe, is that immunity means secrecy, and secrecy means power.
It’s no coincidence that the man who proclaimed “if the president does it, it’s not illegal”—Richard Nixon—was the same man who raised executive secrecy to an art form.
The senators of the Church Committee expressed succinctly the deep flaw in the Nixonian executive: “Abuse thrives on secrecy.” And, in the exhaustive catalogue of their report, they proved it.
In this push for immunity, secrecy is at the center. We find proof in immunity’s original version: a proposal to protect not just the telecoms, but everyone involved in the wiretapping program.
In their original proposal, that is, they wanted to immunize themselves.
Think about that. It speaks to their fear and, perhaps, their guilt: their guilt that they had broken the law, and their fear that in the years to come, they would be found liable or convicted.
They knew better than anyone else what they had done—they must have had good reason to be afraid.
Thankfully, immunity for the executive is not part of the bill before us. But the original proposal tells us something very important: This is, and always has been, a self-preservation bill.
Otherwise, why not have the trial and get it over with? If the proponents of retroactive immunity are right, the corporations would win in a walk.
After all, in the official telling, the telecoms were ordered to help the president spy without a warrant, and they patriotically complied. We’ve even heard on this floor the comparison between the telecom corporations to the men and women laying their lives on the line in Iraq.
But ignore that comparison – which, frankly, I find deeply offensive. Ignore for a moment the fact that in America we obey the laws, not the president’s orders. Ignore that not even the president has the right to scare or bully you into breaking the law, though it seems that tactic has proven surprisingly fruitful.
Ignore that the telecoms were not unanimous; one, Qwest, wanted to see the legal basis for the order, never received it, and so refused to comply.
Ignore that a judge presiding over the case ruled that “AT&T cannot seriously contend that a reasonable entity in its position could have believed that the alleged domestic dragnet was legal.”
Ignore all that: If the order the telecoms received was legally binding, they have an easy case to prove. The corporations only need to show a judge the authority and the assurances they were given, and they’ll be in and out of court in five minutes.
If the telecoms are as defensible as the president says, why doesn’t the president let them defend themselves? If the case is so easy to make, why doesn’t he let them make it?
It can’t be that he’s afraid of leaks. Our federal court system has dealt for decades with the most delicate national security matters, building up expertise in protecting classified information behind closed doors—ex parte, in camera. We can expect no less in these cases.
No intelligence sources need be compromised. No state secrets need be exposed. After litigation at both the district court and circuit court level, no state secrets have been exposed.
In fact, Federal District Court Judge Vaughn Walker, a Republican appointee, has already ruled that the issue can go to trial without putting state secrets in jeopardy.
He reasonably pointed out that the existence of the terrorist surveillance program is a hardly secret at all: “The government has [already] disclosed the general contours of the ‘terrorist surveillance program,’ which requires the assistance of a telecommunications provider.”
As the state secrets privilege is invoked to stall these high-profile cases, it’s useful to consider that privilege’s history. In fact, it was tainted at its birth by a president of my own party, Harry Truman. In 1952, he successfully invoked the new privilege to prevent public exposure of a report on a plane crash that killed three Air Force contractors.
When the report was finally declassified—some fifty years later, decades after anyone in the Truman administration was within its reach—it contained no state secrets at all.
Only facts about repeated maintenance failures that would have seriously embarrassed some important people.
And so the state secrets privilege began its career not to protect our nation—but to protect the powerful.
In his opinion, Judge Walker argued that, even when it is reasonably grounded,
the state secrets privilege [still] has its limits. While the court recognizes and respects the executive’s constitutional duty to protect the nation from threats, the court also takes seriously its constitutional duty to adjudicate the disputes that come before it. To defer to a blanket assertion of secrecy here would be to abdicate that duty, particularly because the very subject matter of this litigation has been so publicly aired.
And that ought to be the epitaph of this Administration: “sacrificing liberty for no apparent enhancement of security.” Worse than selling our soul—giving it away for free!
It is equally wrong to claim that failing to grant this retroactive immunity will make the telecoms less likely to cooperate with surveillance in the future.
The truth is that, since the 1970s, FISA has compelled telecommunications companies to cooperate with surveillance, when it’s warranted—and what’s more, it immunizes them. It’s done that for more than 25 years.
So cooperation in warranted wiretapping is not at stake today, and despite claims by supporters of immunity-- never has been. Collusion in warrantless illegal wiretapping is. And the warrant makes all the difference, because it is precisely the court’s blessing that brings presidential power under the rule of law.
In sum, we know that giving the telecoms their day in court—giving the American people their day in court—would not jeopardize an ounce of our security.
And it could only expose one secret: the extent to which the rule of law has been trampled.
And that is the choice at stake today: Will the secrets of the last years remain closed in the dark? Or will they be open to the generations to come, to our successors in this chamber, so that they can prepare themselves to defend against future outrages of power and usurpations of law from future presidents, of either party?
Thirty years after the Church Committee, history repeated itself. If those who come after us are to prevent it from repeating again, they need the full truth.
And that is why we must not see these secrets go quietly into the night. I am here because the truth is no one’s private property—it belongs to every one of us, and it demands to be heard.
“State secrets,” “patriotic duty”—those, as weak as they are, are the arguments the telecoms’ advocates use when they’re feeling high-minded! When their thoughts turn baser, they make their arguments as amateur economists.
Here’s how Mike McConnell put it: “If you play out the suits at the value they’re claimed, it would bankrupt these companies. So…we have to provide liability protection to these private sector entities.”
To begin with, that’s a clear exaggeration. We are talking about some of the wealthiest, most successful companies in America. Some of them have continued to earn record profits and sign up record numbers of subscribers at the same time as this very public litigation—totally undermining the argument that these lawsuits are doing the telecoms severe “reputational damage.”
Companies of that size couldn’t be completely wiped out by anything but the most exorbitant and unlikely judgment. To assume that the telecoms would lose, and that their judges would then hand down such backbreaking penalties, is already to take several leaps.
Opponents of immunity, including myself, have stated that we would support a reasonable alternative to blanket retroactive immunity.
No one seriously wants to financially cripple our telecommunications industry. The point is to bring checks and balances back to domestic spying. Setting that precedent would hardly require a crippling judgment.
It’s much more troubling, though, that our Director of National Intelligence even bothers to speak to “liability protection for private sector entities.”
This isn’t the Secretary of Commerce we’re talking about, but the head of our nation’s intelligence efforts.
For that matter, how does that even begin to be relevant to letting this case go forward? Since when did we throw entire suits out because the defendant stood to lose too much?
It astounds me that some can speak in the same breath about national security and bottom lines. Approve immunity, and Congress will state clearly: The richer you are, the more successful you are, the more lawless you are entitled to be. A suit against you is a danger to the Republic!
And so, at the rock-bottom of its justifications, the telecoms’ advocates are essentially arguing that immunity can be bought.
The truth is exactly the opposite—and it should be obvious:
The larger the corporation, the greater the potential for abuse.
No one suggests that success should make a company suspect; companies grow large, and essential to our economy, because they are excellent at what they do. But the size and wealth open the realm of possibilities for abuse far beyond the scope of the individual.
After all, if the allegations are true, we are talking about one of the most massive violations of privacy in American history.
Should there not be some retribution or penalty?
If reasonable search and seizure means opening a drug dealer’s apartment, the telecoms’ alleged actions would be the equivalent of strip-searching everyone in the building, ransacking their bedrooms, and prying up all the floorboards.
The scale of these corporations opens unprecedented possibilites for abuse—possibilities far beyond the power of the individual.
What the telecoms have been accused of could not be done by one man or even ten.
It would be inconceivable without the size and resources of a corporate behemoth—the same size that makes Mike McConnell fear the corporations’ day in court. That’s the massive scale we’re talking about—and that massive scale is precisely why no corporation must be above the law.
On that scale, it is impossible to plead ignorance. As Judge Walker ruled, “AT&T cannot seriously contend that a reasonable entity in its position could have believed that the alleged domestic dragnet was legal.”
But the arguments of the president’s allies sink even lower. Listen to the words a House Republican leader spoke on Fox News. They are shameful: “I believe that they deserve immunity from lawsuits out there from typical trial lawyers trying to find a way to get into the pockets of American companies.”
Of course, some of the “typical greedy trial lawyers” bringing these suits actually work for a nonprofit. And the telecoms that some want to portray as pitiable little Davids actually employ hundreds of attorneys, retain the best corporate law firms, and spend multimillion-dollar legal budgets every year. But if the facts actually mattered to immunity supporters, we wouldn’t be here. For some, the pre-written narrative takes precedence far above the mere facts; and here, it is the perennial narrative of the greedy trial lawyers.
With that, some can rest content. They can conclude that we weren’t ever serious about law, or about privacy, or about checks and balances—it was about money all along.
There can no longer be any doubt: One by one, the arguments of the immunity supporters, of the telecoms’ advocates, fail. I’d like to spend a few moments reviewing, in detail, those claims and their failures.
[Slide 1]
One: Immunity supporters argue that granting immunity is a presidential prerogative.
But the fact is that this case belongs in the courts, in cases where the outcome has not been predetermined. The judiciary should be allowed to determine whether the president has exceeded his powers by obtaining from the telecoms wholesale access to the domestic communications of millions of ordinary Americans.
The courts should not simply be in the business of certifying that the companies received some form of documentation. Rather they should be allowed to evaluate the validity of the legal arguments asserted in the document. Was the request legal or not?
Remember also that the administration’s original immunity proposal protected everyone involved in the wiretapping program—not just the telecoms. In their original proposal, that is, they wanted to immunize themselves.
Thankfully, executive immunity is not part of the bill before us. But the origin of immunity tells us a great deal about what’s at stake here: self-preservation.
[Slide 2]
Two: Immunity supporters claim that only foreign communications were targeted—not Americans’ domestic calls.
But the fact is that clear, first-hand evidence, authenticated by the corporations in court, contradicts that claim. “Splitters” at AT&T’s Internet hub in San Francisco diverted into a secret room controlled by the NSA every e-mail, text message, and phone call—foreign or domestic—carried over the massive fiber-optic links of sixteen separate companies.
[Slide 3]
Three: Immunity supporters claim that a lack of immunity will make the telecoms less likely to cooperate with surveillance in the future.
But remember: Since the 1970s, FISA has compelled telecoms to cooperate with warranted surveillance, and it has immunized them. The issue today is not wiretapping—it is warrantless wiretapping. And the warrant is essential, because that is what brings the president’s power under the rule of law.
[Slide 4]
Four: Immunity supporters argue that the telecoms can’t defend themselves without exposing state secrets.
But the fact is that Federal District Court Judge Vaughn Walker has already ruled that the issue can go to trial without putting state secrets in jeopardy. He pointed out that the existence of the warrantless surveillance program is a hardly secret at all: “The government has [already] disclosed the general contours of the ‘terrorist surveillance program,’ which requires the assistance of a telecommunications provider.”
[Slide 5]
Five: Immunity supporters claim that the telecoms are already protected by common law principles.
But the fact is that common law immunities do not trump specific legal duties imposed by statute, such as the specific duties Congress has long imposed on telecoms to protect customer privacy and records. In the pending case against AT&T, the judge already has ruled unequivocally that “AT&T cannot seriously contend that a reasonable entity in its position could have believed that the alleged domestic dragnet was legal.”
Even so, the communications company defendants can and should have the opportunity to present these defenses to the courts, and the courts—not Congress preemptively—should decide whether they are sufficient.
[Slide 6]
Six: Immunity supporters claim that leaks from the trial might damage national security.
But the fact is that our federal court system, in decades of dealing with delicate national security matters, has built up the expertise it takes to secure that information, behind closed doors. If we’re still concerned about national security being threatened as a result of these cases, we can simply get the principals a security clearance.
We can be increasingly confident that these cases will not expose state secrets or intelligence sources—because, after the extensive litigation that has already taken place at both the district court and circuit court level, no sensitive information has leaked out.
[Slide 7]
Seven: Immunity supporters claim that litigation will harm the telecoms by causing them “reputational damage.”
But the fact is that there is no evidence that this litigation has reduced or will reduce the defendant companies’ bottom lines or customer base.
These companies’ reputations can only be harmed if they have done something wrong. If they have not, they have nothing to worry about.
[Slide 8]
Eight: Finally, immunity supporters claim that these lawsuits could bankrupt the telecoms.
But as we’ve seen, such huge corporations could only be wiped out by the most enormous penalties—and also the most unlikely. It takes several leaps to assume that the telecoms will lose, and then that they’ll be slapped with such huge judgments.
But on another level, immunity supporters are staking their claim on a dangerous principle—that a suit can be stopped solely on the basis of how much a defendant stands to lose.
If we accept that premise, we could conceive of a corporation so wealthy, so integral to our economy, that its riches place it outside the law altogether. And that is a deeply flawed argument.
We see, then, that none of the arguments for immunity stand. There is absolutely no reason to halt the legal process, and bar the courthouse door.
And ultimately, Mr. President, that’s all I’m asking for: a fair fight. In any other administration, it would be a humble claim: a day in court, for the companies that have been accused, and for the American citizens who have accused them. To reject immunity would mean to grab hold of the closest thread of lawlessness we have at hand, and to pull until the whole garment unravels.
But ensuring a day in court is not the same as ensuring a verdict. When that day comes, I have absolutely no investment in the verdict, either way. It may be that the federal government broke the law in asking the telecoms to spy, but that the telecoms’ response was innocent. It may be that the government was within the law, and that the telecoms broke it. Maybe both broke the law. Maybe neither did.
But just as it would be absurd for me to declare the telecoms clearly guilty, it is equally absurd to close the case in Congress, without a decision. That is what immunity does. Throughout this debate, the telecoms’ advocates have needed to show not just that they’re right—but that they’re so right, and that we’re so far beyond the pale—that we can shut down the argument right here, today.
That is a burden they have clearly not met. And they cannot expect to meet it when a large majority of the senators who will make the decision have not even seen the secret documents that are supposed to prove the case for retroactive immunity.
Mr. President, my trust is in the courts, in the cases argued openly, in the judges who preside over them, and in the juries of American citizens who decide them. They should be our pride, not our embarrassment. They deserve to do their jobs.
As complex, as diverse, as relentless as the assault on the rule of law has been, our answer to it is a simple one. Far more than any president’s lawlessness, the American way of justice remains deeply rooted in our character.
That, no president can disturb. So I am full of hope, even on this dark day. I have faith that we can unite security and justice—because we have already done it.
My father, Senator Tom Dodd, was the number two American prosecutor at the famous Nuremberg trials. And I have never, never forgotten the example he set.
As Justice Robert Jackson said in his opening statement at Nuremberg: “That four great nations, flushed with victory and stung with injury, stay the hand of vengeance and voluntarily submit their captive enemies to the judgment of the law is one of the most significant tributes that Power has ever paid to Reason.”
Mr. President, what is the tribute that Power owes to Reason?
That America stands for a transcendent idea.
The idea that laws should rule, not men.
The idea that the Constitution does not get suspended for vengeance.
The idea that this nation should never tailor its eternal principles to the conflict of the moment, because if we did, we would be walking in the footsteps of the enemies we despised.
The tribute that Power owes to Reason is due today. I know that we can find the strength to pay it. And if we can’t? We will all have to answer for it.
There’s a famous military recruiting poster that comes to mind. A man is sitting in an easy chair with his son and daughter on his lap, in some future after the war has ended. His daughter is asking him, “Daddy, what did you do in the war?” And his face is shocked and shamed, because he knows he did nothing.
My daughters, Grace and Christina, are six and three. They are growing up in a time of two great conflicts: one between our nation and its enemies, and another, between what is best and worst in our American soul. And someday soon, I know I am going to hear that question: “What did you do?” I want, more than anything else, to give the right answer.
That question is coming for every single one of us in this body. Every single one of us will be judged by a jury from whom there’s no hiding: our sons, our daughters, our grandchildren. Someday soon, they’ll read in their textbooks the story of a great nation, one that threw down tyrants and oppressors for two centuries; one that rid the world of Nazism and Soviet communism; one that proved that great strength can serve great virtue, that right can truly make might.
And then they will read how, in the early years of the 21st century, that nation lost its way.
We do not have the power to strike that chapter. No, Mr. President—we can’t go back.
We can’t un-destroy the CIA’s interrogation tapes. We can’t un-pass the Military Commissions Act. We can’t un-speak Alberto Gonzales’s disgraceful testimony. We can’t un-torture innocent people. And perhaps, sadly, shamefully, we cannot stop retroactive immunity. We can’t un-do anything that has been done in the last six years for the cause of lawlessness and fear.
We cannot blot out that chapter. But we can begin the next one, even today. Let its first words read: “Finally, in June 2008, the Senate said: ‘Enough.’”
I implore my colleagues to write it with me. I implore my colleagues to vote against retroactive immunity and vote against cloture tomorrow morning.
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https://www.mathcounts.org/categorizedresources | 1,627,799,677,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154163.9/warc/CC-MAIN-20210801061513-20210801091513-00032.warc.gz | 907,101,846 | 16,692 | MATHCOUNTS provides many free problem sets, videos, lesson plans and activities that can complement in-person and online learning. We've categorized some of the best resources for several middle school math topics:
• Arithmetic Skills
• Introduction to Counting
• Basic Number Sense
• Exploring Equations
• Properties of Right Triangles
• Multiple Approaches to Problems
• Squares
• Using Similar Figures
• Area and Perimeter
• Sequences, Series and Patterns (Part 1)
• Sequences, Series and Patterns (Part 2)
Arithmetic Skills
Faster Arithmetic Models
Practice Plan
Using the commutative, associative and distributive properties, Mathletes will arrange arithmetic problems in a different order that allows them to be solved more readily.
Order of Operations and Defining New Rules
Practice Plan
After refreshing Mathletes on the order of operations, the video will then focus on how to solve problems where an unfamiliar symbol is defined to be a new type of operations that follows given rules.
The Multiplication Game
National Math Club Game
In The Multiplication Game players take turns chosing factors to obtain a product on the game board. The first player to four squares in a row wins. The game can be used to practice multiplication tables and factor pairs as well as to discuss prime and composite numbers.
Salute
National Math Club Game
In a heads up style game, students use inverse operations to guess the card on their forehead. They may or may not realize they are doing algebra! Register for the free National Math Club to access this game and dozens of others!
A-maze-ing Fractions
National Math Club Exploration
Operations with fractions are often hard for students to conceptualize. With this exploration's dry erase maze boards and four basic arithmetic operations, Mathletes can begin to uncover the secrets of fractions by finding a path that results in the least value or the greatest value. Register for the free National Math Club to access this activity and dozens of others!
Counting Bee
National Math Club Game
Help students improve their basic arithmetic skills by competing in a club counting bee. Given a starting number and counting number, see how far students can count in 15 seconds! Register for the free National Math Club to access this game and dozens of others!
Introduction to Counting
Counting Shapes in a Complex Figure
Practice Plan
This plan will help Mathletes to develop a strategic approach to counting the occurrences of a certain shape in a more complex figure made of multiple intersecting lines.
Counting Paths Along a Grid
Practice Plan
Explore combinatorics by looking at a common type of MATHCOUNTS counting problem – counting paths between two points. End with an extension that connects counting paths to another type of combinatoric problem.
The Fundamental Counting Principle
Practice Plan
This plan will introduce students to The Fundamental Counting Principle – a faster method to determining the total number of possible outcomes of an event without listing them all out!
Counting Possibilities
MATHCOUNTS Mini
This video focuses on using diagrams and organized lists to ensure that each possible outcome is counted once, and only once.
Constructive Counting
MATHCOUNTS Mini
Moving beyond the fundamental counting principle, students will be introduced to the difference between combinations and permutations, and presented with multiple methods for solving these types of problems.
Counting & Combinatorics Stretch
Problem Set
Two sets of ten practice problems from the 2002-2003 and 2015-2016 MATHCOUNTS School Handbook that cover basic counting including some number sense, shapes and paths.
Basic Number Sense
Divisibility Rules
Practice Plan
Students will apply divisibility rules of various integers to simplify computation, better understand number composition and aid in problem solving. In the extension, Mathletes can prove why each of these rules work!
Least Common Multiple
Practice Plan
Calculating the least common multiple is something many students are asked to do, but in this plan they will use their understanding of the least common multiple to stretch themselves to solve more complex problems.
Marble Challenge
National Math Club Game
In the Marble Challenge students will take turns removing marbles with the goal of not taking the last marble. This game encourages students to notice patterns in the numbers and can even be used to introduce modular arithmetic. Register for the free National Math Club to access this game and dozens of others!
Ken Ken
National Math Club Exploration
Using increasingly popular KenKen® puzzles, Mathletes will use teamwork, number sense and logic skills to solve challenges. Register for the free National Math Club to access this activity and dozens of others!
Strategic Guessing Using Divisibility Rules
MATHCOUNTS Mini
Often in MATHCOUNTS you find yourself with a unique problem you don't already have a prescribed method for solving. This mini gives examples of such problems that can be solved with a little logic, number sense and understanding of divisibility rules.
Number Sense Stretches
Problem Sets
In these number sense stretches, there are three problem sets (10 problems each) from old MATHCOUNTS School Handbooks that covers number sense topics such as factoring and divisibility. These are great additional practice in after trying the Practice Plans and MATHCOUNTS Minis.
Exploring Equations
You Don't Have to Solve for x!
Practice Plan
Often the immediate reaction when Mathletes see an algebraic equation is to solve for the unknown but depending on what you are looking for it might be easier to manipulate the equation without solving it.
Mathemagicians
National Math Club Exploration
This exploration is a great way to practice translating word problems into algebraic equations and to develop understanding of the concept of inverse operations. Mathletes will be amazed at first by what appears to be magic, but they will come to understand that the tricks can be explained using algebra. Mathletes can come up with their own magic examples to impress their friends and families and become true mathemagicians! Register for the free National Math Club to access this activity and dozens of others!
Function Battleship
National Math Club Exploration
This exploration lets Mathletes manipulate functions in order to explore and better understand translating, stretching, compressing and other transformations of functions. Through the Desmos platform, with the added twist of similarity to the board game Battleship, Mathletes can graph functions and see the effects of changing coefficients and exponents and adding and subtracting integers. Register for the free National Math Club to access this activity and dozens of others!
Salute
National Math Club Game
In a heads up style game, students use inverse operations to guess the card on their forehead. They may or may not realize they are doing algebra! Register for the free National Math Club to access this game and dozens of others!
Algebraic Equations from Word Problems
MATHCOUNTS Mini
These problems and video focus on translanting the information in word problems into representative algebraic equations.
Seeing Symmetry in Systems of Equations
MATHCOUNTS Mini
When dealing with systems of equations, if you are able to recognize symmetry between the equations, you can simplify the steps to a solution. This Mini will look at some problems and demonstrate how to find and use the symmetry to your advantage.
Properties of Right Triangles
Special Right Triangles
Practice Plan
Mathletes will become familiar with properties of 45-45-90 and 30-60-90 triangles. In this plan, the relationships between the sides of these two special right triangles will be derived. Then, Mathletes will apply these to solve for unknown lengths in geometric figures.
Right Triangles
MATHCOUNTS Mini
From special right triangles to Pythagorean Triples, this video shows how to use properties of right triangles to solve problems.
Proofigami
National Math Club Exploration
This exploration gives Mathletes a brief introduction of the Pythagorean Theorem, then guides them through what we call Proofigami. This fun exploration will feel a lot like origami, but will provide Mathletes with a better understanding of the Pythagorean Theorem and gives club leaders a visual and tactile tool that makes explaining this proof easier. Register for the free National Math Club to access this activity and dozens of others!
30-60-90 Right Triangles
MATHCOUNTS Mini
This MATHCOUNTS Mini will look at ways to use known ratios of 30-60-90 triangles to help solve more complex geometric problems.
Right Triangles Stretch
Problem Set
Practice solving problems by using the Pythagorean Theorem, recognizing Pythagorean triples and applying properties of special right triangles.
Trapezoids and Triangles
MATHCOUNTS Mini
This video explores how we can decompose a figure into trapezoids and triangles to determine its area. The problems associated with this mini will help students determine when and how to apply their right triangle knowledge to solve more complex geometry problems.
Multiple Approaches to Problems
More Than One Way to Solve a Problem
MATHCOUNTS Mini
This video demonstrates multiple problem-solving strategies and emphasizes the importance of solving problems in more than one way to verify that you've solved a problem correctly.
Even More Than One Way to Solve a Problem
MATHCOUNTS Mini
This video reinforces the concept of solving a problem multiple ways to validate your answer.
Fun Problem-Solving Techniques
National Math Club Problem Set
Being able to take multiple different approaches to solve problems is an invaluable skill. In this problem set, students will look at four techniques - creating a model, acting out a situation, drawing a picture and making a list.
Three Tic-Tac-Toes
National Math Club Game
Chances are students are familiar with tic-tac-toe, but these rule variants on the traditional version will challenge students to rethink their strategy. Use this game to talk about symmetry, logic and proof writing.
Draw a Picture
MATHCOUNTS Mini
This video explores how to solve problems by drawing a picture to organize the given information.
Make a Sketch
MATHCOUNTS Mini
This video demonstrates how making a sketch of a given scenario can be a useful strategy when solving problems.
Squares
Recognizing Squares and Solving a Simpler Problem
MATHCOUNTS Mini
This video focuses on recognizing squares and using them to solve a simpler problem.
Using the Difference of Squares to Solve Problems
MATHCOUNTS Mini
This video explores how to use the difference of squares to solve problems and why this method works.
Systems of Equations Stretch
Problem Set
Apply the difference of squares formula in order to solve problems involving systems of equations.
Difference of Squares
Practice Plan
An important formula to know, the difference of squares identity is derived geometrically in the video for this problem set. Mathletes will then try to recognize the difference of squares structure in various expressions and use the identity to find the value.
Perfect Squares/Using a Simpler Case to Solve a Problem
MATHCOUNTS Mini
This video demonstrates how to use perfect squares to find a simpler case to help solve a problem.
Difference of Squares
MATHCOUNTS Mini
This video demonstrates how to solve problems using the difference of squares.
Using Similar Figures
Similar Triangles and Proportional Reasoning
MATHCOUNTS Mini
This video shows how to identify and use similar triangles to solve geometry problems.
Using Similarity to Solve Geometry Problems
MATHCOUNTS Mini
This video explores how to apply properties of similar triangles in solving problems about two-dimensional and three-dimensional figures.
Similarity and Proportional Reasoning Stretches
Problem Sets
Practice with the concept of similarity by answering questions about similar figures, and see how similarity relates to proportional reasoning and geometric transformations.
Using Similarity to Solve Geometry Problems
MATHCOUNTS Mini
This video demonstrates how to use similarity and proportional reasoning to solve difficult geometry problems.
Similarity and Proportional Reasoning
MATHCOUNTS Mini
Sometimes it is necessary to create the similar triangles you'll need in order to solve a problem. This video shows how to look at and build on given diagrams to create similar figures.
Similar Triangles
MATHCOUNTS Mini
This video explores how to use parallel lines and angles to identify similar triangles and solve problems.
Area and Perimeter
Fence Me In
National Math Club Game
After rolling dice to determine the size, in part, of a rectangle, players then use perimeter and area formulas to determine dimensions. The goal is to try to fill up the board first.
Areas of Irregular Convex Polygons
MATHCOUNTS Mini
This video demonstrates two strategies for how to find the area of an irregular convex polygon.
Geometry Stretches
Problem Sets
Find the areas and perimeters of various figures, and see how area and perimeter measurements can be used to solve other types of geometry problems.
Area of Irregular Polygons Reboot
MATHCOUNTS Mini
This video demonstrates how to find the area of an irregular polygon by dividing the figure into smaller regions for which the area is more easily determined.
Trapezoids and Triangles
MATHCOUNTS Mini
This video explores how we can decompose a figure into trapezoids and triangles to determine its area.
Shamrock
Problem of the Week
Practice calculating area and perimeter measurements using the image of a shamrock.
Sequences, Series and Patterns (Part 1)
Number Sense: Looking for Patterns
MATHCOUNTS Mini
This video focuses on techniques for solving problems by looking for patterns that emerge among the digits in large numbers.
Patterns All Around
National Math Club Game
Recognizing patterns in objects in order to express them mathematically is an important skill for students to learn. In this game students will attempt to recognize visual and numeric patterns in a group of cards.
Sequences, Series and Patterns
MATHCOUNTS Mini
This video shows how to find patterns in both visual and numerical sequences and how to use patterns to identify an unknown value in a sequence.
More Sequences, Series and Patterns
MATHCOUNTS Mini
This video demonstrates how to find a pattern in a sequence or series, and prove that it works, to solve problems.
Representing Patterns Numerically
Practice Plan
In this practice plan, Mathletes will recognize visual patterns and practice defining them numerically in order to find the number of elements in the pattern after a large number of repetitions.
Patterns Stretches
Problem Sets
Practice with patterns through problems about visual and numerical sequences and series, the digits of large numbers and other real-world and math topics.
Sequences, Series and Patterns (Part 2)
Arithmetic and Geometric Sequences
MATHCOUNTS Mini
This video explores how to solve problems about arithmetic and geometric sequences.
Relationships Between Arithmetic Sequences, Mean and Median
MATHCOUNTS Mini
This video demonstrates how to use mean and median in solving problems about arithmetic sequences.
Arithmetic Sequences
MATHCOUNTS Mini
This video focuses on techniques for solving problems involving arithmetic sequences, including finding the nth term.
Sequences Stretches
Problem Sets
Practice with standard arithmetic and geometric sequences and series, as well as with other special types of sequences and series, like the Fibonacci sequence.
Patterns, Sequences and Series
MATHCOUNTS Mini
This video shows a few techniques for solving problems using patterns in sequences and series.
Sequences and Central Tendency
MATHCOUNTS Mini
This video demonstrates how the relationship between measures of central tendency and sequences can be used to solve problems. | 3,181 | 16,042 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2021-31 | latest | en | 0.899571 |
http://math.stackexchange.com/questions/64923/why-cant-we-interchange-differentiation-with-taking-a-limit-of-a-series-of-func | 1,469,542,589,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824994.73/warc/CC-MAIN-20160723071024-00250-ip-10-185-27-174.ec2.internal.warc.gz | 162,854,776 | 18,026 | # Why can't we interchange differentiation with taking a limit of a series of functions?
While learning a little Fourier analysis, I ran into this interesting phenomenon:
Consider a series of sawtooth waves such that the height and width of the sawteeth shrinks to zero, but the slope of the sawteeth remains the same. To be specific, let
$$f_n(x) = \frac{nx - \lfloor nx\rfloor}{n}$$
Then define
$$F(x) = \lim_{n\to\infty}f_n(x)$$
It seems intuitively clear that $F(x) = 0$ for all $x$ because the global maximum of $f_n$ is $\frac{1}{n}$.
If $F(x) = 0$, then we should have $F'(x) = 0$ as well. However, if we choose an irrational value of $x$, then $f'_n(x) = 1$ for all $n$, so if $F'(x)$ is found instead by taking
$$F'(x) = \lim_{n\to\infty}f'_n(x)$$
we do not get $F'(x) = 0$.
It seems like the derivative of a limit is not the same as the limit of a derivative, which is pretty counterintuitive to me.
What's going on?
-
This example illustrates the principle well: Functions can be close while their derivatives are not. The standard image to invoke is similar to what you have given here: one function can be constant, while a "nearby" function can stay close to that constant function while oscillating maniacally. Another example to consider would be $\frac{\sin(nx)}{n}\to 0$. – Jonas Meyer Sep 15 '11 at 22:44
That's correct, they are not the same. (In particular, $\lim f_n'(x)$ exists only when $x$ is irrational, and is then equal to $1$ - can you see why?) Since the derivative is itself a limit, this is also a special case of the fact you cannot (in general) interchange two limits. – anon Sep 15 '11 at 22:46
@Mark Eichenlaub : As it has been stated, I'd like to mention that there is a possibility that a lot of interesting things missing when we use convergence under a norm like the $\mathcal{L}^p$. – Rajesh Dachiraju Sep 16 '11 at 8:34
You may want to have a look at this as well, although a different, you might find it interseting. en.wikipedia.org/wiki/Uniform_convergence under the section 'Applications'. It talks about the differentiability and the derivative of the limit function under pointwise convergence. – Rajesh Dachiraju Sep 16 '11 at 8:38 | 613 | 2,192 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2016-30 | latest | en | 0.917103 |
https://us.metamath.org/mpeuni/dfsb2.html | 1,719,104,552,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862425.28/warc/CC-MAIN-20240623001858-20240623031858-00673.warc.gz | 540,471,678 | 4,347 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > dfsb2 Structured version Visualization version GIF version
Theorem dfsb2 2532
Description: An alternate definition of proper substitution that, like dfsb1 2511, mixes free and bound variables to avoid distinct variable requirements. Usage of this theorem is discouraged because it depends on ax-13 2391. (Contributed by NM, 17-Feb-2005.) (New usage is discouraged.)
Assertion
Ref Expression
dfsb2 ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦𝜑) ∨ ∀𝑥(𝑥 = 𝑦𝜑)))
Proof of Theorem dfsb2
StepHypRef Expression
1 sp 2183 . . . 4 (∀𝑥 𝑥 = 𝑦𝑥 = 𝑦)
2 sbequ2 2251 . . . . 5 (𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑𝜑))
32sps 2185 . . . 4 (∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑𝜑))
4 orc 864 . . . 4 ((𝑥 = 𝑦𝜑) → ((𝑥 = 𝑦𝜑) ∨ ∀𝑥(𝑥 = 𝑦𝜑)))
51, 3, 4syl6an 683 . . 3 (∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑 → ((𝑥 = 𝑦𝜑) ∨ ∀𝑥(𝑥 = 𝑦𝜑))))
6 sb4b 2500 . . . 4 (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦𝜑)))
7 olc 865 . . . 4 (∀𝑥(𝑥 = 𝑦𝜑) → ((𝑥 = 𝑦𝜑) ∨ ∀𝑥(𝑥 = 𝑦𝜑)))
86, 7syl6bi 256 . . 3 (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑 → ((𝑥 = 𝑦𝜑) ∨ ∀𝑥(𝑥 = 𝑦𝜑))))
95, 8pm2.61i 185 . 2 ([𝑦 / 𝑥]𝜑 → ((𝑥 = 𝑦𝜑) ∨ ∀𝑥(𝑥 = 𝑦𝜑)))
10 sbequ1 2250 . . . 4 (𝑥 = 𝑦 → (𝜑 → [𝑦 / 𝑥]𝜑))
1110imp 410 . . 3 ((𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑)
12 sb2 2505 . . 3 (∀𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑)
1311, 12jaoi 854 . 2 (((𝑥 = 𝑦𝜑) ∨ ∀𝑥(𝑥 = 𝑦𝜑)) → [𝑦 / 𝑥]𝜑)
149, 13impbii 212 1 ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦𝜑) ∨ ∀𝑥(𝑥 = 𝑦𝜑)))
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 → wi 4 ↔ wb 209 ∧ wa 399 ∨ wo 844 ∀wal 1536 [wsb 2069 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1911 ax-6 1970 ax-7 2015 ax-10 2145 ax-12 2178 ax-13 2391 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-ex 1782 df-nf 1786 df-sb 2070 This theorem is referenced by: dfsb3 2533
Copyright terms: Public domain W3C validator | 1,116 | 1,823 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-26 | latest | en | 0.32952 |
https://www.888casino.com/blog/side-bets/perfect-play-against-the-213-extreme-blackjack-side-bet | 1,726,407,453,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651630.14/warc/CC-MAIN-20240915120545-20240915150545-00267.warc.gz | 581,885,901 | 24,663 | One of the first games I posted about on this blog was the blackjack side bet Twenty One Plus Three (21+3). In this post, I showed that there was very little possible profit against 21+3, even using perfect play. Nevertheless, this post on card counting 21+3 is the most viewed blackjack side bet article in this blog, getting about 70% more hits than this second place post on Lucky Ladies. The 21+3 post ranks 6th in overall views. It continues to be one of the top posts accessed each day. I don’t know why.
For over two years, I have known about a version of 21+3 called “21+3 Extreme” (21+3E). This version pays for the same events as 21+3, but uses a pay table leading to a higher house edge and greater volatility. Because of the extremely low vulnerability of 21+3 to advantage play and the higher house edge of 21+3E, I long ago concluded that 21+3E would be a waste of time to analyze. However, given the continued interest in 21+3 (and a lack of anything better to do), I decided to finally give 21+3E a look. The results of this analysis met my extraordinarily low expectations.
This post covers the six-deck case. With fewer decks, the house edge for 21+3E increases, making it even less vulnerable.
Both 21+3 and 21+3E pay based on the poker value of the three card poker hand consisting of the player’s first two cards and dealer’s up card. The paying hands are Straight Flush, Three of a Kind, Straight and Flush. In the case of 21+3, each of these pays 9-to-1. In the case of 21+3E, the payout for these events is 30-20-10-5.
The following tables give the combinatorial analysis for 21+3 and 21+3E:
In particular the house edge for 21+3 is 3.24%, whereas the house edge for 21+3E is 13.39%. The term “Extreme” certainly describes this side bet: it is “Extremely” good for the casino and “Extremely” bad for the player.
The following Excel spread sheet allows the user to input the exact composition of the shoe and the pay table for the variation of 21+3 in use; it outputs the house edge, hit frequency and standard deviation. I suppose in some universe a spread sheet like this could be used for advantage play against 21+3. Then again, in some universe, a Komodo dragon is the current world chess champion.
Twenty_One_Plus_Three_CA
In investigating card counting 21+3E, the AP can do no better than perfect play (for example, by using the spread sheet given above), where the AP makes a wager on 21+3E if and only if he has the edge. The following table shows the results of perfect play against 21+3E in the six-deck case when the cut card is placed at 260 cards. These results confirmed my intuitive expectation that 21+3E has almost no card counting vulnerability:
The opportunity to make a 21+3E wager with the edge occurs about 3 times per 1000 hands. When the player makes a 21+3E wager, his average edge is about 4.08%. This leads to a gain of about 0.0122 units per 100 hands. If an AP is making a \$100 wager whenever he has the edge over 21+3E, and otherwise does not make a wager, then he will earn about \$1.22 per 100 hands.
The final number given in this table, the desirability index (DI), is discussed for blackjack side bets in this post. The value 0.73 given in this table shows that 21+3E has “minimal vulnerability” to advantage play. (I note that I used the standard deviation from the combinatorial analysis for 21+3E in computing the DI).
To get a different view on the vulnerability of 21+3E, I kept track of the average edge and bet frequency by the location in the shoe at the start of each hand. As noted above, I assumed the cut card was placed at 260 cards, so the simulation did not include values beyond that point. I found that there were essentially no advantageous situations before 175 cards were played from the shoe.
The following plot gives perfect-play results by first-card location, based on a simulation of one million (1,000,000) six-deck shoes with the cut card at 260 cards (click for full-sized image).
The average edge is jagged to begin with. This occurred because there were very few data points being averaged together. As the depth into the shoe increased, there were more instances of advantageous situations, so the average edge smoothed out. The average edge appears to be increasing in direct proportion to the card location, whereas the bet frequency appears to be increasing quadratically. At exactly the 260 card position, the values that the simulation produced were a bet frequency of 5.08% and an average edge of 5.52%.
My conclusion is that 21+3E has a minimal vulnerability to advantage play. No serious AP is going to give 21+3E a second thought without some other vulnerability at the game. For example, if the shuffle is highly trackable and a slug of cards of the same suit can be followed through the shuffle, then it could be very profitable to play 21+3E during the slug.
My recommendations for game protection for 21+3E are as follows: | 1,195 | 4,942 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-38 | latest | en | 0.936127 |
https://studysaga.in/what-is-the-square-root-4/ | 1,725,872,617,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651092.31/warc/CC-MAIN-20240909071529-20240909101529-00701.warc.gz | 519,926,666 | 25,401 | ## what is the square root -4
The square root of a negative number such as -4 is a complex number, and cannot be represented on the real number line. In complex analysis, the square root of -1 is represented by the imaginary unit i, which is defined as i = √(-1). Therefore, the square root of -4 can be written as:
√(-4) = √(4) * √(-1) = 2i
So the square root of -4 is 2i, where i is the imaginary unit. This result is used in many mathematical and engineering applications, including electrical engineering, signal processing, and quantum mechanics.
Calculate other square root value here | 143 | 594 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-38 | latest | en | 0.919079 |
https://forum.nutsvolts.com/viewtopic.php?f=45&t=1862 | 1,611,090,904,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519784.35/warc/CC-MAIN-20210119201033-20210119231033-00322.warc.gz | 366,177,063 | 9,970 | ## Capacitive discharge
This is the place for any magazine-related discussions that don't fit in any of the column discussion boards below.
camino75080
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### Capacitive discharge
Man, am I on a roll tonight. Anyway a am looking for a formula calculate the current of a capacitive dischare, varibles: voltage, current (duh), and resistance. So far I've got this I=(((C*C)*V*R)/2). Please help me.
bodgy
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### Re: Capacitive discharge
From your formula I think you are talking about Joules.<p>W = (C * (V^2)) * 0.5<p>For (dis)charging current the formula is<p>I = C * V /t
OR more correctly<p>Q = C * V<p>I = Q/t<p>Where 't' is the time at a particular instant and Q is the charge in Coulombs.<p>So a 100uF capacitor with 12v across it <p>Q = 1.2mC<p>After 100mS of charging time the current will be<p>I = 1.2mC /100mS = 12mA<p>Colin<p>[ September 06, 2004: Message edited by: bodgy ]</p>
On a clear disk you can seek forever.
camino75080
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### Re: Capacitive discharge
Colin, thanks for the formula, i was way off, but I don't understand how that works without having the resistance. If you can, would you please explain that. Thanks again for the help.
Chris Smith
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### Re: Capacitive discharge
Resistance is everything. Think of a water tower and many different size pipes either feeding it or discharging it. The bigger the pipe size the less resistance, [the longer it is the more resistance] the faster it fills or empties and Vice Versa. <p>The tank size is the capacity, or capacitance, and the height of the tower is pressure [psi], or volts, while the pipe size is a division of the perfect zero ohms, [which really isn’t possible with out super cooling], so everything is more than Zero. One inch is half of two, one ohm is half of two ohms, etc.
camino75080
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### Re: Capacitive discharge
Okay, thanks. So if I plugin time time constant for a given resistor/capacitor combination, for S then I get the current . Thanks again for the help
MrAl
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### Re: Capacitive discharge
Hi there,<p>If you are using a cap being discharged by
a resistor you can use this 'formula' which
by the way includes the resistance and is exact:<p>i=i0*e^(-t/RC)<p>where
i is the instantaneous current at time t,
i0 is the initial current at time t=0,
t is the time in seconds,
RC is the RC time constant = R*C with
R in ohms and C in farads,<p>To get the initial current i0 knowing the
voltage across the cap Vc simply divide
Vc by the resistance R:
i0=Vc/R
where Vc is the voltage across the cap at time
t=0.<p>Also, the voltage across the cap is:
v=v0*e^(-t/RC)
where v0 is the voltage at time t=0.<p>
Take care,
Al<p>PS. The formulas like dv=i*dt/C are only
accurate for very small time intervals--
'small' relative to the time constant.
LEDs vs Bulbs, LEDs are winning.
camino75080
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### Re: Capacitive discharge
Thanks Al, I dint realize there were there were other formulas, I looked on the internet and only for PHD research papers, that sucked, I could make any sense of them!
jwax
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### Re: Capacitive discharge
You'll find that the resistance of an arc discharge in air is a very low value. A thousand volts at a thousand amperes is an ohm, although it's over in a microsecond or so.
bodgy
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### Re: Capacitive discharge
<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>
<p>
Also, the voltage across the cap is:
v=v0*e^(-t/RC)
where v0 is the voltage at time t=0.<p>
<hr></blockquote><p>Of course I shouldn't have been lazy, and should have posted the full formula.<p>But Mr Al shouldn't the voltage across C be -<p>Vc = V0 * [1- e ^(-t/CR)] ?<p>where e is the nat exponent 2.71828<p>Just to add voltage discharge is normally measured at Vc * 0.37 and Current at I * 0.63<p>For charging it is the other way around Vc * 0.63 and I * 0.37<p>Colin
On a clear disk you can seek forever.
camino75080
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### Re: Capacitive discharge
Buy chance, I am calculating the current though an arc. What would be a good aproximation of the arc's resistance?
jwax
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### Re: Capacitive discharge
Take a look at N & V magazine, Sept 2004, p.54-"The Anatomy of a Spark".
There can be a leakage path through air from high voltage of milliamps, or a catastrophic breakdown that carries thousands of amps. The energy available determines the arc properties, along with temperature, humidity, electrode shape, etc.
So the "resistance" of the discharge depends on all those factors.
Chris Smith
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### Re: Capacitive discharge
The [over] general value for an arc is 25k/v per inch. This does not include many factors, but is just a figure for Dry air, general over simplistic conditions such as leakage through a break in insulation to ground.
MrAl
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### Re: Capacitive discharge
Hi again,<p>bodgy:<p>The equation:
v=v0*e^(-t/RC)<p>is actually correct for a *discharging* cap,
where
v is the voltage across the cap and
v0 is the initial voltage across the cap
and the resistor is connected right across
the cap so as to discharge it.<p>
If the cap were *charging* then it would be
v=vp*[1-e^(-t/RC)]
where
v is the voltage across the cap and
vp is the power supply voltage (resistor
connects between it and the cap).<p>
Take care,
Al
LEDs vs Bulbs, LEDs are winning.
camino75080
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### Re: Capacitive discharge
Is the formula for a cap's charging current, I=Io*(1-e^(-t/RC))? Thanks or all the help.<p>[ September 08, 2004: Message edited by: camino75080 ]</p>
MrAl
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### Re: Capacitive discharge
Hi there,<p>Here are all the formula's with the
following resistor connections:<p>When discharging the cap the resistor is connected across the cap.
When charging the cap the resistor is connected
power supply voltage (vp) and the other end
of the cap is grounded (zero volts).<p>Also, the voltage v is across the cap and
the current i is through the cap.<p>CHARGING VOLTS:
v=vp*[1-e^(-t/RC)]<p>CHARGING AMPS:
i=i0*e^(-t/RC), i0=(vp-vc)/R<p>DISCHARGING VOLTS:
v=v0*e^(-t/RC), v0=initial v across cap<p>DISCHARGING AMPS:
i=i0*e^(-t/RC), i0=initial i through cap<p>Note there is only one formula with the "1" in
it and that is for the voltage when charging.<p>If i get a chance i'll post some graphs if you
think that will help?<p>Take care,
Al
LEDs vs Bulbs, LEDs are winning.
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http://barkerhugh.blogspot.com/2015/02/collatz-restated-without-halving.html | 1,534,631,640,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213794.40/warc/CC-MAIN-20180818213032-20180818233032-00390.warc.gz | 46,426,237 | 13,606 | ## Friday, 20 February 2015
### Collatz - restated without the halving.
Here's the way of reformulating Collatz chains that I have been using. This gives us a way to express the conditions for 1) the existence of a loop and 2) the conjecture being true.
Start from n, any odd number
Step 1.
Multiply by 3 and add 1 = 3n+1
Step 2
Multiply by 3 = 9n+ 3
Factor as (2^a)(an odd number)
Add 2^a = 9n+3+2^a
Step 3...
Multiply by 3 = 27n + 9 + 3(2^a)
Factor as (2^b)(an odd number)...
Add 2^b = 27n + 9 + 3(2^a) + 2^b
Repeat until we reach 2^z(n) or 2^z
As the total increases, the cumulative total is
[3^y x n] + [3^(y-1)] + [2^a x 3^(y-2)] + [2^b x 3^(y-3)] …. + [2^c x 3]+ 2^z...
The power of 3 falls by one in each term, until we reach 3^0 in the final term
The power of 2 increases in each term but can rise by more than 1 at a time – so in the example above d>c>b>a.
If the Collatz conjecture is true, we will always reach an equation of the form
[3^y x n] + [3^(y-1)] + [2^a x 3^(y-2)] + [2^b x 3^(y-3)] …. + [2^c x 3]+ 2^d = 2^z
For there to be a 3n+1 loop, which would be a counterexample to the conjecture, we would need this equation to be true:
[3^y x n] + [3^(y-1)] + [2^a x 3^(y-2)] + [2^b x 3^(y-3)] …. + [2^c x 3]+ 2^d = 2^z(n)
We can adapt this method for 5n+1 chains by replacing all the 3s with 5s.
#### 2 comments:
1. do you study math aswell??
2. Wonderful Blog! Do you Matlab to construct these grids or any other program?
Cheers prime | 565 | 1,464 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2018-34 | longest | en | 0.882423 |
http://mathhelpforum.com/algebra/6248-solve-3-2x-1-26-3-x-9-0-a.html | 1,519,369,079,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814493.90/warc/CC-MAIN-20180223055326-20180223075326-00471.warc.gz | 224,943,239 | 9,933 | Thread: solve 3^(2x+1) - 26(3^x) - 9 = 0
1. solve 3^(2x+1) - 26(3^x) - 9 = 0
does anyone know how i make this look like a quadratic? im guessing thats whats uv gotta do. cheers
2. Originally Posted by 24680
does anyone know how i make this look like a quadratic? im guessing thats whats uv gotta do. cheers
3^(2x+1) - 26(3^x) - 9 = 0
3 (3^x)^2 -26 3^x - 9 = 0
so let y=3^x, then we have:
3 y^2 -26 y -9 = 0
RonL
,
,
,
,
,
,
3^(2x 1)-4×3(x 1) 9=0
Click on a term to search for related topics. | 211 | 506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2018-09 | longest | en | 0.793671 |
http://math.stackexchange.com/questions/528375/proof-that-a-subset-of-r-has-same-cardinality-as-r | 1,469,400,453,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824185.14/warc/CC-MAIN-20160723071024-00039-ip-10-185-27-174.ec2.internal.warc.gz | 152,105,696 | 19,770 | # Proof that a subset of R has same cardinality as R
I am sitting with a task where I have to prove the following:
Claim: Every subset of $\mathbb{R}$, that contains an interval $I$ with $a < b$, has the same cardinality as $\mathbb{R}$.
So I think that I should prove that there exist a bijection from $I$ to $\mathbb{R}$? I'm kinda lost, and don't know how to start.
There is a lemma 3 in the book saying: Let $a,b \in \mathbb{R}$ with $-\infty\neq a < b \neq \infty$. There exist a $f\colon ]-1; 1[ \to ]a; b[$ that is bijektiv. The intervals $]-1; 1[$ and $]a; b[$ has same cardinality.
Another lemma 4 says: $f\colon ]0; 1[ \to ]1; \infty[$, $x\to \frac{1}{x}$, is bijective. The intervals $f\colon ]0; 1[ \to ]1; \infty[$ have same cardinality.
(There is an image added to lemma4)
The above interval $]-1; 1[$ has same cardinality as $\mathbb{R}$, and the $f$ is bijective Any help is highly appreciated
-
Let $A$ be a subset that contains a non-trivial interval $I$. Then $\Bbb R=_c I\leq _c A\leq _c\Bbb R$. – Git Gud Oct 16 '13 at 11:44
Are you allowed to use the fact that cardinalities are ordered? That is, do you know that if a subset of $\mathbb{R}$ has a cardinality "not less than" $|\mathbb{R}|$ then it has cardinality equal to $|\mathbb{R}|$? – Eric Stucky Oct 16 '13 at 11:59
Hi. Yes I think I am. Isn't that what the Bernsteins-Schröders sentence says? – user1960836 Oct 16 '13 at 12:02
Hints:
First prove that any two non-empty open intervals have the same cardinality.
Second, pass now to use the nice interval $\;(-\pi/2\,,\,\pi/2)\;$ and a rather nice, simple trigonometric function to show equipotency with $\;\Bbb R\;$
-
There is also a simple rational function that maps $(-1,1)$ to $\mathbb R$ bijectively. – lhf Oct 16 '13 at 11:51
Thanks for the replies people. I am really not good at proofs in maths. That's why I have paid for the course that I am following, so that I can learn more. The posts you have provided so far, doesn't help me more than the lemmas provided, which I can't use to my advantage. Please don't understand this in a negative way, I just don't know how else to say it. So can anyone be a little more specific, or maybe with examples? Thanks – user1960836 Oct 16 '13 at 11:56
For a bijection from $I$ to $\mathbb R$ chose $$f: (a,b) \to (-1,1), \qquad x\mapsto 2\frac{x-a}{b-a} - 1$$ which is (obviously) bijective. Then $$g: (-1,1) \to \mathbb R, \qquad \begin{cases}\frac{1}{x} - 1&x\in(0,1)\\0&x=0\\\frac{1}{x} + 1&x\in(-1,0)\end{cases}$$ (Taken from Lemma 4) $$h := g\circ f : (a,b) \to\mathbb R$$ is bijective as a composition of bijective maps.
-
At this part: f:(a,b)→(−1,1),x↦x−ab−a−1. How did you get to -1 at: x↦x−ab−a−1? Also... What is the conclusion on this: h:=g∘f:(a,b)→R? Why do you do that part? – user1960836 Oct 16 '13 at 13:01
Please use MathJax, your text is very hard to read ($\$\text{\frac\{a\}\{b\}}\$$= \frac a b) Second, f(a) = -1, f(b) = 1 - since f is linear we know by mean-value theorem f((a,b)) = (-1,1). The h is one bijection from (a,b) onto \mathbb R – AlexR Oct 16 '13 at 13:24 I see that you edited the f function and added the number 2 suddenly? And thank you very much for the help and comments. I am thinking. There should be a comment on the interval (a, b), according to Bernsteins-Schröders sentence? won't it clarify the argument why f is bijective? I know that you can just test it with values, but still, for a more formal argument and proof? – user1960836 Oct 16 '13 at 14:06 I am trying to figure out how you got to that formular of the function f. If it is not too much to ask, I'd be really grateful if you could tell me – user1960836 Oct 16 '13 at 14:23 It's a solution of a simple linear equation: Find f(x) = ax + b s.t. f(x_0) = y_0, f(x_1) = y_1. Using the constraints$$f(a) = -1, f(b) = 1$$you get the desired result. The inverse of f can be found by the same method and$$f^{-1}(-1) = a, f^{-1}(1) = b – AlexR Oct 16 '13 at 15:29
No one has pointed out yet that to prove the statement of the OP (where a subset contains an interval), you're also going to need to invoke the theorem that bears the names Cantor, Schroeder, and Bernstein (in some order or other).
The idea is to construct an injection from $\mathbb{R}$ into the subset $S$ by injecting it into an interval $(a, b)$ inside $S$ (the other answers describe how to do this), and of course we have an injection from $S$ into $\mathbb{R}$ given by subset inclusion. Then apply the Cantor-Schroeder-Bernstein theorem, which says that if there is an injective function $f: A \to B$ and an injective function $g: B \to A$, then there exists a bijection $\phi: A \to B$.
- | 1,520 | 4,653 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2016-30 | latest | en | 0.899983 |
http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/liminfsol2directory/LimInfSol2.html | 1,723,310,338,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640810581.60/warc/CC-MAIN-20240810155525-20240810185525-00079.warc.gz | 49,537,599 | 3,608 | ### SOLUTIONS TO LIMITS OF FUNCTIONS AS X APPROACHES PLUS OR MINUS INFINITY
SOLUTION 13 :
=
(This is true because the expression approaches and the expression x + 3 approaches as x approaches . The next step follows from the following simple fact. If A is a positive quantity, then = A . )
=
=
=
(You will learn later that the previous step is valid because of the continuity of the square root function.)
=
(Inside the square root sign lies an indeterminate form. Circumvent it by dividing each term by , the highest power of x inside the square root sign.)
=
=
=
(Each of the three expressions , , and approaches 0 as x approaches .)
=
=
= .
Click HERE to return to the list of problems.
SOLUTION 14 :
=
(This is true because the expression approaches and the expression x + 3 approaches as x approaches . The next step follows from the following simple fact. If A is a negative quantity, then = - A so that = - ( - A ) = A . Please make sure that you think about and understand this before proceeding. )
=
=
=
(You will learn later that the previous step is valid because of the continuity of the square root function.)
=
(Inside the square root sign lies an indeterminate form. Circumvent it by dividing each term by , the highest power of x inside the square root sign.)
=
=
=
(Each of the three expressions , , and approaches 0 as x approaches .)
=
=
= .
Click HERE to return to the list of problems.
SOLUTION 15 :
=
(You will learn later that the previous step is valid because of the continuity of the logarithm function. Note also that the expression leads to the indeterminate form . Circumvent it by dividing each term by , the highest power of x .)
=
=
=
(The term approaches 0 as x approaches .)
=
=
= 0 .
Click HERE to return to the list of problems.
SOLUTION 16 :
=
(You will learn later that the previous step is valid because of the continuity of the cosine function.)
=
=
(The expression leads to the indeterminate form . Circumvent it by dividing each term by , the highest power of x in the expression.)
=
=
=
(Each of the terms and approaches 0 as x approaches .)
=
=
= .
Click HERE to return to the list of problems.
SOLUTION 17 :
(As x approaches each of the expressions and approaches 0. The following steps explain why.)
=
=
=
=
= 0 .
Click HERE to return to the list of problems.
SOLUTION 18 :
=
(Circumvent this indeterminate form by dividing each term in the expression by . Division by also works . You might want to try it both ways to convince yourself of this. Also, BEWARE of making one of the following common MISTAKES : = or \ = .)
=
=
=
(Since approaches 0 and approaches as x approaches , we get the following resultant limit.)
=
= .
(Thus, the limit does not exist.)
Click HERE to return to the list of problems.
SOLUTION 19 :
= `` '' truein truein (BEWARE of making the following common MISTAKE : = . Realize also that the form `` '' is an indeterminate one ! It is not equal to 1 ! Circumvent it in the following algebraic ways.)
=
=
(Factor out the term . If you have time, try factoring out the term to convince yourself that it DOESN'T seem to help !)
=
=
=
=
=
(The expressions and approach 0 as x approaches .)
=
= .
= 9 .
Click HERE to return to the list of problems.
Duane Kouba
Wed Apr 2 10:10:40 PST 1997 | 787 | 3,347 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-33 | latest | en | 0.939758 |
https://cynohub.com/vtu-b-tech-aeronatical-engineering-3rd-sem-syllabus-for-complex-analysis-probability-and-statistical-methods-pdf-2022/ | 1,660,876,089,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573540.20/warc/CC-MAIN-20220819005802-20220819035802-00239.warc.gz | 186,507,660 | 22,115 | # VTU B.TECH AERONATICAL-Engineering 3rd SEM Syllabus For Complex analysis, probability and statistical methods PDF 2022
### Get Complete Lecture Notes for Complex analysis, probability and statistical methods on Cynohub APP
You will be able to find information about Complex analysis, probability and statistical methods along with its Course Objectives and Course outcomes and also a list of textbook and reference books in this blog.You will get to learn a lot of new stuff and resolve a lot of questions you may have regarding Complex analysis, probability and statistical methods after reading this blog. Complex analysis, probability and statistical methods has 5 units altogether and you will be able to find notes for every unit on the CynoHub app. Complex analysis, probability and statistical methods can be learnt easily as long as you have a well planned study schedule and practice all the previous question papers, which are also available on the CynoHub app.
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### Complex analysis, probability and statistical methods Unit One
#### Module-1
Calculus of complex functions: Review of function of a complex variable, limits, continuity, and differentiability. Analytic functions: Cauchy-Riemann equations in Cartesian and polar forms and consequences. Construction of analytic functions: Milne-Thomson method-Problems.
### Complex analysis, probability and statistical methods Unit Two
#### Module-2
Conformal transformations: Introduction. Discussion of transformations:ᡵ=ᡒ⡰,ᡵ=ᡗこ,ᡵ=ᡸ+⡩こ,䙦ᡸ≠0䙧.Bilinear transformations- Problems.Complex integration: Line integral of a complex function-Cauchy’s theorem and Cauchy’s integral formula and problems.
### Complex analysis, probability and statistical methods Unit Three
#### Module-3
Probability Distributions: Review of basic probability theory. Random variables (discrete and continuous), probability mass/density functions. Binomial, Poisson, exponential and normal distributions- problems (No derivation for mean and standard deviation)-Illustrative examples.
### Complex analysis, probability and statistical methods Unit Four
#### Module-4
Statistical Methods: Correlation and regression-Karl Pearson’s coefficient of correlation and rank correlation -problems. Regression analysis- lines of regression –problems.Curve Fitting: Curve fitting by the method of least squares- fitting the curves of the form-ᡷ=ᡓᡶ+ᡔ,ᡷ=ᡓᡶ〩ᡓᡦᡖᡷ=ᡓᡶ⡰+ᡔᡶ+ᡕ.
### Complex analysis, probability and statistical methods Unit Five
#### Module-5
Joint probability distribution: Joint Probability distribution for two discrete random variables, expectation and covariance. Sampling Theory: Introduction to sampling distributions, standard error, Type-I and Type-II errors. Test of hypothesis for means, student’s t-distribution, Chi-square distribution as a test of goodness of fit.
### Complex analysis, probability and statistical methods Course Objectives
To provide an insight into applications of complex variables, conformal mapping and special functions arising in potential theory, quantum mechanics, heat conduction and field theory. •To develop probability distribution of discrete, continuous random variables and joint probability distribution occurring in digital signal processing, design engineering and microwave engineering.
### Complex analysis, probability and statistical methods Course Outcomes
At the end of the course the student will be able to:•Use the concepts of analytic function and complex potentials to solve the problems arising in electromagnetic field theory. •Utilize conformal transformation and complex integral arising in aerofoil theory, fluid flow visualization and image processing. •Apply discrete and continuous probability distributions in analyzing the probability models arising inengineering field. •Make use of the correlation and regression analysis to fit a suitable mathematical model for the statistical data. • Construct joint probability distributions and demonstrate the validity of testing the hypothesis.
### Complex analysis, probability and statistical methods Text Books
Higher Engineering Mathematics-B. S. Grewal
Engineering Mathematics-Srimanta Pal et al
### Complex analysis, probability and statistical methods Reference Books
Advanced Engineering Mathematics-C. Ray Wylie, Louis C.Barrett
Introductory Methods of Numerical Analysis-S.S.Sastry
Higher Engineering Mathematics-B. V. Ramana
A Text Book of Engineering Mathematics-N. P. Bali and Manish Goyal
### Scoring Marks in Complex analysis, probability and statistical methods
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https://prime-numbers.fandom.com/wiki/677 | 1,566,650,334,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027320734.85/warc/CC-MAIN-20190824105853-20190824131853-00322.warc.gz | 603,292,272 | 34,548 | ## FANDOM
836 Pages
677 is a prime number from 601-700. 677 has 2 factors, 1 and 677. It is the 123rd prime number, and the fourteenth prime number from 601-700.
## Proofs
• — 677 can be divided by 1.
• — 677 cannot be divided by 2-676.
• — 677 can be divided by 677.
## Relationship with other odd numbers
### The numbers before
• The previous prime number is 673.
• 673 and 677 are four numbers apart; therefore, they are cousin primes.
• The previous prime number before 673 is 661.
• Notably, 667 is the product of 23 and 29, while 671 is divisible by 11. They are both composite numbers.
### The numbers after
• The next prime number is 683.
• 677 and 683 are six numbers apart; therefore, they are sexy primes.
• 679 is divisible by 7, therefore not a prime number.
• 691 is the next prime number after 683.
• 689 is a composite number, being the product of 13 and 53.
## Trivia
Community content is available under CC-BY-SA unless otherwise noted. | 269 | 965 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2019-35 | latest | en | 0.883042 |
https://physics.stackexchange.com/questions/164583/why-is-work-done-on-the-system-considered-when-calculating-the-work-output | 1,713,128,219,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816893.9/warc/CC-MAIN-20240414192536-20240414222536-00488.warc.gz | 425,631,334 | 39,003 | # Why is work done on the system considered when calculating the work output?
The thermodynamic efficiency $\eta$ is calculated by $\eta= \frac{W_{out}}{Q_{in}}$
Using the first law of thermodynamics we usually say that $W_{out}$ is $Q_c+Q_h$, where $Q_c$ is the heat dissipated into a cold reservoir, and $Q_h$ is the heat absorbed by a hot reservoir. Both are measured within the system, such that $Q_c<0$ and $Q_h>0$
However I object to that. $W_{out}$ is not $Q_c+Q_h$. That calculation is simply the magnitude of net energy in the process due to work. Namely, its considering the $W_{input}$, done by the surroundings on the gas, to calculate the $W_{output}$.
Consider the a Carnot engine. I would say that the work output is the area underneath the expansion isotherm, and the expansion adiabat. All other works are done on the system and are not "outputs".
You might say that the other works are "negative outputs". But although this makes sense in a mathematical sense, it doesn't make practical sense that this should be a $W_{output}$.
• For starters, you can tell us what $Q_c$ and $Q_h$ are. Feb 11, 2015 at 18:33
• Ok, I've done it on my 2nd paragraph. thx
– DLV
Feb 11, 2015 at 18:40
• Hmm. Not sure I follow ... Is the problem the terminology, and what exactly is meant by "work output"? Does it help to consider $W_{out}$ as being the net work, which for an engine is "output"? I think that's the intended meaning. Feb 11, 2015 at 23:04
• So I'm saying Qc+Qh is not the work coming out of the system. This calculation is considering work done on the system. The work coming out of the system or "output", shouldn't have anything to do with the work coming into it if it really is an "output".
– DLV
Feb 11, 2015 at 23:36
• Well, it is the net output, and I think that's the sense of the word that is intended. Feb 12, 2015 at 2:45
The Carnot cycle is ... well ... a cycle. In each turn of the cycle, the system returns to exactly the same state over and over again. Pick a point, call it the beginning of the cycle. After one turn of the cycle, the system is where it started, with exactly the same energy it had when it started. During the cycle, it has done some work. Where has that energy come from? It can't have come from the system, the system has exactly the same energy it had when it started. The energy can only have come from the heat reservoirs. $Q_h$ is removed from the hot side, $Q_c$ is deposited the cold side. The net energy, $Q_h + Q_c$ has been converted to work. The Carnot engine provides the mechanism for the transfer of the energy from heat to work.
• Hey, I don't think this is my question. My question regards the definition of $\eta$ and why $Qh+Qc$ is not a valid calculation of $W_{out}$ Thanks. | 737 | 2,750 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2024-18 | latest | en | 0.950951 |
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Chapter 3: Answers to Questions and Problems
1.
a. When P = \$12, R = (\$12)(1) = \$12. When P = \$10, R = (\$10)(2) = \$20. Thus, the price decrease results in an \$8 increase in total revenue, so demand is elastic over this range of prices.
b. When P = \$4, R = (\$4)(5) = \$20. When P = \$2, R = (\$2)(6) = \$12. Thus, the price decrease results in an \$8 decrease total revenue, so demand is inelastic over this range of prices.
c. Recall that total revenue is maximized at the point where demand is unitary elastic. We also know that marginal revenue is zero at this point. For a linear demand curve, marginal revenue lies halfway between the demand curve and the vertical axis. In this case, marginal revenue is a line starting at a price of \$14 and intersecting the quantity axis at a value of Q = 3.5. Thus, marginal revenue is 0 at 3.5 units, which corresponds to a price of \$7 as shown below.
Price
\$14
\$12
\$10
\$8
\$6
\$4
\$2
\$0
MR
0123456
Figure 3-1
Demand
Quantity
2.
3.
a.
b.
c.
At the given prices, quantity demanded is 700 units:
Q =−
x
d
1000
2 ()154
+
.02 ()400
the elasticity formula gives:
E
Q
x
=
,
P
x
700
.
=− 2
Substituting the relevant information into
P
x
Q
x
=− 2 154 =− 0.44 . Since this is less 700
than one in absolute value, demand is inelastic at this price. If the firm charged a lower price, total revenue would decrease.
At the given prices, quantity demanded is 300 units:
Q =−
x
d
1000
2 ()354
+
.02 ()400
the elasticity formula gives:
E
Q
x
=
,
P
x
300 . Substituting the relevant information into
=−
2
P
x
Q
x
=−
2
⎜ ⎝
354
300
⎟ ⎠
=− 2.36
. Since this is
greater than one in absolute value, demand is elastic at this price. If the firm increased its price, total revenue would decrease.
At the given prices, quantity demanded is 700 units:
Q =−
x
d
1000
2 ()154
+
.02 ()400
the elasticity formula gives:
E
Q
x
=
,
P
Z
700 . Substituting the relevant information into
⎟ ⎠
P
Z
⎜ ⎝
400
700
== .02 ⎜ ⎟
Q
x
.02
=
0.011
. Since this
number is positive, goods X and Z are substitutes.
a.
b.
c.
d.
The own price elasticity of demand is simply the coefficient of ln P x , which is –
0.5. Since this number is less than one in absolute value, demand is inelastic.
The cross-price elasticity of demand is simply the coefficient of ln P y , which is –
2.5. Since this number is negative, goods X and Y are complements.
The income elasticity of demand is simply the coefficient of ln M, which is 1. Since this number is positive, good X is a normal good.
The advertising elasticity of demand is simply the coefficient of ln A, which is 2.
4.
a. Use the own price elasticity of demand formula to write %
ΔQ
d
x
5
= −
2
. Solving,
we see that the quantity demanded of good X will decrease by 10 percent if the
price of good X increases by 5 percent.
b. Use the cross-price elasticity of demand formula to write %
ΔQ
d
x
10
= −
6 . Solving,
we see that the demand for X will decrease by 60 percent if the price of good Y
increases by 10 percent.
c. Use the formula for the advertising elasticity of demand to write %
ΔQ
d
x
2
=
4 .
Solving, we see that the demand for good X will decrease by 8 percent if
d. Use the income elasticity of demand formula to write %
ΔQ
d
x
3
= 3 . Solving, we
see that the demand of good X will decrease by 9 percent if income decreases by
3 percent.
5. Using the cross price elasticity formula,
50
% Δ P
y
=− 5
. Solving, we see that the price
of good Y would have to decrease by 10 percent in order to increase the consumption of good X by 50 percent.
6. Using the change in revenue formula for two products, ΔR = [\$30,000(1 2.5)+ \$70,000()1.1 ](.01) = \$320 . Thus, a 1 percent increase in the price of good X would cause revenues from both goods to increase by \$320.
7.
Table 3-1 contains the answers to the regression output.
SUMMARY OUTPUT
Regression Statistics
Multiple R R Square Adjusted R Square Standard Error Observations 0.62 0.39 0.37 190.90 100.00 ANOVA degrees of freedom SS MS F Significance F Regression 2.00 2,223,017.77 1,111,508.88 30.50 0.00 Residual 97.00 3,535,019.49 36,443.50 Total 99.00 5,758,037.26 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept Price of X Income 187.15 534.71 0.35 0.73 -880.56 1,254.86 -4.32 0.69 6.26 0.00 -5.69 -2.96 0.09 0.02 4.47 0.00 0.05 0.14
Table 3-1
a. b. Q d x 4.32 =−+PM . 187.15 .09 x Only the coefficients for the Price of X and Income are statistically significant at the 5 percent level or better. c. The R-square is fairly low, indicating that the model explains only 39 percent of the total variation in demand for X. The adjusted R-square is only marginally lower (37 percent), suggesting that the R-square is not the result of an excessive number of estimated coefficients relative to the sample size. The F-statistic, however, suggests that the overall regression is statistically significant at better than the 5 percent level.
8. The approximate 95 percent confidence interval for a is
ˆ ±
a
2
σ
a ˆ
=
10
±
2
. Thus, you
can be 95 percent confident that a is within the range of 8 and 12. The approximate
±
95 percent confidence interval for b is
percent confident that b is within the range of –3.5 and –1.5.
ˆ
b
2
σ
ˆ
b
=− 2.5 ±
1
. Thus, you can be 95
9.
a. The t statistics are as follows:
t ˆ
a
9369.45
=
11067.07
= 0.848
;
t
ˆ =
b
1.36
0.56
= 2.429
; and
b.
t
c ˆ
0.14
=
0.05
=− 2.80
.
Since
However, since
statistically different from zero.
t
a ˆ
< 2
the coefficient estimate, aˆ , is not statistically different from zero.
t
ˆ >
b
2
and
t
c ˆ
> 2
, the coefficient estimates b ˆ and cˆ are
c. The R-square and adjust R-square tell us the proportion of variation explained by the regression. The R-square tells us that 24 percent of the variability in the dependent variable is explained by price and income. The adjusted R-square confirms that fact and the R-square is not the result of estimating too many coefficients (i.e. few degrees of freedom).
10. a. The own-price elasticity of demand is -1.36, so demand is elastic. b. The income elasticity of demandis-0.14, so X is an inferior good. 11. The result is not surprising. Given the available information, the own price elasticity
=
137
of demand for major cellular telephone manufacturer is
17 this number is greater than one in absolute value, demand is elastic. By the total revenue test, this means that a reduction in price will increase revenues.
E
=− 8.06
. Since
Q P
,
12.
The regression output is as follows:
SUMMARY OUTPUT
Regression Statistics
Multiple R R Square Adjusted R Square Standard Error Observations 0.97 0.94 0.94 0.00 49 ANOVA df SS MS F Significance F Regression 2 0.00702 0.004 370.38 0.0000 Residual 46 0.00044 0.000 Total 48 0.00745 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept 1.29 0.41 3.12 0.00 0.46 2.12 LN Price -0.07 0.00 -26.62 0.00 -0.08 -0.07 LN Income -0.03 0.09 -0.33 0.74 -0.22 0.16
Table 3-2
Thus, the demand for your batteries is given by ln Q =−1.29
. Since this is a log-linear demand equation, the best estimate of the income elasticity of demand for your product is -.03: Your batteries are an inferior good. However, note the estimated income elasticity is very close to zero (implying that a 3 percent reduction in global incomes would increase the demand for your product by less than one tenth of one percent). More importantly, the estimated income elasticity is not statistically different from zero (the 95 percent confidence interval ranges from a low of -.22 to a high of .16, with a t-statistic that is well below 2 in absolute value). On balance, this means that a 3 percent decline in global incomes is unlikely to impact the sales of your product. Note that the R-square is reasonably high, suggesting the model explains 94 percent of the total variation in the demand for this product. Likewise, the F-test indicates that the regression fit is highly significant.
0.07 ln PM0.03ln
13. Based on this information, the own price elasticity of demand for Big G cereal is 3
E
Q P
,
=
2
=− 1.5
. Thus, demand for Big G cereal is elastic (since this number is
greater than one in absolute value). Since Lucky Charms is one particular brand of cereal for which even more substitutes exist, you would expect the demand for Lucky
Charms to be even more elastic than the demand for Big G cereal. Thus, since the demand for Lucky Charms is elastic, one would predict that the increase in price of Lucky Charms resulted in a reduction in revenues on sales of Lucky Charms.
% Δ Q
d
14. Use the income elasticity formula to write
4 purchases are expected to decrease by 7 percent.
= 1.75 . Solving, we see that coffee
15.
To maximize revenue, Toyota should charge the price that makes demand unit elastic. Using the own price elasticity of demand formula,
E Q P
,
(
= −
1.25
)
⎝ ⎜
P
⎠ ⎟
1.25 P
100,000
=−
1 . Solving this equation for P implies that the
revenue maximizing price is P = \$40,000 .
16. Using the change in revenue formula for two products, ΔR = [\$600(1 2.5)+ \$400()0.2 ]× (.01) = \$9.8 million , so revenues will increase by \$9.8 million.
17. The estimated demand function for residential heating fuel is
Q
d
RHF = 136.96 91.69
P
RHF
+ 43.88
P
NG
11.92
P
E
0.05
M , where
P
RHF
is the price
of residential heating fuel,
electricity, and M is income. However, notice that coefficients of income and the price of electricity are not statistically different from zero. Among other things, this means that the proposal to increase the price of electricity by \$5 is unlikely to have a statistically significant impact on the demand for residential heating fuel. Since the
coefficient of
reduction in the consumption of residential heating fuel (since (-91.69)(\$2) = - 183.38
units). Since the coefficient of P
P
NG
is the price of natural gas,
P
E
is the price of
P
RHF
is -91.69, a \$2 increase in
NG
P
RHF
would lead to a 183.38 unit
P
NG
is 43.88, a \$1 reduction in
43.88 unit reduction in the consumption of residential heating fuel (since (43.88)(-\$1) = -43.88). Thus, the proposal to increase the price of residential heating fuel by \$2 would lead to the greatest expected reduction in the consumption of residential heating fuel.
18. The regression output is as follows:
SUMMARY OUTPUT
Regression Statistics
Multiple R 0.97 R Square 0.94 Adjusted R Square 0.94 Standard Error 0.06 Observations 41 ANOVA df SS MS F Significance F Regression 1 2.24 2.24 599.26 0.00 Residual 39 0.15 0.00 Total 40 2.38 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept 4.29 0.12 37.17 0.00 4.06 4.53 ln (Price) -1.38 0.06 -24.48 0.00 -1.50 -1.27
Table 3-3
Thus, the least squares regression line is ln Q = 4.29 1.38ln P . The own price elasticity of demand for broilers is –1.38. From the t-statistic, this is statistically different from zero (the t-statistic is well over 2 in absolute value). The R-square is relatively high, suggesting that the model explains 94 percent of the total variation in the demand for chicken. Given that your current revenues are \$750,000 and the elasticity of demand is –1.38, we may use the following formula to determine how much you must change price to increase revenues by \$50,000:
Δ R =
[
P
x
Q
x
( 1
+
E
Q
x
,
P
x
)]
×
Δ
P
x
P
x
\$50,000 = \$750,000 1 1.38
[
(
)]
ΔP
x
P
x
Solving yields
Δ
P
x
\$50,000
= =
P
x
\$285,000
0.175 . That is, to increase revenues by \$50,000,
you must decrease your price by 17.5 percent.
19. The regression output (and corresponding demand equations) for each state are presented below:
ILLINOIS
SUMMARY OUTPUT
Regression Statistics
Multiple R R Square Adjusted R Square Standard Error Observations 0.29 0.09 0.05 151.15 50 ANOVA degrees of freedom SS MS F Significance F Regression 2 100540.93 50270.47 2.20 0.12 Residual 47 1073835.15 22847.56 Total 49 1174376.08 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept -42.65 496.56 -0.09 0.93 -1041.60 956.29 Price 2.62 13.99 0.19 0.85 -25.53 30.76 Income 14.32 6.83 2.10 0.04 0.58 28.05
Table 3-4
The estimated demand equation is Q =−42.65 + 2.62PM+ 14.32
that demand slopes upward, note that coefficient on price is not statistically different from zero. An increase in income by \$1,000 increases demand by 14.32 units. Since the t-statistic associated with income is greater than 2 in absolute value, income is a significant factor in determining quantity demanded. The R-square is extremely low, suggesting that the model explains only 9 percent of the total variation in the demand for KBC microbrews. Factors other than price and income play an important role in determining quantity demanded.
. While it appears
INDIANA
SUMMARY OUTPUT
Regression Statistics
Multiple R R Square Adjusted R Square Standard Error Observations 0.87 0.76 0.75 3.94 50 ANOVA degrees of freedom SS MS F Significance F Regression 2 2294.93 1147.46 73.96 0.00 Residual 47 729.15 15.51 Total 49 3024.08 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept 97.53 10.88 8.96 0.00 75.64 119.42 Price -2.52 0.25 -10.24 0.00 -3.01 -2.02 Income 2.11 0.26 8.12 0.00 1.59 2.63
Table 3-5
The estimated demand equation is Q = 97.53 2.52P + 2.11M . This equation says that increasing price by \$1 decreases quantity demanded by 2.52 units. Likewise, increasing income by \$1,000 increases demand by 2.11 units. Since the t-statistics for each of the variables is greater than 2 in absolute value, price and income are significant factors in determining quantity demanded. The R-square is reasonably high, suggesting that the model explains 76 percent of the total variation in the demand for KBC microbrews.
MICHIGAN
SUMMARY OUTPUT
Regression Statistics
Multiple R R Square Adjusted R Square Standard Error Observations 0.63 0.40 0.37 10.59 50 ANOVA degrees of freedom SS MS F Significance F Regression 2 3474.75 1737.38 15.51 0.00 Residual 47 5266.23 112.05 Total 49 8740.98 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept 182.44 16.25 11.23 0.0000 149.75 215.12 Price -1.02 0.31 -3.28 0.0020 -1.65 -0.40 Income 1.41 0.35 4.09 0.0002 0.72 2.11
Table 3-6
The estimated demand equation is Q = 182.44 1.02P + 1.41M . This equation says that increasing price by \$1 decreases quantity demanded by 1.02 units. Likewise, increasing income by \$1,000 increases demand by 1.41 units. Since the t-statistics associated with each of the variables is greater than 2 in absolute value, price and income are significant factors in determining quantity demanded. The R-square is relatively low, suggesting that the model explains about 40 percent of the total variation in the demand for KBC microbrews. The F-statistic is zero, suggesting that the overall fit of the regression to the data is highly significant.
MINNESOTA
SUMMARY OUTPUT
Regression Statistics
Multiple R R Square Adjusted R Square Standard Error Observations 0.64 0.41 0.39 16.43 50 ANOVA degrees of freedom SS MS F Significance F Regression 2 8994.34 4497.17 16.67 0.00 Residual 47 12680.48 269.80 Total 49 21674.82 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept 81.70 81.49 1.00 0.32 -82.23 245.62 Price -0.12 2.52 -0.05 0.96 -5.19 4.94 Income 3.41 0.60 5.68 0.00 2.20 4.62
Table 3-7
The estimated demand equation is Q =−81.70
that increasing price by \$1 decreases quantity demanded by 0.12 units. Likewise, a \$1,000 increase in consumer income increases demand by 3.41 units. Since the t- statistic associated with income is greater than 2 in absolute value, it is a significant factor in determining quantity demanded; however, price is not a statistically significant determinant of quantity demanded. The R-square is relatively low, suggesting that the model explains 41 percent of the total variation in the demand for KBC microbrews.
0.12PM+ 3.41
. This equation says
MISSOURI
SUMMARY OUTPUT
Regression Statistics
Multiple R R Square Adjusted R Square Standard Error Observations 0.88 0.78 0.77 15.56 50 ANOVA degrees of freedom SS MS F Significance F Regression 2 39634.90 19817.45 81.81 0.00 Residual 47 11385.02 242.23 Total 49 51019.92 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept 124.31 24.23 5.13 0.00 75.57 173.05 Price -0.79 0.58 -1.36 0.18 -1.96 0.38 Income 7.45 0.59 12.73 0.00 6.27 8.63
Table 3-8
The estimated demand equation is Q = 124.31 −+0.79PM7.45
that increasing price by \$1 decreases quantity demanded by 0.79 units. Likewise, a \$1,000 increase in income increases demand by 7.45 units. The t-statistic associated with price is not greater than 2 in absolute value; suggesting that price does not statistically impact the quantity demanded. However, the estimated income coefficient is statistically different from zero. The R-square is reasonably high, suggesting that the model explains 78 percent of the total variation in the demand for KBC microbrews.
. This equation says
OHIO
SUMMARY OUTPUT
Regression Statistics
Multiple R R Square Adjusted R Square Standard Error Observations 0.99 0.98 0.98 10.63 50 ANOVA degrees of freedom SS MS F Significance F Regression 2 323988.26 161994.13 1434.86 0.00 Residual 47 5306.24 112.90 Total 49 329294.50 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept 111.06 23.04 4.82 0.0000 64.71 157.41 Price -2.48 0.79 -3.12 0.0031 -4.07 -0.88 Income 7.03 0.13 52.96 0.0000 6.76 7.30
Table 3-9
The estimated demand equation is Q = 111.06 −+2.48PM7.03
that increasing price by \$1 decreases quantity demanded by 2.48 units. Likewise, increasing income by \$1,000 increases demand by 7.03 units. Since the t-statistics associated with each of the variables is greater than 2 in absolute value, price and income are significant factors in determining quantity demanded. The R-square is very high, suggesting that the model explains 98 percent of the total variation in the demand for KBC microbrews.
. This equation says
WISCONSIN
SUMMARY OUTPUT
Regression Statistics
Multiple R R Square Adjusted R Square Standard Error Observations 0.999 0.998 0.998 4.79 50 ANOVA degrees of freedom SS MS F Significance F Regression 2 614277.37 307138.68 13369.30 0.00 Residual 47 1079.75 22.97 Total 49 615357.12 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept 107.60 7.97 13.49 0.00 91.56 123.65 Price -1.94 0.25 -7.59 0.00 -2.45 -1.42 Income 10.01 0.06 163.48 0.00 9.88 10.13
Table 3-10
The estimated demand equation is Q = 107.60 −+1.94PM10.01
that increasing price by \$1 decreases quantity demanded by 1.94 units. Likewise, increasing income by \$1,000 increases demand by 10.01 units. Since the t-statistics associated with price and income are greater than 2 in absolute value, price and income are both significant factors in determining quantity demanded. The R-square is very high, suggesting that the model explains 99.8 percent of the total variation in the demand for KBC microbrews.
. This equation says
20. Table 3-11 contains the output from the linear regression model. That model indicates that R 2 = .55, or that 55 percent of the variability in the quantity demanded is explained by price and advertising. In contrast, in Table 3-12 the R 2 for the log-linear model is .40, indicating that only 40 percent of the variability in the natural log of quantity is explained by variation in the natural log of price and the natural log of advertising. Therefore, the linear regression model appears to do a better job explaining variation in the dependent variable. This conclusion is further supported by comparing the adjusted R 2 s and the F-statistics in the two models. In the linear regression model the adjusted R 2 is greater than in the log-linear model: .54 compared to .39, respectively. The F-statistic in the linear regression model is 58.61, which is larger than the F-statistic of 32.52 in the log-linear regression model. Taken together these three measures suggest that the linear regression model fits the data better than the log-linear model. Each of the three variables in the linear regression model is statistically significant; in absolute value the t-statistics are greater than two. In contrast, only two of the three variables are statistically significant in the log-linear model; the intercept is not statistically significant since the t-statistic is less than two in absolute value. At P = \$3.10 and A = \$100, milk consumption is 2.029 million
gallons per week ( Q
d
milk
= 6.52 1.61(3.10)+ .005()100 = 2.029)
.
SUMMARY OUTPUT LINEAR REGRESSION MODEL
Regression Statistics
Multiple R R Square Adjusted R Square Standard Error Observations 0.74 0.55 0.54 1.06 100.00 ANOVA df SS MS F Significance F Regression 2.00 132.51 66.26 58.61 2.05E-17 Residual 97.00 109.66 1.13 Total 99.00 242.17 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept 6.52 0.82 7.92 0.00 4.89 8.15 Price -1.61 0.15 -10.66 0.00 -1.92 -1.31 Advertising 0.005 0.0016 2.96 0.00 0.00 0.01
Table 3-11
SUMMARY OUTPUT LOG-LINEAR REGRESSION MODEL
Regression Statistics
Multiple R R Square Adjusted R Square Standard Error Observations 0.63 0.40 0.39 0.59 100.00 ANOVA df SS MS F Significance F Regression 2.00 22.40 11.20 32.52 1.55E-11 Residual 97.00 33.41 0.34 Total 99.00 55.81 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept -1.99 2.24 -0.89 0.38 -6.44 2.46 ln(Price) -2.17 0.28 -7.86 0.00 -2.72 -1.62 ln(Advertising) 0.91 0.37 2.46 0.02 0.18 1.65
21.
Table 3-12
Given the estimated demand function and the monthly subscriptions prices, demand is
172,000 subscribers( Q
\$8.6 million, which are not sufficient to cover costs. Revenues are maximized when
demand is unit elastic
d
= 152.5 0.9()50 + 1.05()30 + 1.10()30 ) . Thus, revenues are
P
sat
217
.9
P
sat
=
1
: Solving yields
P
sat
= \$120.56
. Thus, the
sat
⎛ ⎛ ⎜ ⎜ .9 ⎜ ⎜
⎝ ⎝
maximum revenue News Corp. can earn is \$13,080,277.76
(TR = P × Q = 120.56 × (217 .9 ×120.56)×1000). News Corp. cannot cover its costs in the current environment.
22.
23.
The manager of Pacific Cellular estimated that the short-term price elasticity of demand was inelastic. In the market for cellular service, contracts prevent many customers from immediately responding to price increases. Therefore, it is not surprising to observe inelastic in the short-term. However, as contracts expire and customers have more time to search for alternatives, quantity demanded is likely to drop off much more. Given a year or two, the demand for cellular service is much more elastic. The price increase has caused Pacific to lose more customers than they initially estimated.
The owner is confusing the demand for gasoline for the entire U.S. with demand for the gasoline for individual gasoline stations. There are not a great number of substitutes for gasoline, but in large towns there are usually a very high number of substitutes for gasoline from an individual station. In order to make an informed decision, the owner needs to know the own price elasticity of demand for gasoline from his stations. Since gas prices are posted on big billboards, and gas stations in cities are generally close together, demand for gas from a small group of individual stations tends to be fairly elastic. | 6,929 | 23,587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2020-40 | latest | en | 0.850199 |
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