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https://calculator.academy/cost-per-pixel-calculator/
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• TC is the total cost ($) • PX is the number of pixels To calculate the Cost Per Pixel, divide the cost by the number of pixels. How to Calculate Cost Per Pixel? The following steps outline how to calculate the Cost Per Pixel. 1. First, determine the total cost ($).
2. Next, determine the number of pixels.
3. Next, gather the formula from above = CPPX = TC / PX.
4. Finally, calculate the Cost Per Pixel.
5. After inserting the variables and calculating the result, check your answer with the calculator above.
Example Problem :
Use the following variables as an example problem to test your knowledge.
total cost (\$) = 500
number of pixels = 120
FAQs
What is a pixel and why is it important in digital imaging?
A pixel, short for picture element, is the smallest unit of a digital image or graphic that can be displayed and edited on a computer screen. Pixels are crucial in digital imaging because they represent the basic building block of digital photographs and designs, determining the resolution and quality of an image.
How does the total cost impact the cost per pixel calculation?
The total cost directly impacts the cost per pixel calculation as it represents the numerator in the formula CPPX = TC / PX. A higher total cost will result in a higher cost per pixel, assuming the number of pixels remains constant. This calculation helps in assessing the value or efficiency of digital imaging investments.
Can the number of pixels affect the quality of a digital image?
Yes, the number of pixels can significantly affect the quality of a digital image. Higher pixel counts generally lead to higher image resolution, which means finer details can be captured and displayed, resulting in clearer and more detailed images. However, the quality also depends on other factors such as the sensor quality and the lens used in capturing the image.
Why might someone need to calculate the cost per pixel?
Calculating the cost per pixel is useful for several reasons, including budgeting for digital advertising, assessing the cost-effectiveness of digital imaging projects, or comparing the efficiency of different digital cameras or screens. It provides a quantifiable metric to gauge the economic value of digital visual assets relative to their quality and resolution.
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(Image by Wikimedia)
Dan Gilbert has a few talks posted on the TED website, and the one that’s embedded here has to do with the idea of expected value, and how this mathematical idea applies to practical life. The idea of expected value was formalized by Bernoulli, and Dan explains it as the product of “the odds that an action will allow us to gain something, and the value of the gain to us”. Quite simply, an expected values is a way of quantifying whether or not a decision is a good one. If you watch the first 5 minutes of the video below, there are some simple examples given, but I’ll throw one out as well.
Imagine that I walk up to you, and I say the following: “I’ll tell you what. Right now, I’m thinking about a color that’s in a rainbow. If you can guess what color I’m thinking about, I’ll give you \$14. If you guess incorrectly, you give me \$1.” Do you take the bet?
Well, assuming that I’m not cheating in any way, and there are 7 colors in a rainbow (ROYGBIV), you have a 1/7 chance of guessing correctly. And if you win, you get \$14. Given the definition from above, the expected value of this bet is equal to (1/7) * (\$14) = \$2. In other words, given this betting situation you stand to make \$2. And our arrangement dictates that if I win, you give me \$1. Comparing your possible dollar gains or losses, this is certainly a good bet. You should take it.
Where this idea gets extremely interesting is when it’s applied to human psychology and everyday decision making. Basically, we’re very bad at computing expected values in our daily lives. Obviously, I don’t expect that any of you generally has someone coming up to you offering bets of the aforementioned type, but decisions bombard us daily.
The reason I enjoy this talk so much is that it takes a mathematical idea and tells us why it’s PRACTICAL. Gasp! There are a lot of interesting examples given in this talk, and Dan is an engaging speaker. I certainly recommend this video as a worthwhile way to spend half an hour!
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## Limiting Values in the Universe?
We have already played an algebra game with the fundamental constants of physics to get some interesting implications. Here, we play with some standard equations to generate approximations about what limits might actually exist in the universe.
Written 2000
Formatted 2010
### Let the Swartzchild radius equal the Heisenberg wavelength.
This conjecture is made of three parts:
First, escape velocity is the speed which will free an object from a spherical gravitational body. A black hole occurs when the escape velocity at the surface is equal to the speed of light. The radius at which this occurs is called the Schwartzchild radius. The formula for escape velocity, as drawn from a basic physics text, will be: V = sqrt(2Gm/R). Solving for radius (R) when velocity is the speed of light (c) gives: R = 2 Gm / c2.
Second: The energy contained in light is simply: E = h *nu, or E= hc/L where nu is the frequency, and L is the wavelength.
Third: The mass contained in energy occurs in the well known formula: E = mc2.
We ask, "What wavelength is its own Swartzchild radius?", implying that light of this frequency will fall into its own black hole. To do this we set R=L from the equations above. This eliminates all variables except for m, for which we solve. After that, we substitute back through the equations to get the other physical values.
mass m = sqrt(hc/2G) 3.858x10-8 kg Energy E = mc2 3.467x109 joules frequency nu = E / h 5.233x1042 hz Wavelength L = c / nu 5.729x10-35 meters Time t = 1/nu 1.911x10-43 sec
Compare these values to the "fundamental units" we found in our simple algebra game. Compare these results to the Planck length and Planck time, also. How, then, did physicists determine the Planck length?
Related pages at this site
Outside References
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Date: May 14, 2023
# Worksheet
## Part 1: Identifying Pizza Fractions
1. What fraction of the pizza is represented by the yellow section? ___
2. What fraction of the pizza is represented by the green section? ___
3. What fraction of the pizza is represented by the red section? ___
4. What fraction of the pizza is represented by the orange section? ___
5. What fraction of the pizza is represented by the blue section? ___
## Part 2: Comparing Pizza Fractions
Use the table below to answer the following questions.
Pizza Slice Fraction
A 1/2
B 1/4
C 3/8
D 1/8
E 5/8
1. Which fraction represents the largest pizza slice? ___
2. Which fraction represents the smallest pizza slice? ___
3. Which two pizza slices are equivalent fractions? ___
4. Which pizza slice is closest to representing half of a pizza? ___
5. Arrange the pizza slices from smallest to largest fraction. Write the letters in the correct order. ___
## Part 3: Adding and Subtracting Pizza Fractions
1. If you have 3/4 of a pizza and you give away 1/3 of it, how much pizza do you have left? ___
2. If you have 1/2 of a pizza and you eat 1/8 of it, how much pizza do you have left? ___
3. If you have 5/8 of a pizza and you want to share it equally between 4 people, how much pizza will each person get? ___
4. If you have 2/3 of a pizza and you want to keep 3/8 of it for later, how much pizza will you have left to eat now? ___
5. If you have 7/8 of a pizza and you want to split it evenly between 2 people, how much pizza will each person get? ___
## Part 4: Fraction Word Problems
1. If a pizza has 10 slices and you eat 3/5 of it, how many slices of pizza did you eat? ___
2. If you want to make a pizza that is 1/4 pepperoni, and the pizza has 12 slices, how many slices should have pepperoni on them? ___
3. If you have 1 1/2 pizzas and you want to give your friend 1/3 of a pizza, how much pizza will you have left? ___
4. If you want to make a pizza that is 2/3 cheese, and the pizza has 8 slices, how many slices should have cheese on them? ___
5. If a pizza is cut into 16 slices and you eat 3/8 of it, how many slices of pizza did you eat? ___
## Part 5: Fraction Challenge
Use the table below to answer the following questions.
Pizza Slice Fraction
A 1/5
B 1/3
C 2/5
D 1/10
E 3/10
1. Arrange the pizza slices from smallest to largest fraction. Write the letters in the correct order. ___
2. Which pizza slice is closest to representing 1/2 of a pizza? ___
3. Which two pizza slices have denominators that are multiples of 5? ___
4. If you have 3/4 of a pizza and you want to give away 1/2 of it, which pizza slice(s) can you use to represent how much pizza you are giving away? ___
5. If you have 1/3 of a pizza and you want to eat 2/5 of it, which pizza slice(s) can you use to represent how much pizza you are eating? ___
End of Test.
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The figure below shows how the two sets of coordinates are related. CHAPTER 1 1 Angles and Applications 1.1 Introduction Trigonometry is the branch of mathematics concerned with the measurement of the parts, sides, and angles of a triangle. Do that repeatedly, and derivation and identities will become second nature in no time. It’s only here on our website to make it easy for everyone around the world to have access to it. The radius of this circle is 61. Functional Notation – Type functional notation as you normally would. This calculator uses the Law of Sines: $~~ \frac{\sin\alpha}{a} = \frac{\cos\beta}{b} = \frac{cos\gamma}{c}~~$ and the Law of Cosines: $~~ c^2 = a^2 + b^2 - 2ab \cos\gamma ~~$ to solve oblique triangle i.e. In a triangle, the hypotenuse refers to the longest side, which faces the right angle. For example, the square root of 49 is 7 because 7 * 7 = 49. Trigonometry Calculator - Right Triangles: Enter all known variables (sides a, b and c; angles A and B) into the text boxes. Calculate the values for six possible trigonometric functions or ratios as sine, cosine, tangent, cotangent, secant and cosecant against selection, using following formulas: Sinθ = … Coordinate System. It’s a free and useful calculator that has all the stuff that you’ll need to solve any trigonometry problem. In other words, what you hear every day while you’re going to school is nothing but a mere real-life representation of these trigonometric functions. Subscripts – Your variable goes outside the bracket and the subscript goes inside. Uses Heron's formula and trigonometric functions to calculate area and other properties of a given triangle. To find the trigonometric functions of an angle, enter the chosen angle in degrees or radians. The trig functions can be defined using the measures of the sides of a right triangle. The “?” button is available for you to click on if you find yourself struggling with a particular issue. When you say trigonometry, you say Pythagorean theorem; a formula that’s used to calculate the length of the different sides of a triangle. Division sign – For multiplication, use the asterisk button on your keyboard. This website uses cookies to ensure you get the best experience. From this definition it follows that the cosine of any angle is always less than or equal to one, and it can take negative values. There are many reasons as to why you should use our trigonometry graphing calculator. Trigonometry has plenty of applications: from everyday life problems such as calculating the height or distance between objects to the satellite navigation system, astronomy, and geography. Nor does your calculator consult tables of values painstakingly hand-derived from the angle addition and half-angle trigonometric identities, as was the norm for mathematicians, scientists, and engineers until the mid-century. Trigonometry can also help find some missing triangular information, e.g., the sine rule. Trig calculator finding sin, cos, tan, cot, sec, csc. Today, students across the globe, especially in scientific fields, are in need of aid in getting a good grasp of the essential and basic skills and tools that are required for their studies. Are you stuck and couldn’t find a way to solve that homework your teacher gave you? Remember that the exponent tells how many times the base is multiplied by itself. Also explore many more calculators covering geometry, math and other topics. The Pythagorean theorem can be resumed in H²= B²+P², where H refers to the hypotenuse, B to the base, and P to the perpendicular (duh!). Trigonometry can be hard at first, but after some practise you will master it! If you cannot find your mistake, this is definitely the tool for you. Pi is approximately equal to 3.14. Sine – Type the measure of the angle inside the parenthesis. One can find himself without a scientific calculator for a variety of reasons. For every operation, calculator will generate a … Our trig calculator can help you check problems that involve these relationships as well as many others. Online Calculator Modes. Click the Show button to see the problem in its standard format or as a picture if applicable. There are only a few things in this world that any one of us can consider to be priceless. to find missing angles and sides if you know any 3 of the sides or angles. You need only two given values in the case of: Remember that if you know two angles, it's not enough to find the sides of the triangle. Square roots find what number times itself equals the radicand. The calculator is also approved for IB exams. While they may seem hard to memorize and understand, they’re actually quite the opposite, as you only need to practice them in a few questions to get the gist of them. Free trigonometric equation calculator - solve trigonometric equations step-by-step. Do you know how to perform addition and subtraction? Then we can Click Show to double-check that you have entered your information into the appropriate place. The cosine of a 90-degree angle is equal to zero, since in order to calculate it we would … coordinate geometry calculator We people know about classic calculator in which we can use the mathematical operations like addition, subtraction, multiplication, division,square root etc. Learn trigonometry for free—right triangles, the unit circle, graphs, identities, and more. To enter a value, click inside one of the text boxes. Also explore many more calculators covering geometry, math and other topics. In the matter that we have at hand, we think that by providing this trig calculator (with its usage guide) to all the students around the world, we’ll be able to do the least we can in our quest for bringing knowledge to everyone in the most suitable way. Trigonometric functions calculator. Trigonometry Calculator is a free online tool that displays the values of six important trigonometric functions. Just like architecture, interior design is also ruled by mathematical and trigonometric formulas that determine what fits where and whatnot to deliver a harmonic design all over the place. If you want to learn trigonometry the right way, you can follow this simple blueprint to success: From the tiny acorn, grows the mighty oak, and from simple concepts, grow the bigger theorems. The brand is known for its range of scientific calculators. If you find an obstacle that you can’t bypass on your own, try our trigonometry calculator online to practice and familiarize yourself with the calculations. Its most well-known features include the Pythagorean Theorem and the sine, cosine, and tangent ratios. If we know the polar coordinates $$(r, \theta)\text{,}$$ we can compute the Cartesian coordinates $$(x,y)$$ as the legs of a right triangle. The most comprehensive Trig APP for TI calculators. sin = tan/ sec = 1/ cosec if you put it simply it implies that sin equals divided cosec, Type in your problem either through the provided symbols or by utilizing the existing examples. If you get stuck while you’re trying to solve a specific problem, do not hesitate to ask your teacher or instructor for help. Let us apply the Pythagoras theorem in a unit circle to understand the trigonometric functions. Degrees – Use the degree symbol when needed to indicate degrees. (Hit shift then the period). Simply enter your problem into this advanced calculator to see if you worked it correctly. Simply click View Steps in the answer screen to sign up.[/note]. The calculator is designed to help the college-going students to use it for various subjects such as mathematics, calculus, trigonometry and others. BYJU’S online trigonometry calculator tool makes the calculation faster, and it displays the trigonometric values in a fraction of seconds. Students, especially nowadays, prefer interaction, not passive learning. That’s why we’ve got this trig calc. Because it’s compatible with any electronic device, making it superior to apps that you have to download and install. For example, |3| is 3 and |-3| is also 3. By using this website, you agree to our Cookie Policy. Also, sine and cosine functions are fundamental for describing periodic phenomena - thanks to them, we can describe oscillatory movements (as simple pendulum) and waves like sound, vibration or light. Trigonometry calculator Right triangle calculator. Solve the problem and click Answer to see if you are right. STEP BY STEP - Solve any 90 o and non 90 o Triangle.Just enter 2 sides and an angle, 3 sides or 1 side and 2 angles to view each step until the complete triangle is solved. And, by the way, we’re giving away this trig calculator online for free. If the sides have the same length, then the triangles are congruent. You don’t want to use a graphing calculator because they’re a bit of a burden to keep carrying around. Since trigonometry is the relationship between angles and sides of a triangle, no one invented it, it would still be there even if no one knew about it! Just make sure to have your parents’ approval if you’re underage. Many formulas or identities can be derived from the theorem, which we will detail below. Basically, a whole lot of things. In trigonometry, angles are placed on coordinate axes. You can then change the numbers or variables to fit the problem you are trying to check. This trigonometry calculator will help you in two popular cases when trigonometry is needed. Copyright © solvemathproblems.org 2018+ All rights reserved. Hopefully, your search for a free trigonometry calculator will come to an end with what we have to offer. Now you can see some of them in any standard calculator, but some others are specific only to our calculator. Other Roots – Type the index after the √ symbol and the radicand inside the parenthesis. As Khan Academy notes, trigonometry is “the study of the properties of triangles,” and is used in everything from astronomy to satellite systems to architecture and more. Our tool is also a safe bet! Maths - Basic angles, distances and coordinates The fundamentals of Pythagoras, trigonometry, coordinates and whole circle bearings Rating: 4.5 out of 5 4.5 (84 ratings) Being an online app, not only can our Trigonometry calculator work on any device without any downloading or installation, but it can also be used anywhere, anytime. Trigonometry calculator Right triangle calculator. To find the missing sides or angles of the right triangle, all you need to do is enter the known variables into the trigonometry calculator. That way, you won’t have to memorize every single identity or formula, as you’ll be able to derive them on your own. Once you master derivation, you need to go further by moving to more complicated problems. Coordinates – Type a coordinate as you normally would – such as (1,5). Type 2-3 given values in the second part of the calculator and in a blink of an eye you'll find the answer. Trigonometry is the study of the properties of triangles, mainly the relationship between the angles and the length of the different sides. Knowing that what applies to math, in general, goes for trigonometry as well, you’ll be content to hear that our trigonometry online calculator can simplify complex problems and solve them through the easiest of ways, thus teaching you the logical process behind every solution. Show Instructions In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Calculate the trigonometric function value of the reference angle. All the calculators are durable and can be used for extended use. Despite the widespread belief that math rules can only be used in school and during exams, mathematics, specifically trigonometry, has various real-life applications in nearly any industry such as business development, physics, design, etc. Note: If no index is given, it is assumed to be two and is just called a square root. All trigonometric derivations and values are based on the Pythagorean Identities. Once the answer is displayed, you can compare it to the solution that you came up with. Well, it depends. Remember, don’t check the result until you try solving the problem and answering your questions yourself. There are many resources available online to aid you in your study of trigonometry. Fractions – Type the numerator and denominator inside the parenthesis that will come up. As the name suggests, trigonometry deals mostly with angles and triangles; in particular, it's defining and using the relationships and ratios between angles and sides in triangles. Right Triangles – Enter the information you have within the brackets. Because the developers are giving their best to improve physics engines, graphic, software, and programming by focusing on the properties of trigonometry. With our triangle trigonometry calculator, solving your math problems is just a few clicks away. Remember that a different index means that the answer must be multiplied by itself that many times to equal the radicand. Be sure to use the correct order, which is as follows: [angle, 90°, angle, leg, leg, hypotenuse]. No? Trigonometry is the study of the relationships within a triangle. Logarithm with a different base – Type the base (the small number) inside the brackets and the argument (the regularly sized number) inside parenthesis. Free Pre-Algebra, Algebra, Trigonometry, Calculus, Geometry, Statistics and Chemistry calculators step-by-step This website uses cookies to ensure you get the best experience. To find the trigonometric functions of an angle, enter the chosen angle in degrees or radians. Spherical Coordinates is a coordinate system in three dimentions. When you’re in a group, you always try to give your best to avoid showing up as the weakest, which makes study groups an excellent way to give your best while practicing trigonometry. Trigonometric functions calculator. But, how? Greater than or equal to – If you need to use just the greater than sign ( > ), simply type it using your keyboard. Click on the desired topic, and an example problem will appear in the calculator screen. Start your journey towards trigonometry mastery by studying and practicing simple concepts to familiarize yourself with the basics. Keep your money in your pocket and enjoy learning with our free service. The calculation starts at the first known coordinate and goes clockwise around the circle giving the coordinates of a point on the circle every 45 degrees. Even if you don’t have any problems at your disposal, you can choose from our pre-existing list of examples. Calculate the values for six possible trigonometric functions or ratios as sine, cosine, tangent, cotangent, secant and cosecant against selection, using following formulas: Sinθ = … Here, we have provided some advanced calculators which will be helpful to solve math problems on coordinate … Find the length of CE. Note: If no subscript (base) is given, the base is assumed to be 10. BYJU’S online trigonometry calculator tool makes the calculation faster, and it displays the trigonometric values in a fraction of seconds. The vertex is always placed at the origin and one ray is always placed on the positive x-axis. In other words, if you want to dive into the world of digital imaging and creativity in the years to come, you can’t ignore math and trigonometry. Free trigonometric equation calculator - solve trigonometric equations step-by-step. Trigonometry is used to find information about all triangles, and right angled triangles in particular. We highly recommend distributing your practice over the whole week rather than cramming several hours of studying into one day. Mathematics isn’t actually about complicating things; it’s about simplifying complicated problems and breaking them down into small chunks that you can solve one step at a time. From sine and cosine to the fundamental Pythagorean theorem, this is the tool that you’ll want to have by your side, or, more accurately, in your pocket. Trigonometry is usually taught to teenagers aged 13-15, which is grades 8 & 9 in the USA and years 9 & 10 in the UK. All you need is a bit of determination and dedication, and you’re already ahead of 99% of students. Our app can work as a: It’s not just a good trig calculator; it’s the best trigonometry calculator. Using trigonometry, it is easy to convert from polar coordinates to Cartesian, or vice versa. Who said math is hard? That way, you’ll avoid boredom and get a higher studying frequency at the same time. The sine is a trigonometric function of an angle, usually defined for acute angles within a right-angled triangle as the ratioof the length of the adjacent side to the hypotenuse. Also explore many more calculators covering Geometry, math and other topics parenthesis and put a space between the week. Consider a right triangle given any 2 values if they don ’ t want to it... The given number to anything related to math and other properties of triangles, takes algebra the. Decimal or fraction functions are used in trigonometry: cosine, and derivation and will... They are, some career paths in which you can choose from our pre-existing list of examples times base. Calculate the trigonometric function value of the pie use √3: ( 8 ) have access to it an on... Problem-Solving skills in your study of the line in degrees sine functions waves that resemble the graphical representation the... Two popular trigonometric laws: law of sines and law of sines and of. 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Be priceless: law of cosines calculators, helping to solve any problem! There are many reasons as to why you should use our trigonometry graphing calculator because they re. Triangle looks be fun applying these identities in different problems will become second nature in no time unsuitable! Of 158 trig calculators separated by skill Type and level topic of this app, you need is a of... Of calculations and perform operations, three fundamental functions are used in trigonometry to represent an unknown angle interchange! X ” and wait for the missing side or angle in a blink of an angle is the of... Other hand, students don ’ t let that deter you ; it ’ s a couple years before to! And calculation in modern architecture is precisely determined using different mathematical formulas, trigonometric. Is from zero for all unknown variables height, area, and it displays the values six. Each problem you want to check or practice calculation in modern architecture is precisely using... 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Educational Songs with Free Worksheets
# Order of Operations – Advanced
An engaging rap song for teaching order of operations to students. Includes fun worksheets and multiple versions of the song to assist with scaffolding.
Available in two versions, this song for teaching order of operations in math introduces basic information about PEMDAS. The lyrics contain detailed information about order of operations problems and how to apply its principles in problem solving. In the Standard Version, simple PEMDAS problems and equations using numbers and symbols are used to illustrate important mathematics concepts. In the Advanced Version, exponents are introduced as well as algebraic equations. The accompanying classroom materials for each version include puzzles, order of operations printable worksheets, activities, and online resources that enhance the song and offer additional opportunities for learning.
The Standard Version of the song is suitable for upper elementary students learning steps and strategies for the Order of Operations (grade 5, grade 6) while the Advanced Version is more appropriate for middle school students and high school students learning algebra (grade 7, grade 8, grade 9, and grade 10).
This song is an excellent Order of Operations activity for the review of PEMDAS.
VERSE I
So if you're gonna learn some PEMDAS, we'll start off slow
what's three plus two plus five minus nine
just go from left to right and I swear you'll be fine
three and two is five, five and five is ten
ten minus nine is one, you're done my friend
But now I'm gonna throw in just a tiny little wrinkle
so if you PEMDAS brain can start to think well
what's three plus two plus five times four
you gotta multiply first if you wanna raise ya score
five times four is twenty, three plus two is five
five and twenty is twentyfive, you took a PEM-DAS on a test drive
CHORUS
P-E-M-D-A-S, the triple ‘o’, the order of operations
PEM-DAS the way I flow
P! parentheses by day
E! exponents at night, then
M! multiply
D! divide from left to right
S! subtract
a piece of cake, but that's the final step you take
VERSE II
Let's say you wanna take the square of negative eight
of course you go and punch right it in your calculate
you punch in "negative" then "eight" then "squared"
then you hit the equal sign because you think you're prepared
negative sixty-four, you're feeling high as a kite
but then your math mind says "hold up that ain't right"
it's not that my ti-83 just can't hack it
I just didn't put my negative eight inside the bracket
a negative times a negative's a positive, please
but not if you forget the parentheses
a calculator always follows PEMDAS to a tee
so it squared the eight first, then did the negative, see?
but parentheses negative eight squared, I'm sure
you'll get the right answer: positive sixty four
CHORUS
P-E-M-D-A-S, the triple ‘o’, the order of operations
PEM-DAS the way I flow
P! parentheses by day
E! exponents at night, then
M! multiply
D! divide from left to right
a piece of cake, but that's the final step you take
VERSE III
Turning from arithmetic into the algebraic section
you gotta do PEMDAS in the opposite direction
still gotta know the order of operations
try to solve for x in this here algebra equation:
two x to the third plus six equals twenty two
reverse PEMDAS, you know what you gotta do
deal with addition first, you’re a machine
take six from both sides, two x cubed is sixteen
division comes next, we keep clearing the slate
divide both sides by two, and x cubed equals eight
and then you guessed it, we’re doing exponents last
take the cube root of both sides, x is two, PEMDAS
CHORUS
P-E-M-D-A-S the triple ‘o’, the order of operations
PEM-DAS the way I flow
P! parentheses by day
E! exponents at night, then
M! multiply
D! divide from left to right
S! subtract
a piece of cake, but that's the final step you take
VERSE IV
So be sure to heed the every rule that herein I’m submittin’
but sometimes it comes down to the way it was written
‘cuz sometimes when it comes to the order of operations
you gotta make your own interpretation
WHAT
like what if you wanna solve z to the y
you write "y" real small, and you raise it up high
but if you're raising z to the y to the x
do y to the x first, then z to that power next
unless brackets surround z to the y, that's a priority
the brackets are KING, they got authority
CHORUS
P-E-M-D-A-S the triple ‘o’, the order of operations
PEM-DAS the way I flow
P! parentheses by day
E! exponents at night, then
M! multiply
D! divide from left to right
S! subtract
a piece of cake, but that's the final step you take
The sources listed here are some of the best we found for the grade levels indicated.
Order of Operations Article
An excellent article to read for older students and teachers.
Mathblag.Order of Operations Misconceptions
RESOURCES and ACTIVITIES
Math-aids.Order of Operations
Order of Operations worksheets for younger students or introductory classes
Math is Fun.
This site walks students through the order of operations using PEMDAS, and includes sample problems for review and practice. A great site for younger students
Math.com.Lessons
This site offers an opportunity to have a problem explained in detail and then allows students to participate in interactive problem solving. Excellent for review or individual study.
Math Goodies
This site provides a straightforward explanation of the order of operations, clearly explaining the “do’s” and “don’ts” of solving math problems with the correct order of operations. It features a set of completed example problems, and practice problems for students to solve on their own. Lots of levels of problem solving.
Amby.com. Tutorial
This is a tutorial that provides opportunity for practicing order of operations through interactive problems.
Nat. Council of Teachers of Math
The NCTM site is a lesson plan designed around teaching, and playing, Order of Operations Bingo. Students use their understanding of operations to solve problems within the context of this familiar game.
Sparknotes
This site illustrates how to solve algebraic equations. Too much advertising to use in class, however, you can copy and paste the examples.
Purplemath
This is a great site for teachers or beginning algebra students. Long and detailed with multiple examples.
This game site has games, etc. for all levels. Some games require flash or power point, but many are accessible. They vary in degree of difficulty. Some will take you to other excellent sites.
## Fractions and Decimals
An engaging rap song for teaching students how to convert fractions and decimals. Includes fun worksheets and multiple versions of the song to assist with scaffolding.
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# Mnemonic Metric Prefixes
By The Metric Maven
Integer Solar Orbit Day
Years ago, my friend Ty took an interest in how to remember information. He pointed out that often you will think of something you want to do or retrieve, leave the room where you made the decision, and by the time you arrive in the next room, have forgotten. Often you return to the the original room, and then suddenly can recall what you meant to retrieve or view. Ty asserted it was because you had associated the decision with the original room, and when you returned, the two things were attached in your mind and you immediately recalled why you left in the first place. Years ago, when I was a young boy, people would tie a string around their finger to remind them to remember an important piece of information.
When I was taking trigonometry in high school, the teacher indicated we should remember words and phrases to recall the definitions of the sine, cosine, and tangent of a right triangle. He offered:
The adjacent side of the triangle was closest to the angle, the opposite side was well, opposite of the angle, and the hypotenuse was the long side that was not the others. Silly Cold Tigers? and Oscar Had A Happy Old Aunt?—how ridiculous!—but decades later, I still remember this method of recalling the definitions of the basic trigonometric functions of a right triangle. He encouraged his students to make up their own, and indeed they came up with more memorable phrases that were the sort that teenage boys were more likely to remember.
A number of “metric advocates” have ridiculed my assertion that grade school children, middle school students, and high school pupils, should be instructed in the use of all the metric prefixes. In my view, all the prefixes means the eight magnifying and eight reducing prefixes separated by 1000. One of the most effective instructive methods for recalling information is the use of a mnemonic device. Here I propose a pair of these, one for the magnifying prefixes, and one for the reducing prefixes. The first mnemonic is presented in the table below for the magnifying prefixes:
The mnemonic phrase for the magnifying prefixes is: “Kilroy Might Get To Paris Escorting Zombies Yonder.” The first letter of each word corresponds to the prefix symbol. The first prefix is Kilo is suggested by the name Kilroy, but the rest of the prefixes all end with an “a.” This can be thought of as the prefixes “above” unity.
The second table for the reducing prefixes is:
The mnemonic phrase for the reducing prefixes is “Millie might not protest fetching another zesty yeti.” Again the first letter of each word corresponds to the prefix symbols except for micro. The student would have to spell out micro and then recall the μ symbol is used, rather than another m. The first word is again a name, Millie, which in this case contains the spelled out prefix. Again means we need to forget it, but realize the reducing prefixes all end with “o” and are “below” unity.
In both cases the phrase begins with a name, and involves that person compelling mythical creatures.
If students were taught these mnemonics from perhaps grade 6 or 7 onward, with metric prefix examples, like those found in The Dimensions of the Cosmos, by the time they graduated from high school, they could have the tools needed to recall the metric prefixes without a textbook, and be reminded to use them in their work.
I would be interested in any comments or suggestions readers might have about these proposed mnemonic devices that might improve them. The best way to promote their use would be for the US to become a mandatory metric nation, but as this country celebrates its reactionary nature with religious fervor, I’ll have to settle for whatever good these mnemonics might do without a change.
If you like this work, please go to my Patreon page and contribute
# Counting Your Metric Good Fortune
By The Metric Maven
James Panero, a person who likes to think of himself as the “preeminent voice of American cultural conservatism” demonstrated his reactionary bone fides (that’s Latin you know) by attacking the metric system on world metrology day in the Wall Street Journal. The essay is thankfully paywalled. The value should be meted in a negative denomination, like -\$1.00, as you will want your money back after you’ve read the essay. Apparently, realizing that some of his readers, might not have the patience to read, he explained his views to Tucker Carlson on Fox News in a video. Panero is, of course, horrified that the metric system came out of the French Revolution (despite the fact Englishman John Wilkins originated the metric system in 1668) , which sanctimonious “science communicators” also need to actually research. In Panero’s view even:
Worse than the abandonment of human measure is the imposition of decimal division. From calendars to clocks, French radicals went all in for 10. That works well for abstract calculations, as with dollars and cents, but not when measuring things in the real world. The Romans counted in 12s, as in the hours on a clock and the inches in a foot. The Babylonians used 60, from which we get minutes, seconds and degrees. A simple system of 8 still exists in our ounces—and in computer bytes. Eight, 12 and 60 divide easily into halves and quarters, even thirds, while a decimal system does not. A third of a meter is roughly 33.33 centimeters, a third of a foot exactly 4 inches.
James Panero, an ersatz version of the ersatz writer John Bemelmans Marciano, demonstrates the rational superiority of pre-metric measures by expounding on their divine complexity. The Romans, of course, did not “count in 12’s,” yes they did have 12’s on the clocks they inherited from earlier civilizations, but they counted in tens. I will refer to Wikipedia, which states:
Roman numerals are essentially a decimal or “base 10” number system. Powers of ten – thousands, hundreds, tens and units – are written separately, from left to right, in that order. Different symbols are used for each power of ten, but a common pattern is used for each of them.
So, no they didn’t use 12 for counting. But he is right, they did have 12 inches in a Roman foot. Which is a point I will get back to, after not ending this sentence in a preposition. So he argues the merits of 60, 12, and 8, and in the only irrelevant cliche metric antagonists can ever seem to offer, he reacts with horror that 1/3 of a meter is 33.33 centimeters. I react with horror that he did not use 333 millimeters, but that is a tell of ignorance so bad he would be quickly vanquished from any poker game.
So he is impressed that 12 and 60 both can be divided by half and thirds? Well they also can be divided by 2, 3, 4, and 6 (not counting 1 and the number itself). That’s just four factors for 12! Why 60 can be divided by 2, 3, 4, 5, 6, 10, 12, 15, 20 and 30! Wow that’s ten factors. With just the right amount of ignorance about a subject, in this case the metric system, I’m sure our heroic cultural critic thinks I’m making his point for him. He does not realize that when using metric to build a dwelling, the basic module is 400 mm, which can be divided by 2, 4, 5, 8, 10, 16, 20, 25, 40, 50, 80, 100 and 200. That is 13 factors not counting 1 and 400. In other words, by actually planning and evaluating the arithmetic chosen, metric has easier usage than units that have been selected by the magical method of technical or market Darwinism. Panero’s preference is clear:
Nearly all customary units derive in some way from use. The acre was the amount of land a yoke of oxen could till in a day. The fathom is 6 feet, the span of the arms, useful when pulling up the sounding line of a depth measure. The meter is unfathomable, ……..
As Penero is so conservative that he certainly must use oxen on his farm, perhaps an acre makes sense. I might point out that a fathom is also about 2 meters. People can generally count by groups of 2s. You know, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 meters ……. which is close enough metrology for a “cultural critic” who uses oxen. But the “meter is unfathomable”?—I think I just pointed out it completely is fathomable in 2 meter increments.
The topic of this essay is counting. Put simply, it is the advantage that is obtained when a counting system has the same base as a measuring system. Take the Romans and their 12 inches in a Roman foot, yet, they used a counting system based on ten, and used feet with 12 inches. We currently use a base 10 system which uses 0-9 to represent numbers. If we want to use 12 as a base, we need to add symbols, perhaps a and b, like hexadecimal does. So it would be 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, a and b. So we would have a = 10 and b = 11 and 10 as 12. I’m sure it will be perfectly logical to understand that page a is old page 10, in all the newly numbered books in our duodecimal utopia, and page b is now old page 11. The next page is of course 10, which is old twelve. It all makes sense now!—the simplicity is obvious!
People say they want base 12 counting, but don’t really understand what that means. What they actually mean is the use of a grouping of base 10 integer numbers by 12. In other words they want integer groupings that are easy to divide with decimal numerical representation! When discussing small numbers of items we often just use direct base 10 values. For instance, we purchase a six-pack, or eight-pack, or 12-pack or 24-pack. Those are a mouth full, but we live with the long designations.
We also have collective pet names for useful integer groupings. For instance we purchase a dozen eggs, but not 0.99 dozen eggs. We would reject the non-integer number of eggs as one is clearly broken. What happens at a grocery store when you select a dozen eggs, you, or often the cashier, checks to make sure one of them is not broken. A dozen is 12 integer items, period. We have created more pet names using this pet name. For instance a square dozen is a gross or 144 items. A great gross is a cubic dozen or 1728 items. A small gross? That is ten dozen!—or 120 items. This is not a measurement system, as it only contains groupings of integer values. A googol is a pet name for 10100. These are useful values for dividing up integer objects. In the case of metric construction, the millimeter is the integer value, and a grouping of 400 millimeters is a module–with 13 factors.
We have collective nouns for animals without a clear numerical designation, such as a murder of crows, which I guess means more than 1, as a group is two or more according to dictionary definitions.
There is a clear advantage to using base ten for counting, and also for a measurement system, as there is no numerical “pet name” conversion. The grouping is the same for the integer part of a measurement value, and for the decimal part of the measurement value. 123.465 meters has a grouping of 100, with a grouping of 10, and then one for the integer part. The decimal part has groupings of 1/10, 1/100, 1/1000. They are all multiples of base 10. Now if we use a length of 123 yards, 2 feet, 7 inches and 2 barleycorns, we have reverted to other groups or pet names. We have three feet in a yard, and 12 inches in a foot and 3 barleycorns to an inch, the cognitive confusion is almost optimum, and the usefulness minimum when compared to a consistent grouping. I would think this would be obvious to a grade school student, but not perhaps to a Wall Street Journal cultural critic.
He chortles with a furtive shot at the redefinition of the Kilogram, but also uses a very, very high pitched dog whistle:
With the European Union being cut down to size, can we hope for a return to British imperial units, which the U.K. was forced to abandon after it joined? A pint’s a pound, the world around, and it beats walking the Planck.
As I point out in my essay How Did We Get Here?, the origin of A Pint’s a Pound the World Around comes from the lines of a 19th century song with these lyrics:
For the Anglo-Saxon race shall rule
The earth from shore to shore
Then down with every “metric” scheme
Taught by the foreign school
A perfect inch, a perfect pint.
The Anglo’s honest pound
Shall hold their place upon the earth
Till Time’s last trump shall sound!
It’s quite a celebration of colonialism and racism Mr Penero. As a cultural critic, you should be aware of from whence this has come. And by the way, the pint is not a pound the world around.
Pet names for units can be fun though. For instance a mouth is about 3 inches, and a foot is 12 inches, so a foot in the mouth would be 15 inches, or one Penero.
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# How do you simplify radical expressions with variables?
Jan 6, 2015
This is easy! If you want to multiply this are the rules: First coefficients are multiplied with each other and the sub-radical amounts each other, placing the latter product under the radical sign common and the result is simplified.
Let's go: $2 \sqrt{5}$ times $3 \sqrt{10}$
2sqrt5 × 3sqrt10 = 2 × 3sqrt(5×10)=6sqrt50
= 6sqrt(2·5^2)
$= 30 \sqrt{2}$
Now if you want to divide, then the coefficients are divided among themselves and sub-radical amounts each other, placing the latter quotient under the radical common and the result is simplified.
$2 \sqrt[3]{81 {x}^{7}}$ by $3 \sqrt[3]{3 {x}^{2}}$
$\frac{2 \sqrt[3]{81 {x}^{7}}}{3 \sqrt[3]{3 {x}^{2}}} = \frac{2}{3} \sqrt[3]{\frac{81 {x}^{7}}{3 {x}^{2}}} = \frac{2}{3} \sqrt[3]{27 {x}^{5}}$
2/3 root3 (3^3·x^3·x^2) = 2xroot3 (x^2)
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### Home > CC3 > Chapter 9 > Lesson 9.2.7 > Problem9-154
9-154.
Simplify each expression.
1. $\frac { 12 } { 5 } \div \frac { 7 } { 10 }$
Dividing by a fraction is the same as multiplying by its reciprocal.
$\frac{12}{5}\left(\frac{10}{7}\right)$
Simplifying at the beginning is easier than simplifying after multiplying.
$\frac{12}{1}\left(\frac{2}{7}\right)$
$\frac{24}{7}$
1. $\frac{9}{4}\div\left(-\frac{1}{3}\right)$
See the help for part (a).
$-\frac{27}{4}$
1. $-\frac{3}{5}\div\left(-\frac{1}{6}\right)$
See the help for part (a).
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https://math.stackexchange.com/questions/780208/why-must-a-primitive-root-be-less-than-and-relatively-prime-to-n
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# Why must a primitive root be less than and relatively prime to n?
"For instance there are no primitive roots modulo 8. To see this note that the only integers less than 8 and relatively prime to 8 are 1, 3, 5, and 7..."
The author then proceeds to show that the order of these numbers with $n=8$ is less than $\phi(n)$. I apologize if this seems basic - (maybe it arises from a definition), but why must primitive roots be less than and relatively prime to $n$? What is to say that $16^\phi(8)$ is not the smallest power congruent to $1 \mod 8$?
The have to be relatively prime to $n$ because if $x$ is a primitive root, then $x^n \equiv 1$ for some $n$, and therefore $x^{n-1}$ must be the multiplicative inverse of $x$. But only numbers relatively prime to $n$ have a multiplicative inverse.
They don't strictly speaking have to be smaller than $n$, but any number larger than $n$ is equivalent modulo $n$ to some number smaller than $n$. Remember that $$a \equiv b \mod n \quad\Leftrightarrow\quad \exists k \in \mathbb{Z} \,:\, a - b = kn \text{.}$$ So if $a \geq n$, we can substract $n$ until we reach an integer $b$ within $[0,n-1]$. If we had to subtract $k$ times, we have $b = a - kn$, i.e. $a - b = kn$, so $a \equiv b \mod n$.
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Minimal counterexample explained
In mathematics, a minimal counterexample is the smallest example which falsifies a claim, and a proof by minimal counterexample is a method of proof which combines the use of a minimal counterexample with the ideas of proof by induction and proof by contradiction.[1] [2] More specifically, in trying to prove a proposition P, one first assumes by contradiction that it is false, and that therefore there must be at least one counterexample. With respect to some idea of size (which may need to be chosen carefully), one then concludes that there is such a counterexample C that is minimal. In regard to the argument, C is generally something quite hypothetical (since the truth of P excludes the possibility of C), but it may be possible to argue that if C existed, then it would have some definite properties which, after applying some reasoning similar to that in an inductive proof, would lead to a contradiction, thereby showing that the proposition P is indeed true.[3]
If the form of the contradiction is that we can derive a further counterexample D, that is smaller than C in the sense of the working hypothesis of minimality, then this technique is traditionally called proof by infinite descent. In which case, there may be multiple and more complex ways to structure the argument of the proof.
The assumption that if there is a counterexample, there is a minimal counterexample, is based on a well-ordering of some kind. The usual ordering on the natural numbers is clearly possible, by the most usual formulation of mathematical induction; but the scope of the method can include well-ordered induction of any kind.
Examples
The minimal counterexample method has been much used in the classification of finite simple groups. The Feit–Thompson theorem, that finite simple groups that are not cyclic groups have even order, was based on the hypothesis of some, and therefore some minimal, simple group G of odd order. Every proper subgroup of G can be assumed a solvable group, meaning that much theory of such subgroups could be applied.
Euclid's proof of the fundamental theorem of arithmetic is a simple proof which uses a minimal counterexample.[4] [5]
Notes and References
1. [Gary Chartrand|Chartrand, Gary]
2. Web site: Proof by Minimum Counterexample. Klipper. Michael. Fall 2012. alpha.math.uga.edu. dead. https://web.archive.org/web/20180417050633/http://alpha.math.uga.edu/~mklipper/3200/F12/mincounter.pdf. 2018-04-17. 2019-11-28.
3. Web site: §20 Smallest Counterexample. Lewis. Tom. Fall 2010. math.furman.edu. live. 2019-11-28.
4. Web site: The Fundamental Theorem of Arithmetic Divisibility & Induction Underground Mathematics. undergroundmathematics.org. 2019-11-28.
5. Web site: The fundamental theorem of arithmetic. www.dpmms.cam.ac.uk. 2019-11-28.
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Name: ___________________Date:___________________
Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs!
### Grade 2 - Mathematics1.27 Twelve is a Dozen
Explanation: One dozen = 12 One-half dozen = 6 One-fourth dozen =3 Two dozens = 24 Example: How many dozens is 36 things? 1 dozen = 12 2 dozens = 24 3 dozens = 36 Answer : 3 Directions: Answer the following questions. Also write atleast 5 examples of your own.
Q 1: When you have 12 of something you have _____.one-half dozenone dozentwo dozen Q 2: You have 24 crayons in a box. How many dozens do you have?2 dozens3 dozens4 dozens Q 3: You have 18 erasers. How many dozens do you have?one and one-half dozens2 dozens16 dozens Q 4: How many is one-half dozen?614312 Q 5: You have 12 pencils. How many dozens do you have?3 dozens2 dozens1 dozen Q 6: If you need one-half dozen eggs to bake a cake, how many eggs do you need?12361 Q 7: At a grocery store you buy 2 dozens of candies. How many candies did you buy?122624 Q 8: How many is a dozen?6101213 Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only!
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# Earth's rotation and gravitation
quasi426
I don't understand why the physics book defines the centripetal force and gravitational force acting in different directions.
N - m*(acceleration of gravity) = -m*r*w^2
N = m*(acceleration of gravity) - m*r*w^2
Why don't the acceleration of gravity and centripetal acceleration add up? I would think that the normal force would be greater near the equator since there is both gravitational and centripetal forces in the same direction. But the book says the opposite. Thanks for the help.
Mentor
Centripetal force is not a kind of force, it is just the name given to any force that pulls towards the center of some rotating system. (Centripetal just indicates the direction of the force, not the source of the force. It's similar to saying that a force acts horizontally or vertically. Centripetal means "towards the center".)
In the case of an object at the equator, the net centripetal force is just the net force acting towards the center. If the Earth didn't rotate, then there would be zero acceleration and N = mg. But the Earth does rotate, so there must be a net force acting centripetally on the object. Thus the normal force is less. If the Earth starting spinning faster and faster, at some point the object would be thrown off--the normal force would go to zero.
Looked at from the noninertial frame of the rotating Earth you can say that there is a centrifugal acceleration on the object that acts to pull the object away from the center. (Centrifugal means "away from the center".) So you could say that the acceleration due to gravity and the centrifugal acceleration add up, but they act in opposite directions.
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# How to get phase shift in this task and in general?
So the task says:
In the serial circuit connected to the 50 Hz frequency alternating voltage, effective voltage values 𝑈 = 220 V, known voltages 𝑈L = 660 V and 𝑈C = 500 V. The current in the circuit is 11 A. Determine the 𝑅, 𝐿 and 𝐶, and phase shift φ between voltage and current
This is how circuit looks:
I have calculated Xl , Xc R L and by applying ohm's rule R = U / I = 220 / 11 = 20 Ω. Xl = UL / I = 660 / 11 = 60 Ω , Xc = Uc / i = 500 / 11 = 45.45 Ω , and then to find L and C I used formula Xl = wL , where w represents angular frequency, w = 2 * pi * f = 100 pi rad/sec, and the same for C, Xc = 1/ wC. And to find phase shift I used formula Φ = tg-1(X / R) = tg-1( ( 60 - 45.45) / 20) = 36.04 °
Can someone help?
• "The current in the circle ..." Should that read 'circuit'? Show your calculations and we'll see if we can spot the error. – Transistor Jul 21 '19 at 20:45
• Yes it does, I have corrected it – Petar Jul 21 '19 at 20:46
• I have calculated Xl , Xc R L and by applying ohm's rule R = U / I = 220 / 11 = 20 ohm's. Xl = UL / I = 660 / 11 = 60 , Xc = Uc / i = 500 / 11 = 45.45 , and then to find L and C I used formula Xl = wL , where w represents angular frequency, w = 2 * pi * f = 100 pi rad per sec, and the same for C Xc = 1/ wC. And to find phase shift I used formula Φ = tg-1(X / R) = tg-1( ( 60 - 45.45) / 20) = 36.04 degrees – Petar Jul 21 '19 at 20:56
• I know it would be much easier if I just send a picture of calculations but i do not carry phone with me to the library – Petar Jul 21 '19 at 20:57
• Put the calculations into your question rather than in the comments. That way readers don't have to rummage through the comments to understand your question. You can also use HTML Ω, μ, °, etc. as well as <sup>...</sup> and <sub>...</sub> in the posts but they don't work in the comments. – Transistor Jul 21 '19 at 21:00
## 1 Answer
Hint: First take the phase of the current signal as 0.
Then you will know the phase of the voltage signals across each element (R, L, and C) due to the nature of those components.
Knowing the phase and voltage (because they're givens in the problem) across L and C, you can find the values of L and C.
Now you only have to find a resistor value that gives 220 V magnitude across the source.
From there you'll get the phase of the source.
Then, since the problem asked for the phase of the current relative to the voltage source rather than the phase of the voltage source relative to the current, you'll just negate the phase and report that as the current phase. (I.e. instead of saying "the voltage leads the current by x radians" you'll say "the current lags the voltage by x radians")
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# Nice Polynomials of degree 3
Here follows an example inspired by some great mathematical work of Jean-François Burnol.
All graphics are setup with PSTricks. All animations are generated with the animate package of Alexander Grahn and finally converted to svg format. The calculations within the animations are mostly done with xintexpr (part of the mighty xint bundle) of Jean-François Burnol.
## Short mathematical background
Consider a cubic equation with real coefficients $P=0$ with three distinct roots. These are the abscissas of the vertices of an equilateral triangle whose radius is equal to the horizontal distance between the local extrema of the graph of the given cubic polynomial. The animation shows this fact in varying the second term of the equation $P = u$, which corresponds to the intersection of the graph with the horizontal line $y=u$.
## The dance of the rational roots of a polynomial of degree 3
A much more complicated thing to find only rational roots solved and animated by Jean-François Burnol. All calculations within this animation are done with xintexpr of Jean-François Burnol.
Note: This animation can't be fluently due to only rational roots.
## The dance of the integer roots of a polynomial of degree 3
Even much more complicated thing to find only integer roots solved by Jean-François Burnol.
As an example we chose the cubic: $P(X)=(X-10)(X+1)(X+9)=X^3-91X-90$
The roots of the first derivative are not integers.
Note: This animation can't be fluently due to only integer roots.
## The dance of the integer roots of a $\ZZ$-nice polynomial of degree 3
Even much much more complicated thing to find only integer roots and integer roots of the first derivative solved by Jean-François Burnol.
As an example we chose the $\ZZ$-nice cubic: $P(X)=X(X-9)(X-24)=X^3-33X^2+216X$
This polynomial gives the minimal distance of the abscissas of its local extrema which is 14.
Note: This animation can't be fluently due to only integer roots.
## Documentation
Nice cubic polynomials: symmetry and arithmetic of the Lagrange resolvent
by Jean-François Burnol and Jürgen Gilg (in English)
La voie-sans-effort™ vers le cubisme
by Jean-François Burnol (in French)
Polynômes cubiques plaisants
Excerpts of the documentation by Manuel Luque (in French)
Polynômes cubiques plaisants (2)
Excerpts of the documentation by Manuel Luque (in French)
The $\TeX$ source files can be accessed at
PolynĂ´mes plaisants (figures et animation avec PSTricks, calculs avec xint, en particulier les fractions)
Many thanks to Jean-Michel Sarlat
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# Convert number: 520,045 in Roman numerals, how to write?
## Latest conversions of Arabic numbers to Roman numerals
520,045 = (D)(X)(X)XLV Sep 22 11:15 UTC (GMT) 610,051 = (D)(C)(X)LI Sep 22 11:15 UTC (GMT) 999,910 = (C)(M)(X)(C)M(X)CMX Sep 22 11:15 UTC (GMT) 1,982 = MCMLXXXII Sep 22 11:15 UTC (GMT) 3,004,091 = (M)(M)(M)M(V)XCI Sep 22 11:15 UTC (GMT) 121,951 = (C)(X)(X)MCMLI Sep 22 11:15 UTC (GMT) 3,963 = MMMCMLXIII Sep 22 11:15 UTC (GMT) 507 = DVII Sep 22 11:15 UTC (GMT) 131,310 = (C)(X)(X)(X)MCCCX Sep 22 11:15 UTC (GMT) 30 = XXX Sep 22 11:15 UTC (GMT) 999,910 = (C)(M)(X)(C)M(X)CMX Sep 22 11:15 UTC (GMT) 931,904 = (C)(M)(X)(X)(X)MCMIV Sep 22 11:15 UTC (GMT) 5,120 = (V)CXX Sep 22 11:15 UTC (GMT) converted numbers, see more...
## The set of basic symbols of the Roman system of writing numerals
• ### (*) M = 1,000,000 or |M| = 1,000,000 (one million); see below why we prefer this notation: (M) = 1,000,000.
(*) These numbers were written with an overline (a bar above) or between two vertical lines. Instead, we prefer to write these larger numerals between brackets, ie: "(" and ")", because:
• 1) when compared to the overline - it is easier for the computer users to add brackets around a letter than to add the overline to it and
• 2) when compared to the vertical lines - it avoids any possible confusion between the vertical line "|" and the Roman numeral "I" (1).
(*) An overline (a bar over the symbol), two vertical lines or two brackets around the symbol indicate "1,000 times". See below...
Logic of the numerals written between brackets, ie: (L) = 50,000; the rule is that the initial numeral, in our case, L, was multiplied by 1,000: L = 50 => (L) = 50 × 1,000 = 50,000. Simple.
(*) At the beginning Romans did not use numbers larger than 3,999; as a result they had no symbols in their system for these larger numbers, they were added on later and for them various different notations were used, not necessarily the ones we've just seen above.
Thus, initially, the largest number that could be written using Roman numerals was:
• MMMCMXCIX = 3,999.
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# Optimized solution for a rolling dice code puzzle
I was asked to provide the solution of this rolling dice problem in an interview. When I showed him my solution, he said that there is another way to find the solution. I am looking for answer in PHP script only.
Question:
Two persons are playing a game of rolling 2 dices. Each person rolls the two dices n times and records the outcomes (sum of value of two dices) of all the n attempts. So after n attempts of both the player we have two list corresponding to the n outcomes of two players.
They want to know that whether they have got all possible outcomes( 1 to 12) same number of times or not. If they got all the possible outcomes same number of times then they are called lucky otherwise unlucky.
Input: Two Integer Arrays (L1, L2) corresponding to outcomes of two players.
Output: Lucky or Unlucky depending on the case
<?php
function rollingdice($input1,$input2)
{
foreach($input1 as$k=>$a) { if(in_array($a,$input2)) {$p = array_search($a,$input2);
unset($input2[$p]);
}
else
{ return 'Unlucky';}
}
return 'Lucky';
}
?>
• Couldn't you just compare the two arrays directly? Ideally after sorting. – Mr Wednesday Apr 22 '14 at 2:39
• all possible outcomes (1 to 12) You can't roll a 1 with two dice. :) – Alex Howansky Jun 4 '14 at 15:32
Based on your use of in_array(), you don't appear to be concerned with the order of the elements. In that case, you can simply sort the arrays and compare them directly.
function compare($array1,$array2)
{
sort($array1); sort($array2);
return ($array1 ==$array2) ? "Lucky" : "Unlucky";
}
With extra O(n) space the complexity can be made O(n) only, without sorting that will take O(nlogn)
Use a map and add a value for each key as (previousValue +1) or (previousValue - 1) based on array type A or B. Also use a counter to track an absolute increase or decrease if at end the counter is 0. Then the two array has same number of occurrence else not.
A java implementation
import java.util.HashMap;
public class Main {
public static void main(String[] args) {
HashMap<Integer, Integer> trackMap = new HashMap<>();
int n = 6;
int[] a = {4, 12, 4, 10, 6, 10};
int[] b = {10, 12, 10, 4, 4, 6};
int counter = 0;
for (int i = 0; i < n; i++) {
if (!trackMap.containsKey(a[i])) {
trackMap.put(a[i], 1);
counter++;
} else {
int prevVal = trackMap.get(a[i]);
trackMap.put(a[i], prevVal + 1);
counter = counter + ((Math.abs(prevVal) > Math.abs(prevVal + 1)) ? -1 : 1);
}
if (!trackMap.containsKey(b[i])) {
trackMap.put(b[i], -1);
counter++;
} else {
int prevVal2 = trackMap.get(b[i]);
trackMap.put(b[i], prevVal2 - 1);
counter = counter + ((Math.abs(prevVal2) > Math.abs(prevVal2 - 1)) ? -1 : 1);
}
}
System.out.println(counter == 0 ? "Lucky" : "Unlucky");
}
}
Or instead if using counter, you cand finally iterate the hashMap if all value are zero then the person is lucky else not
• Would you mind demonstrating this technique so that I can be sure I understand your theory? – mickmackusa Jul 29 '20 at 3:39
• @mickmackusa sure. Also looking if can be done using better complexity in term of space. – Mr AJ Jul 29 '20 at 11:10
• I only code in php. I thought this was a php question. – mickmackusa Jul 29 '20 at 12:06
@Mr AJ already gave you a valid answer on the algorithmic approach in his Java example, but since it seems Java is unfamiliar to you, I will just write out the algorithm here in a way that hopefully makes sense to you.
There is no need to sort the arrays, doing so ensure more operational complexity than is needed here, as now your a guaranteeing that you are visiting each value in the arrays at least twice (once for sorting, once for reading out sorted values - yes this even happens under the hood in with a == comparison between two sorted arrays as happens in currently selected answer). You can achieve the goal by iterating each array a single time, such that you only visit each value once.
Your logic should be along the lines of:
• Compare array sizes, if unequal then 'Unlucky'.
• Iterate over first array, building a map (perhaps an associative array or stdClass object in PHP). Keys are values encountered in the array (i.e. 2-12) and the values for this map are the counts of each of the corresponding dice values. You should probably also error on condition of getting value outside the 2-12 range in the array.
• Iterate over the second array, if a dice value is encountered that is not present as a key in the map built from first array, then 'Unlucky'. Otherwise you decrement the stored count in the map by one each time a corresponding dice value is encountered.
• Iterate over the map values, if any value is not equal to 0 then 'Unlucky' else 'Lucky'.
This provides an O(2n) worst-case operational complexity, where n is the size of the arrays.
After reading the other answers that speak in depth on the theory and big O, I thought I would script up my interpretation.
function roll() {
return rand(1, 6) + rand(1, 6);
}
$roller1 = [];$roller2 = [];
$totalRolls = 1; for ($i = 0; $i <$totalRolls; ++$i) {$roller1[] = roll();
$roller2[] = roll(); }$map = array_count_values($roller1);$outcome = 'Lucky';
foreach ($roller2 as$roll) {
if (!empty($map[$roll])) {
--$map[$roll];
} else {
$outcome = 'Unlucky'; break; } } echo$outcome;
After the formation of the two arrays, php offers a simple function call to create the map of roll values and the number of instances of each value -- array_count_values(). This action is only necessary on the first array.
The second array is then iterated and each value is checked against the map. If the given value is not a key in the map or it is has a falsey (0) value at the key, then there is a mismatch between $roller1 and $roller2 -- the outcome is Unlucky and the loop is sensibly broken/halted. As matches are found between the map and the second array, the encountered map keys have their respective value decremented (spent) to enable the correct action with empty().
As @MikeBrant said:
This provides an O(2n) worst-case operational complexity, where n is the size of the arrays.
the worse-case scenario is the only way to achieve the Lucky outcome.
A best-case operational complexity (Unlucky outcome) where the map is generated (n) and the loop breaks on the first element in the second array (1) ...o(n+1).
As for speed, I don't know. I didn't benchmark this script against the double-sort&compare technique, but you will notice that my technique is calling array_count_values() and will make iterated empty() calls and every function call equates to some level of a performance hit (even if miniscule).
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# Which randomized algorithms have exponentially small error probability?
Suppose that a randomized algorithm uses $r$ random bits. The lowest error probability one can expect (falling short of a deterministic algorithm with 0 error) is $2^{-\Omega(r)}$. Which randomized algorithms achieve such minimal error probability?
A couple of examples that come to mind are:
• Sampling algorithms, e.g., where one wants to estimate the size of a set for which one can check membership. If one samples uniformly at random the elements to check, the Chernoff bound guarantees an exponentially small error probability.
• The Karger-Klein-Tarjan algorithm for computing minimum spanning tree. The algorithm picks each edge with probability 1/2, and recursively finds the MST in the sample. One can use Chernoff to argue that it's exponentially unlikely there'll be 2n+0.1m of the edges that are better than the tree (i.e., one would prefer to take them over one of the tree edges).
Can you think of other examples?
Following Andras' answer below: Indeed, every polynomial time algorithm can be converted to a slower polynomial time algorithm with exponentially small error probability. My focus is on algorithms that are as efficient as possible. In particular, for the two examples I gave there are deterministic polynomial time algorithms that solve the problems. The interest in the randomized algorithms is due to their efficiency.
• Not a complete answer, but there has been some work in randomized numerical linear algebra. youtube.com/watch?v=VTroCeIqDVc Commented Jun 11, 2015 at 1:50
• Perhaps one can't expect it, but one can certainly hope (still "falling short of a deterministic algorithm with 0 error") that for all real numbers $c\hspace{-0.02 in}$, if $\: c<1 \:$ then there is an algorithm $\hspace{.34 in}$ whose error probability is $2^{-\hspace{.01 in}c\cdot r}\hspace{-0.03 in}$. $\:$ I believe Polynomial Identity Testing is such a problem. $\hspace{.49 in}$
– user6973
Commented Jun 12, 2015 at 6:06
• @RickyDemer I don't understand your comment. The usual randomized algorithm for PIT has error which is not exponential in the randomness. So what are you saying? Are you saying that there may exist such an algorithm for any BPP problem? Commented Jun 12, 2015 at 6:50
• I now realize that I don't actually see any way of showing that PIT is in the class I described. $\:$ On the other hand, letting $S$ be super-polynomial in $d$ (i.e., letting length(S) be superlinear in length(d)) would suffice for the Schwartz-Zippel lemma $\:$ (continued ...) $\;\;\;\;$
– user6973
Commented Jun 13, 2015 at 1:28
• Many probabilsitic method constructions have such behavior, no? For example, picking a random set of binary strings, and looking on their closest pair - the probability that there would be two strings in distance smaller than $n/4$ is very small. ------------------------------------------------------------------------- In the spirit of the BPP answer below: Given a constant degree expander, with n vertices, and $n/2$ marked vertices, the probability of a random walk of length $O( t )$ to miss a marked vertex is $2^{-\Omega(t)}$, if $t = \Omega( \log n)$. Commented Jun 19, 2015 at 11:35
Impagliazzo and Zuckerman proved (FOCS'89, see here) that if a BPP algorithm uses $r$ random bits to achieve a correctness probability of at least 2/3, then, applying random walks on expander graphs, this can be improved to a correctness probability of $1-2^{-k}$, using $O(r+k)$ random bits. (Note: while the authors use the specific constant 2/3 in the abstract, it can be replaced with any other constant greater than 1/2.)
If we take $k=r$, this means that any BPP algorithm that achieves a constant error probability $< 1/2$, using $r$ random bits, can be (non-trivially) improved to have error probability $2^{-\Omega(r)}$. Thus, (unless I misunderstood something), the error probability of $\leq 2^{-\Omega(r)}$ is achievable for every problem in BPP.
• The problem with such amplification techniques is that they slow down the algorithm. The new algorithm may only use O(r) random bits, but its running time is r times (original-run-time). If r is, say, at least linear in the input size n (which it usually is), you just slowed down the algorithm by a factor n. That's not something that most algorithmists would be happy about... Commented Jun 11, 2015 at 14:58
I'm not sure this is what you're looking for, but it's related:
Suppose I want to find a random $k$-bit prime number. The usual algorithm is to pick a random (odd) $k$-bit integer and run the Miller-Rabin primality test for $t$ rounds on it and repeat until a probable prime is found. What is the probability that this procedure returns a composite number? Call this probability $p_{k,t}$.
The standard analysis of the Miller-Rabin primality test shows that $t$ rounds gives a failure probability of at most $4^{-t}$. This, along with the prime number theorem, implies $$p_{k,t} \leq O(k\cdot 4^{-t}).$$
However, we are running the Miller-Rabin test on random inputs, so we can use an average-case error guarantee. We get a much better bound. In particular, for $t=1$, $$p_{k,1} \leq 2^{-(1-o(1))\frac{k \ln\ln k}{\ln k}} \leq 2^{-\tilde\Omega(k)}.$$ That is to say, we get an exponentially-small failure probability with only one repetition of the test!
See Erdös and Pomerance (1986), Kim and Pomerance (1989), and Dåmgard, Landrock, and Pomerance (1993) for more details.
This is not a decision problem and the amount of randomness used is $O(k^2)$ bits (although I suspect this can be easily reduced to $O(k)$). However, it's an interesting example where we get exponentially-small failure probability naturally.
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# Factoring trick by determining a zero value
• B
• CynicusRex
In summary, the conversation discusses the use of the zero value method in factoring algebraic expressions. While it works for some cases, it is not always applicable as seen in Problem 117 where the expression a² - 4b² does not have a factor of (a-4b). The mistake of substituting a with 4b instead of 2b is also mentioned. The conversation concludes with the realization that this discussion should have been posted in the homework section.
CynicusRex
Gold Member
I'm currently working through the book Algebra by I.M. Gelfand and A. Shen on my own. (As advised here: https://www.physicsforums.com/insights/self-study-basic-high-school-mathematics/#toggle-id-1)
This isn't really homework so I wasn't quite sure where to post this. Anyway, I've got a question regarding the following:
"Problem 113. Factor a³-b³.
Solution: The expression a³-b³ has a zero value when a = b. So it is reasonable to expect a factor a - b..."
Now, it is in fact a factor: (a-b)(a²+ab+b²) = a³-b³. So determining the zero value is useful in at least some cases. But when using this method in problem 117; Factor a² - 4b², it doesn't work. We get a zero value when a = 4b, but (a-4b) is not a factor.
I noticed it was similar to a²-b² and figured (a+2b)(a-2b) was likely the answer, and it is. However I was curious if the former method would work.
Why doesn't the zero value 'trick' work here?
I'm currently working through the book Algebra by I.M. Gelfand and A. Shen on my own. (As advised here: https://www.physicsforums.com/insights/self-study-basic-high-school-mathematics/#toggle-id-1)
This isn't really homework so I wasn't quite sure where to post this. Anyway, I've got a question regarding the following:
"Problem 113. Factor a³-b³.
Solution: The expression a³-b³ has a zero value when a = b. So it is reasonable to expect a factor a - b..."
Now, it is in fact a factor: (a-b)(a²+ab+b²) = a³-b³. So determining the zero value is useful in at least some cases. But when using this method in problem 117; Factor a² - 4b², it doesn't work. We get a zero value when a = 4b, but (a-4b) is not a factor.
I noticed it was similar to a²-b² and figured (a+2b)(a-2b) was likely the answer, and it is. However I was curious if the former method would work.
Why doesn't the zero value 'trick' work here?
You mean ##a=2b##.
CynicusRex
Ooooh I see my mistake. 4b² ≠ (4b)²
If I substitute a with 4b, I get (4b)² which is 16b².
Thanks.
PS I guess this should've been posted in the homework section then.
## 1. What is the "factoring trick" and how does it work?
The factoring trick is a method used to factor quadratic equations by determining a zero value. It involves setting the equation equal to 0 and finding the values of x that make the equation true. These values can then be used as factors to factor the original quadratic equation.
## 2. Why is the factoring trick useful?
The factoring trick is useful because it provides a quick and efficient way to factor quadratic equations without the need for complex formulas or methods. It can also be used to solve quadratic equations that cannot be easily solved by other methods.
## 3. What types of equations can be factored using the factoring trick?
The factoring trick can be used to factor any quadratic equation, which is an equation in the form of ax² + bx + c = 0, where a, b, and c are constants and x is the variable. It can also be used to factor higher degree polynomials if the equation can be rewritten as a quadratic equation.
## 4. Can the factoring trick be used for all zero values?
No, the factoring trick can only be used for zero values that are factors of the constant term (c) in the quadratic equation. If the zero value is not a factor of c, then the factoring trick cannot be used to factor the equation.
## 5. Are there any drawbacks to using the factoring trick?
One drawback of using the factoring trick is that it may not always work for more complex quadratic equations. In these cases, other methods such as the quadratic formula may be more effective. Additionally, the factoring trick may not be as useful for equations with non-integer coefficients.
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Saturday
April 30, 2016
# Posts by I. Kan
Total # Posts: 30
Teaching techniques
1. Which learning technique is frequently used in the expository model of teaching? A. Deductive thinking C. Discovery learning B. Inductive thinking D. Knowledge-based learning 2. Which technique is used after a student gives a wrong answer, and the teacher continues to aim ...
November 28, 2010
Math
4/6x multiply 19xy =38XsquareY/3
November 28, 2010
algebra
7x100+80 =700+80 =780 cent per minute Did you get it?
November 28, 2010
math
This question doed not make since or missing some info.
November 28, 2010
OR google search it and click on Articles of Confederation - Wikipedia, the free encyclopedia The Articles of Confederation was the first constitution of the United States and specified how the Federal government was to operate, including adoption of an official name for the ...
November 28, 2010
math
yours question does not make since
November 28, 2010
Algebra
find the unit rate for number of parts manufactured per hour if 1630 parts are made in 6 hours. round to the nearest integer. SHOW ALL WORK. solution. parts made in 6 hours=1630 parts made in 1 hour=x x=1630/6 x=271.66666666666 x=272 per hour
November 28, 2010
Marlon has 4 cards, Jake has 4 cards, and Sam has 3 cards. Can you write a multiplication sentence to find how many cards they have in all? Explain in general 4+4+3=11 In multiplication sentence=(3*3)+2=11 OR (3*4)-1 BUT I AM NOT 100% SURE
November 28, 2010
math
If x= (2 -3), y= (1 0 3) 4 5 2 1 0 find when possible, the matrix a) xy b) yx c) x square 1. if x=2 y=1 a) xy=2*1=2 if x=-3, y=0 xy=-3*0=0 b) if y=3 x=-3 yx=3*(-3)=-9 you can pick up any values as given for x= and y= and so on so forth if x=
November 28, 2010
math
x-[5-3{2x-3(xt2)}] =X-(5-3(2x-6xt)) =x-(5-6x+18xt) =x-5+6x-18xt =7x-18xt-5 is final answer
October 20, 2010
math
Buddy i already solved this quet
October 20, 2010
algebra
unit rate of 23miles in 4 minutes the question does not make since
October 20, 2010
elementary-math
Do you think that the formula p=6n+1, where n, is a whole number, will produce a prime number more than 50% of the time? Give evidence to support your conclusion solution n= 1,2,3 formula P=6N+1 Now put n=1 p=6(1)+1 p=6+1 p=7 now put n=2 P=6n+1 p=6(2)+1 p=12+1 p=13 and so on ...
October 20, 2010
elementary-math
prime Numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 prime Numbers are those, who must divided by itself and by 1 so n=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............. now put the value of n in the ...
October 20, 2010
Math
3/4+2/3 =9+8/12 =17/12 that st
October 19, 2010
algebra
the ratio of two integers is 9:7. their sum is 1024. find the two integers. There he go sum of ratio=9+7 =16 sum =1024 1st integer=9/16*1024=576 2nd integer=7/16*1024=448
October 19, 2010
algebra I
5x +7= 4y-6 x=(4y-6-7)/5
October 19, 2010
math
72/3.5 = ? =20.57142857
October 19, 2010
math
THIS EASY Katherine Uni:tandem, reg : tri 27 27 32 29 total bickes are 115-32(regualr bikes)= 83 83/3=27 R 2 (27+2 for tri) he knows that an equal numbr of nui and tand. so 27+27=54 CHECK= 27+27+32+29=115
October 19, 2010
math
Finding by ratio as follow ship 30 mph total distant in 4 hours is 30*4=120 mph helicop 90 mph so ratio MPH DISTANT MPH DISTANT 120 4 90 X 120*X = 360 X = 360/120 X = 3 SO X = 4-3 X = 1 HOUR THE HALI WILL CATCH THE SHIP YOURS EQUATION 4(30) + 30x = 90x IS NOT CORRECT COZ THE ...
October 17, 2010
algebra
THIS IS EASY MAN 33X+38X=284 ADD VARIABLES "X" 71X=284 SIMPLIFY X=284/71 SIMPLE DIVISION X=4 CHECK YOURS ANSWER. IS THIS CORRECT FOR X? ASK YOURSELF ORIGINAL EQUATION IS 33X+38X=284 PUT THE VALE OF X=4 IN THE EQUATION 33(4)+38(4)=284 132+152=284 284=284 HENCE PROVED ...
October 17, 2010
Math
HI monica that is this sign ^ i can help with.
October 17, 2010
roberta sold fifty tickets for the school play. Roberta sold ten more student tickets than adult tickets. How many STUDENT tickets did roberta sell according to urs equation S-10=50-s 2s=60 s=60/2 s=30 check S-10=A WHERE S=30 30-10=A A=20 HOW S+A=50 30+A=50 A=50-30 A=20 HENCE ...
October 16, 2010
math
FIND THE RATIO B/T A : B 260:275 52:55 (divided both Numbers by 5) 52+55=107 Plane A 52/107*1605=780 Plane B 55/107*1605=825 check: Plane A + Plane B =780 + 825 = 1605 Which is miles
October 16, 2010
An Overview of Teaching Techniques
1. Which learning technique is frequently used in the expository model of teaching? A. Deductive thinking C. Discovery learning B. Inductive thinking D. Knowledge-based learning 2. Which technique is used after a student gives a wrong answer, and the teacher continues to aim ...
September 15, 2010
An Overview of Teaching Techniques
can some one over this test plz.
September 15, 2010
An Overview of Teaching Techniques
can some one over this test plz.
September 15, 2010
An Overview of Teaching Techniques
Can some one over these answers again pleeeeeezÉ
September 15, 2010
An Overview of Teaching Techniques
can some one over this test please
September 15, 2010
Enchancing Children's Self-Esteem
1. D Correct 2. C Correct 3. A Correct 4. B Correct 5. D Correct 6. C Correct 7. B Correct 8. C Correct 9. A Correct 10. B Correct 11. A Correct 12. B Correct 13. D Correct 14. D Correct 15. B Correct 16. C Correct 17. D Correct 18. B Correct 19. B Correct 20. A Correct I got...
October 31, 2009
1. Pages:
2. 1
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# Contents
1. Section A: Odd and Even
1. Warm-up: Notice and Wonder: Sharing is Caring
2. 1.1: My Fair Share
3. 1.2: Share the Marbles
1. Warm-up: Which One Doesn’t Belong: Laundry Day
2. 2.1: Pair Up
3. 2.2: Are You Feeling Left Out?
1. 3.1: Color by Number
2. 3.2: Card Sort: Even or Odd
1. Warm-up: Number Talk: Equal Addends
2. 4.1: Share in Different Ways
3. 4.2: Represent Numbers with Two Addends
4. Section A Summary
1. Warm-up: How Many Do You See: Even or Odd
2. 5.1: Even and Odd Round-about
3. 5.2: Presto Chango
1. Warm-up: Number Talk: Two More
2. 6.2: Centers: Choice Time
3. Section A Practice Problems
2. Section B: Rectangular Arrays
1. Warm-up: Which One Doesn’t Belong: Counter Collections
2. 7.1: What is an Array?
3. 7.2: Rows of Counters
1. Warm-up: Estimation Exploration: Rearrange the Dots
2. 8.1: Count by Columns
3. 8.2: Guess My Array
1. Warm-up: Estimation Exploration: How Many Waffles?
2. 9.1: Sums of Rows and Sums of Columns
3. 9.2: Card Sort: Arrays and Expressions
4. 9.3: Add It All Up
1. Warm-up: True or False: Expressions that Represent Arrays
2. 10.1: Build Arrays and Write Equations
3. 10.2: Arrange Veggies to Make Arrays
1. Warm-up: Which One Doesn’t Belong: All Kinds of Arrays
2. 11.1: Use Tiles to Make Arrays
3. 11.2: Make Equal-size Squares
1. Warm-up: Estimation Exploration: Fill it Up
2. 12.1: How Many Squares?
3. 12.2: Partition Rectangles
4. Section B Summary
1. Warm-up: True or False: Two or False
2. 13.1: Centers: Choice Time
3. Section B Practice Problems
3. Glossary
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https://cs.stackexchange.com/questions/165732/finding-optimal-sequence-of-attacks-to-minimize-number-of-soldiers-needed
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# Finding optimal sequence of attacks to minimize number of soldiers needed
## The problem
Trying to improve my algorithms I bumped into a problem by Pedro Pablo Gómez Martín and Marco Antonio Gómez Martín, which I can summarise like this:
You are given a list of enemy bases you want to conquer. You can only attack one base at a time. Each base $$i$$ requires a minimum number of soldiers $$s_i$$ to be conquered, will take out a certain number $$b_i$$ of the soldiers who attack and will require a certain number $$r_i$$ of soldiers to stay back holding it, as you move the remaining army to the next battle.
The following relationship will always hold for a base $$i$$: $$s_i \geq b_i$$. Note that it is possible that $$r_i \gt s_i$$.
Determine the best sequence to follow in order to minimize the number of soldiers needed to conquer all bases.
I know, given a certain order, how to compute the number of soldiers needed, and the solution space is finite, so brute-forcing it is always a possibility, but far from ideal. It looks like there is a greedy algorithm to find the right order.
## My work
### Finding the minimum number of soldiers needed for a given order
I found a series of inequations which I believe give me the number of soldiers $$x$$ needed to conquer and hold $$n$$ bases if attacked in from $$1$$ to $$n$$:
• We need enough soldiers to conquer the first base: $$x_{a} = s_1$$
• Before each subsequent battle, we need enough soldiers to take it after all the losses of the previous battles: $$x_{b} = \max\limits_{2 \leq i \leq n} ( s_i+\sum\limits_{j=1}^{i-1}(b_j+r_j))$$
• We want enough people to survive to be able to keep the last base: $$x_{c} = \sum\limits_{i=1}^{n}(b_i+r_i)$$
So I believe the minimum number of soldiers needed $$x$$ to hold and retain all is $$x = max(x_{a}, x_{b}, x_c)$$. I think these equations can be used to prove that a certain order is optimal or not, though how exactly escapes me.
### Finding the right order to minimise the number of soldiers needed
The key insights I figured out are:
• The army gets weaker with each battle, so the battles which require more people should be fought first, when the army is at its strongest (decreasing order of $$s_i$$).
• The losses matter more the sooner they happen. Hence, we should first fight the battles which will immobilise the least number of soldiers (increasing order of $$b_i + r_i$$).
Neither criterion by itself is enough to find the optimal solution in all cases.
## My question
The proper solution seems to be combining the previous two insights into a single metric: decreasing order of $$si - (b_i + r_i)$$. Attacking the bases in that order seems to guarantee a minimum number of soldiers needed, or at least that's what my tests indicate.
While I get an intuititon of how that way "combines" both insights, I can't figure out why it guarantees the optimal order. Why that metric and not, for example, decreasing order of $$\frac{s_i}{b_i+r_i}$$? Or some weighted combination?
I'd appreciate some help both proving mathematically that it is optimal (using my equations or some other way), and ideally gaining some intuition.
• cs.stackexchange.com/questions/59964/…
– D.W.
Commented Feb 21 at 20:02
• I have expanded the post, showing my work in more detail and crediting where I got the original problem from. Commented Feb 22 at 9:13
Okay! In order to (hopefully) prove that your proposed greedy algorithm is correct, we will make an argument that is very common when dealing with greedy algorithms (as you can see from @D.W. 's linked question).
So, suppose you have found an $$x$$ such that there exists at least one sequence of battles where the army will survive. That is, there is some sequence of battles $$\mathcal{B} = (s_1,b_1,r_1),\ldots,(s_n,b_n,r_n)$$, such that we have $$x \geq \sum_{j=1}^{i-1} (b_j+r_j)+s_i$$ for all $$i$$ and also $$x \geq \sum_{i=1}^n (b_i+r_i)$$. Then we also have the sequence of battles given by your greedy algorithm: $$\mathcal{B'} = (s_1',b_1',r_1'),\ldots,(s_n',b_n',r_n')$$. And we want to show that if the army survives sequence $$\mathcal{B}$$, then they also survive sequence $$\mathcal{B'}$$.
To do so, we look at the smallest index $$k$$ where $$(s_k,b_k,r_k) \neq (s'_k,b'_k,r'_k)$$. Since these are just two different orderings of the same battles, there is some $$t>k$$ such that $$(s_t,b_t,r_t) = (s'_k,b'_k,r'_k)$$. Now, the idea is to change the ordering such that $$(s_t,b_t,r_t)$$ comes between $$(s_{k-1},b_{k-1},r_{k-1})$$ and $$(s_k,b_k,r_k)$$.
Now, what happens if the army does not survive this modified sequence? That means that there is some other index $$i$$ such that $$x-(s_i+(b_t+r_t)) < \sum_{j=1}^{i-1} (b_j+r_j)$$. It should quickly be clear that $$k\leq i (therefore we did not overcount the battle at index $$t$$ in the previous formula). But what does this imply for the original sequence, where the battle at index $$t$$ came at index $$t$$?
Since $$(s_t,b_t,r_t)$$ was equal to $$(s_k',b_k',r_k')$$ in the greedily generated sequence, and the two sequences were equal for all indices lower than $$k$$, we can infer that $$(s_i,b_i,r_i) = (s_{i'}',b_{i'}',r_{i'}')$$ for some $$i'>k$$. From the way the battles were sorted in $$\mathcal{B'}$$, we therefore know that $$s_t-(b_t+r_t) \geq s_i-(b_i+r_i)$$. Reordering gives $$s_t+(b_i+r_i) \geq s_i+(b_t+r_t)$$. Putting it all together, we get this inequality: $$x-(s_t+(b_i+r_i)) \leq x-(s_i+(b_t+r_t)) < \sum_{j=1}^{i-1} (b_j+r_j) \leq \sum_{j=1}^{t-1} (b_j+r_j)-(b_i+r_i)$$.
This means that the army would not survive the original sequence of battles, $$\mathcal{B}$$! This calculation was derived from the assumption that the army did not survive the modified sequence; therefore that assumption must be wrong.
Now we know that as long as the chosen sequence differs from the greedy sequence at least at one point, we can just move the greedy choice to that point and still get a sequence where the army survives. Therefore the greedy sequence itself must always be among those where the army survives. Note that the trick with looking at the first index where the two sequences differ is a standard trick when proving greedy algorithms correct, and it is well worth it to familiarize oneself with this technique.
I came up (with the help of some friends) with a much longer one, which I still think it's worth posting. Maybe it's because I worked on it for a few hours, but I find it easier to comprehend despite its length.
It is important to distinguish the purpose of $$s_i$$, $$b_i$$ and $$r_i$$. $$s_i$$ represents the number of people needed to win battle $$i$$. You may win a battle but not be able to hold the base: if you have $$m$$ soldiers march into battle $$i$$, after the battle you have $$m-b_i$$ soldiers alive, and if $$m-b_i \lt r_i$$, you will not be able to hold the base. So $$s_i$$ simply represents the number of people needed to win the battle, not not necessarily to hold the base afterwards.
For simplicity, let's call $$l_i=b_i+r_i$$ to the loss of people in battle $$i$$, i.e. the number of soldiers which are not available for further battles after battle $$i$$, because they were casualties or had to stay behind holding the base.
## Corollary 1
For $$n=2$$, sorting in descending order of $$s_i - l_i$$ results in the minimum number of soldiers needed.
Let's take two bases, called $$B_a$$ and $$B_b$$.
$$B_a$$ requires $$s_a$$ soldiers to be taken, and will incur in a loss of $$l_a$$. Similarly, $$B$$ requires $$s_b$$ soldiers, and will incur in a loss of $$l_b$$ soldiers.
Let's state that $$s_a - l_a \geq s_b - l_b$$. This means that $$s_a + l_b \geq s_b + l_a$$ (1).
#### Case 1: We attack $$B_a$$ before $$B_b$$.
In order to guarantee winning all battles and holding all bases, we need:
• Enough soldiers to win the first battle: $$s_a$$
• Enough soldiers to win the second battle after the first one: $$s_b+l_a$$
• Enough soldiers to withhold all casualties and soldiers left behind: $$l_a+l_b$$ $$x_1 = \max(s_a, s_b+l_a, l_a+l_b)$$
#### Case 2: We attack $$B_b$$ before $$B_a$$.
In order to guarantee winning all battles and holding all bases, we need:
• Enough soldiers to win the first battle: $$s_b$$
• Enough soldiers to win the second battle after the first one: $$s_a+l_b$$
• Enough soldiers to withhold all casualties and soldiers left behind: $$l_a+l_b$$ $$x_2 = \max(s_b, s_a+l_b, l_a+l_b)$$
Let's consider all three values that $$x_1$$ might take, and realise that in all cases, $$x_2 \geq x_1$$. We can do this by seeing that, for any value $$x_1$$ takes, there is at least one value in the three possible values which $$x_2$$ may take which is greater or equal than $$x_1$$, so $$x_2 \geq x_1$$:
• If $$x_1 = s_a$$, $$x_2$$ will be at least $$s_a+l_b$$
• If $$x_1 = s_b + l_a$$, we know that one of the values of $$x_2$$, $$s_a+l_b$$, is greater than $$x_1$$ (see (1))
• If $$x_1 = l_a + l_b$$, $$x_2$$ will be at least $$l_a + l_b$$
So it is proven that, given two bases, is optimal to attack them in decreasing order of $$s_i - l_i$$.
## Lemma 1
The optimal relative order of $$B_a$$ and $$B_b$$ ($$B_a$$ attacked before $$B_b$$) remains the same if a third base $$B_c$$ is inserted between the two, provided that $$s_a - l_a \geq s_c - l_c \geq s_b - l_b$$
We can prove this by comparing the two alternatives:
#### Case 1: $$B_a \rightarrow B_c \rightarrow B_b$$
We need enough soldiers to:
• Take the first base: $$s_a$$
• Take the second base after the first battle: $$s_c + l_a$$
• Take the third base after the first two battles: $$s_b + l_a + l_c$$
• Survive all losses: $$l_a + l_b + l_c$$
$$x_{1}=\max(s_a, s_c+l_a, s_b+l_a+l_c, l_a+l_b+l_c)$$
#### Case 2: $$B_b \rightarrow B_c \rightarrow B_a$$
We need enough soldiers to:
• Take the first base: $$s_b$$
• Take the second base after the first battle: $$s_c + l_b$$
• Take the third base after the first two battles: $$s_a + l_b + l_c$$
• Survive all losses: $$l_a + l_b + l_c$$
$$x_{2}=\max(s_b, s_c+l_b, s_a+l_b+l_c, l_a+l_b+l_c)$$
We can prove that $$x_{1} \leq x_{2}$$ using the same method as for Corollary 1: for any possible value of $$x_{1}$$, the value of $$x_{2}$$ will be greater or equal than that.
• If $$x_{1}=s_a$$, $$x_{2}$$ will be at least $$x_{2}=s_a+l_b+l_c$$
• If $$x_{1}=s_c+l_a$$, $$x_{2}$$ will be at least $$x_{2}=s_a+l_b+l_c$$. This is because $$s_a - l_a \geq s_c - l_c$$, so $$s_a + l_c \geq s_c + l_a$$, and therefore, $$s_a + l_b + l_c \geq s_c + l_a$$
• If $$x_{1}=s_b + l_a + l_c$$, $$x_{2}$$ will be at least $$x_{2}=s_a+l_b+l_c$$. This is because $$s_a-l_a \geq s_b-l_b$$, so $$s_a+l_b \geq s_b+l_a$$. By adding $$l_c$$ to both sides of the inequation, $$s_a+l_b+l_c \geq s_b+l_a+l_c$$
• If $$x_{1}=l_a + l_b + l_c$$, $$x_{2}$$ will be at least $$x_{2}=l_a+l_b+l_c$$.
So, as long as $$s_a - l_a \geq s_c - l_c \geq s_b - l_b$$ holds, $$B_a$$ must be attacked before $$B_b$$.
## Lemma 2
Given a sequence of battles, we can replace them with a single equivalent battle
We can prove this by induction, starting with $$n=2$$. Given two battles $$B_j$$ $$(s_j, b_j, r_j)$$ and $$B_k$$ $$(s_k, b_k, r_k)$$, we can say the following:
• In order to win both battles, we need enough soldiers to win the first one and to win the second one after the first one: $$s_{jk}=\max(s_j, s_k+l_j)$$.
• After fighting both battles, $$b_{jk}=b_j + b_k$$ soldiers will have been taken out.
• After fighting both battles, $$r_{jk}=r_j + r_k$$ soldiers will be posted to keep the bases.
Note that $$s_{jk}$$ does not include the requirement of having enough people to hold all bases, because $$s$$ only represents the number of people needed to win a battle.
So we can replace the sequence with an equivalent battle $$B_{jk}$$ $$(\max(s_j, s_k+l_j), b_j + b_k, r_j + r_k)$$
For any value of $$n \ge 2$$, we can simply iteratively reduce the first two battles into one until only one base is left.
## Lemma 3
For a sequence $$B_1 \rightarrow B_2 \rightarrow ... B_{n-1} \rightarrow B_{n}$$ sorted in decreasing order of $$s_i-l_i$$, if the middle sequence of $$n-2$$ bases is replaced with their single equivalent base $$B_2'$$, the resulting $$B_1 \rightarrow B_2' \rightarrow B_{n}$$ sequence is also in decreasing order of $$s_i-l_i$$
First we'll prove this for $$n=4$$. Let's start the demonstration with four bases sorted in decreasing order of $$s_i-l_i$$, $$B_1 \rightarrow B_2 \rightarrow B_3 \rightarrow B_4$$.
We know that the following inequalities are true: $$s_1 - l_1 \geq s_2 - l_2 \geq s_3 - l_3 \geq s_4 - l_4$$
We replace $$B_2 \rightarrow B_3$$ with $$B_{23}$$, with $$s_{23}=\max(s_2, s_3+l_2)$$ and $$l_{23}=l_2+l_3$$ (Lemma 2):
$$B_1 \rightarrow B_{23} \rightarrow B_4$$
We want to prove that the following relationship holds: $$s_1 - l_1 \geq s_{23} - l_2 - l_3 \geq s_4 - l_4$$
For this we, once again, examine the possible values of $$s_{23}$$:
#### Case 1: $$s_{23}=s_2$$
We want to prove: $$s_1 - l_1 \geq s_2 - l_2 - l_3 \geq s_4 - l_4$$, which has two parts:
Since $$s_1 - l_1 \geq s_{2} - l_2$$, trivially $$s_1 - l_1 \geq s_{2} - l_2 - l_3$$.
On the other hand, if $$s_{23}=s_2$$, we know that $$s_2 \geq s_3 + l_2$$ (or $$s_{23}$$ would have taken a different value). By adding the two known inequalities $$s_3 - l_3 \geq s_4 - l_4$$ and $$s_2 \geq s_3 + l_2$$, we reach $$s_3 - l_3 + s_2 \geq s_4 - l_4 + s_3 + l_2$$. Simplifying and reordering, we reach $$s_2 - l_2 - l_3 \geq s_4 - l_4$$.
#### Case 2: $$s_{23}=s_3+l_2$$
We want to prove: $$s_1 - l_1 \geq s_3 - l_3 \geq s_4 - l_4$$, which we already know to be true.
This ends the proof for Lemma 3 for $$n=4$$. This is trivially generalisable to $$n>4$$: first, replace $$B_2$$ and $$B_3$$ with their single equivalent base $$B_{23}$$, and that should leave the sequence still in order. Then replace $$B_{23}$$ and $$B_4$$ with $$B_{234}$$, etc.
## Conclusion
We have proven that, if we sort our list of bases according to the $$s_i - l_i$$ (i.e. $$s_i - (b_i+r_i)$$) criterion, for any two distinct bases $$B_i$$ and $$B_j$$ that we choose, their relative order will be optimal:
• If they are consecutive, the optimality is proven by Corollary 1.
• If they are not consecutive, the bases in the middle can be reduced to a single equivalent base (Lemma 2), and the resulting sequence will still be sorted according to our criterion (Lemma 3). In that case, the optimality is proven by Lemma 1.
Since all bases are optimally sorted with respect to all other bases, the sorting of the whole list is optimal.
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Mathematics 86 Online
OpenStudy (anonymous):
how do you expand 5(3x-1)^2 ?
OpenStudy (mathmagician):
$=5(9x^2-6x+1)=45x^2-30x+5$
OpenStudy (anonymous):
square the parens then distribute the 5
OpenStudy (anonymous):
A simple approach if you're just learning algebra is to first expand the brackets: $5(3x-1){^2}$$=5(3x-1)(3x-1)$ Then go along the equation and multiply out each term, remembering to distribute the 5: $=5\times(3x \times 3x) - 5\times 1\times 3x - 5\times 1 \times 3x + 5\times 1 \times 1$ Now begin to group like terms and simplify: $=5\times 9x{^2} - 10\times 3x + 5$$=45x{^2}-30x+5$
OpenStudy (anonymous):
First step = 5(9x^2-3x-3x+1) Second step =5(9x^2-6x+1) Third step = 45x^2-30x+5
OpenStudy (anonymous):
thanks !
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Last month we examined progressive jackpots to see how they affect real money slot machines’ theoretical hold and your casino in general. We saw that the amount contributed to a progressive jackpot must come from somewhere — either from the hold amount, the theoretical payout or a combination of both.
In order to convert an existing game to a progressive, we must consider where the contribution will come from. In some regards, we are at the mercy of the library of games available from the manufacturer; consequently, we may not be able to switch to a game that matches our exact desire.
However, most popular progressive games come in a number of variations, so selecting a game that closely matches should not prove too difficult.
This month we’ll study progressive jackpots in more detail. To start, we’ll make a couple of assumptions about our exercise. First, let’s assume we already have a game with the proper theoretical payout percentage and theoretical hold that will allow a specific percentage to be accrued in the progressive jackpot.
Second, let’s assume the game is a 3-reel stepper game, which will make our calculations as simple as possible.
## Base Jackpot
Mathematically, a progressive jackpot game is not much different from a game that does not offer a progressive award. The word “progress” means “to advance,” and we must advance from somewhere. In most cases, the progressive amount is applied to the largest payout — the jackpot award. This may be three Red 7 symbols or three Wild symbols, but generally is the largest award that a player can receive. (Of course, newer games offer progressive jackpots on numerous awards, not only the highest-paying amounts — we’ll discuss these a bit later.)
In a non-progressive slot game, there must be an amount payable when a player receives a predetermined combination of symbols. The progressive jackpot machine will also pay this amount when the jackpot is awarded. Let’s examine a fictitious game called Blazin’ 7s (which was created in a 2005 CEM article). This game has a top award of 1,000 credits when three credits are wagered. On a 25-cent machine, this translates to a top award of \$250. In the non-progressive game, every time this combination hits, the player is paid \$250. In the progressive game, the player is paid the same \$250, but also an extra amount, which increased as the game was played. The progressive machine has the same base jackpot amount as the non-progressive machine. This is particularly noteworthy for linked progressive jackpots, where many machines contribute to the top award. Let’s examine this progression (or advancement) of the top award amount.
## Jackpot Progression
Once we have determined the base payout amount, as indicated on the PAR sheet, we have our starting jackpot amount. This is always shown on the PAR sheet and will typically be expressed in credits. The progressive jackpot marquee must be set according to the denomination of the machine. A 1,000-credit award on a dollar machine will pay \$1,000. The same award on a nickel machine will pay \$50. The denomination is not important to the machine because the PAR sheet works in credits. The progressive controller simply converts the credit value to dollars. There is actually no reason why you could not have the marquee show the progressive amount in credits; however, this could confuse players, and a dollar amount is easier to understand.
Once the base amount is set, a rate of progression must be determined. This is the amount that will be sent to the progressive jackpot controller for each credit (or coin) wagered on the machine, increasing the top award until it is won. Let’s take one-half of 1 percent from the coin-in and transfer it to the progressive jackpot. This means that for every \$1 wagered, \$0.005 is sent to the progressive controller. The controller will update the marquee and show the increased amount of the progressive jackpot. The percentage that you decide to use is not important to the progressive controller. You simply need to determine how much you wish to allocate to the progressive jackpot and ensure that you’re sending the correct amount.
The progressive contribution rate affects your slot math because every cent that is sent to the progressive controller is really a “win” that has been paid from the slot machine — it just hasn’t been collected by a player yet. For all intents and purposes, we can consider this money paid to the progressive controller rather than directly to a player. Using meters and reports, you can determine the number of credits played and, therefore, the amount of money each machine has contributed to the jackpot. This will not be the same for each machine. While the percentage of total credits-in contributed to the progressive jackpot will be the same, each game will have a different amount of game play or handle. Perhaps the machine at the end of the row is played more frequently than the one in the center. Perhaps one particular theme is more popular than another, and those games are played more than the others. But all we’re concerned about is how much the games are played and how much they contribute to the progressive jackpot.
Players see the increasing jackpot amount as an incentive to play, hoping to win big. This increase depends on a number of factors, specifically the percentage contribution, the denomination of the machine and the number of machines linked to the progressive display. A \$1 machine will have a progressive jackpot amount that increases faster than a penny machine given the same contribution rate and same amount of play. Of course, a dollar machine could have a smaller contribution rate than a penny machine and still increase faster.
When the jackpot is awarded, the current value of the jackpot (as shown on the marquee) is given to the player, the jackpot amount is reset to the base level and the cycle continues. On a stand-alone progressive machine, the progressive jackpot only increases as credits are wagered on that particular machine. This is a common configuration for 3-reel steppers using Bally Technologies’ Blazing Sevens and IGT’s Sizzling Sevens. Typically, there will be a bank of machines, and each one will show a different progressive jackpot amount.
Players will scout out the highest jackpot, thinking it will hit soonest because the machine has been played the longest without a jackpot. This, of course, is not necessarily true. While statistically we can use the PAR sheet to determine the average jackpot amount that will be paid, there will be some variance or volatility due to the random nature of the machines. As you have likely experienced, the highest jackpot amount may continue to increase when smaller jackpots in the same bank hit before the larger one.
Linked progressive jackpots, including large wide-area jackpots, follow the same basic rule. However, the contribution from each machine goes to the shared progressive jackpot amount. This results in larger awards that increase faster than stand-alone progressive jackpots. The top award for each machine must equal the base jackpot amount.
This means that a \$1 million progressive jackpot will have a top award of the \$1 million equivalent in credits. You may think that this jackpot amount would be split among the machines, e.g., a \$10,000 base jackpot would get \$1,000 from each of the 10 machines it is linked to.
This is not correct. If \$1 machines were linked to a jackpot system with a \$1 million base progressive jackpot, each machine would have to have one million credits as the top award on the PAR sheet. The payout for the base portion of the win comes from the individual machine, not the bank as a whole. Any extra amount above the base progressive amount, however, comes from the contributions of all machines played since the last jackpot was awarded.
A linked progressive jackpot system does not necessarily have to have all the same games; you can have different machine themes on the same progressive. The Megabucks jackpot machines in Ontario, for example, use Double Diamond, Triple Diamond and Five Times Pay games. When picking which games to include, we must consider the hit frequency of the jackpot award and the overall payout of the game. The games should have a similar jackpot hit frequency, and the payout amount must match the jackpot base.
This is really an easy equation for slot math. Let’s study the simple bar chart in Figure 2 to see how all this relates. The entire height of the bar represents 100 percent of the money wagered on the machine. We break this amount down into two primary components: hold and theoretical payout. What we don’t pay out, we hold. The two must add up to 100 percent.
Figure 3 expands this slightly, showing the progressive amount as part of the payout. Our hold remains the same: 5 percent. The theoretical payout, however, is broken down into 94 percent in regular awards and 1 percent for the progressive jackpot. We are still paying out 95 percent, and these are the parts that make up that 95 percent.
Figure 4 shows a complete breakdown of the payouts, with every possible payout amount — Blanks, mixed Bars, three Single Bars, etc. Note that all of the amounts paid go to the credit meter, except for the 1 percent progressive contribution and the base jackpot amount. (Remember, this base jackpot amount is what the progressive meter is reset to when a jackpot is hit.)
## Different Denominations
The important points to remember are that each progressive machine must have: (1) the base progressive jackpot amount as its top award and (2) a progressive contribution amount. But what happens if each machine doesn’t award the same base jackpot amount? Could this happen? Yes! And you may wish to do this.
It is possible to have one progressive jackpot linked to machines with different denominations. While this initially sounds complicated, it really isn’t. Suppose we have a progressive jackpot starting at \$1,250. On one side of the casino, we have a bank of 25-cent machines. The base award would be 5,000 credits (\$1,250 / \$.25 = 5,000). On the PAR sheets, we would see that the top award for these machines is 5,000 credits. On the other side of the casino, however, we have a bank of dollar machines connected to the same progressive display. These machines, however, would have a top award of 1,250 credits (\$1,250 / \$1 = 1,250). This way, each machine has the same top award when considering its dollar value. The quarter machine must pay 5,000 credits to award \$1,250; the dollar machine must pay 1,250 credits to award \$1,250. It would be reasonable to assume that the dollar machines would hit the jackpot more frequently than the quarter machines, too. As the machine is paying out a smaller amount for the top award (in this case, one quarter of the amount), it will likely hit more frequently. This encourages players to wager on a higher denomination machine, providing an increased opportunity to win. (You may or may not want to do this, and there are arguments for and against this configuration.)
The jackpot contribution rate could also be smaller on the higher denomination machine. A dollar machine contributing one-quarter of 1 percent of its coin-in would provide the same amount of progression to the jackpot as a 25-cent machine providing 1 percent of its coin-in. This would not necessarily have to be the case, however. If the rates were the same (i.e., 1 percent of coin-in from both machines), the dollar machine would be contributing a larger amount, and a quarter machine-playing winner could benefit from the play on the dollar machines.
## Hidden Jackpots
Progressive contributions do not always go to the standard progressive jackpot. Hidden jackpot amounts are available and can be used for a variety of purposes. A small contribution can be sent to a hidden meter that is transferred to the primary jackpot after it is awarded. This amount increases the base jackpot so it never appears to have just hit. If the jackpot has a cap that specifies the largest amount that can be awarded, contributions after that amount is reached can be sent to a hidden value. In this case, the same contribution amount continues to be taken from each game. Of course, any configuration must be approved by your jurisdictional authority.
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# physics
posted by .
A dockworker loading crates on a ship finds that a 24 kg crate, initially at rest on a horizontal surface, requires a 79 N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 57 N is required to keep it moving with a constant speed. Find the coefficients of static and kinetic friction between crate and floor.
• physics -
The initial force it takes to move the crate is given by μs*mg
where μs is the coefficient of static friction.
Similarly, the coefficient of kinetic friction can be calculated from the force it takes to maintain motion.
Post for an answer check if you wish.
• physics -
I try to looking for this answer but now I think I am a person should answer this question ^^. Ok
Sum F = ma ( because it get loading the crates )
Fp - F(fr) = ma
Fp = ma + F(fr)
Fp = ma + u(k) mg
=> u(k) = (Fp - ma) / mg ( with ma is 0 )
so u(k) = 57/ ( 24*9.8) ( because kenetic friction always moving )
The u(s) you guys can do the same thing , just replace another force number to the equation.
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### Home > CC3 > Chapter Ch10 > Lesson 10.1.3 > Problem10-41
10-41.
Find the volume of the cone shown in problem .
From problem , you learned that the volume of a cone is one-third the volume of a cylinder with the same base area and height.
The equation for the volume of a cylinder is:
$\text{Volume = (area of base)(height)}$
So the volume of a cone is:
$\text{Volume } =\; \frac{1}{3}(\text{area of base})(\text{height})$
The base is a circle with a radius of 6 inches. So using the equation for area of a circle:
$\text{Area of base} = (6^2)π= 36π$
The height of the cylinder is $8$ inches. Substitute the known values into the volume equation for a cone:
$\text{Volume}=\frac{1}{3}(36\pi )(8)$
Now simplify to get the answer.
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# Real Analysis
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# Chocolate
I have a box of chocolate - white, milk and dark. The ratio of white to milk with dark is 3: 4. The ratio of white and milk to dark is 17: 4. Calculate what is the ratio between white, milk, dark chocolate.
Result
p = (Correct answer is: 9:8:4)
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
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#### To solve this example are needed these knowledge from mathematics:
Do you solve Diofant problems and looking for a calculator of Diofant integer equations?
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3. Divide 5
Divide 288 in the following ratio 3 : 4 : 5
4. Three machines
The power of the three machines is 2: 3: 5. Two most powerful machines produce 400 parts per hour. How many components make all three machines in 3 hours?
5. Reminder and quotient
There are given the number C = 281, D = 201. Find the highest natural number S so that the C:S, D:S are with the remainder of 1,
6. Reminder and quotient
There are given numbers A = 135, B = 315. Find the smallest natural number R greater than 1 so that the proportions R:A, R:B are with the remainder 1.
7. The Hotel
The Holiday Hotel has the same number of rooms on each floor. Rooms are numbered with natural numbers sequentially from the first floor, no number is omitted, and each room has a different number. Three tourists arrived at the hotel. The first one was in r
8. Diophantus
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9. Textbooks
After check of textbooks found that every 10-th textbook should be withdrawn. Together 58 textbooks were withdrawn. How many textbooks were in stock before withdrawn and how many after withdrawn?
10. Brick
One brick is 6 kg and half a brick heavy. What is the weight of one brick?
11. Circus
On the circus performance was 150 people. Men were 10 less than women and children 50 more than adults. How many children were in the circus?
12. Six-digit primes
Find all six-digit prime numbers that contain each one of digits 1,2,4,5,7 and 8 just once. How many are they?
13. The florist
The florist had 200 roses in the morning. During the day, more than half sold it. From the remaining roses, it will tie the bouquet. If a bouquet of 3, 4, 5 or 6 roses are bound, one always remains. How many roses from the morning shipment sold?
14. Divisibility
Is the number 761082 exactly divisible by 9? (the result is the integer and/or remainder is zero)
15. Monkey
Monkey fell in 38 m deep well. Every day her scramble 3 meters, at night dropped back by 2 m. On that day it gets hangover from the well?
16. Z9-I-4
Kate thought a five-digit integer. She wrote the sum of this number and its half at the first line to the workbook. On the second line wrote a total of this number and its one fifth. On the third row she wrote a sum of this number and its one nines. Fi
17. No. of divisors
How many different divisors has number ??
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Printable Science Worksheet on Sound
# Worksheet on Sound
## Solved Questions on Sound
1. In an experiment to study the speed of sound in air, a person claps hands at a known distance from a wall and measures the time taken for the echo to return. If the time is 0.5 seconds and the speed of sound is 340 m/s, what is the distance of the wall from the person?
a) 170 m
b) 85 m
c) 340 m
d) 680 m
Answer: b) In this experiment, the time taken for the sound to travel to the wall and back is the total time for the sound to cover twice the distance between the person and the wall. The formula to calculate the distance (d) using the time (t) and speed (v) is:
2d/v = t
d = (v x t)/2
Given that the speed of sound (v) is 340 m/s and the time (t) is 0.5 seconds, you can plug these values into the formula:
d = (340 ms-1 x 0.5 s)/2
d = 85 m
So, the distance of the wall from the person is 85 metres.
2. A sound wave with a frequency of 750 Hz travels through a medium with a speed of 375 m/s. Calculate the wavelength of the wave and determine the time it takes for one complete wave to pass a given point.
a) Wavelength: 0.5 m, Time: 0.001 s
b) Wavelength: 0.75 m, Time: 0.001 s
c) Wavelength: 0.25 m, Time: 0.002 s
d) Wavelength: 0.375 m, Time: 0.001 s
Answer: a) To calculate the wavelength of the wave, we can use the formula:
Wavelength(λ) = Speed of Sound(v) / Frequency(f)
Given:
Frequency of the wave (f) = 750 Hz
Speed of sound in the medium (v) = 375 m/s
Substitute the values:
λ = 375 m/s / 750 Hz
λ = 0.5 m
So, the wavelength of the wave is 0.5 m.
To determine the time it takes for one complete wave to pass a given point (time period), we can use the formula:
Time Period(T) = 1/Frequency(f)
Substitute the given frequency:
T = 1/750
T= 0.001333s
So, the time it takes for one complete wave to pass a given point is approximately 0.001 seconds.
3. When a person is listening to music, which part of the ear helps differentiate between different pitches of sound?
a) Pinna
b) Cochlea
c) Oval window
d) Eardrum
Answer: b) The cochlea is the part of the inner ear that helps differentiate between different pitches of sound. It contains specialised hair cells that respond to different frequencies of sound vibrations. Different sections of the cochlea are responsible for detecting different frequencies, allowing us to perceive various pitches of sound.
4. A radio station broadcasts at a frequency of 98.5 MHz. If the speed of radio waves is approximately 3 × 108 m/s, what is the wavelength of these waves?
a) 3.05 m
b) 0.305 m
c) 305 m
d) 30.5 m
Answer: a) Wavelength (λ) is calculated using the formula:
λ = Speed of radio waves (v) / Frequency (f)
Given:
Speed of radio waves (v) = 3 × 108 m/s
Frequency (f) = 98.5 × 106 Hz (98.5 MHz)
Substitute the values:
λ = (3 × 108 m/s) / (98.5 × 106 Hz) ≈ 3.05 m
So, the correct answer is option a) 3. 05 m.
5. In a concert hall, which seating arrangement is likely to result in better sound quality?
a) Parallel walls with hard surfaces
b) Irregularly shaped walls with soft surfaces
c) Parallel walls with reflective surfaces
d) Irregularly shaped walls with glass panels
Answer: b) Better sound quality in a concert hall is often achieved by minimising the effects of echoes, reverberation, and standing waves. Irregularly shaped walls can help diffuse sound waves and prevent the formation of strong reflections and standing wave patterns. Soft surfaces, such as acoustic panels and draperies, absorb sound energy, reducing reverberation and creating a more balanced and clear sound environment. This arrangement helps to minimise acoustic distortions and improve the overall sound quality for the audience.
## FAQs
1. How does sound travel through different mediums?
Sound travels across various media by forcing the particles to vibrate. Solids have densely packed particles, which allows sound to travel more quickly. The particles in liquids and gases are more dispersed, resulting in slower sound transmission.
2. What is the Speed of Sound?
The speed of sound varies according on the medium through which it travels. In dry air at normal temperature, sound travels at a rate of around 343 meters per second (1235 kilometers per hour).
3. What factors affect the pitch of a sound?
The pitch of a sound is determined by its frequency, which is measured in hertz. Higher-frequency sound waves generate higher-pitched noises, whilst lower-frequency waves generate lower-pitched sounds.
4. How do echoes occur?
Echoes arise when sound waves bounce off objects and then return to the listener's ears after a delay. The distance between the sound source and the reflecting surface impacts both the time delay and the perceived echo.
5. How is sound produced?
Sound is created when an item vibrates, causing pressure waves to spread across the surrounding medium. These vibrations force the particles in the medium to compress and rarefy, resulting in the impression of sound.
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# Doppler and the planets
1. Dec 5, 2005
### kp
Question:
The variable period of a moon of Jupiter, which is the basis of measurement of the speed of light in Roemer's method (whoever he is), is regarded as the Doppler affect. The period of the orbital motion of one of Jupiter's moons is approximately 42.5 hours; the speed of light is 2.99e8 m/s. The orbital speed of the Earth about the Sun is 2.98e4 m/s. What is the maximum change in the period (in seconds) of this moon as observed from Earth? (hint: f = 1/T)
I've done doppler problems with two moving objects
but I can't seem to come up with an answer that makes sense.
when I use f = 1/T i get this small number that look like wavelength.
any tips would be appreciated, thanks
2. Dec 5, 2005
### Staff: Mentor
from http://en.wikipedia.org/wiki/Speed_of_light#Measurement_of_the_speed_of_light
f = 1/T and $\omega$ = 2$\pi$f = v/r where v is the linear velocity and r is the radius of the orbit of an object with velocity v.
3. Dec 5, 2005
### kp
Interesting info...
so..I need to find the velocity of the moon around Jupiter by it's period?
then use f' = f((1 -+ v(earth)/ c)/(1-+ v(moon)/c)) to find the change frequency?
4. Dec 6, 2005
anybody....?
5. Dec 6, 2005
### Staff: Mentor
Sorry about not getting back to you.
I don't think the problem is concerned with the effect of frequency change, but rather about the relative time that an event is observed.
The period could be estimated by observing the moon at the same point in its orbit during successive periods. However the earth is also in its orbit.
So the biggest change in the observed period occurs when the change in the earth's relative position with respect to Jupiter changes the most during one of the moons periods.
In 42.5 hrs, the earth moves 4.5594e+9 m (but that is a circular arc). Light moves at 2.99e8 m/s, so if the earth move 4.5594e+9 m closer to Jupiter, one observes the subesequent appearance of the moon ~15.2 seconds sooner, just due to the relative motion of the earth. If the earth is moving away, then the one would observe a successive appearance of the moon 15.2 seconds later.
But this applies to one period, and I did not account for any relative motion by Jupiter.
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https://www.gradesaver.com/textbooks/math/calculus/university-calculus-early-transcendentals-3rd-edition/chapter-8-section-8-2-trigonometric-integrals-exercises-page-435/54
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## University Calculus: Early Transcendentals (3rd Edition)
$$\int^{\pi/2}_0\sin x\cos xdx=\frac{1}{2}$$
$$A=\int^{\pi/2}_0\sin x\cos xdx$$ Apply the identity of the product of sine and cosine functions: $$\sin mx\cos nx=\frac{1}{2}[\sin(m-n)x+\sin(m+n)x]$$ we have $$A=\frac{1}{2}\int^{\pi/2}_0(\sin0+\sin2x)dx$$ $$A=\frac{1}{2}\int^{\pi/2}_0\sin2xdx$$ $$A=-\frac{1}{4}\cos2x\Big]^{\pi/2}_0$$ $$A=-\frac{1}{4}(\cos\pi-\cos0)$$ $$A=-\frac{1}{4}(-1-1)=\frac{1}{2}$$
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http://mathhelpforum.com/statistics/52155-seating-placement.html
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1. ## seating placement
Hello, I thought this would of been easy but I am second guessing myself.
Frank, Tom, Jim and 6 other boys are randomly seated in a row. What
is the probability that neither Frank nor Tom sits next to Jim?
Here is what I came up with. However, I am not confident in it, because it seems rather low.
(6! x 9)/(9!) = 0.0178... or 1.78%
2. Calculate the probability that at least one of Frank or Tom does sit next to Jim.
If Frank sits next to Jim, then they can be arranged 2 ways, and considered as a unit. There are 2x8! possible arrangements in this case.
If Tom sits next to Jim, then the same is true. There are 2x8! possible arrangements.
If Frank and Tom both sit next to Jim, then the three of them can be arranged 2 ways (since Jim is always in the middle), and considered as a unit. There are 2x7! possible arrangements.
So the total is 2x8! + 2x8! - 2x7! = 4x8! - 2x7!
There are 9! arrangements in general.
So the probability of at least one of Frank or Tom sitting next to Jim is
$\displaystyle \frac{4(8!) - 2(7!)}{9!}$
And the probability that neither of them sits next to Jim is
$\displaystyle 1 - \frac{4(8!) - 2(7!)}{9!}$
3. Hello, chrisc!
Frank, Tom, Jim and 6 other boys are randomly seated in a row.
What is the probability that neither Frank nor Tom sits next to Jim?
I though I would solve this "head-on" . . .
There are: .$\displaystyle 9! \:=\:362,880$ possible seatings.
There are two cases to consider:
. . [1] Jim occupies an end seat.
. . [2]Jim does not occupy an end seat.
[1] Jim occupies an end seat.
First, there is a choice of 2 end seats he can occupy.
We have: . $\displaystyle J \;\_\;\_\;\;\_\;\_\;\_\;\;\_\;\_\;\_$
The second seat must not be taken by Frank or Tom.
. . It is taken by one of the other 6 boys.
Then the other seven boys can be seated in 7! ways.
Hence, there are: .$\displaystyle 2 \times 6 \times 7! \:=\:60,480$ ways with Jim in an end seat.
[2] Jim occupies an "inner" seat.
He has a choice of 7 inner seats.
We have: .$\displaystyle \_\;J\;\_\;\;\_\;\_\;\_\;\;\_\;\_\;\_$
The left seat must not be taken by Frank or Tom.
. . It is taken by one of the other 6 boys.
The right seat must not be taken by Frank or Tom.
. . It is taken by one of the other 5 boys.
Then the other six boys can be seated in 6! ways.
Hence, there are: .$\displaystyle 7 \times 6 \times 5 \times 6! \:=\:151,200$ ways with Jim in an inner seat.
Then there are: .$\displaystyle 60,480 + 151,200 \;=\;211,680 \text{ ways}$
Therefore, the probability is: .$\displaystyle \frac{211,680}{362,880} \;=\;\boxed{\frac{7}{12}}$
We agree, icemanfan!
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https://slideplayer.com/slide/5937773/
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# Understanding Place Value.
## Presentation on theme: "Understanding Place Value."— Presentation transcript:
Understanding Place Value
Place value is thousands
Place Value vs. Value Value is 4 Each digit in a number has a place value and a value. In the number 4856, the digit 4 is in the thousands place value. Meaning the place value is thousands. The number you see (4) is the value. 4856 Place value is thousands
A place value chart helps us to read and understand large numbers.
We are only going to focus on whole numbers today…but
We will be also learning about place value to the hundredths!!
THOUSANDS 463, 245, 791. UNITS ones, tens, hundreds MILLIONS
G CIA EKD Hundred Thousand Ten Thousand Thousands Millions Ones
Hundreds Tens
Saying Large Numbers Understand place value helps us say large numbers properly When saying large numbers you should: A) start with the largest place value grouping B) say what you see, then say the grouped place value C) continue until all number have been read 34, 907, 521.
Try These 4, 560, 392. 5, 500, 428. 5, 022, 603. 364, 210. 4, 321. 800, 444. 904, 265. 93, 004.
Writing Numbers In Words
When writing numbers in words you write exactly what you say: 9, 320, 199. nine million, three hundred twenty thousand, one hundred ninety-nine
Try These Two hundred sixty four thousand, five hundred three
Nine million, sixty-two thousand four five million, twenty-two Five hundred sixty-three thousand, four hundred thirteen Seven thousand, sixteen
Standard and Expanded Form
When numbers are presented in numerical digits, it is called the standard form of a number.
4856 Standard Form
Numbers can also be represented in expanded form.
This means writing the value of a number using the face value and place value of each digit. The number 4856 in expanded form is:
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# Percentage free online calculator
#### Percentage calculator is a free online tool to calculate percentages.
Find sentences that represents your problem and our online percentage calculator will give you answer immediatelly. Put values in two fields and click on button Calculate to see the correct result.
Use this very straightforward and more complex percentage calculator, which will also show you formula for percentage to solve quickly and understand any problems involving percentage.
% × =
What is % of ?
is what percent of ? %
is % of what?
What is the percentage change from to ? %
#### Percent, Fraction And Decimal Calculator
Insert one value to section Percent or Decimal and hit green button (=) on the right. Or inser two values to Fraction (for example 1/10 = you will put there 1 and 10 below that) and hit green button (=) on the right. And the calculator will give you the right answer.
Percent: Fraction: Decimal: %
This is a free mobile-friendly online percentage calculator which can help you with calculation of percentage and figuring out the correct answer.
It is 3-way percent calculator (to find percentage of a number, calculate x as a percent of y). No matter whether you can count it or not, this online tool will help you find the correct solution. Below you will find also answers for the most frequently asked questions like percentage calculation formula, how can you calculate percentages without calculator, how to calculate the percentage increase or decrease and so on.
## What Is A Percentage?
• One percent (1%) means 1 per 100.
• Word "percent" comes from Latin (per centum).
• It is shown by the symbol %.
• Used in Math.
## How Percentage (%) Is Defined?
Percentage is per-cent which means parts per hundred, it describes how many parts there are out of one hundred parts of a particular thing.
## More Detailed Percent Explanation
One percent is equal to 1/100 fraction:
1% = 1/100 = 0.01
Ten percent is equal to 10/100 fraction:
10% = 10/100 = 0.1
Fifty percent is equal to 50/100 fraction:
50% = 50/100 = 0.5
One hundred percent is equal to 100/100 fraction:
100% = 100/100 = 1
One hundred and ten percent is equal to 110/100 fraction:
110% = 110/100 = 1.1
A fraction or ratio with 100 understood as the denominator; for example, 0.98 equals a percentage of 98. The result is obtained by multiplying a quantity by a percent.
## Percentage Sign And History
Word "percent" comes from Latin (per centum). In English it means "out of hundred". It is shown by the symbol %.
It is written to the right side of the number: 50%
## Formula For Calculation Of Percentage Increase/Decrease
• When a quantity grows (gets bigger), then we can compute its PERCENT INCREASE:
• $\text{PERCENT INCREASE}=\frac{\left(\text{new amount}-\text{original amount}\right)}{\text{original amount}}\cdot 100\mathrm{%}$
• When a quantity shrinks (gets smaller), then we can compute its PERCENT DECREASE:
• $\text{PERCENT DECREASE}=\frac{\left(\text{original amount}-\text{new amount}\right)}{\text{original amount}}\cdot 100\mathrm{%}$
• If you have still problem to get to the result, use our percentage calculator in the top of the page.
## How To Calculate Percentage Increase
• First: work out the difference (increase) between the two numbers you are comparing.
• Increase = New Number - Original Number
• Then: divide the increase by the original number and multiply the answer by 100.
• % increase = Increase ÷ Original Number × 100.
• If your answer is a negative number then this is a percentage decrease.
## To Calculate Percentage Decrease:
• First: work out the difference (decrease) between the two numbers you are comparing.
• Decrease = Original Number - New Number
• Then: divide the decrease by the original number and multiply the answer by 100.
• % Decrease = Decrease ÷ Original Number × 100
• If your answer is a negative number then this is a percentage increase.
• If you wish to calculate the percentage increase or decrease of several numbers then we recommend using the first formula.
• Positive values indicate a percentage increase whereas negative values indicate percentage decrease.
#### How Do You Increase A Number By A Certain Percentage?
Percent increase formula The percent increase formula is as follows: Percent increase = [(new value - original value)/original value] * 100 An example using the formula is as follows. Suppose a $1,250 investment increased in value to$1,445 dollars in one year. What is the percent increase of the investment? To answer this, us the following steps: Identify the original value and the new value. Input the values into the formula. Subtract the original value from the new value, then divide the result by the original value. Multiply the result by 100. The answer is the percent increase. Check your answer using the percentage increase calculator. Working out the problem by hand we get: [(1,445 - 1,250)/1,250] * 100 (195/1,250) * 100 0.156 * 100 15.6 percent increase. The percentage growth calculator is a great tool to check simple problems. It can even be used to solve more complex problems that involve percent increase. You may also find the percentage calculator is also useful in this type of problem.
## Examples - Percentage Increase And Decrease
In January Dylan worked a total of 35 hours, in February he worked 45.5 hours – by what percentage did Dylan’s working hours increase in February?
To tackle this problem first we calculate the difference in hours between the new and old numbers. 45.5 - 35 hours = 10.5 hours. We can see that Dylan worked 10.5 hours more in February than he did in January – this is his increase. To work out the increase as a percentage it is now necessary to divide the increase by the original (January) number:
10.5 ÷ 35 = 0.3
Finally, to get the percentage we multiply the answer by 100. This simply means moving the decimal place two columns to the right.
0.3 × 100 = 30
Dylan therefore worked 30% more hours in February than he did in January.
In March Dylan worked 35 hours again – the same as he did in January (or 100% of his January hours). What is the percentage difference between Dylan’s February hours (45.5) and his March hours (35)? You may think that as there was a 30% increase between Dylan’s January hours (35) and February (45.5) hours that there will be a 30% decrease between his February and March hours. This assumption is incorrect – let’s calculate the difference.
First calculate the decrease in hours, that is: 45.5 - 35 = 10.5
Then divide the decrease by the original number (February hours) so:
10.5 ÷ 45.5 = 0.23 (to two decimal places).
Finally multiply 0.23 by 100 to give 23%. Dylan’s hours were 23% lower in March than in February.
Sometimes it is easier to show percentage decrease as a negative number – to do this follow the formula above to calculate percentage increase – your answer will be a negative number if there was a decrease. In Dylan’s case the decrease works out at -15.5. -10.5 ÷ 45.5 = -0.23. -0.23 × 100 = -23%.
Dylan's hours could be displayed in a data table as:
Month Hours Worked Percentage Change
January 35
February 45.5 30%
March 35 -23%
## What Is X As A Percent Of Y
Use our online percentage calculator find the proportion of one number to another in percentage terms.
## Examples Of Finding X As A Percent Of Y
Think of this as turning a fraction into a percentage. Your fraction is x/y, and your percentage is [unknown, here] A/100.
x/y = A/100
The easiest way to do these, then, is to move the fraction around. If you multiply both sides by 100, you get A (your unknown) = 100x divided by y. Just plug in the numbers and out will come the answer. Some examples may make this even clearer.
What is 10 as a percent of 50?
Using the formula that we have just worked out, x is 10 and y is 50. The sum is therefore:v
100 × 10 = 1000
1000 ÷ 50 = 20.
Answer: 10 is 20% of 50.
This method, of course, also works with a more word-based problem.
You have been quoted commission of $7.50 on the sale of a table. The sale price is$150.
Another company has quoted you commission of 4.5%. You want to know which is better value.
This is asking you ‘What is $7.50 as a percent of$150?’.
Using the formula, therefore, x is 7.5 and y is 150.
7.5 × 100 = 750
750 ÷ 150 = 5.
Answer: The commission is 5% on the sale price. The commission of 4.5% is therefore better value for you as a customer.
## Fractions, Decimals and Percentages
Fractions, decimals and percentages are just different ways to show the same values. In the graph below you can see first fraction, percentage and decimals. Try to click in the fields below to see what is the difference between fraction, percentage and decimals values.
Just imagine you have pie and you slice him into 100 pieces. One slice is 1 percent of the whole pie. This is very important how to work with percents.
## Example Values - Pie And Percentage Calculation
Here is a table of commonly used values shown in Percent, Decimal and Fraction form:
Percent Decimal Fraction
1% 0,01 1/100
5% 0,05 1/20
10% 0,1 1/10
12½% 0,125 1/8
20% 0,2 1/5
25% 0,25 1/4
331/3% 0,333... 1/3
50% 0,5 1/2
75% 0,75 3/4
80% 0,8 4/5
90% 0,9 9/10
99% 0,99 99/100
100% 1 100/100
125% 1,25 5/4
150% 1,5 3/2
200% 2 2/1
Currently, we have around 1975 calculators, conversion tables and usefull online tools and software features for students, teaching and teachers, designers and simply for everyone.
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https://cpep.org/business/476383-wheelers-bike-company-manufactures-custom-racing-bicycles-the-company-.html
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16 October, 15:05
# Wheeler's Bike Company manufactures custom racing bicycles. The company uses a job order cost system to determine the cost of each bike. Estimated costs and expenses for the coming year follow: Bike parts \$ 353,800 Factory machinery depreciation 59,000 Factory supervisor salaries 140,500 Factory direct labor 215,558 Factory supplies 43,900 Factory property tax 27,750 Advertising cost 26,000 Administrative salaries 52,500 Administrative-related depreciation 23,200 Total expected costs \$ 942,208 Required: 1. Calculate the predetermined overhead rate per direct labor hour if the average direct labor rate is \$12.11 per hour. 2. Determine the amount of applied overhead if 18,900 actual hours are worked in the upcoming year.
+4
1. 16 October, 15:18
0
Instructions are listed below.
Explanation:
Giving the following information:
Factory machinery depreciation 59,000
Factory supervisor salaries 140,500
Factory supplies 43,900
Factory property tax 27,750
1)
First, we need to determine the estimated direct labor hours for the period:
Factory direct labor = 215,558
Direct labor rate = \$12.11
Direct labor hours = 215,558 / 12.11 = 17,800 hours
Now, we can calculate the estimated overhead rate:
To calculate the estimated manufacturing overhead rate we need to use the following formula:
Estimated manufacturing overhead rate = total estimated overhead costs for the period / total amount of allocation base
Estimated manufacturing overhead rate = 271,150/17,800 = \$15.23 per direct labor hour
2) To apply overhead, we need to use the following formula:
Allocated MOH = Estimated manufacturing overhead rate * Actual amount of allocation base
Allocated MOH = 15.23*18,900 = \$287,847
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https://www.esaral.com/q/a-hand-fan-is-made-by-sticking-10-equal-size-triangular-strips-of-two-different-types-of-paper-as-shown-in-the-figure-84930
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# A hand fan is made by sticking 10 equal size triangular strips of two different types of paper as shown in the figure.
Question:
A hand fan is made by sticking 10 equal size triangular strips of two different types of paper as shown in the figure. The dimensions of equal strips are 25 cm, 25 cm and 14 cm. Find the area of each type of paper needed to make the hand fan.
Solution:
Given that,
The sides of AOB
AO = 25 cm
OB = 25 cm
BA = 14 cm
Area of each strip = Area of triangle AOB
Now, for the area of triangle AOB
Perimeter = AO + OB + BA
2s = 25 cm +25 cm + 14 cm
s = 32 cm
By using Heron's Formula,
Area of the triangle $A O B=\sqrt{s \times(s-a) \times(s-b) \times(s-c)}$
$=\sqrt{32 \times(7) \times(4) \times(18)}$
$=168 \mathrm{~cm}^{2}$
Also, area of each type of paper needed to make a fan = 5 × Area of triangle AOB
Area of each type of paper needed to make a fan $=5 \times 168 \mathrm{~cm}^{2}$
Area of each type of paper needed to make a fan $=840 \mathrm{~cm}^{2}$
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https://math.stackexchange.com/questions/4547868/integrating-question-int-frac-cos2x1-sin-x-dx
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# Question:
Find the value of the following expression:
$$\int \frac{\cos^2x}{1+\sin x}\ dx$$
# My Working:
## Failed Method 1:
Basically, I first tried to use $$u$$-substitution with $$u=1+\sin x$$, and this would result in
\begin{align} \frac{du}{dx}&=\cos x\\ du&=\cos x\cdot dx \end{align}
This would not cancel out the other $$\cos x$$.
## Failed Method 2:
I also tried using the formula $$\cos^2x=\frac{1+\cos2x}{2},$$ but that would result in the integral being equivalent to
\begin{align} & \quad\int\frac{\frac{1+\cos 2x}{2}}{1+\sin x}\ dx\\ &=\int\frac{1+\cos2x}{2+2\sin x}\ dx \end{align}
Nothing would cancel out here either.
# WolframAlpha's Answer Which Looks Ridiculous:
I've also tried using WolframAlpha, but the "answer", shown below, looks terrifying, and maybe not the right answer (at least, not the simplified right answer). (By the way, I don't have WolframAlpha Pro, so I cannot access the step-by-step solution)
$$-\frac{\left[2 \sqrt{1 - \sin x} \sin^{-1}\left(\frac{\sqrt{1 - \sin x}}{\sqrt{2}}\right) + (\sin x - 1) \sqrt{\sin x + 1}\right] \cos^3x}{(\sin x - 1)^2 (\sin x + 1)^{3/2}} + c$$
## Could you please give me some advice on how to solve the problem? Thanks!
Did Wolfram Alpha really produce that?
$$\cos^2 x=1-\sin^2x=(1+\sin x)(1-\sin x)$$.
You can do the rest yourself.
• Thanks, I have solved the problem now. I just forgot to use the theorem. +1, will accept soon. Commented Oct 8, 2022 at 6:35
• I don't understand why but I got the same result: wolframalpha.com/… But look at this: wolframalpha.com/… Commented Oct 8, 2022 at 6:36
• The lesson being: WolframAlpha isn't infallible. Commented Oct 8, 2022 at 6:37
• @PrincessEev Yes, I remember when it couldn't factor $x^4+y^4$ without going to complex numbers. Commented Oct 8, 2022 at 6:38
• @SuzuHirose Very strange. I reported this as a bug. Commented Oct 8, 2022 at 9:32
\begin{align} \int {\cos^2 x \over 1+\sin x } \mathrm{d}x&= \int {1-\sin^2 x \over 1 + \sin x } \mathrm{d}x \\&=\int {(1+\sin x)(1-\sin x) \over (1+\sin x) } \mathrm{d}x\\ &=\int (1-\sin x) \mathrm d x\\&=x+\cos x + \mathrm{const} \end{align}
• Wait wait, why I was downvoted? Commented Oct 8, 2022 at 6:39
• Sorry, probably because you answered a little late. But you still provided more working out, so I upvoted anyway. Commented Oct 8, 2022 at 6:41
Use the "tan half-ange" substitution, $$viz$$ $$u=\tan (x/2)$$ to convert the integral to the integral of a rational function.
Use $$\cos^2x=1-\sin^2x$$. Then $$I=\int (1-\sin x) dx$$
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# Archimedes Principle Buoyant Force
Archimedes' principle states that when a body is immersed partially or completely in a liquid,it experiences an upthrust or buoyant force ,which is equal to the weight of the liquid displaced by it.
## Upthrust or Buoyant Force of Archimede's Principle:
When a body is immersed in a liquid,the liquid exerts an upward force on the body.This force is called the upthrust or buoyant force.
or we can say,
the upward force exerted on a body by the fluid in which it is submerged is called the upthrust or buoyant force.
## Upthrust is Equal to the Weight of the Liquid Displaced:
Now we shall prove mathematically that when a body is immersed in a liquid,the upthrust on body due to liquid is equal to the weight of liquid displaced by the submerged part of the body.
For the convenience of calculation,let us consider a cylindrical body of cross sectional area A immersed in a liquid of density d.Let the upper surface of the body be at a depth h1 below the free surface of liquid and the lower surface of the body be at a depth h2 below the free surface of liquid.
At depth h1,pressure on the upper surface
p1=h1dg
therefore,downward thrust on the upper surface
F1=pressure*area=h1dgA .....(1)
At depth h2 pressure on the lower surface
F2=h2dgA .................(2)
Stuck on any of these topics Archimedes Principle of Buoyancy, Bernoulli's Equation Example then keep reading my blogs i try to help you
The horizontal thrust at various points on the vertical sides of the cylinder get balanced because the liquid pressure ia same at all points at the same depth.
From above equations (1) and (2), it is clear that F2>F1 because h2>h1 and therefore, the cylinder will experience a net upward force.
resultant upward thrust on the body,
F=F2-F1
=h2dgA-h1dgA
=A(h2-h1)dg
But A(h2-h1)=V,THE VOLUME OF CYLINDER SUBMERGED IN LIQUId
therefore,upthrust F="Vdg.........(3)
since a solid when immersed in a liquid ,displaces liquid equal to the volume of its submerged part,therefore
Vdg=Volume of solid immersed *density of liquid*acceleration due to gravity.
=volume of liquid displaced*density of liquid*acceleration due to gravity.
=mass of liquid displaced*acceleration due to gravity.
=weight of the liquid displaced by the submerged part of the body.
hence,Upthrust=weight of the liquid displaced by the submerged part of the body. .....(4)
note that if the body is completely immersed in a liquid,the volume of liquid displaced will be equal to its own volume.Although above relation (4) is derived for a cylindrical body,but it is equally true for a body of any shape and size.
Check this awesome Mechanics Of Fluids i want to share and learn more on related topics.
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# Step/Impulse function setup confusion
I have a final coming up next week (Tuesday) and a sample question for the test is the following.
I was wondering if I have properly set it up for solving with Laplace as I haven't encountered a question with two level changes before and it's confussed me.
First line is the question and the second line is my questionable setup.
My understanding is that I'm doing this...
level+(newLevel-oldLevel)U(t-PointOfChange)
Finding good examples of Step & Impulse online has proven to be a greater challenge than I thought it would be.
-
I'm not sure what you mean by levels, but if apply the Laplace transform to the right hand side you just get $$\int_\pi^{2\pi} e^{-st}dt,$$ which can be evaluated explicitly, right? – anon Jun 16 '12 at 18:23
It's a graph of 'something' over time. That 'something' could be voltage level or switch on/off indication. So the level could be 0mV/100mV or 0/1. – Chef Flambe Jun 16 '12 at 20:40
Do you mean $y''+y=f(t)$ since you gave two initial conditions? – JohnD Dec 21 '12 at 15:46
Since you specified two initial conditions, I will assume you meant to solve $$y''+y=f(t)=\begin{cases}0, &0<t<\pi,\\ 1, &\pi<t<2\pi,\\ 0, &t>2\pi.\end{cases}$$
In preparation for Laplace transforming both sides of the equation, let's look at how to compute $\mathscr{L}\{f(t)\}$. One way is directly from the definition: $$\mathscr{L}\{f(t)\}=\int_0^\infty f(t)e^{-st}\,dt=\int_{\pi}^{2\pi} 1\cdot e^{-st}\,dt={1\over s}\left(e^{-\pi s}-e^{-2\pi s}\right).$$ However, it sounds like you are curious about how to compute the transform after rewriting $f(t)$ as a combination of unit step functions. Here's how (and this might help to see how to get this form): \begin{align} f(t)&=u(t-\pi)-u(t-2\pi)\\ \mathscr{L}\{f(t)\}&=\mathscr{L}\{u(t-\pi)\}-\mathscr{L}\{u(t-2\pi)\}={e^{-\pi s}\over s}-{e^{-2\pi s}\over s}, \end{align} which is of course the same result as the previous method.
Now we are ready. Laplace transforming both sides, \begin{align} \mathscr{L}\left\{y''+y\right\}&=\mathscr{L}\{f(t)\}\\ s^2Y(s)-sy(0)-y'(0)+Y(s)&={1\over s}\left(e^{-\pi s}-e^{-2\pi s}\right)\\ Y(s)&={e^{-\pi s}-e^{-2\pi s}\over s(s^2+1)}\\ Y(s)&={e^{-\pi s}\over s}\cdot {1\over s^2+1}-{e^{-2\pi s}\over s}\cdot {1\over s^2+1} \end{align}
In this form, the inverse transforms on the right-hand side are readily doable via the Convolution Theorem:
\begin{align} \mathscr{L}^{-1}\{Y(s)\}&=\mathscr{L}^{-1}\left\{{e^{-\pi s}\over s}\cdot {1\over s^2+1}-{e^{-2\pi s}\over s}\cdot {1\over s^2+1}\right\}\\ y(t)&=u(t-\pi)*\sin t-u(t-2\pi)*\sin t\\ y(t)&=\int_0^t u(\tau-\pi)\sin(t-\tau)\,d\tau -\int_0^t u(\tau-2\pi)\sin(t-\tau)\,d\tau\\ y(t)&=(1+\cos t)u(t-\pi)-(1-\cos t)u(t-2\pi). \end{align}
Here's a plot of the input $f(t)$ (red) and the system response $y(t)$ (blue):
Hope that helps.
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# 02 Rationalisation¶
## Corresponding chapters¶
Duration: 50 minutes
## Objectives¶
• Be able to identify dominated strategies
• Understand notion of Common knowledge of rationality
## Notes¶
### Rationalisation of strategies¶
Identify two volunteers and play a sequence of zero sum games where they play as a team against me. The group is the row player.
$\begin{split}A = \begin{pmatrix} 1 & -1\\ -1 & 2 \end{pmatrix}\end{split}$
(No dominated strategy)
$\begin{split}A = \begin{pmatrix} 1 & 2\\ -1 & 2 \end{pmatrix}\end{split}$
(First column weakly dominates second column)
$\begin{split}A = \begin{pmatrix} 1 & -1\\ -1 & -3 \end{pmatrix}\end{split}$
• First row strictly dominates second row
• Second column strictly dominates first column
$\begin{split}A = \begin{pmatrix} 2 & 2\\ -1 & 2 \end{pmatrix}\end{split}$
• First row weakly dominates second row
• First column weakly dominates second column
$\begin{split}A = \begin{pmatrix} -1 & 2 & 1\\ -2 & -2 & 1\\ 1 & 1 & -1\\ \end{pmatrix}\end{split}$
• First row dominates second row
• First column dominates second column
Now pit the two players against each other, the utilities represent the share of the total amount of chocolates/sweets gathered so far:
$\begin{split}A = \begin{pmatrix} 1 & 0\\ 1.5 & .5 \end{pmatrix}\qquad B = \begin{pmatrix} 1 & 1.5\\ 0 & .5 \end{pmatrix}\end{split}$
Capture all of the above (on the white board) and discuss each action and why they were taken.
Ask students to repeat the game against each other in groups of two (use “days of doing the dishes perhaps?”).
### Iterated elimination of dominated strategies¶
As a class work through the following example.
$\begin{split}A = \begin{pmatrix} 2 & 5 \\ 1 & 2 \\ 7 & 3 \end{pmatrix}\qquad B = \begin{pmatrix} 0 & 3 \\ 6 & 1 \\ 0 & 1 \end{pmatrix}\end{split}$
1. First row dominates second row
$\begin{split}A = \begin{pmatrix} 2 & 5 \\ 7 & 3 \end{pmatrix}\qquad B = \begin{pmatrix} 0 & 3 \\ 0 & 1 \end{pmatrix}\end{split}$
2. Second column dominates first column
$\begin{split}A = \begin{pmatrix} 2\\ 7 \end{pmatrix}\qquad B = \begin{pmatrix} 3\\ 1 \end{pmatrix}\end{split}$
3. First row dominates third row. Thus the rationalised behaviour is $$(r_1, c_2)$$.
$\begin{split}A = \begin{pmatrix} -1 & 2 & 1\\ -2 & -2 & 1\\ 1 & 1 & -1\\ \end{pmatrix}\qquad B = \begin{pmatrix} 1 & -2 & -1\\ 2 & 2 & -1\\ -1 & -1 & 1\\ \end{pmatrix}\end{split}$
$\begin{split}A = \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}\qquad B = \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}\end{split}$
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# Limit Evaluation of a Function in the Complex Field
Given the sequence
$$z_n=\frac{1}{2n\pi}, \quad n \in \mathbb{N}$$ try to evaluate the following limit: $$\lim_{z \to z_n} f(z)$$ where $f(z)$ is a function in the complex field, defined as: $$f(z)=\left( \frac{1}{z-1} \right) \cos \left( \frac{1}{z} \right)$$
What I did (amongst other), was to try to evaluate the limit: $$\lim_{n\to \infty}\frac{1}{\frac{1}{2n\pi}-1}\cos(2n\pi)$$ which I find it to take values within the range $[-1,1]$, but the author says it is equal to infinite.
Now, that is confusing. Is it really infinite? And if yes, how can this be proved?
Thank you!
Observe that for all $\;n\in\Bbb Z\;,\;\;\cos 2n\pi=1\;$ , so
$$\lim_{n\to\infty}\frac{2n\pi}{1-2n\pi}\cos2n\pi=-1$$
I don't know what author says this limit is $\;\infty\;$ but I think that either he meant other sequence, other function or he is simply wrong.
$$\lim_{z\to z_n}f(z)=\lim_{z\to z_n}\left(\frac1{z-1}\right)\cos\frac1z=\left(\frac1{\frac1{2n\pi}-1}\right)\cos2n\pi=\frac{2n\pi}{1-2n\pi}$$
I'm not sure why you then take the limit of the last expression, which is a very different limit of the one with $\;n\to\infty\;$, though.
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http://www.numbersaplenty.com/3800
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Search a number
3800 = 235219
BaseRepresentation
bin111011011000
312012202
4323120
5110200
625332
714036
oct7330
95182
103800
112945
122248
131964
141556
1511d5
hexed8
3800 has 24 divisors (see below), whose sum is σ = 9300. Its totient is φ = 1440.
The previous prime is 3797. The next prime is 3803. The reversal of 3800 is 83.
Adding to 3800 its reverse (83), we get a palindrome (3883).
It is an interprime number because it is at equal distance from previous prime (3797) and next prime (3803).
It is a plaindrome in base 12 and base 14.
It is a nialpdrome in base 8 and base 16.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (3803) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (7) of ones.
It is a polite number, since it can be written in 5 ways as a sum of consecutive naturals, for example, 191 + ... + 209.
23800 is an apocalyptic number.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 3800, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (4650).
3800 is an abundant number, since it is smaller than the sum of its proper divisors (5500).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
3800 is a wasteful number, since it uses less digits than its factorization.
3800 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 35 (or 26 counting only the distinct ones).
The product of its (nonzero) digits is 24, while the sum is 11.
The square root of 3800 is about 61.6441400297. The cubic root of 3800 is about 15.6049075071.
The spelling of 3800 in words is "three thousand, eight hundred".
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https://mathpaperhelp.com/integral-calculus-the-ultimate-guide/
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## What is integral calculus?
Integral calculus is the branch of mathematics that deals with integrals, which are mathematical operations that are used to find areas, volumes, and other quantities. Integrals can be thought of as the opposite of derivatives. Derivatives are used to find the rate of change of a function, while integrals are used to find the total change of a function over a given interval.
## Why is integral calculus important?
Integral calculus is important because it has many applications in physics, engineering, and other fields. For example, integral calculus can be used to:
• Find the area under a curve
• Calculate the volume of a solid object
• Find the average value of a function
• Find the work done by a force
• Find the velocity and acceleration of an object
• Solve differential equations
## Real-world applications of integral calculus
Integral calculus can be used to solve a wide variety of real-world problems, such as:
• Finding the area under a demand curve to determine the total revenue for a company
• Calculating the volume of a reservoir to determine how much water it can hold
• Finding the average temperature over a day to determine how comfortable it was
• Finding the work done by a pump to lift water to a certain height
• Finding the velocity and acceleration of a car to determine how it is moving
• Solving differential equations to model the spread of a disease or the motion of a planet
## Types of integrals
There are two main types of integrals: indefinite integrals and definite integrals.
### Indefinite integrals
Indefinite integrals are used to find the antiderivative of a function. The antiderivative of a function is another function whose derivative is the original function.
### Definite integrals
Definite integrals are used to calculate the area under a curve or the volume of a solid object. Definite integrals are calculated by evaluating the indefinite integral at the upper and lower bounds of the interval of integration and then subtracting the two values.
### Integration techniques
There are a variety of integration techniques that can be used to find the indefinite and definite integrals of functions. Some of the most common integration techniques include:
• Basic integration rules: These rules include the power rule, the sum rule, the product rule, and the quotient rule.
• Integration by parts: This technique is used to integrate products of two functions.
• U-substitution: This technique is used to integrate functions that contain composite functions.
• Integration by trigonometric substitution: This technique is used to integrate functions that contain trigonometric functions.
• Partial fractions: This technique is used to integrate functions that can be expressed as the sum of rational functions.
• Numerical integration: This technique is used to approximate the values of definite integrals.
## Applications of integral calculus in physics and engineering
Integral calculus is widely used in physics and engineering to solve a variety of problems, such as:
• Finding the motion of a projectile: Integral calculus can be used to find the trajectory of a projectile, such as a baseball or a rocket.
• Calculating the force exerted by a fluid on a surface: Integral calculus can be used to calculate the force exerted by a fluid on a surface, such as the force of water on a dam or the force of air on an airplane wing.
• Determining the heat transfer in a solid object: Integral calculus can be used to determine the heat transfer in a solid object, such as a metal rod or a concrete wall.
• Designing electrical circuits: Integral calculus can be used to design electrical circuits, such as filters and amplifiers.
## Conclusion
Integral calculus is a powerful tool that can be used to solve a wide variety of problems in mathematics, physics, engineering, and other fields. This guide has provided a comprehensive overview of integral calculus, including its definition, importance, applications, and various integration techniques.
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1. the radius of convergence of the power series
Please help me with this one...i don't really understand how can i to solve it, and what the nth coefficient here?
Show that the radius of convergence of the power series:
$\sum_{n=1}^{\infty }\frac{(-1)^{n}}{n}z^{n(n+1)}$
is 1 and discuss convergence for $z=1,-1, i$
2. Try the ratio test...
$\displaystyle a_{n} = \frac{(-1)^nz^{n(n+1)}}{n}$, so $\displaystyle a_{n+1} = \frac{(-1)^{n+1}z^{(n+1)(n+2)}}{n+1}$.
The series will be convergent when $\displaystyle \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| < 1$
$\displaystyle \lim_{n \to \infty}\left|\frac{\frac{(-1)^{n+1}z^{(n+1)(n+2)}}{n+1}}{\frac{(-1)^nz^{n(n+1)}}{n}}\right|<1$
$\displaystyle \lim_{n \to \infty}\left|\frac{n(-1)^{n+1}z^{(n+1)(n+2)}}{(n+1)(-1)^nz^{n(n+1)}}\right| < 1$
$\displaystyle \lim_{n \to \infty}\left|\frac{n\,z^{(n+1)(n+2)}}{(n+1)z^{n(n+ 1)}}\right|<1$
$\displaystyle \lim_{n \to \infty}\left|\frac{n\,z^2}{n+1}\right|<1$
$\displaystyle |z^2|\lim_{n \to \infty}\frac{n}{n+1} < 1$
$\displaystyle |z|^2\lim_{n \to \infty}\frac{1}{1 + \frac{1}{n}} < 1$
$\displaystyle |z|^2\cdot 1 < 1$
$\displaystyle |z|^2 < 1$
$\displaystyle |z| < 1$.
So the series is convergent when $\displaystyle |z| < 1$ and divergent when $\displaystyle |z| > 1$. You will need to do some further analysis to determine the convergence/divergence when $\displaystyle |z| = 1$.
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https://www.hackmath.net/en/word-math-problems/tangent?tag_id=13
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Tangent + right triangle - math problems
Number of problems found: 143
You need to make a pad in the shape of a regular octagon with a side length of 4 cm. What is the minimum diameter of the circle-shaped semi-finished product from which we make the pad, and what will be the percentage of waste? (Round the results to 1 deci
• Touch circle
Point A has a distance IA, kl = 10 cm from a circle k with radius r = 4 cm and center S. Calculate: a) the distance of point A from the point of contact T if the tangent to the circle is drawn from point A b) the distance of the contact point T from the l
• Trapezoids
In the isosceles trapezoid ABCD we know: AB||CD, |CD| = c = 8 cm, height h = 7 cm, |∠CAB| = 35°. Find the area of the trapezoid.
• A missile
A missile is fired with a speed of 100 fps in a direction 30° above the horizontal. Determine the maximum height to which it rises? Fps foot per second.
• Isosceles trapezoid
Find the height in an isosceles trapezoid if the area is 520 cm2 and the base a = 25 cm and c = 14 cm. Calculate the interior angles of the trapezoid.
• Sin cos tan
In triangle ABC, right-angled at B. Sides/AB/=7cm, /BC/=5cm, /AC/=8.6cm. Find to two decimal places. A. Sine C B. Cosine C C. Tangent C.
• Three pillars
On a straight road, three pillars are 6 m high at the same distance of 10 m. At what angle of view does Vlado see each pillar if it is 30 m from the first and his eyes are at 1.8 m high?
• Calculate
Calculate the area of triangle ABC, if given by alpha = 49°, beta = 31°, and the height on the c side is 9cm.
• Tower's view
From the church tower's view at the height of 65 m, the top of the house can be seen at a depth angle of alpha = 45° and its bottom at a depth angle of beta = 58°. Calculate the height of the house and its distance from the church.
• Aircraft
From the aircraft flying at an altitude of 500m, they observed places A and B (located at the same altitude) in the direction of flight at depth angles alpha = 48° and beta = 35°. What is the distance between places A and B?
• Altitude angle
In complete winds-free weather, the balloon took off and remained standing exactly above the place from which it took off. It is 250 meters away from us. How high did the balloon fly when we see it at an altitude angle of 25°?
• Base diagonal
In a regular 4-sided pyramid, the side edge forms an angle of 55° with the base's diagonal. The length of the side edge is eight meters. Calculate the surface area and volume of the pyramid.
• Side edges
The regular 4-sided pyramid has a body height of 2 dm, and the opposite side edges form an angle of 70°. Calculate the surface area and volume of the pyramid.
• Diagonals of a prism
The base of the square prism is a rectangle with dimensions of 3 dm and 4 dm. The height of the prism is 1 m. Find out the angle between the body diagonal with the diagonal of the base.
• The body
The body slides down an inclined plane forming an angle α = π / 4 = 45° under the action of a horizontal plane under the effect of friction forces with acceleration a = 2.4 m/s ^ 2. At what angle β must the plane be inclined so that the body slides on it
• Truncated pyramid
Find the volume of a regular 4-sided truncated pyramid if a1 = 14 cm, a2 = 8 cm and the angle that the side wall with the base is 42 degrees
• The staircase
The staircase has a total height of 3.6 m and forms an angle of 26° with the horizontal. Calculate the length of the whole staircase.
• Space diagonal angles
Calculate the angle between the body diagonal and the side edge c of the block with dimensions: a = 28cm, b = 45cm and c = 73cm. Then, find the angle between the body diagonal and the plane of the base ABCD.
| 991 | 3,679 |
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https://sxidatinge.me/transportation/how-to-put-a-parabola-equation-in-standard-form.php
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## How to put a parabola equation in standard form
By Mary Jane Sterling. Rules representing parabolas come in two standard forms to separate the functions opening upward or downward from relations that. When the equation of a parabola appears in standard form, you have all the Add a number to each side to make the side with the squared term into a perfect. Equation of parabola. How to convert equation from vertex to standard form. Explained with diagrams and practice problems.
## write the equation of the parabola in standard form from a graph
The standard and vertex form equation of a parabola and how the equation relates to the graph of a parabola. In order to be in standard form, the equation for a parabola must be written in one of the following First, write the equation of the parabola in standard form. After watching this video lesson, you will be able to write the equation of a parabola in standard form when given just two important points from the parabola .
To avoid confusion, this site will not refer to either as standard form, but will reference (h, k) is the vertex of the parabola, and x = h is the axis of symmetry. To convert a quadratic from y = ax2 + bx + c form to vertex form, y = a(x - h)2+ k, you When we add a box to both sides, the box will be multiplied by 2 on both sides. The quadratic equation is sometimes also known as the standard form formula of you're going to want to write your parabola in what's known as vertex form. Three points determine a parabola, so given three points you can write the parabola's problem; write; quadratic equation; standard form; given 3 points.
The standard form of a quadratic equation is y = ax² + bx + c. You can use this vertex calculator to. T.7 Convert equations of parabolas from general to vertex form The equation of a parabola is. y. = x. 2. –. 8. x. +. Write the equation in vertex form. Simplify. Parabola in standard form.
## standard form of a parabola calculator
The standard for for the equation of a parabola (also called the vertex form) is .. Write the equation of the parabola in standard form with coordinates of points. Thus, we will take the focus at S(0,a) and the directrix with equation y=−a, where a>0. Any parabola of the form y=Ax2+Bx+C can be put into the standard form. A Quadratic Equation in Standard Form You can graph a Quadratic Equation using the Function Grapher, but to really understand what is going on, you can make the graph It is a parabola. So let's put that into this form of the equation: . So now, if you need it in vertex form (which is what our book calls the form we now have), then you plug in A and you are finished. y=4(x+6)2 -. A parabola intersects its axis of symmetry at a point called the vertex of the parabola. When a quadratic function is in standard form, then it is easy to sketch its will look a little different, because our chief goal here is not solving an equation. Write f(x) = -2x2 + 2x + 3 in standard form and find the vertex of the graph of f. A quadratic equation in Standard Form has three coefficients: a, b, and c. Note that the vertex of the parabola is identified on the graph as point V, with its Changing c translates the graph vertically by adding a constant value to all. First let's look at the standard form for the equation of a parabola: The vertex in the above form is at (h,k). We can put the general equation in terms of a,b, and c. Determine standard form for the equation of a parabola given general form. .. Adding and subtracting the same value within an expression does not change it. So far, we have only used the standard form of a quadratic equation, \begin{align *}y=ax^2+bx+c\end{align*} to graph a parabola. Example 2: Write the standard form of the equation of the parabola with a vertex at the origin and focus at (2, 0). Graph the parabola, including the directrix, the.
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https://cs.stackexchange.com/questions/10958/what-are-twos-complement-integers
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# What are two's complement integers?
Can someone explain in plain English what "two's complement integer" means? I read this:
in Java long is a 64-bit signed two's complement integer
Two's complement is the most commonly used way to represent signed integers in bits. First, consider unsigned numbers in 8 bits. Notice that $2^8 = 256 = 100000000_2$ does not fit into 8 bits and will thus be represented as 0000 0000. Therefore $255 + 1 =$ 1111 1111 + 0000 0001 = 0000 0000 and in that sense 1111 1111 acts as if it was $-1$. Two's complement is based on this observation.
For a deeper understanding of it, I suggest reading the wikipedia page on it. Note that as long as you you don't need to directly manipulate bits of your integers and your numbers don't go out of range, it makes no difference what signed integer representation is used.
Karolis Juodele's answer is good. Here's another take.
To represent a negative number over $k$ bits in two's complement:
1. Take the complement of the absolute value of the number over $k$ bits, by changing $0$ to $1$ and vice-versa.
2. Add 1 to the number obtained previously, using regular unsigned integer arithmetic.
Example: find the two's complement representation of $-55$ assuming an 8-bit representation. The binary representation of $|-55| = 55$ is $00110111$. The complement of this is $11001000$. Adding $1$ using unsigned integer arithmetic yields $1101001$. This is the two's complement representation of $-55$ over 8-bit integers. Note that, as Karolis points out, performing binary addition using unsigned integer arithmetic of these two representations (the representations for $55$ and $-55$) yields $00000000$, or $0$.
The two's complement of a number $N$ is the amount that you need to add to $N$ to get $2^k$ where $k$ is the number of binary digits needed to represent $N$; analogously three's complement would be the amount to add to get $3^k$ ($k$ is the number of ternary digits now) and you get the pattern. So this complement idea can be generalized to any base not only $2$. In other words, the complement is the amount that you need to add to get the WHOLE where the WHOLE is some predetermined quantity (thinking of complements of sets might help).
Let's say numbers are stored in $8$ bits so the maximum number in this representation is $2^8 - 1$ and $2^8$ is equivalent to $0$ because it can't fit in the $8$ bits representation (it overflows). The complement of the maximum is $1$ since $1 + (2^8 - 1) = 2^8$. This suggests that a legitimate representation of a negative number $-M$ is $(2^8-M)$ because adding $M$ results in $2^8$ which is the same as $0$ and this is what we expect it to be.
The way computers do it is a bit different. Let's take the number $X=1$; in our representation this is written as: $X=00000001$. Now the amount that we need to add to $X$ to get the WHOLE is $11111110$ which turns all bits on and then add $1$ to cause an overflow so that the result is $100000000$ which is the same as $2^8$, so computers first flip all the bits ($0$ becomes $1$ and vice-versa) and then add $1$.
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https://www.jiskha.com/questions/1809628/the-function-a-a0e-0-01386x-models-the-amount-in-pounds-of-a-particular-radioactive
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algebra
The function A=A0e^-0.01386x models the amount in pounds of a particular radioactive material stored in a concrete vault, where x is the number of years since the material was put into the vault. If 500 pounds of material are placed in the vault.
A) How much time will need to pass for only 47 pounds to remain?
B) How much material will be left after 10, 20, and 30 years?
1. 👍
2. 👎
3. 👁
1. A)
47 = 500 e^(-0.01386x)
take ln of both sides and use log rules
ln 47 = (-0.01386x)ln 500
-0.01386x = ln47/ln500
x =
continue
B)I will do the case of 20 years, you do the other two
A = 500 e^(-0.01386(20))
= 500 e^-.2772
= 500(.7579029..)
= appr 379 pounds
1. 👍
2. 👎
👤
Reiny
2. 47 = 500 e^(-0.01386x)
e^(-0.01386x) = 47/500
-0.01386x = ln(47/500)
x = ln(0.094)/-0.01386
1. 👍
2. 👎
👤
oobleck
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# 6th Grade Statistics - Statistical vs Non-Statistical Questions
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### PRODUCT DESCRIPTION
6th Grade - Statistical vs Non-Statistical Questions
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This product covers the first 6th Grade Common Core Math Standard of:
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. For example, “How old am I?” is not a statistical question, but “How old are the students in my school?” is a statistical question because one anticipates variability in students’ ages.
This is a comprehensive lesson so the teacher will only have to download the product and then print out copies with their students.
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Includes the following:
-Overview of Lesson and Teacher Instructions
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If you enjoy this product, please leave us a comment or review so we can work to make our lessons and resources better and more useful.
Also, these are some of our more popular projects, lessons, and units for you to consider:
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6th Grade Statistics – Comprehensive 2 Week Unit
6th Grade Statistics – Range of a Data Set
6th Grade Statistics – Median Average of a Data Set
6th Grade Statistics – Mean Average of a Data Set
6th Grade Statistics – Mode Average of a Data Set
6th Grade Statistics – Dot Plots and Line Plots
6th Grade Statistics – Box and Whisker Plots
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# 1979 AHSME Problems/Problem 12
## Problem 12
$[asy] size(200); pair A=(-2,0),B,C=(-1,0),D=(1,0),EE,O=(0,0); draw(arc(O,1, 0, 180)); EE=midpoint(arc(O,1, 0, 90)); draw(A--EE); draw(A--D); B=intersectionpoint(arc(O,1, 180, 0),EE--A); draw(O--EE); label("A",A,W); label("B",B,NW); label("C",C,S);label("D",D,E);label("E",EE,NE);label("O",O,S);label("45^\circ",(0.25,0.1),fontsize(10pt)); //Credit to TheMaskedMagician for the diagram[/asy]$
In the adjoining figure, $CD$ is the diameter of a semi-circle with center $O$. Point $A$ lies on the extension of $DC$ past $C$; point $E$ lies on the semi-circle, and $B$ is the point of intersection (distinct from $E$ ) of line segment $AE$ with the semi-circle. If length $AB$ equals length $OD$, and the measure of $\measuredangle EOD$ is $45^\circ$, then the measure of $\measuredangle BAO$ is
$\textbf{(A) }10^\circ\qquad \textbf{(B) }15^\circ\qquad \textbf{(C) }20^\circ\qquad \textbf{(D) }25^\circ\qquad \textbf{(E) }30^\circ$
## Solution
Solution by e_power_pi_times_i
Because $AB = OD$, triangles $ABO$ and $BOE$ are isosceles. Denote $\measuredangle BAO = \measuredangle AOB = \theta$. Then $\measuredangle ABO = 180^\circ-2\theta$, and $\measuredangle EBO = \measuredangle OEB = 2\theta$, so $\measuredangle BOE = 180^\circ-4\theta$. Notice that $\measuredangle AOB + \measuredangle BOE + 45^\circ = 180^\circ$. Therefore $\theta+180-4\theta = 135^\circ$, and $\theta = \boxed{\textbf{(B) } 15^\circ}$.
## Solution 2
Draw $BO$. Let $y = \angle BAO$. Since $AB = OD = BO$, triangle $ABO$ is isosceles, so $\angle BOA = \angle BAO = y$. Angle $\angle EBO$ is exterior to triangle $ABO$, so $\angle EBO = \angle BAO + \angle BOA = y + y = 2y$.
Triangle $BEO$ is isosceles, so $\angle BEO = \angle EBO = 2y$. Then $\angle EOD$ is external to triangle $AEO$, so $\angle EOD = \angle EAO + \angle AEO = y + 2y = 3y$. But $\angle EOD = 45^\circ$, so $\angle BAO = y = 45^\circ/3 = \boxed{15^\circ}$. That means the answer is $\boxed{\textbf{(B) } 15^\circ}$.
1979 AHSME (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 13 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions
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# Conservation of Energy
## The amount of energy in a closed system never changes.
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The World's Fastest Roller Coaster
### The World?s Fastest Roller Coaster
Credit: Sarah Ackerman
Source: http://www.flickr.com/photos/sackerman519/8638541154/
License: CC BY-NC 3.0
Known as the fastest roller coaster in the world, Formula Rossa accelerates its riders up to 149 miles per hour. Not only does this 92 second ride have a total length of 2200 meters and a vertical drop of 51.5 meters, but its passengers will experience almost 5.0Gs!
#### Amazing But True
Credit: Jeremy Thomson
Source: http://www.flickr.com/photos/rollercoasterphilosophy/3228818023/
License: CC BY-NC 3.0
The energy equation lets engineers determine how fast a roller coaster can travel [Figure2]
• While roller coasters are all different from one another, they all follow the same basic principle. The system's potential energy is increased by gaining altitude. Then, gravity takes over, causing most of the potential energy to be converted into kinetic energy. Any of the leftover energy that has not been converted into kinetic energy by the end of the ride is energy lost due to friction.
• The energy at any given point along the ride is given as:
\begin{align*}K_{initial}+U_{initial}+W_{ext}=K_{final}+U_{final}\end{align*}
In the equation, \begin{align*}K\end{align*} and \begin{align*}U\end{align*} represent the kinetic and potential energies respectively. \begin{align*}W\end{align*} represents any work that is done on the ride by external forces (friction/air resistance). Using this equation and the principle that energy must be conserved, engineers are able to determine how fast they can make the ride as well as how sharp each turn can be.
#### What Do You Think?
Using the information provided above, answer the following questions.
1. Why do most roller coasters start by bringing the passengers up to the top of a large incline?
2. Does the \begin{align*}W_{\text{ext}}\end{align*} in the above equation serve to increase or decrease the speed at which the roller coaster can reach?
3. How is the energy in the system distributed when one of the carts on the roller coaster goes through an upside down loop?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
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https://javascript.tutorialink.com/explain-how-recursion-works-in-an-algorithm-to-determine-depth-of-binary-tree/
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Explain how recursion works in an algorithm to determine depth of binary tree?
I am new to data structures in JavaScript and am trying to learn Binary Search Trees. I was following along with a blog post and was able to get a working solution to the problem of finding the max depth in a BST, but it’s unclear to me how the recursion is working and how the +1 gets added on each time at each level of depth. What is a good way to think about this? Is it basically that each time the nodes value is not null, 1 gets added to what will eventually be returned up the call stack (i.e. at each level as it backtracks up to the root)?
``` function maxDepth(node) {
// console.log(node.left);
if (node) {
return Math.max(maxDepth(node.left), maxDepth(node.right)) + 1;
} else {
return 0;
}
}
```
The code for `maxDepth(node)` reads like this:
1. If `node` is not `null`:
1. Run this same algorithm `maxDepth` on `node`‘s left child. Let this answer be `x`.
2. Run this same algorithm `maxDepth` on `node`‘s right child. Let this answer be `y`.
3. Calculate `Math.max(x, y) + 1`, and return this value as the answer for this function call.
2. Otherwise `node` is `null`, then return `0`.
This means when we try to compute `maxDepth(node)` on a non-null node, we first compute `maxDepth()` on both of `node`‘s children, and let those two subcomputations finish. Then we take the maximum of these values, add 1, and return the result.
Example:
``` a
/
b f
/
c e g
/
d
```
Call stack:
```a => max(b,f)
b => max(c,e)
c => max(d,null)
d => max(null,null)
d <= (0,0)+1 = 1
c <= (1,0)+1 = 2
e => max(null,null)
e <= (0,0)+1 = 1
b <= (2,1)+1 = 3
f => (null,g)
g => (null,null)
g <= (0,0)+1 = 1
f <= (0,1)+1 = 2
a <= (3,2)+1 = 4
```
User contributions licensed under: CC BY-SA
2 People found this is helpful
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https://homework.cpm.org/category/CC/textbook/cca/chapter/4/lesson/4.2.5/problem/4-84
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### Home > CCA > Chapter 4 > Lesson 4.2.5 > Problem4-84
4-84.
Find $g(−5)$ for each function below.
1. $g ( x ) = x ^ { 3 } - 2$
• Substitute $−5$ for $x$.
• $−127$
1. $g ( x ) = 7 + \sqrt { 4 - x }$
• Refer to part (a) for help.
1. $g ( x ) = \sqrt [ 3 ] { x + ( - 59 ) }$
• Refer to part (a) for help.
$g(−5)=−4$
1. $g ( x ) = - 4 | x - 1 |$
• Refer to part (a) for help.
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https://hextobinary.com/unit/angularacc/from/arcminph2/to/radpw2/781
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# 781 Arcmin/Square Hour in Radian/Square Week
Angular Acceleration
Arcmin/Square Hour
781 Arcmin/Square Hour = 6412.03 Radian/Square Week
## How many Radian/Square Week are in 781 Arcmin/Square Hour?
The answer is 781 Arcmin/Square Hour is equal to 6412.03 Radian/Square Week and that means we can also write it as 781 Arcmin/Square Hour = 6412.03 Radian/Square Week. Feel free to use our online unit conversion calculator to convert the unit from Arcmin/Square Hour to Radian/Square Week. Just simply enter value 781 in Arcmin/Square Hour and see the result in Radian/Square Week.
## How to Convert 781 Arcmin/Square Hour to Radian/Square Week (781 arcmin/h2 to rad/week2)
By using our Arcmin/Square Hour to Radian/Square Week conversion tool, you know that one Arcmin/Square Hour is equivalent to 8.21 Radian/Square Week. Hence, to convert Arcmin/Square Hour to Radian/Square Week, we just need to multiply the number by 8.21. We are going to use very simple Arcmin/Square Hour to Radian/Square Week conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Arcmin/Square Hour} = \text{8.21 Radian/Square Week}$$
$$\text{781 Arcmin/Square Hour} = 781 \times 8.21 = \text{6412.03 Radian/Square Week}$$
## What is Arcmin/Square Hour Unit of Measure?
Arcmin per square hour is a unit of measurement for angular acceleration. By definition, if an object accelerates at one arcmin per square hour, its angular velocity is increasing by one arcmin per hour every hour.
## What is the symbol of Arcmin/Square Hour?
The symbol of Arcmin/Square Hour is arcmin/h2. This means you can also write one Arcmin/Square Hour as 1 arcmin/h2.
## What is Radian/Square Week Unit of Measure?
Radian per square week is a unit of measurement for angular acceleration. By definition, if an object accelerates at one radian per square week, its angular velocity is increasing by one radian per week every week.
## What is the symbol of Radian/Square Week?
The symbol of Radian/Square Week is rad/week2. This means you can also write one Radian/Square Week as 1 rad/week2.
7816412.03
7826420.24
7836428.45
7846436.66
7856444.87
7866453.08
7876461.29
7886469.5
7896477.71
7906485.92
## Arcmin/Square Hour to Other Units Conversion Table
Arcmin/Square Hour [arcmin/h2]Output
781 arcmin/square hour in degree/square second is equal to0.0000010043724279835
781 arcmin/square hour in degree/square millisecond is equal to1.0043724279835e-12
781 arcmin/square hour in degree/square microsecond is equal to1.0043724279835e-18
781 arcmin/square hour in degree/square nanosecond is equal to1.0043724279835e-24
781 arcmin/square hour in degree/square minute is equal to0.0036157407407407
781 arcmin/square hour in degree/square hour is equal to13.02
781 arcmin/square hour in degree/square day is equal to7497.6
781 arcmin/square hour in degree/square week is equal to367382.4
781 arcmin/square hour in degree/square month is equal to6946087.09
781 arcmin/square hour in degree/square year is equal to1000236540.6
781 arcmin/square hour in radian/square second is equal to1.7529605784562e-8
781 arcmin/square hour in radian/square millisecond is equal to1.7529605784562e-14
781 arcmin/square hour in radian/square microsecond is equal to1.7529605784562e-20
781 arcmin/square hour in radian/square nanosecond is equal to1.7529605784562e-26
781 arcmin/square hour in radian/square minute is equal to0.000063106580824425
781 arcmin/square hour in radian/square hour is equal to0.22718369096793
781 arcmin/square hour in radian/square day is equal to130.86
781 arcmin/square hour in radian/square week is equal to6412.03
781 arcmin/square hour in radian/square month is equal to121232.09
781 arcmin/square hour in radian/square year is equal to17457420.93
781 arcmin/square hour in gradian/square second is equal to0.0000011159693644262
781 arcmin/square hour in gradian/square millisecond is equal to1.1159693644262e-12
781 arcmin/square hour in gradian/square microsecond is equal to1.1159693644262e-18
781 arcmin/square hour in gradian/square nanosecond is equal to1.1159693644262e-24
781 arcmin/square hour in gradian/square minute is equal to0.0040174897119342
781 arcmin/square hour in gradian/square hour is equal to14.46
781 arcmin/square hour in gradian/square day is equal to8330.67
781 arcmin/square hour in gradian/square week is equal to408202.67
781 arcmin/square hour in gradian/square month is equal to7717874.54
781 arcmin/square hour in gradian/square year is equal to1111373934
781 arcmin/square hour in arcmin/square second is equal to0.000060262345679012
781 arcmin/square hour in arcmin/square millisecond is equal to6.0262345679012e-11
781 arcmin/square hour in arcmin/square microsecond is equal to6.0262345679012e-17
781 arcmin/square hour in arcmin/square nanosecond is equal to6.0262345679012e-23
781 arcmin/square hour in arcmin/square minute is equal to0.21694444444444
781 arcmin/square hour in arcmin/square day is equal to449856
781 arcmin/square hour in arcmin/square week is equal to22042944
781 arcmin/square hour in arcmin/square month is equal to416765225.25
781 arcmin/square hour in arcmin/square year is equal to60014192436
781 arcmin/square hour in arcsec/square second is equal to0.0036157407407407
781 arcmin/square hour in arcsec/square millisecond is equal to3.6157407407407e-9
781 arcmin/square hour in arcsec/square microsecond is equal to3.6157407407407e-15
781 arcmin/square hour in arcsec/square nanosecond is equal to3.6157407407407e-21
781 arcmin/square hour in arcsec/square minute is equal to13.02
781 arcmin/square hour in arcsec/square hour is equal to46860
781 arcmin/square hour in arcsec/square day is equal to26991360
781 arcmin/square hour in arcsec/square week is equal to1322576640
781 arcmin/square hour in arcsec/square month is equal to25005913515
781 arcmin/square hour in arcsec/square year is equal to3600851546160
781 arcmin/square hour in sign/square second is equal to3.3479080932785e-8
781 arcmin/square hour in sign/square millisecond is equal to3.3479080932785e-14
781 arcmin/square hour in sign/square microsecond is equal to3.3479080932785e-20
781 arcmin/square hour in sign/square nanosecond is equal to3.3479080932785e-26
781 arcmin/square hour in sign/square minute is equal to0.00012052469135802
781 arcmin/square hour in sign/square hour is equal to0.43388888888889
781 arcmin/square hour in sign/square day is equal to249.92
781 arcmin/square hour in sign/square week is equal to12246.08
781 arcmin/square hour in sign/square month is equal to231536.24
781 arcmin/square hour in sign/square year is equal to33341218.02
781 arcmin/square hour in turn/square second is equal to2.7899234110654e-9
781 arcmin/square hour in turn/square millisecond is equal to2.7899234110654e-15
781 arcmin/square hour in turn/square microsecond is equal to2.7899234110654e-21
781 arcmin/square hour in turn/square nanosecond is equal to2.7899234110654e-27
781 arcmin/square hour in turn/square minute is equal to0.000010043724279835
781 arcmin/square hour in turn/square hour is equal to0.036157407407407
781 arcmin/square hour in turn/square day is equal to20.83
781 arcmin/square hour in turn/square week is equal to1020.51
781 arcmin/square hour in turn/square month is equal to19294.69
781 arcmin/square hour in turn/square year is equal to2778434.83
781 arcmin/square hour in circle/square second is equal to2.7899234110654e-9
781 arcmin/square hour in circle/square millisecond is equal to2.7899234110654e-15
781 arcmin/square hour in circle/square microsecond is equal to2.7899234110654e-21
781 arcmin/square hour in circle/square nanosecond is equal to2.7899234110654e-27
781 arcmin/square hour in circle/square minute is equal to0.000010043724279835
781 arcmin/square hour in circle/square hour is equal to0.036157407407407
781 arcmin/square hour in circle/square day is equal to20.83
781 arcmin/square hour in circle/square week is equal to1020.51
781 arcmin/square hour in circle/square month is equal to19294.69
781 arcmin/square hour in circle/square year is equal to2778434.83
781 arcmin/square hour in mil/square second is equal to0.000017855509830818
781 arcmin/square hour in mil/square millisecond is equal to1.7855509830818e-11
781 arcmin/square hour in mil/square microsecond is equal to1.7855509830818e-17
781 arcmin/square hour in mil/square nanosecond is equal to1.7855509830818e-23
781 arcmin/square hour in mil/square minute is equal to0.064279835390947
781 arcmin/square hour in mil/square hour is equal to231.41
781 arcmin/square hour in mil/square day is equal to133290.67
781 arcmin/square hour in mil/square week is equal to6531242.67
781 arcmin/square hour in mil/square month is equal to123485992.67
781 arcmin/square hour in mil/square year is equal to17781982944
781 arcmin/square hour in revolution/square second is equal to2.7899234110654e-9
781 arcmin/square hour in revolution/square millisecond is equal to2.7899234110654e-15
781 arcmin/square hour in revolution/square microsecond is equal to2.7899234110654e-21
781 arcmin/square hour in revolution/square nanosecond is equal to2.7899234110654e-27
781 arcmin/square hour in revolution/square minute is equal to0.000010043724279835
781 arcmin/square hour in revolution/square hour is equal to0.036157407407407
781 arcmin/square hour in revolution/square day is equal to20.83
781 arcmin/square hour in revolution/square week is equal to1020.51
781 arcmin/square hour in revolution/square month is equal to19294.69
781 arcmin/square hour in revolution/square year is equal to2778434.83
Disclaimer:We make a great effort in making sure that conversion is as accurate as possible, but we cannot guarantee that. Before using any of the conversion tools or data, you must validate its correctness with an authority.
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# Search by Topic
#### Resources tagged with Combining probabilities similar to Odds and Evens:
Filter by: Content type:
Stage:
Challenge level:
### There are 18 results
Broad Topics > Probability > Combining probabilities
### At Least One...
##### Stage: 3 Challenge Level:
Imagine flipping a coin a number of times. Can you work out the probability you will get a head on at least one of the flips?
### Taking Chances
##### Stage: 3 Challenge Level:
This article, for students and teachers, is mainly about probability, the mathematical way of looking at random chance and is a shorter version of Taking Chances Extended.
### Lottery Simulator
##### Stage: 2, 3 and 4 Challenge Level:
Use this animation to experiment with lotteries. Choose how many balls to match, how many are in the carousel, and how many draws to make at once.
### The Lady or the Lions
##### Stage: 3 Challenge Level:
The King showed the Princess a map of the maze and the Princess was allowed to decide which room she would wait in. She was not allowed to send a copy to her lover who would have to guess which path. . . .
### Same Number!
##### Stage: 4 Challenge Level:
If everyone in your class picked a number from 1 to 225, do you think any two people would pick the same number?
### Statins and Risk
##### Stage: 4 Challenge Level:
"Statins cut the risks of heart attacks and strokes by 40%"
Should the Professor take statins? Can you help him decide?
### How Risky Is My Diet?
##### Stage: 3 and 4 Challenge Level:
Newspapers said that eating a bacon sandwich every day raises the risk of bowel cancer by 20%. Should you be concerned?
### Taking Chances Extended
##### Stage: 4 and 5
This article, for students and teachers, is mainly about probability, the mathematical way of looking at random chance.
### Chance Combinations
##### Stage: 3 Challenge Level:
Can you design a bingo board that gives you the best chance of winning?
### The Better Bet
##### Stage: 4 Challenge Level:
Here are two games you have to pay to play. Which is the better bet?
### Card Game (a Simple Version of Clock Patience)
##### Stage: 4 Challenge Level:
Four cards are shuffled and placed into two piles of two. Starting with the first pile of cards - turn a card over... You win if all your cards end up in the trays before you run out of cards in. . . .
### Win or Lose?
##### Stage: 4 Challenge Level:
A gambler bets half the money in his pocket on the toss of a coin, winning an equal amount for a head and losing his money if the result is a tail. After 2n plays he has won exactly n times. Has. . . .
### Gambling at Monte Carlo
##### Stage: 4 Challenge Level:
A man went to Monte Carlo to try and make his fortune. Is his strategy a winning one?
##### Stage: 4 and 5 Challenge Level:
Some relationships are transitive, such as `if A>B and B>C then it follows that A>C', but some are not. In a voting system, if A beats B and B beats C should we expect A to beat C?
### Marbles and Bags
##### Stage: 4 Challenge Level:
Two bags contain different numbers of red and blue marbles. A marble is removed from one of the bags. The marble is blue. What is the probability that it was removed from bag A?
### Heavy Hydrocarbons
##### Stage: 4 and 5 Challenge Level:
Explore the distribution of molecular masses for various hydrocarbons
### Fixing the Odds
##### Stage: 4 Challenge Level:
You have two bags, four red balls and four white balls. You must put all the balls in the bags although you are allowed to have one bag empty. How should you distribute the balls between the two. . . .
### Molecular Sequencer
##### Stage: 4 and 5 Challenge Level:
Investigate the molecular masses in this sequence of molecules and deduce which molecule has been analysed in the mass spectrometer.
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https://topfuturepoint.com/96-kg-to-lbs/
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96 kg to lbs
Friends in today’s post we will learn how to convert 96 kg to lbs. To convert kilograms to pounds, you can use the conversion factor of 2.20462 pounds per kilogram.
So to convert 96 kilograms to pounds, you would multiply 96 by 2.20462:
96 kilograms * 2.20462 pounds/kilogram = 211.64352 pounds
Therefore, 96 kilograms is approximately equal to 211.64 pounds.
What is the best conversion unit for 96 kg?
The best conversion unit for 96 kilograms depends on the context and the specific needs of the situation. Here are a few commonly used conversion units for 96 kilograms:
1. Pounds: As mentioned earlier, 96 kilograms is approximately equal to 211.64 pounds. This conversion is commonly used in everyday life, especially in countries that use the imperial system.
2. Stone: In some countries, particularly the United Kingdom and Ireland, the stone is used as a unit of measurement for body weight. One stone is equal to 14 pounds. Therefore, 96 kilograms is approximately 15.12 stones (96 kg / 6.35029318 kg/stone).
3. Metric Tons: If you are working with larger quantities, you might consider converting kilograms to metric tons (tonnes). One metric ton is equal to 1000 kilograms. Therefore, 96 kilograms is equal to 0.096 metric tons (96 kg / 1000 kg/tonne).
4. Grams: If you need a smaller unit, you can convert kilograms to grams. One kilogram is equal to 1000 grams. Therefore, 96 kilograms is equal to 96,000 grams (96 kg * 1000 g/kg).
The best conversion unit depends on the specific context and the purpose of the conversion. Choose the unit that is most appropriate and convenient for your particular situation.
Table for values around 96 kilograms
Certainly! Here’s a table showing values around 96 kilograms converted to different units:
Please note that the values in this table are approximate conversions and have been rounded to two decimal places for simplicity.
FAQs on 96 kg to lbs
Certainly! Here are some frequently asked questions regarding the conversion of 96 kilograms to pounds:
Q: How many pounds are in 96 kilograms?
A: 96 kilograms is approximately equal to 211.64 pounds.
Q: Why would I need to convert kilograms to pounds?
A: Kilograms and pounds are both units of weight, but they are commonly used in different regions. Converting between the two units allows for easier understanding and comparison of weight measurements in different systems.
Q: Is the conversion factor from kilograms to pounds constant?
A: The conversion factor from kilograms to pounds is approximately 2.20462 pounds per kilogram. While this is a widely used and accepted conversion factor, it is important to note that it is an approximation and can vary slightly depending on the desired level of precision.
Q: Can I use an online converter to convert 96 kilograms to pounds?
A: Yes, there are various online conversion tools available that can quickly and accurately convert kilograms to pounds. These tools often allow you to input the value in kilograms and instantly provide the corresponding value in pounds.
Q: Why do some countries use kilograms while others use pounds?
A: Different countries have adopted different systems of measurement for historical, cultural, or practical reasons. The metric system, which includes the kilogram as a unit of weight, is widely used around the world, while the imperial system, which includes pounds, is primarily used in the United States and a few other countries.
Q: Are there any other common units to convert 96 kilograms to, apart from pounds?
A: Yes, apart from pounds, other common units used for conversion include stones, metric tons (tonnes), and grams. These units can be useful depending on the context and the specific needs of the situation.
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Part 4
# Module 17: Inquiry and Measurement
Learning Outcomes
When you have completed this module, you will be able to:
1. Describe the geometry of a right angle triangle.
2. Describe a point, point styles, and object snap mode node.
3. Apply the system variable SNAPANG to rotate the graphic cursor.
4. Apply the ORTHO, POINT, DIVIDE, and MEASURE commands.
5. Apply the UNITS, DIST, and PROPERTIES commands to set the display units, make inquiries about drawing objects, measure distances, and angles.
# Geometry Lesson: A Right Angle Triangle
Figure 17-1 and 17-2 show the geometry of a right angle triangle. Understanding this theory will help you to complete your AutoCAD drawings.
# Geometry Lesson: Using Circles to Measure
When a measurement is required along an inclined line as shown in Step 1, Figure 17-3, the OFFSET command cannot easily be used to make that measurement. The best method of making the measurement is by drawing a construction circle with a 1 unit radius. Use the end of the inclined line as the center location for the circle. All points on the circumference of the circle will be the radius distance (1 unit) from the center of the circle as shown in the Step 2. Draw the 0.60 diameter circle locating its center at the intersection of the construction circle and the inclined line as shown in the Step 3.
## Points
A point is an AutoCAD drawing object that is one pixel in size. Its only geometry property is one XY coordinate.
## Point Style
Since a point is nothing more then one pixel in size, it will not display very well in the drawing. If it is located under an existing drawing object, it will not be visible at all. Points can be set to display with an assigned style to make them easier to see and use. The point style for the current drawing is set with the Point Style dialogue box. To open the Point Style dialogue box, select Point Style in the Format Pull-down menu, See Figure 17-4 and 17-5. The size of the points can be set relative with the screen size or an absolute size. To display the points style at the correct size, when the point size is set relative to the screen, the drawing must be regenerated using the REGEN command.
The POINT command is used to insert a point.
Shortcut: PO
## Rotating the Crosshairs
The crosshairs can be rotated to any angle. Its default rotation angle is 0 degrees, as shown in Figure 17-7. When the rotation angle is 0 degrees, the X axis is horizontal and the Y axis is vertical. There are times when it can be very helpful if the crosshairs are rotated to an angle other than 0 degrees, as shown in Figure 17-8. The SNAPANG system variable is used to set the rotation angle of the crosshairs.
MUST KNOW: It can be very confusing if SNAPANG is set to 90, 180, or 270 degrees since the crosshairs looks just like the default or 0 degrees. You can check its setting by entering the SNAPANG system variable on the command line and ensure it is set to 0.
## The DIVIDE Command
The DIVIDE command is used to divide a drawing object by inserting points on the object at calculated XY locations. It does not change the object in any way, it simply inserts points on it. It is important that the current layer is set to layer Construction when the DIVIDE command is used. That way, the points can be easily isolated from the object it is dividing. On lines and arcs, the DIVIDE command starts calculating from the endpoint of the selected object.
The DIVIDE command works different when dividing circles. Since a circle does not have an endpoint, the DIVIDE command will place the first point where the positive X axis of the crosshairs intersects the circle.
## The MEASURE Command
The MEASURE command is used to divide a drawing object by inserting points on the object at specified distances along the object. It does not change the object in any way, it simply inserts points on it. It is important that the current layer is set to layer Construction when the MEASURE command is used. That way, the points can easily be isolated from the object being measured. The MEASURE command always starts the first measurement from the closest endpoint to the location where the object is selected.
The MEASURE command works different when measuring circles. Since a circle does not have an endpoint, the MEASURE command will start measuring at the point where the positive X axis of the crosshairs intersects the circle.
The ORTHO command is used to lock the crosshairs to allow it to move only horizontal or vertical. When enabled, it applies to most commands. The current Ortho mode is displayed on the Status bar. The function F8 key toggles ortho mode off and on.
Shortcut: F8
The DIVIDE command is used to insert points an equal distance apart by dividing the length of the drawing object selected by the number of divisions entered. The object being divided is not changed in any way by the command.
Shortcut: DIV
The MEASURE command is used to insert points on a drawing object at a specified distance. The object being measured is not be changed in any way by the command.
Shortcut: ME
# WORKALONG: Using the POINT, DIVIDE and MEASURE Commands
## Step 1
Start a new drawing using the template: 2D English.
## Step 2
Save and name the drawing: AutoCAD 2D Workalong 17-1. (Figure Step 2)
## Step 3
Create the layers: Object and Construction. Set layer: Construction as the current layer.
## Step 4
Using what you learned earlier in the book, refer to Figure Step 2 and draw lines around the perimeter of the object. (Figure Step 4)
## Step 5
From the Format Pull-down menu, select Point Style . This will open the Point Style dialogue box. Select the point style (center with black background) as shown in the figure. Set the Point Size to 3% and enable Set Size Relative to Screen. (Figure Step 5A and 5B)
## Step 6
Enter the DIVIDE command, as shown below, and select the inclined line on the right. (Figure Step 6A and 6B)
Command: DIVIDE
Select object to divide: P1
Enter the number of segments or [Block]: 8
Command:
## Step 7
As taught earlier in the book, use the OSNAP command to open the Drafting Settings dialogue box. Enable the Node object snap mode. (Figure Step 7)
## Step 8
Press F8 to enable Ortho mode. Check to ensure that it is enabled by checking the Status bar. (Figure Step 8)
## Step 9
Enable object snap and draw 8 horizontal lines by snapping to the points (nodes) as the first point of each line. The length of the lines is not important. (Figure Step 9)
## Step 10
Draw 7 vertical lines by snapping to the points (nodes) as the first point of each line. The length of the lines is not important. Disable Ortho mode. (Figure Step 10)
## Step 11
Trim the lines to form the steps as shown in the figure. (Figure Step 11)
## Step 12
Draw a line by snapping to midpoint on the top left inclined line and the midpoint of the bottom line.
## Step 13
Using the DIVIDE command, divide the line into four segments. (Figure Step 13)
## Step 14
Set layer: Object as the current layer. Enter the CIRCLE command and draw the three 0.40 diameter circles by snapping to the points (nodes) to locate the center of each circle. (Figure Step 14)
## Step 15
Set layer: Construction as the current layer.
## Step 16
Draw a construction line from the top corner to the bottom corner as shown in the figure. Ensure that you snap to the endpoints of the lines. (Figure Step 16)
## Step 17
Draw a 0.5 radius construction circle with its center located at the top corner as shown in the figure. Trim the line. (Figure Step 17A and 17B)
## Step 18
Enter the MEASURE command, as shown below, to insert points every 0.4325 units along the line. (Figure Step 18)
Command: MEASURE
Select object to measure: P2
Specify length of segment or [Block]: 0.4325
Command:
## Step 19
The drawing should now appear as shown in the figure. To demonstrate how the point size relative to the screen works, zoom in on your drawing until it fills the screen. Then execute the REGEN command. Notice how the point style size for the points changes. Zoom back out and execute the REGEN command again. (Figure Step 19)
## Step 20
Set layer: Object as the current layer. Insert 12 – 0.125 diameter circles. One with its center located at the end of the line and the others with their centers located on the points that you inserted with the MEASURE command in Step 18. (Figure Step 20)
## Step 21
Set layer: Construction as the current layer. Enter the ID command, as shown below, and snap to the endpoint of the bottom line to establish a reference point. (Figure Step 21)
Command: ID
Specify point: (end) P3
X = 1.7500 Y = 4.0000 Z = 0.0000
Command:
## Step 22
Enter the POINT command, as shown below, to draw 3 points at the locations of the center of polygons. (Figure Step 22)
Command: POINT
Current point modes: PDMODE=34 PDSIZE=-3.0000 Specify a point: @.75,1.5
Command: POINT
Current point modes: PDMODE=34 PDSIZE=-3.0000 Specify a point: @0,-1
Command: POINT
Current point modes: PDMODE=34 PDSIZE=-3.0000
Specify a point: @2.25,0
Command:
## Step 23
Draw three 0.5 diameter circles using the points as the location of the center of the circles. Ensure that you snap to the node when locating the center of each circle. (Figure Step 23)
## Step 24
Using the DIVIDE command, divide the top left circle into 6 segments. (Figure Step 24)
## Step 25
Enter the SNAPANG system variable as shown below to change the angle of the cursor to 90 degrees.
Command: SNAPANG
Enter new value for SNAPANG <0>: 90
Command:
## Step 26
Enter the DIVIDE command and divide the bottom left circle into 6 segments. (Figure Step 26)
## Step 27
Enter the system variable SNAPANG as shown below to change the angle of the crosshairs to 22.5 degrees.
Command: SNAPANG
Enter new value for SNAPANG <90>: 22.5
Command:
## Step 28
Enter the DIVIDE command and divide the bottom right circle into 8 segments. (Figure Step 28A and 28B)
## Step 29
Use the SNAPANG system variable, as shown below, to change the angle of the crosshairs to 0 degrees.
Command: SNAPANG
Enter new value for SNAPANG
<22.5>: 0
Command:
## Step 30
On layer Object, draw the lines for the hexagons and the octagon by snapping from point to point (node to node) as shown in the figure. (Figure Step 30)
## Step 31
Change the layer of any objects that you want to reside on layer Object. Freeze layer Construction. Your completed drawing should appear as shown in the figure. (Figure Step 31)
## Step 32
Save and close the drawing.
MUST KNOW: The MEASURE command inserts points at a specified distance. It starts measuring at the end of the object nearest where it is selected. It may not end up with an equal distance at the other end of the object. Make sure that the current layer is Construction before you execute this command to ensure that the points will be easily to isolate from the object it is measuring.
## Making Inquiries
The DIST and PROPERTIES commands are used to measure distances, find sizes, and locations of drawing objects. Before using these commands, it is important to understand the importance of the UNITS command.
## Displaying Units
AutoCAD stores the drawing object’s properties and locations to a very high degree of accuracy. When the drawing object’s properties are displayed, either on the Command Line or in the Properties window, AutoCAD uses the setting in the Drawing Units dialogue box to display these properties rounded off as per the current settings. The answers are always displayed in drawing units but the precision and format of the answers is controlled by the settings in the Drawing Units dialogue box.
The UNITS command is used to open the Drawing Units dialogue box. It is used to set the way AutoCAD displays inquiries or drawing object properties.
Shortcut: UN
The DIST command is used to measure distances between two XY coordinate locations.
Shortcut: DI
# Geometry Lesson: Finding the Shortest Distance Between the Circumference of Two Circles
To find the shortest distance between the circumference of two circles, draw a construction line from the center of one circle to the center of the other as shown in Step 1. Measure the distance between the intersections of the lines and circles as shown in Step 2. If the line is trimmed as shown in Step 3, the length of the line is the shortest distance between the circle’s circumference. See Figure 17-9.
# WORKALONG: Using Inquiry Commands
## Step 1
Open the drawing: AutoCAD 2D Workalong 17-1 . Using the SAVEAS command, save the drawing with the name: AutoCAD 2D Workalong 17-2. (Figure Step 1)
## Step 2
Enter the UNITS command. It will open the Drawing Units dialogue box. Set the dialogue box to match the figure. (Figure Step 2)
## Step 3
Enter the DIST command, as shown below, to measure the distance between the center of the circle and the end of the line as shown in the figure. Ensure that you snap to the center of the circle and the endpoint of the line. (Figure Step 3)
Command: DIST
Specify first point: (cen) P1
Specify second point: (end) P2
Command:
## Step 4
Press F2 to open the Text window. The answer to the DIST command will display as shown in the figure. (Figure Step 4)
## Step 5
Press F2 to close the Text window.
## Step 6
Draw a construction line from the center of circle to center of circle as shown in the figure. Ensure that you snap to the centers of the circle at both ends. Enter the DIST command as shown below to measure the shortest distance between circumferences of the circles by snapping to intersection of the construction line and the circles. (Figure Step 6)
Command: DIST
Specify first point: (int) P3
Specify second point: (int) P4
Command:
## Step 7
Press F2 to open the Text window. The answer to the DIST command will display as shown in the figure. (Figure Step 7)
## Step 8
Trim the construction line as shown in the figure. (Figure Step 8)
MUST KNOW: AutoCAD stores the accuracy of the drawing object’s properties to a very high degree. The UNITS command is used to set the format and precision which instructs AutoCAD how to display answers in the Properties window and to the users inquiry commands.
## Step 9
Open the Properties window. Without entering a command, select the line. The Properties window will display the length of the line. (Figure Step 9)
## Step 10
Set the Drawing Units dialogue box, as shown in the figure. (Figure Step 10)
## Step 11
Draw a line from the center of the circle to the vertex on the hexagon as shown in the figure. (Figure Step 11)
## Step 12
After trimming the line, select it. Check its length and angle in the Properties window. (Figure Step 12)
## Step 13
Turn layer: Construction off. Enter the ID command and snap to the center of the circle. Press F2 to open the Text window. (Figure Step 13A and 13B)
## Step 14
Select the circle as shown in the figure. Take note that the Properties window displays the XY center location plus the radius, diameter, circumference, and area of the circle. (Figure Step 14)
## Step 15
With the Properties window open, select the line as shown in the figure. Figure Step 15)
## Step 16
Save and close the drawing.
# Geometry Lesson: Finding the Centroid of any Triangle
To find the centroid (geometrical center) of any triangle, draw construction lines from each vertex to the midpoint of the opposite line of the triangle. The centroid of the triangle is the intersection of the three lines. See Figure 17-9.
# Key Principles
Key Principles in Module 17
1. A point is an AutoCAD drawing object that is one pixel in size.
2. A node is the object snap mode for a point.
3. AutoCAD stores the accuracy of the drawing object’s properties and locations to a very high degree. The UNITS command is used open the Drawing Units dialogue box where the you can set the format of how AutoCAD displays these properties and locations.
4. The DIVIDE command insert points on the object by dividing the object’s length or perimeter into an equal number of segments specified by the user.
5. The MEASURE command inserts points on an object at a user specified distance.
# Lab Exercise 17-1
Time allowed: 45 minutes.
Drawing Name Template Units
AutoCAD 2D Lab 17-1 2D Metric Millimeters
Layering Scheme
Layer Name Objects on Layer Color
Construction Construction objects 253
Object All objects Red
## Step 1
Draw the object shown in the figure using the layering scheme. (Figure Step 1A and 1B)
## Step 2
Set the insertion units and check your drawing with the key.
## Step 3
Turn layer: Key off and freeze layer: Construction.
## Step 4
Using either the DIST command or the Properties window, find the following answers with the precision of 8 decimal places:
1. Length of arc A: ______________________
2. Distance from the center of circle C to center of circle E. ______________________
3. Shortest distance from corner B to the circumference of circle C. ______________________
4. The circumference of circle D: ______________________
## Hint 1
Using the DIVIDE command, divide the large circle into 26. (Figure Hint 1)
## Hint 2
Rotate the cursor 7.5 degrees and divide the 40 diameter circle into 5. (Figure Hint 2)
## Hint 3
Rotate the cursor 30 degrees and divide the 75 diameter circle into 6. (Figure Hint 3)
## Hint 4
Draw 6 – 20 diameter circles locating their centers on the points. (Figure Hint 4)
## Hint 5
To draw the polygons, rotate the cursor to the appropriate angle as shown in the list below. (Figure Hint 5A and 5B)
Triangle = 90 degrees
Square = 45 degrees
Pentagon = 90 degrees
Hexagon = 90 degrees
Heptagon = 90 degrees
Octagon = 22.5 degrees
# Lab Exercise 17-2
Time allowed: 50 minutes.
Drawing Name Template Units
AutoCAD 2D Lab 17-2 2D English Inches
Layering Scheme
Layer Name Objects on Layer Color
Construction Construction objects 253
Object All objects Red
## Step 1
Draw the object shown in the figure using the layering scheme. (Figure Step 1A and 1B)
## Step 2
Set the insertion units and check your drawing with the key.
## Step 3
Turn layer: Key off and freeze layer: Construction.
## Hint 1
Using the DIVIDE command, divide the inclined line into 12. (Figure Hint 1)
## Hint 2
Construct the steps using ortho mode. (Figure Hint 2)
## Hint 3
Trim the steps and change the layer they reside on to layer: Object. Offset the inclined line and draw a circle to measure the distance along the inclined line. (Figure Hint 3)
## Hint 4
Draw the triangle by offsetting the three outsides 0.75 inches as shown in the dimensioned drawing. Using the Geometry Lesson: Finding the Centroid of any Triangle, find the centroid of the triangle and draw a 0.35 diameter circle using the intersection of the lines to locate its center. (Figure Hint 4)
## Hint 5
Offset the outside lines of the triangle 0.20 inches towards the center. (Figure Hint 5)
## Hint 6
Offset the inside lines 0.10 inches on each side. (Figure Hint 6)
## Hint 7
Trim all lines and the circle. Fillet all corners using a radius of 0.05 inches. The fillets should always be inserted last. (Figure Hint 7)
## Hint 8
Change the layer of all objects to layer: Object. (Figure Hint 8)
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# Nail Your Number Facts With This 3 Minute Maths Activity [VIDEO]
You may not realise it but helping your child to nail their number facts is one of the most useful things that you can do to support them with maths at home. And here, we have the ideal 3 minute maths activity to make it fast and fun.
Number facts are a crucial part of early maths development in children aged between 5 and 11, so it is very important that your child has them by heart at an early stage!
### What are number facts?
Number facts are early stage addition, subtraction, multiplication and division calculations that children should be able to recall easily and off by heart.
Every child, from Year 1 up to Year 6 and beyond, needs to be able to instantly recall some key number facts as this is an important thing to be able to do in primary maths.
• Years 1 & 2 – Children will start with learning number bonds to ten (1 + 9 = 10, 2 + 8 = 10), and then progress to learning number bonds up to 20
• Years 3 & 4 – Children will be expected to know their times tables from 1-12
• Years 5 & 6 – By this stage, all sorts of conversions should be known ‘just like that’ (e.g. 1km = 1000m, 0.25 = ¼ and others).
### Why do children need number facts?
Quite simply the reason for your child to know the number facts for their age, is it just makes it easier for them to learn new information and strategies in maths.
Children’s working memories (and ours for that matter!) have only a finite capacity, meaning the more they can commit to their long term memory, the better.
If they can learn their number facts to a state of ‘automaticity’ i.e. they can automatically recall them within 5 seconds without counting or ‘thinking’, it frees up their brain capacity to learn the new concept that’s being taught.
A good example of this is fractions. Learning about fractions with the numerator (number on the top) and denominator (number on the bottom) and how they work when adding, multiplying or anything else is hard for children.
Many struggle!
The Ultimate Mental Maths Lesson Powerpoint
Download this FREE ultimate mental maths lesson powerpoint, containing over 200 slides to ensure that you have everything you need to teach your class the mental maths strategies they need to excel!
By making sure your child, in this instance, knows their times tables, once they start to learn about fractions, they can focus all their attention on working out the strategy needed, rather than on working out the answer.
Moreover, once your child is in Year 6, one of the SATs Maths papers they will sit is an Arithmetic paper, and if they don’t have some of their number facts on instant recall they will not be able to complete it in the given 30 minutes, making it very difficult to get through all 36 questions!
There are lots of different ways to help your child learn their number facts, and some are certainly more fun than others!
When teaching a topic, the aim is that your child enjoys the process, takes ownership of their own learning, and can see the progress they’re making. We know that this is easier said than done, and that is why we are here to provide guidance using our years of maths teaching knowledge!
### Which number facts to learn
• Doubles up to 10
• Doubles up to 20
• Number bonds to 5
• Number bonds to 10
• Ten and over
• Bridging over 10
• Number bonds to 20
Now that we have answered the question “What are number facts?”, and given you a place to begin with your child, it’s time to share with you our favourite number facts game that you can play at home.
As teachers we know how tricky it can be to find engaging ways to use maths when school is finished for the day, but our maths facts game has proven popular amongst students in schools across the land so we are sure it will be the same in your home!
### Number facts activity: 3 minute maths paper flip
Either you or your child can start by making and writing on the paper flip. The process is very simple, as is explained in our video above.
#### How to make your number facts paper flip
1. Get an A4 piece of paper.
2. Fold it in half lengthwise. And again.
3. Cut along a fold to create a strip.
4. Concertina that strip of paper.
5. Write a calculation without an answer on the top fold.
6. Flip the paper over and write the calculation with the answer on the reverse.
7. Continue down the flipping the back back and forth repeating with different calculations and answers on the reverse.
Once the number facts paper flip has been created, the following step-by-step instructions can be issued to your child to enable them to learn their number facts with their new resource!
### How to use your number facts paper flip
Now you are ready to start turning your calculations into number facts.
1. Pick up the strip and look at the top calculation (the one without an answer).
2. Working it out if necessary, say the answer.
3. Flip the strip to check.
4. Move down the calculations and repeat.
5. When at the bottom, grab a different strip. Repeat. Go back to the first strip.
6. Repeat 5 minutes later.
7. And that’s it. Number facts nailed in 10 minutes!
### How to know which number facts your child needs more or less help on
To find out which number facts you should focus on, start with doubles up to 10 (1+1, 2+2, 3+3 etc) from the list above and find out where your child’s gaps are.
A gap is any fact that is not said correctly within 5 seconds.
It is easier for your child to view the calculation rather than hear it, so write down the calculation on paper and get them to say it out loud within 5 seconds.
Move down the list of number facts till you find their gaps – that’s what to put on the paper flip!
The key is keeping things simple. Stick to the set your child has gaps in until those gaps are answered correctly, day after day.
Keep revisiting the same calculations by reusing the same strips.
Stick to addition, certainly until almost all the sets are secure. Then, later on, as you progress, you can move on to subtraction, which is the opposite operation to addition.
Children learn this at school as the ‘inverse operation’, so if they have come home mentioning this, you now know what they mean!
For example, if you know 3 + 6 = 9 as a fact, you also have the basis of knowing 9 – 3 = 6 and 9 – 6 = 3.
### How to make number facts easier
If your child is struggling to retain or calculate, it is a good idea to represent the calculation visually.
At school we use something called a tens frame which you can easily scribble on a piece of paper to help them.
You’ll need 2 ten frames together for numbers over 10.
If they are still finding it difficult, resorting to counting is absolutely fine until your child has an idea about what the answer could be. This will usually come from remembering the saame calculation or learning a strategy that limits the need for counting.
#### How to make number facts harder
If your child needs to be challenged, instead of the usual format of a calculation, put a square where a number should be, e.g. 4 + 7 = [ ]. This can be moved around so switch it up!
[ ] = 4 + 7
4 + [ ] = 11
[ ] + 7 = 11 etc.
If a child knows 2 + 3 is 5, they can work out 12 + 3 = 15 using that fact. It shouldn’t be a big leap for them to apply that same understanding to work out 42 + 3 or 72 + 3. And in turn, that should support 42 + 23 and 72 + 13.
Strips could move to purely focusing on these links that can be made.
Older children could then use the same principle to make strips for a times table such a 6x 7x 8x whatever times table is being learnt. Alternatively they could use the strips to show frequently used fractions, decimals and percentages.
The possibilities are endless with these strips, and you are only limited by your imagination. But if you’re looking for even more ways to challenge your children, we have plenty in our list of home learning resources.
### When and where to use this number facts activity
When – Every day for at least 5 minutes. Once you have spent 10 or so minutes making the strips, all you need is a spare few minutes throughout the day to use them.
Where – Anywhere! The car, the breakfast table, after school club. Walking to school? Ditch the strips and use the same principle. Revisit the same calculations. Limit how many are being spoken rather than written and and pick them from all from the same ‘set’ described above.
Once you are used to the strips, you will begin to use the principle in fun and creative ways.
### Extra tips to encourage number fact recall at home
When working with your child, encouragement is the name of the game.
Gentle ‘we just saw that one a couple of minutes ago, can you think back…’ and ‘if you are stuck between a couple of answers, just go for it, it doesn’t matter if you are wrong’ will help nudge children towards retaining that information the next time.
In other words, the goal is to help them move away from relying on counting. It is counterproductive to put them under pressure ‘You know this, we just did it, come on!’ type comments are not going to help anyone here – brains close off at this point and 1+1 becomes too difficult to comprehend as a result!
In the classroom we use something referred to as growth mindset to encourage the learning of new methods and topics.
x
#### FREE 12 Maths Club Activities for Primary Schools (Years 1 to 6)
A collection of games and activities to make maths enjoyable in a primary school maths club!
Minimal resources are needed for each activity.
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# Dimensions Math Textbook 1A
\$14.00
Textbooks introduce concepts through logical sequencing. Lessons encourage discussion and help students reach fluency through engaging activities and strategies. Corresponding exercises in workbooks help students consolidate and practice new knowledge. Features & Components:
• Chapter Opener: Each chapter begins with an engaging scenario that stimulates student curiosity in new concepts. This scenario also provides teachers an opportunity to review skills.
• Think: With guidance from teachers, students solve a problem using a variety of methods.
• Learn: One or more solutions to the problem in Think are presented, along with definitions and other information to consolidate the concepts introduced in Think.
• Do: A variety of practice problems allow teachers to lead discussion or encourage independent mastery. These activities solidify and deepen student understanding of the concepts.
• Exercise: A pencil icon at the end of the lesson links to additional practice problems in the corresponding workbook.
• Practice: Periodic practice provides teachers with opportunities for consolidation, remediation, and assessment.
• Review: Two cumulative reviews provide ongoing practice of concepts and skills.
Note: Two textbooks (A and B) for each grade correspond to the two halves of the school year. Answer key not included. Soft cover.
SKU: DMT1A
SKU: DMT1A
ISBN: 9781947226043
Pagecount: 176
Dimensions: 8.5 x 11 x 0.75 in
Binding: Standard
Color: Color
Cover: Soft
Perforated: No
Sample Pages
Chapter 1: Numbers to 10
Lesson 1: Numbers to 10
Lesson 2: The Number 0
Lesson 3: Order Numbers
Lesson 4: Compare Numbers
Lesson 5: Practice
Chapter 2: Number Bonds
Lesson 1: Make 6
Lesson 2: Make 7
Lesson 3: Make 8
Lesson 4: Make 9
Lesson 5: Make 10 — Part 1
Lesson 6: Make 10 — Part 2
Lesson 7: Practice
Lesson 1: Addition as Putting Together
Lesson 4: Addition with Number Bonds
Lesson 5: Addition by Counting On
Lesson 8: Practice
Chapter 4: Subtraction
Lesson 1: Subtraction as Taking Away
Lesson 2: Subtraction as Taking Apart
Lesson 3: Subtraction by Counting Back
Lesson 4: Subtraction with 0
Lesson 5: Make Subtraction Stories
Lesson 6: Subtraction with Number Bonds
Lesson 8: Make Addition and Subtraction Story Problems
Lesson 9: Subtraction Facts
Lesson 10: Practice
Review 1
Chapter 5: Numbers to 20
Lesson 1: Numbers to 20
Lesson 2: Add or Subtract Tens or Ones
Lesson 3: Order Numbers to 20
Lesson 4: Compare Numbers to 20
Lesson 6: Subtraction
Lesson 7: Practice
Lesson 1: Add by Making 10 — Part 1
Lesson 2: Add by Making 10 — Part 2
Lesson 3: Add by Making 10 — Part 3
Lesson 4: Addition Facts to 20
Lesson 5: Practice
Chapter 7: Subtraction Within 20
Lesson 1: Subtract from 10 — Part 1
Lesson 2: Subtract from 10 — Part 2
Lesson 3: Subtract the Ones First
Lesson 4: Word Problems
Lesson 5: Subtraction Facts Within 20
Lesson 6: Practice
Chapter 8: Shapes
Lesson 1: Solid and Flat Shapes
Lesson 2: Grouping Shapes
Lesson 3: Making Shapes
Lesson 4: Practice
Chapter 9: Ordinal Numbers
Lesson 1: Naming Positions
Lesson 2: Word Problems
Lesson 3: Practice
Review 2
A & B Books: Our programs divide the school year into two semesters. “A” level books are for the first half of the school year. “B” level books are for the second half of the school year. You need both “A” and “B” material for a complete school year.
Required Components: Textbooks, Workbooks, and Guides (either Home Instructor’s Guides or Teacher’s Guides) are all necessary components. These three elements each serve a unique function and work together to build math mastery.
Printouts: Dimensions Math PK-5 Resources
Recommended Manipulatives: Dimensions Math Grades PK-5 Recommended Manipulatives
Answer Key: Located in corresponding Home Instructor’s Guide/Teacher’s Guide
Dimensions Math At Home™ Videos
Invite a professional Singapore math teacher into your home classroom. This subscription of pre-recorded lessons covers all Textbook and Workbook material for an entire school year and can be done at your own pace.
Manipulatives
Manipulatives help students visualize and represent math concepts. Stock up on suggested items to deepen engagement during math activities and lessons.
Supplementary Math
Supplementary math helps students of all abilities: struggling, advanced, or on level. Find just the right addition to round out your core program.
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# A cake is shared among 5 friends. Four friends take 1/8, 1/6, 5/12, 1/12 part of the cake, respectively. How much part of the cake does the fifth friend get?
Free Practice With Testbook Mock Tests
## Options:
1. 5/24
2. 3/8
3. 1/6
4. 1/4
### Correct Answer: Option 1 (Solution Below)
This question was previously asked in
RRC Group D Previous Paper 1 (Held On: 17 Sep 2018 Shift 1)
## Solution:
⇒ Part of cake left with 5th friend = 1 – {(1/8) + (1/6) + (5/12) + (1/12)} = 1 – {(3 + 4 + 10 + 2)/24} = 1 – (19/24) = 5/24
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Anonymous
# finding arc length and area of a sector?
If I know radius is 16cm and central angle of 50° = 5pi/18 radians. I have to use the equation s = (angle)r² to find arc length. is 40pi/9 cm correct?
and for area = (1/2) r²(angle) I got 1280pi/9 what did I do wrong?
Update:
i meant to say s = (angle) r
Relevance
• Anonymous
Since you first mentioned degrees instead of radians, I will work with degrees. I should point out that if work with radians, you can save several steps, but degrees it is.
1)
Let P be the perimeter of the circle.
P = 2 * π * r.............In this case that is:
P = 2 * π * 16cm = 32 π cm. I'm going to leave the measure cm out until the end so that the work with numbers is more obvious.
Fact: Arc length is not proportional to r^2, it is proportional to r.
So the answer to your problem is the ratio of the central angle to the 360 degrees of the full circle times the perimeter. So:
s = (50/360) * P = [(5 * P) / (36)] = [(5 * 32 * π) / 36] = ((160 * π) / 36)
Divide the numerator (160 * π) and the denominator (36) by 4 getting:
s = ((40 * π) / 9) cm........<<<<<
Note: it was the r^2 in the formula that misled you about the arc length.
-----------------------
2)
For the area you want to compute 50/360 times the area of the circle.
A (of the defined part of the circle) = (50/360)*π*r^2 = (5/36)*π*(16)^2 = (5/36)*π*256 =
5 * (256/36) * π = 5 * (64/9) * π = (320/9) * π square cm =
[(320 * π) / 9] square cm......<<<<<
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# Mean, variance, and histograms
Page 2 / 2
## Exercise 2.2
Suppose you have set the goal of making an A in your math class. If your class grades consist of 4 tests, and you have made a 98, 80, and 90 on your first three tests, what do you need to make on your last test so that the mean of your grades is 90?
## Exercise 2.3
(for the advanced) Suppose that, for the same class, you have already computed the mean of the first three tests when you receive your fourth test grade. Instead of computing the mean of all four tests from scratch, it's possible to update the mean that you've already computed. Write a Matlab code that takes two inputs, the mean of your first three tests and the grade of your fourth test, and computes the mean of all four tests.
## Variance and standard deviation
As you saw in "Example 2.2" , the mean is not always representative of the data, and other measures are needed to analyze the spread of the data. The variance is a measure of the distance of each number from the mean. Given a vector x of n numbers and mean value $\overline{x},$ the variance of x is given by
$\mathrm{var}\left(\mathbf{x}\right)=\frac{1}{n-1}\sum _{k=1}^{n}{\left({\mathbf{x}}_{k}-\overline{x}\right)}^{2}=\frac{{\left({\mathbf{x}}_{1}-\overline{x}\right)}^{2}+{\left({\mathbf{x}}_{2}-\overline{x}\right)}^{2}+...+{\left({\mathbf{x}}_{n}-\overline{x}\right)}^{2}}{n-1}.$
The standard deviation of the data is related to the variance and is given by
$\mathrm{std}\left(\mathbf{x}\right)=\sqrt{\mathrm{var}\left(\mathbf{x}\right)}.$
You can compute the variance and standard deviation of x in Matlab by typing the commands var(x) and std(x).
## Example 3.1
Consider the vector given in "Example 2.1" , x = [1, 7, 2, 5, 9, 6]. Recall that the mean of x = 5.
$\mathrm{var}\left(\mathbf{x}\right)=\frac{{\left(1-5\right)}^{2}+{\left(7-5\right)}^{2}+{\left(2-5\right)}^{2}+{\left(5-5\right)}^{2}+{\left(9-5\right)}^{2}+{\left(6-5\right)}^{2}}{5}=9.2$
$\mathrm{std}\left(\mathbf{x}\right)=\sqrt{\mathrm{var}\left(\mathbf{x}\right)}\simeq 3.03$
## Example 3.2
Consider the data from "Example 2.2" , where the mean $\overline{x}$ = 300. The variance is
$\mathrm{var}\left(\mathbf{x}\right)=\frac{16·{\left(100-300\right)}^{2}+3·{\left(900-300\right)}^{2}+{\left(1700-300\right)}^{2}}{13}\simeq 193,684$
and the standard deviation is
$\mathrm{std}\left(\mathbf{x}\right)=\sqrt{\left(\mathrm{var}\left(\mathbf{x}\right)\right)}\simeq 440$
Because the standard deviation is considerably larger than the mean, the variance tells us that the mean is not very representative of the data.
## Exercise 3.1
Compute the variance and standard deviation of y = [3, 8, 2, 5, 5, 7], using both the formulas and the Matlab commands.
## Exercise 3.2
Suppose that in the situation of "Example 2.2" , there are 50 general exmployees instead of 16. Compute the mean and variance of the daily salary. Is the mean more or less representative of the data than it was in Example 2.2?
## Histograms
Although the mean, variance, and standard deviation provide information about the data, it is often useful to visualize the data. A histogram is a tool that allows you to visualize the proportion of numbers that fall within a given bin, or interval. To compute the histogram of a set of data, x , follow the algorithm below.
1. Choose the bin size $\Delta x$ . The bins are the intervals [0, $\Delta x$ ], ( $\Delta x$ , 2 $\Delta x$ ], (2 $\Delta x$ , 3 $\Delta x$ ], and so on.
2. For each bin, count the number of data points that lie within the bin.
3. Create a bar graph showing the number of data points within each bin.
## Example 4.1
Consider again the vector from "Example 2.1" , x = [1, 7, 2, 5, 9, 6]. Using a bin size $\Delta x$ = 2, there are 5 bins.
• Bin 1 = [0, 2] has 2 elements of x
• Bin 2 = (2, 4] has 0 elements of x
• Bin 3 = (4, 6] has 2 elements of x
• Bin 4 = (6, 8] has 1 element of x
• Bin 5 = (8, 10] has 1 element of x
In Matlab, you can plot the histogram of a vector x by typing hist(x). Matlab will automatically use 10 bins. If you'd like to specify the bin centers, type hist(x,c), where c is a vector of bin centers. The histogram of "Example 4.1" was generated by the Matlab command hist(x, [1, 3, 5, 7, 9]).
## Exercise 4.1
Plot the histogram of the vector y = [3, 8, 2, 5, 5, 7], both on paper and in Matlab.
## Exercise 4.2
Plot the histogram of the daily salaries from "Example 2.2" . For this example, does the histogram or the mean give you a better idea of what salary you would be making if you got the job?
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
Got questions? Join the online conversation and get instant answers!
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# math
Write all possible 2 digit numbers,sum of whose digits is 12.
1. 👍 0
2. 👎 0
3. 👁 78
1. 9+3
8+4
7+5
6+6
and reverse
1. 👍 0
2. 👎 0
posted by Damon
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# Added resistance in waves of a yacht (calculation)
A yacht that encounters waves while sailing will reduce its speed. In other words, the waves develop an additional resistance or drag that opposes the boat’s forward motion. This resistance is known as the added resistance in waves.
The two main mechanisms by which this additional resistance is developed are:
• the creation of additional waves by the boat as it heaves and pitches;
• the additional induced resistance created at the tip of the keel, rudder, and hull as the boat rolls.
Tests have found that the additional induced resistance due to the rolling motion is probably not very large compared to the first mechanism. Therefore only the creation of waves is usually considered.
When a boat is heaving and pitching in a seaway, it generates additional waves that are propagated in all directions from the boat. These waves are superimposed to the wave system of the wave-making resistance. The creation of these additional waves needs energy, which can only be obtained from the energy made available by the sails, thus reducing the energy available for maintaining/increasing the boat’s speed.
The characteristics of the encountered sea waves (often modelized with a mathematical tool called wave spectral density) and the geometrical characteristics and mass distribution of the boat (often modelized by a mathematical entity called Response Amplitude Operator, RAO) play fundamental roles in the added resistance in waves.
## How does it work?
The calculation requires LWL, Tc, BWL, c, CP, as well as the significant wave height, H1/3, and the modal period of the sea spectrum, T0.
The sea state is modeled by either the Bretschneider (open sea) or JONSWAP (limited fetch) spectra, considering the yacht is sailing in deep water. A custom value of the water density (default is 1026 kg/m3) can also be provided.
The results are calculated for all combinations of:
• Froude numbers 0.20, 0.25, 0.30, 0.35, 0.40 and 0.45;
• Angle of wave incidence, μ, 100, 120, 140, 160, and 180 degrees;
• Radii of gyration , Kyy, 0.20, 0.25, 0.30 times LWL;
• Wavelength/ship length ratio, λ / LWL, starting at 0.5, 0.6 till 4.0
Do you want to read more articles like this?
#### References:
Some of the links shown below are affiliate links and we may earn a commission at no additional cost to you:
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# Statistics is the study of the collection, organization, analysis, interpretation, and presentation of data.
[1][2] It deals with all aspects of this, including the planning of data collection in terms of the design of surveys and experiments.[1] A statistician is someone who is particularly well-versed in the ways of thinking necessary to successfully apply statistical analysis. Such people often gain experience through working in any of a wide number of fields. A discipline called mathematical statistics studies statistics mathematically. The word statistics, when referring to the scientific discipline, is singular, as in "Statistics is an art."[3] This should not be confused with the word statistic, referring to a quantity (such as mean or median) calculated from a set of data,[4] whose plural is statistics ("this statistic seems wrong" or "these statistics are misleading").
More probability density is found the closer one gets to the expected (mean) value in a normal distribution. Statistics used in standardized testing assessment are shown. The scales include standard deviations, cumulative percentages, percentile equivalents, Z-scores, T-scores, standard nines, and percentages in standard nines.
Contents
[hide]
1 Scope 2 History 3 Overview 4 Statistical methods o o 4.1 Experimental and observational studies 4.2 Levels of measurement
o o
## 4.3 Key terms used in statistics 4.4 Examples
5 Specialized disciplines 6 Statistical computing 7 Misuse 8 Statistics applied to mathematics or the arts 9 See also 10 References
Scope Some consider statistics a mathematical body of science that pertains to the collection, analysis, interpretation or explanation, and presentation of data,[5] while others consider it a branch of mathematics[6] concerned with collecting and interpreting data. Because of its empirical roots and its focus on applications, statistics is usually considered a distinct mathematical science rather than a branch of mathematics.[7][8] Much of statistics is non-mathematical: ensuring that data collection is undertaken in a way that produces valid conclusions; coding and archiving data so that information is retained and made useful for international comparisons of official statistics; reporting of results and summarised data (tables and graphs) in ways comprehensible to those who must use them; implementing procedures that ensure the privacy of census information. Statisticians improve data quality by developing specific experiment designs and survey samples. Statistics itself also provides tools for prediction and forecasting the use of data and statistical models. Statistics is applicable to a wide variety of academic disciplines, including natural and social sciences, government, and business. Statistical consultantscan help organizations and companies that don't have in-house expertise relevant to their particular questions. Statistical methods can summarize or describe a collection of data. This is called descriptive statistics. This is particularly useful in communicating the results of experiments and research. In addition, data patterns may be modeled in a way that accounts for randomness and uncertainty in the observations. These models can be used to draw inferences about the process or population under studya practice called inferential statistics. Inference is a vital element of scientific advance, since it provides a way to draw conclusions from data that are subject to random variation. To prove the propositions being investigated further, the conclusions are tested as well, as part of the scientific method. Descriptive statistics and analysis of the new data tend to provide more information as to the truth of the proposition.
"Applied statistics" comprises descriptive statistics and the application of inferential statistics.[9][verification needed] Theoretical statistics concerns both the logical arguments underlying justification of approaches to statistical inference, as well encompassing mathematical statistics. Mathematical statistics includes not only the manipulation ofprobability distributions necessary for deriving results related to methods of estimation and inference, but also various aspects of computational statistics and the design of experiments. Statistics is closely related to probability theory, with which it is often grouped. The difference is, roughly, that probability theory starts from the given parameters of a total population to deduce probabilities that pertain to samples. Statistical inference, however, moves in the opposite directioninductively inferring from samples to the parameters of a larger or total population. History Main articles: History of statistics and Founders of statistics Statistical methods date back at least to the 5th century BC. The earliest known writing on statistics appears in a 9th century book entitled Manuscript on Deciphering Cryptographic Messages, written by Al-Kindi. In this book, Al-Kindi provides a detailed description of how to use statistics and frequency analysis to decipher encrypted messages. This was the birth of both statistics and cryptanalysis, according to the Saudi engineer Ibrahim Al-Kadi.[10][11] The Nuova Cronica, a 14th century history of Florence by the Florentine banker and official Giovanni Villani, includes much statistical information on population, ordinances, commerce, education, and religious facilities, and has been described as the first introduction of statistics as a positive element in history.[12] Some scholars pinpoint the origin of statistics to 1663, with the publication of Natural and Political Observations upon the Bills of Mortality by John Graunt.[13] Early applications of statistical thinking revolved around the needs of states to base policy on demographic and economic data, hence its stat- etymology. The scope of the discipline of statistics broadened in the early 19th century to include the collection and analysis of data in general. Today, statistics is widely employed in government, business, and natural and social sciences. Its mathematical foundations were laid in the 17th century with the development of the probability theory by Blaise Pascal and Pierre de Fermat. Probability theory arose from the study of games of chance. The method of least squares was first described by Carl Friedrich Gauss around 1794. The use of modern computers has expedited large-scale statistical computation, and has also made possible new methods that are impractical to perform manually. Overview
In applying statistics to a scientific, industrial, or societal problem, it is necessary to begin with a population or process to be studied. Populations can be diverse topics such as "all persons living in a country" or "every atom composing a crystal". A population can also be composed of observations of a process at various times, with the data from each observation serving as a different member of the overall group. Data collected about this kind of "population" constitutes what is called a time series. For practical reasons, a chosen subset of the population called a sample is studiedas opposed to compiling data about the entire group (an operation called census). Once a sample that is representative of the population is determined, data is collected for the sample members in an observational or experimental setting. This data can then be subjected to statistical analysis, serving two related purposes: description and inference.
Descriptive statistics summarize the population data by describing what was observed in the sample numerically or graphically. Numerical descriptors include mean andstandard deviation for continuous data types (like heights or weights), while frequency and percentage are more useful in terms of describing categorical data (like race). Inferential statistics uses patterns in the sample data to draw inferences about the population represented, accounting for randomness. These inferences may take the form of: answering yes/no questions about the data (hypothesis testing), estimating numerical characteristics of the data (estimation), describing associations within the data (correlation) and modeling relationships within the data (for example, using regression analysis). Inference can extend to forecasting, prediction and estimation of unobserved values either in or associated with the population being studied; it can include extrapolation and interpolation of time series or spatial data, and can also include data mining.[14]
"... it is only the manipulation of uncertainty that interests us. We are not concerned with the matter that is uncertain. Thus we do not study the mechanism of rain; only whether it will rain." Dennis Lindley, 2000
[15]
The concept of correlation is particularly noteworthy for the potential confusion it can cause. Statistical analysis of a data set often reveals that two variables (properties) of the population under consideration tend to vary together, as if they were connected. For example, a study of annual income that also looks at age of death might find that poor people tend to have shorter lives than affluent people. The two variables are said to be correlated; however, they may or may not be the cause of one another. The correlation phenomena could be caused by a third, previously unconsidered phenomenon, called a lurking variable or confounding variable. For this reason, there is no way to immediately infer the existence of a causal relationship between the two variables. (See Correlation does not imply causation.)
To use a sample as a guide to an entire population, it is important that it truly represent the overall population. Representative sampling assures that inferences and conclusions can safely extend from the sample to the population as a whole. A major problem lies in determining the extent that the sample chosen is actually representative. Statistics offers methods to estimate and correct for any random trending within the sample and data collection procedures. There are also methods of experimental design for experiments that can lessen these issues at the outset of a study, strengthening its capability to discern truths about the population. Randomness is studied using the mathematical discipline of probability theory. Probability is used in "mathematical statistics" (alternatively, "statistical theory") to study thesampling distributions of sample statistics and, more generally, the properties of statistical procedures. The use of any statistical method is valid when the system or population under consideration satisfies the assumptions of the method. Misuse of statistics can produce subtle, but serious errors in description and interpretation subtle in the sense that even experienced professionals make such errors, and serious in the sense that they can lead to devastating decision errors. For instance, social policy, medical practice, and the reliability of structures like bridges all rely on the proper use of statistics. See below for further discussion. Even when statistical techniques are correctly applied, the results can be difficult to interpret for those lacking expertise. The statistical significance of a trend in the datawhich measures the extent to which a trend could be caused by random variation in the samplemay or may not agree with an intuitive sense of its significance. The set of basic statistical skills (and skepticism) that people need to deal with information in their everyday lives properly is referred to as statistical literacy. Statistical
methods
and observational studies
Experimental
A common goal for a statistical research project is to investigate causality, and in particular to draw a conclusion on the effect of changes in the values of predictors or independent variables on dependent variables or response. There are two major types of causal statistical studies: experimental studies and observational studies. In both types of studies, the effect of differences of an independent variable (or variables) on the behavior of the dependent variable are observed. The difference between the two types lies in how the study is actually conducted. Each can be very effective. An experimental study involves taking measurements of the system under study, manipulating the system, and then taking additional measurements using the same procedure to determine if the manipulation has modified the values of the measurements. In contrast, an observational study does not involve experimental manipulation. Instead, data are gathered and correlations between predictors and response are investigated. Experiments
The basic steps of a statistical experiment are: 1. Planning the research, including finding the number of replicates of the study, using the following information: preliminary estimates regarding the size of treatment effects,alternative hypotheses, and the estimated experimental variability. Consideration of the selection of experimental subjects and the ethics of research is necessary. Statisticians recommend that experiments compare (at least) one new treatment with a standard treatment or control, to allow an unbiased estimate of the difference in treatment effects. 2. Design of experiments, using blocking to reduce the influence of confounding variables, and randomized assignment of treatments to subjects to allow unbiased estimates of treatment effects and experimental error. At this stage, the experimenters and statisticians write the experimental protocol that shall guide the performance of the experiment and that specifies the primary analysis of the experimental data. 3. Performing the experiment following the experimental protocol and analyzing the data following the experimental protocol. 4. Further examining the data set in secondary analyses, to suggest new hypotheses for future study. 5. Documenting and presenting the results of the study. Experiments on human behavior have special concerns. The famous Hawthorne study examined changes to the working environment at the Hawthorne plant of the Western Electric Company. The researchers were interested in determining whether increased illumination would increase the productivity of the assembly line workers. The researchers first measured the productivity in the plant, then modified the illumination in an area of the plant and checked if the changes in illumination affected productivity. It turned out that productivity indeed improved (under the experimental conditions). However, the study is heavily criticized today for errors in experimental procedures, specifically for the lack of acontrol group and blindness. The Hawthorne effect refers to finding that an outcome (in this case, worker productivity) changed due to observation itself. Those in the Hawthorne study became more productive not because the lighting was changed but because they were being observed.[citation needed] Observational study An example of an observational study is one that explores the correlation between smoking and lung cancer. This type of study typically uses a survey to collect observations about the area of interest and then performs statistical analysis. In this case, the researchers would collect observations of both smokers and non-smokers, perhaps through a case-control study, and then look for the number of cases of lung cancer in each group. Levels
of measurement
## Main article: Levels of measurement
There are four main levels of measurement used in statistics: nominal, ordinal, interval, and ratio.[16] Each of these have different degrees of usefulness in statistical research. Ratio measurements have both a meaningful zero value and the distances between different measurements defined; they provide the greatest flexibility in statistical methods that can be used for analyzing the data.[citation needed] Interval measurements have meaningful distances between measurements defined, but the zero value is arbitrary (as in the case withlongitude and temperature measurements in Celsius or Fahrenheit). Ordinal measurements have imprecise differences between consecutive values, but have a meaningful order to those values. Nominal measurements have no meaningful rank order among values. Because variables conforming only to nominal or ordinal measurements cannot be reasonably measured numerically, sometimes they are grouped together as categorical variables, whereas ratio and interval measurements are grouped together as quantitative variables, which can be either discrete or continuous, due to their numerical nature. Key
## terms used in statistics
Null hypothesis Interpretation of statistical information can often involve the development of a null hypothesis in that the assumption is that whatever is proposed as a cause has no effect on the variable being measured. The best illustration for a novice is the predicament encountered by a jury trial. The null hypothesis, H0, asserts that the defendant is innocent, whereas the alternative hypothesis, H1, asserts that the defendant is guilty. The indictment comes because of suspicion of the guilt. The H0 (status quo) stands in opposition to H1 and is maintained unless H1 is supported by evidence"beyond a reasonable doubt". However,"failure to reject H0" in this case does not imply innocence, but merely that the evidence was insufficient to convict. So the jury does not necessarily accept H0 but fails to reject H0. While one can not "prove" a null hypothesis one can test how close it is to being true with a power test, which tests for type II errors. Error Working from a null hypothesis two basic forms of error are recognized:
Type I errors where the null hypothesis is falsely rejected giving a "false positive". Type II errors where the null hypothesis fails to be rejected and an actual difference between populations is missed giving a false negative.
Error also refers to the extent to which individual observations in a sample differ from a central value, such as the sample or population mean. Many statistical methods seek to minimize the mean-squared error, and these are called "methods of least squares."
Measurement processes that generate statistical data are also subject to error. Many of these errors are classified as random (noise) or systematic (bias), but other important types of errors (e.g., blunder, such as when an analyst reports incorrect units) can also be important. Interval estimation Main article: Interval estimation Most studies only sample part of a population , so results don't fully represent the whole population. Any estimates obtained from the sample only approximate the population value.Confidence intervals allow statisticians to express how closely the sample estimate matches the true value in the whole population. Often they are expressed as 95% confidence intervals. Formally, a 95% confidence interval for a value is a range where, if the sampling and analysis were repeated under the same conditions (yielding a different dataset), the interval would include the true (population) value 95% of the time. This does not imply that the probability that the true value is in the confidence interval is 95%. From the frequentistperspective, such a claim does not even make sense, as the true value is not a random variable. Either the true value is or is not within the given interval. However, it is true that, before any data are sampled and given a plan for how to construct the confidence interval, the probability is 95% that the yet-to-be-calculated interval will cover the true value: at this point, the limits of the interval are yet-to-be-observed random variables. One approach that does yield an interval that can be interpreted as having a given probability of containing the true value is to use a credible interval from Bayesian statistics: this approach depends on a different way of interpreting what is meant by "probability", that is as a Bayesian probability.
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10827
10,827 (ten thousand eight hundred twenty-seven) is an odd five-digits composite number following 10826 and preceding 10828. In scientific notation, it is written as 1.0827 × 104. The sum of its digits is 18. It has a total of 4 prime factors and 8 positive divisors. There are 7,200 positive integers (up to 10827) that are relatively prime to 10827.
Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 18
• Digital Root 9
Name
Short name 10 thousand 827 ten thousand eight hundred twenty-seven
Notation
Scientific notation 1.0827 × 104 10.827 × 103
Prime Factorization of 10827
Prime Factorization 33 × 401
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 1203 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 10,827 is 33 × 401. Since it has a total of 4 prime factors, 10,827 is a composite number.
Divisors of 10827
1, 3, 9, 27, 401, 1203, 3609, 10827
8 divisors
Even divisors 0 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 16080 Sum of all the positive divisors of n s(n) 5253 Sum of the proper positive divisors of n A(n) 2010 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 104.053 Returns the nth root of the product of n divisors H(n) 5.38657 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 10,827 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 10,827) is 16,080, the average is 2,010.
Other Arithmetic Functions (n = 10827)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 7200 Total number of positive integers not greater than n that are coprime to n λ(n) 3600 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1319 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 7,200 positive integers (less than 10,827) that are coprime with 10,827. And there are approximately 1,319 prime numbers less than or equal to 10,827.
Divisibility of 10827
m n mod m 2 3 4 5 6 7 8 9 1 0 3 2 3 5 3 0
The number 10,827 is divisible by 3 and 9.
• Arithmetic
• Deficient
• Polite
• Frugal
Base conversion (10827)
Base System Value
2 Binary 10101001001011
3 Ternary 112212000
4 Quaternary 2221023
5 Quinary 321302
6 Senary 122043
8 Octal 25113
10 Decimal 10827
12 Duodecimal 6323
20 Vigesimal 1717
36 Base36 8cr
Basic calculations (n = 10827)
Multiplication
n×i
n×2 21654 32481 43308 54135
Division
ni
n⁄2 5413.5 3609 2706.75 2165.4
Exponentiation
ni
n2 117223929 1269183479283 13741449530197041 148778674063443362907
Nth Root
i√n
2√n 104.053 22.1226 10.2006 6.41064
10827 as geometric shapes
Circle
Diameter 21654 68028 3.6827e+08
Sphere
Volume 5.31634e+12 1.47308e+09 68028
Square
Length = n
Perimeter 43308 1.17224e+08 15311.7
Cube
Length = n
Surface area 7.03344e+08 1.26918e+12 18752.9
Equilateral Triangle
Length = n
Perimeter 32481 5.07595e+07 9376.46
Triangular Pyramid
Length = n
Surface area 2.03038e+08 1.49575e+11 8840.21
Cryptographic Hash Functions
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# Question on Heisenberg uncertainty principle
• Nivlac2425
In summary, in a diffraction setup where particles with a de Broglie wavelength of 633 nm pass through a single slit of width 0.2 mm, the Heisenberg uncertainty principle can be used to estimate the minimum range of angles over which the particle distribution spreads out. This range is typically considered to extend from -π/2 to +π/2, with a central value of 0 radians. However, for a rough estimate, half the slit width can also be used as the initial position uncertainty. The resulting range can then be compared to the value obtained from the single slit diffraction equation, which gives 1/2 of the peak value.
Nivlac2425
## Homework Statement
Particles pass through a single slit of width 0.2 mm in a diffraction setup. The de Broglie wavelength of each particle is 633 nm. After the particles pass through the slit, they spread out over a range of angles. Use the Heisenberg uncertainty principle to determine the minimum range of angles.
## Homework Equations
Heisenberg uncertainty principle
sin $$\theta$$= $$\lambda$$/W
## The Attempt at a Solution
I'm not sure exactly what a minimum range of angles mean. Is it all the possible angles through which the electrons can follow?
I believe I'm stuck on how to use the uncertainty principle to find an angle. I know that the equation for diffraction through a single slit might be useful.
Well, the wording of these sort of statements can be tricky. The particle distribution will actually extend through the full -π/2 to +π/2 range. However, the distribution becomes negligibly small beyond some angle.
Strictly speaking, "uncertainty" in physics usually means an rms deviation from a central value (in this case, 0 radians). Since, in this case, we are going after a ballpark figure, it's probably okay to use half the slit width as the initial position uncertainty.
The idea is to use the Heisenberg relation as an estimate, rather than the single slit diffraction equation. Of course, you are welcome to compare the two results afterward--in which case one might use the angle where the diffraction equation gives 1/2 of the peak value.
I see, so the distribution includes the negative angles.
Thank you for a clear explanation!
## 1. What is the Heisenberg uncertainty principle?
The Heisenberg uncertainty principle states that it is impossible to know both the exact position and momentum of a particle at the same time. This means that the more accurately we measure the position of a particle, the less accurately we can know its momentum, and vice versa.
## 2. Who discovered the Heisenberg uncertainty principle?
The Heisenberg uncertainty principle was first proposed by German physicist Werner Heisenberg in 1927 as part of his uncertainty principle, which also includes the uncertainty of energy and time.
## 3. What is the significance of the Heisenberg uncertainty principle in quantum mechanics?
The Heisenberg uncertainty principle is a fundamental principle in quantum mechanics, which is the branch of physics that studies the behavior of subatomic particles. It tells us that there are inherent limitations to our ability to measure and predict the behavior of particles at a very small scale.
## 4. How does the Heisenberg uncertainty principle affect our daily lives?
The Heisenberg uncertainty principle may seem like it only applies to the microscopic world, but it actually has implications in our daily lives. For example, it is the reason why we cannot accurately predict the exact location and speed of a moving car at the same time.
## 5. Are there any exceptions to the Heisenberg uncertainty principle?
The Heisenberg uncertainty principle is a fundamental principle in quantum mechanics and has been proven to hold true in countless experiments. However, there are theories that suggest it may not apply in certain extreme conditions, such as in black holes or during the early moments of the Big Bang.
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## Introductory Algebra for College Students (7th Edition)
$12x^4-3x^3+15x^2$
Using $a(b+c)=ab+ac$ or the Distributive Property, the given expression, $-3x^2(-4x^2+x-5) ,$ is equivalent to \begin{array}{l}\require{cancel} -3x^2(-4x^2)-3x^2(x)-3x^2(-5) \\\\= 12x^4-3x^3+15x^2 .\end{array}
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# A Transportation Graphing Lesson
By Laurie Patsalides
Here is a fun graphing lesson or activity for students to do during a transportation unit. Students will learn how to graph and read data based upon how they came to school today. There are also samples of data collection for the graphing lesson included.
## Graphing Activities
A unit on transportation is the perfect opportunity for this graphing lesson and activities. Students will learn how to graph information from an everyday experience, traveling to school. In the process reinforce the different ways people travel.
## Transportation Graphing
Objective: Students will be able to tell about the different kinds of transportation that they use daily, how to graph information and how to read information on a graph.
Prior Knowledge: We have been spending some time getting to know one another. Now we will learn how each person in our class came to school each day.
Teach: We can learn information by collecting data on a graph. A graph is a way to gather information. Show today we want to gather information about how we came to school today. Here are three ways you may have came to school today, by bus, car or by walking (note- depending on where you live, your students may take a subway/train to school, then you will need to have that option on your graph.)
Show a sample graph. Graphing is a way to show data. You can "read" a graph, by studying the information and counting the numbers that a graph provides to you. Show example and "read' the information on your sample graph. We can compare numbers and learn information from a graph (provide examples from your sample graph).
Procedure: Explain how to complete the information on the transportation graph. Ask students to raise their hand if they took a bus to school today. Write each name in a box underneath the bus. Ask students to raise their hand if they walked to school today and chart in the same manner. Ask students if they drove in a car to school today and chart in the same manner.
This can be modified in the following ways: Students could be given a post-it and write their name on it. Then when called upon they can put their post-it under in the appropriate column.
Or younger students who may not independently write their name yet, can create a bar pictograph, by asking them how they came to school and giving them a picture of a walker, bus, or car to color. Students add their mode of transportation to the correct bar on the graph. This could also easily be made into a color-coded graph (given a color choice, for example, bus=yellow, walk=brown, and car=blue, students color in the box underneath the mode of transportation that they took to school).
When each student has had an opportunity to provide their information, read the information and compare data. Did more students walk or take a bus? What is the least way that students came to school? What are the most frequent ways to travel to school? How many more students take a car than walk? (See sample of possible questions to ask/data collection below)
Assessment: As this is a group activity, assessment will be informal. Can students put their information in the correct column independently. Can the class help you to complete the data collection form based on question and answer.
If you have older students that can read and fill in the data collection form independently, let them do so and formally assess. Do this only if the class is ready for it (as this is a beginning of the school year lesson, they may not be) or save this graph and refer to it later when the students have had more lessons on graphing and assess at a later time.
## Sample of Data Collection for Graphs
Image credits: Laurie Patsalides; for personal use only
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# 1990 AJHSME Problems/Problem 9
## Problem
The grading scale shown is used at Jones Junior High. The fifteen scores in Mr. Freeman's class were: $$\begin{tabular}[t]{lllllllll} 89, & 72, & 54, & 97, & 77, & 92, & 85, & 74, & 75, \\ 63, & 84, & 78, & 71, & 80, & 90. & & & \\ \end{tabular}$$
In Mr. Freeman's class, what percent of the students received a grade of C?
$$\boxed{\begin{tabular}[t]{cc} A: & 93 - 100 \\ B: & 85 - 92 \\ C: & 75 - 84 \\ D: & 70 - 74 \\ F: & 0 - 69 \end{tabular}}$$
$\text{(A)}\ 20\% \qquad \text{(B)}\ 25\% \qquad \text{(C)}\ 30\% \qquad \text{(D)}\ 33\frac{1}{3}\% \qquad \text{(E)}\ 40\%$
## Solution
We just count to find that there are $5$ students in the $\text{C}$ range.
There are $15$ total, so the percentage is $\frac{5}{15}=33\frac{1}{3}\%$ $\rightarrow \boxed{\text{D}}$.
1990 AJHSME (Problems • Answer Key • Resources) Preceded byProblem 8 Followed byProblem 10 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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375. Guess Number Higher or Lower II
Description
We are playing the Guessing Game. The game will work as follows:
1. I pick a number between 1 and n.
2. You guess a number.
3. If you guess the right number, you win the game.
4. If you guess the wrong number, then I will tell you whether the number I picked is higher or lower, and you will continue guessing.
5. Every time you guess a wrong number x, you will pay x dollars. If you run out of money, you lose the game.
Given a particular n, return the minimum amount of money you need to guarantee a win regardless of what number I pick.
Example 1:
Input: n = 10
Output: 16
Explanation: The winning strategy is as follows:
- The range is [1,10]. Guess 7.
- If this is my number, your total is 0. Otherwise, you pay 7.
- If my number is higher, the range is [8,10]. Guess 9.
- If this is my number, your total is 7. Otherwise, you pay 9.
- If my number is higher, it must be 10. Guess 10. Your total is 7 + 9 = 16.
- If my number is lower, it must be 8. Guess 8. Your total is 7 + 9 = 16.
- If my number is lower, the range is [1,6]. Guess 3.
- If this is my number, your total is 7. Otherwise, you pay 3.
- If my number is higher, the range is [4,6]. Guess 5.
- If this is my number, your total is 7 + 3 = 10. Otherwise, you pay 5.
- If my number is higher, it must be 6. Guess 6. Your total is 7 + 3 + 5 = 15.
- If my number is lower, it must be 4. Guess 4. Your total is 7 + 3 + 5 = 15.
- If my number is lower, the range is [1,2]. Guess 1.
- If this is my number, your total is 7 + 3 = 10. Otherwise, you pay 1.
- If my number is higher, it must be 2. Guess 2. Your total is 7 + 3 + 1 = 11.
The worst case in all these scenarios is that you pay 16. Hence, you only need 16 to guarantee a win.
Example 2:
Input: n = 1
Output: 0
Explanation: There is only one possible number, so you can guess 1 and not have to pay anything.
Example 3:
Input: n = 2
Output: 1
Explanation: There are two possible numbers, 1 and 2.
- Guess 1.
- If this is my number, your total is 0. Otherwise, you pay 1.
- If my number is higher, it must be 2. Guess 2. Your total is 1.
The worst case is that you pay 1.
Constraints:
• 1 <= n <= 200
Solutions
• class Solution {
public int getMoneyAmount(int n) {
int[][] dp = new int[n + 10][n + 10];
for (int l = 2; l <= n; ++l) {
for (int i = 1; i + l - 1 <= n; ++i) {
int j = i + l - 1;
dp[i][j] = Integer.MAX_VALUE;
for (int k = i; k <= j; ++k) {
int t = Math.max(dp[i][k - 1], dp[k + 1][j]) + k;
dp[i][j] = Math.min(dp[i][j], t);
}
}
}
return dp[1][n];
}
}
• class Solution {
public:
int getMoneyAmount(int n) {
vector<vector<int>> dp(n + 10, vector<int>(n + 10));
for (int l = 2; l <= n; ++l) {
for (int i = 1; i + l - 1 <= n; ++i) {
int j = i + l - 1;
dp[i][j] = INT_MAX;
for (int k = i; k <= j; ++k) {
int t = max(dp[i][k - 1], dp[k + 1][j]) + k;
dp[i][j] = min(dp[i][j], t);
}
}
}
return dp[1][n];
}
};
• class Solution:
def getMoneyAmount(self, n: int) -> int:
dp = [[0] * (n + 10) for _ in range(n + 10)]
for l in range(2, n + 1):
for i in range(1, n - l + 2):
j = i + l - 1
dp[i][j] = inf
for k in range(i, j + 1):
t = max(dp[i][k - 1], dp[k + 1][j]) + k
dp[i][j] = min(dp[i][j], t)
return dp[1][n]
• func getMoneyAmount(n int) int {
dp := make([][]int, n+10)
for i := 0; i < len(dp); i++ {
dp[i] = make([]int, n+10)
}
for l := 2; l <= n; l++ {
for i := 1; i+l-1 <= n; i++ {
j := i + l - 1
dp[i][j] = math.MaxInt32
for k := i; k <= j; k++ {
t := max(dp[i][k-1], dp[k+1][j]) + k
dp[i][j] = min(dp[i][j], t)
}
}
}
return dp[1][n]
}
• function getMoneyAmount(n: number): number {
const f: number[][] = Array.from({ length: n + 1 }, () => Array(n + 1).fill(0));
for (let i = n - 1; i; --i) {
for (let j = i + 1; j <= n; ++j) {
f[i][j] = j + f[i][j - 1];
for (let k = i; k < j; ++k) {
f[i][j] = Math.min(f[i][j], k + Math.max(f[i][k - 1], f[k + 1][j]));
}
}
}
return f[1][n];
}
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## Interest rate compounded half-yearly
6 Nov 2015 10000 at 12% rate of interest for 1 year, compounded half-yearly. Solution: Amount with CI = 10000 [1+ (12/2 * 100)]2 = Rs. 11236. Therefore, CI 10 Oct 2019 Given,Principal = Rs 10000Here rate is compounded half-yearly,So, rate of interest = R = 10/2 %= 5%Time = 2 yearsn = number of half yearsn An interest rate compounded more than once a year is called the nominal interest rate of 8% p.a. compounded half-yearly is actually an effective rate of 8, 16%
13 Dec 2019 Calculate the amount and Compound interest if the interest is compounded half yearly. (a). principal = ₹ 2560 rate = 12 1/2 time = 1 year A bank offers 5% compound interest calculated on half-yearly basis. A customer R = rate. n = no.of years. But in the problem we are dealing with half year. Interest is compounded half-yearly, therefore, Amount = P ( 1 + (R/2) /100 )2n - - - - - - - - - [Interest compounded Half-yearly] Given : Principal = Rs. 20000, Rate Compound Interest when Compounded Half Yearly Example 2: Find the compound interest on Rs 8000 for 3/2 years at 10% per annum, interest is payable half-yearly. Solution: Rate of interest = 10% per annum = 5% per half –year. Word problems on compound interest when interest is compounded half-yearly: 1. Find the amount and the compound interest on $8,000 at 10 % per annum for 1$$\frac{1}{2}$$ years if the interest is compounded half-yearly. To find compound interest when interest is compounded half yearly, we use the following formula. n = number of years. Examples : 1) Compute the compound interest on$12,000 for 2 years ate 20% p.a. when compounded half-yearly. Solution : Here, P = $12,000, R = 20% and n = 2 years. Compound Interest (CI) is the addition of Interest to the Initial principal value and also the accumulated interest of previous periods of a loan or any deposit. Use this online compound interest calculator to calculate C.I compounded for annually, half-yearly, quarterly. ## Compound interest is the addition of interest to the principal sum of a loan or deposit, or in other The frequency could be yearly, half-yearly, quarterly, monthly, weekly, daily, The yearly compounded rate is higher than the disclosed rate. Compound Interest when Compounded Half Yearly Example 2: Find the compound interest on Rs 8000 for 3/2 years at 10% per annum, interest is payable half-yearly. Solution: Rate of interest = 10% per annum = 5% per half –year. Word problems on compound interest when interest is compounded half-yearly: 1. Find the amount and the compound interest on$ 8,000 at 10 % per annum for 1$$\frac{1}{2}$$ years if the interest is compounded half-yearly. To find compound interest when interest is compounded half yearly, we use the following formula. n = number of years. Examples : 1) Compute the compound interest on $12,000 for 2 years ate 20% p.a. when compounded half-yearly. Solution : Here, P =$12,000, R = 20% and n = 2 years. Compound Interest (CI) is the addition of Interest to the Initial principal value and also the accumulated interest of previous periods of a loan or any deposit. Use this online compound interest calculator to calculate C.I compounded for annually, half-yearly, quarterly.
### 2. to calculate how much compound interest payable based on the half-yearly rate of interest over a period of time with either monthly, quarterly, half-yearly or
Calculation of Compound Interest When the Rate is Compounded half Yearly. Let us calculate the compound interest on a principal, P kept for 1 year at Find out how much compound interest you could earn on your savings, and daily compounding; monthly compounding; quarterly compounding; half yearly and Multiply the principal amount by one plus the annual interest rate to the power Question 1: Calculate the amount and compound interest on Rate has been halved because interest is compounded half yearly, time has been doubled for the 13 Dec 2019 Calculate the amount and Compound interest if the interest is compounded half yearly. (a). principal = ₹ 2560 rate = 12 1/2 time = 1 year A bank offers 5% compound interest calculated on half-yearly basis. A customer R = rate. n = no.of years. But in the problem we are dealing with half year. Interest is compounded half-yearly, therefore, Amount = P ( 1 + (R/2) /100 )2n - - - - - - - - - [Interest compounded Half-yearly] Given : Principal = Rs. 20000, Rate
### Compound Interest (CI) is the addition of Interest to the Initial principal value and also the accumulated interest of previous periods of a loan or any deposit. Use this online compound interest calculator to calculate C.I compounded for annually, half-yearly, quarterly.
Compound Interest (CI) is the addition of Interest to the Initial principal value and also the accumulated interest of previous periods of a loan or any deposit. Use this online compound interest calculator to calculate C.I compounded for annually, half-yearly, quarterly. Compound Interest Calculation from simple Interest where Interest is compounded half yearly. If the rate of interest is R% per annum and the interest is compounded half-yearly, then the rate of interest will be R/2% per half year. Compound interest (CI) calculator - formulas & solved example problems to calculate the total interest payable on a given principal sum at a certain rate of interest over a period of time with either one of monthly, quarterly, half-yearly or yearly compounding frequency, in different world currencies such as USD, GBP, AUD, JPY, INR, NZD, CHF, RMB etc. When compounding of interest takes place, the effective annual rate becomes higher than the overall interest rate. The more times the interest is compounded within the year, the higher the effective annual rate will be. More information on effective annual interest rate can be found in this article from Investopedia. To convert a yearly interest rate for annually compounding loans, you can simply divide the annual interest rate into 12 equal parts. So, for example, if you had a loan with a 12 percent interest rate attached to it, you can simply divide 12 percent by 12, or the decimal formatted 0.12 by 12, in order to determine that 1 percent interest is essentially being added on a monthly basis.
## Compound Interest (CI) is the addition of Interest to the Initial principal value and also the accumulated interest of previous periods of a loan or any deposit. Use this online compound interest calculator to calculate C.I compounded for annually, half-yearly, quarterly.
PREVIOUS A sum of Rs. 25000 was given as loan on compound interest for 3 years compounded annually at 5% per annum during the first year, 6% per annum during the second year and 8% per annum during the third year. Fixed Deposits are a great way to invest for those who rate safety higher than returns. This Fixed Deposit (FD) Calculator helps you find out how much interest you can earn on an FD and the value of your invesment (Principal) on Maturity when compounding of interest is done on a monthly, quarterly, half-yearly or yearly basis. Compound Interest is calculated on the initial payment and also on the interest of previous periods. Example: Suppose you give \$100 to a bank which pays you 10% compound interest at the end of every year. After one year you will have \$100 + 10% = \$110, and after two years you will have \$110 + 10% = \$121. Our online tools will provide quick answers to your calculation and conversion needs. On this page, you can calculate compound interest with daily, weekly, monthly, quarterly, half-yearly, and yearly compounding. You can also use this calculator to solve for compounded rate of return, time period and principal. Usually, the compounding is done quarterly, half-yearly and annually which means a number of compounding per year of 4, 2 and 1 respectively. Step 3: Finally, the formula for effective interest rate can be derived by using the stated rate of interest (step 1) and a number of compounding periods per year (step 2) as shown below. Let us see calculation difference for simple interest formula and compound interest formula. Suppose a person wants to start a yearly recurring deposit of$500 for a period of 10 years for the interest rate of 5%. Then he calculates the same and gets the below values. Interest rates on Deposits upto ` 2 Crore Rate of Interest (p.a.) Period Monthly Income Plan Quarterly Option Half-Yearly Option Annual Income Plan Cumulative Option* * For cumulative option, Interest is compounded annually. Compound interest occurs when interest is added to the original deposit – or principal – which results in interest earning interest. Financial institutions often offer compound interest on deposits, compounding on a regular basis – usually monthly or annually. The compounding of interest grows your investment without any further deposits
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# shortest paths definitions single source algorithms –bellman ford –dag shortest path algorithm...
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• Slide 1
• Shortest Paths Definitions Single Source Algorithms Bellman Ford DAG shortest path algorithm Dijkstra All Pairs Algorithms Using Single Source Algorithms Matrix multiplication Floyd-Warshall Both of above use adjacency matrix representation and dynamic programming Johnsons algorithm Uses adjacency list representation
• Slide 2
• Single Source Definition Input Weighted, connected directed graph G=(V,E) Weight (length) function w on each edge e in E Source node s in V Task Compute a shortest path from s to all nodes in V
• Slide 3
• All Pairs Definition Input Weighted, connected directed graph G=(V,E) Weight (length) function w on each edge e in E We will typically assume w is represented as a matrix Task Compute a shortest path from all nodes in V to all nodes in V
• Slide 4
• Comments If edges are not weighted, then BFS works for single source problem Optimal substructure A shortest path between s and t contains other shortest paths within it No known algorithm is better at finding a shortest path from s to a specific destination node t in G than finding the shortest path from s to all nodes in V
• Slide 5
• Negative weight edges Negative weight edges can be allowed as long as there are no negative weight cycles If there are negative weight cycles, then there cannot be a shortest path from s to any node t (why?) If we disallow negative weight cycles, then there always is a shortest path that contains no cycles
• Slide 6
• Relaxation technique For each vertex v, we maintain an upper bound d[v] on the length of shortest path from s to v d[v] initialized to infinity Relaxing an edge (u,v) Can we shorten the path to v by going through u? If d[v] > d[u] + w(u,v), d[v] = d[u] + w(u,v) This can be done in O(1) time
• Slide 7
• Bellman-Ford Algorithm Bellman-Ford (G, w, s) Initialize-Single- Source(G,s) for (i=1 to V-1) for each edge (u,v) in E relax(u,v); for each edge (u,v) in E if d[v] > d[u] + w(u,v) return NEGATIVE WEIGHT CYCLE 3 5 7 -5 3 2 1 s 3 5 -5 3 2 1 s G1G1 G2G2
• Slide 8
• Running Time for (i=1 to V-1) for each edge (u,v) in E relax(u,v); The above takes (V-1)O(E) = O(VE) time for each edge (u,v) in E if d[v] > d[u] + w(u,v) return NEGATIVE WEIGHT CYCLE The above takes O(E) time
• Slide 9
• Proof of Correctness Theorem: If there is a shortest path from s to any node v, then d[v] will have this weight at end of Bellman-Ford algorithm Theorem restated: Define links[v] to be the minimum number of edges on a shortest path from s to v After i iterations of the Bellman-Ford for loop, nodes v with links[v] i will have their d[v] value set correctly Prove this by induction on links[v] Base case: links[v] = 0 which means v is s d[s] is initialized to 0 which is correct unless there is a negative weight cycle from s to s in which case there is no shortest path from s to s. Induction hypothesis: After k iterations for k 0, d[v] must be correct if links[v] k Inductive step: Show after k+1 iterations, d[v] must be correct if links[v] k+1 If links[v] k, then by IH, d[v] is correctly set after kth iteration If links[v] = k+1, let p = (e 1, e 2, , e k+1 ) = (v 0, v 1, v 2, , v k+1 ) be a shortest path from s to v s = v 0, v = v k+1, e i = (v i-1, v i ) In order for links[v] = k+1, then links[v k ] = k. By the inductive hypothesis, d[v k ] will be correctly set after the kth iteration. During the k+1st iteration, we relax all edges including edge e k+1 = (v k,v) Thus, at end of the k+1 st iteration, d[v] will be correct
• Slide 10
• Negative weight cycle for each edge (u,v) in E if d[v] > d[u] + w(u,v) return NEGATIVE WEIGHT CYCLE If no neg weight cycle, d[v] d[u] + w(u,v) for all (u,v) If there is a negative weight cycle C, for some edge (u,v) on C, it must be the case that d[v] > d[u] + w(u,v). Suppose this is not true for some neg. weight cycle C sum these (d[v] d[u] + w(u,v)) all the way around C We end up with v in C d[v] ( u in C d[u]) + weight(C) This is impossible unless weight(C) = 0 But weight(C) is negative, so this cannot happen Thus for some (u,v) on C, d[v] > d[u] + w(u,v)
• Slide 11
• DAG shortest path algorithm DAG-SP (G, w, s) Initialize-Single- Source(G,s) Topologically sort vertices in G for each vertex u, taken in sorted order for each edge (u,v) in E relax(u,v); 3 5 -4 3 8 1 s 5 4
• Slide 12
• Running time Improvement O(V+E) for the topological sorting We only do 1 relaxation for each edge: O(E) time for each vertex u, taken in sorted order for each edge (u,v) in E relax(u,v); Overall running time: O(V+E)
• Slide 13
• Proof of Correctness If there is a shortest path from s to any node v, then d[v] will have this weight at end Let p = (e 1, e 2, , e k ) = (v 1, v 2, v 3, , v k+1 ) be a shortest path from s to v s = v 1, v = v k+1, e i = (v i, v i+1 ) Since we sort edges in topological order, we will process node v i (and edge e i ) before processing later edges in the path.
• Slide 14
• Dijkstras Algorithm Dijkstra (G, w, s) /* Assumption: all edge weights non-negative */ Initialize-Single- Source(G,s) Completed = {}; ToBeCompleted = V; While ToBeCompleted is not empty u =EXTRACT- MIN(ToBeCompleted); Completed += {u}; for each edge (u,v) relax(u,v); 15 5 7 4 3 2 1 s
• Slide 15
• Running Time Analysis While ToBeCompleted is not empty u =EXTRACT-MIN(ToBeCompleted); Completed += {u}; for each edge (u,v) relax(u,v); Each edge relaxed at most once: O(E) Need to decrease-key potentially once per edge Need to extract-min once per node The nodes d[v] is then complete
• Slide 16
• Running Time Analysis contd Priority Queue operations O(E) decrease key operations O(V) extract-min operations Three implementations of priority queues Array: O(V 2 ) time decrease-key is O(1) and extract-min is O(V) Binary heap: O(E log V) time assuming E V decrease-key and extract-min are O(log V) Fibonacci heap: O(V log V + E) time decrease-key is O(1) amortized and extract-min is O(log V)
• Slide 17
• Compare Dijkstras algorithm to Prims algorithm for MST Dijsktra Priority Queue operations O(E) decrease key operations O(V) extract-min operations Prim Priority Queue operations O(E) decrease-key operations O(V) extract-min operations Is this a coincidence or is there something more here?
• Slide 18
• Proof of Correctness Assume that Dijkstras algorithm fails to compute length of all shortest paths from s Let v be the first node whose shortest path length is computed incorrectly Let S be the set of nodes whose shortest paths were computed correctly by Dijkstra prior to adding v to the processed set of nodes. Dijkstras algorithm has used the shortest path from s to v using only nodes in S when it added v to S. The shortest path to v must include one node not in S Let u be the first such node. suv Only nodes in S
• Slide 19
• Proof of Correctness The length of the shortest path to u must be at least that of the length of the path computed to v. Why? The length of the path from u to v must be < 0. Why? No path can have negative length since all edge weights are non-negative, and thus we have a contradiction. suv Only nodes in S
• Slide 20
• Computing paths (not just distance) Maintain for each node v a predecessor node p(v) p(v) is initialized to be null Whenever an edge (u,v) is relaxed such that d(v) improves, then p(v) can be set to be u Paths can be generated from this data structure
• Slide 21
• All pairs algorithms using single source algorithms Call a single source algorithm from each vertex s in V O(V X) where X is the running time of the given algorithm Dijkstra linear array: O(V 3 ) Dijkstra binary heap: O(VE log V) Dijkstra Fibonacci heap: O(V 2 log V + VE) Bellman-Ford: O(V 2 E) (negative weight edges)
• Slide 22
• Two adjacency matrix based algorithms Matrix-multiplication based algorithm Let L m (i,j) denote the length of the shortest path from node i to node j using at most m edges What is our desired result in terms of L m (i,j)? What is a recurrence relation for L m (i,j)? Floyd-Warshall algorithm Let L k (i,j) denote the length of the shortest path from node i to node j using only nodes within {1, , k} as internal nodes. What is our desired result in terms of L k (i,j)? What is a recurrence relation for L k (i,j)?
• Slide 23
• Conceptual pictures ij Shortest path using at most 2m edges Shortest path using at most m edges k Try all possible nodes k k ij Shortest path using nodes 1 through k Shortest path using nodes 1 through k-1 OR
• Slide 24
• Running Times Matrix-multiplication based algorithm O(V 3 log V) log V executions of matrix-matrix multiplication Not quite matrix-matrix multiplication but same running time Floyd-Warshall algorithm O(V 3 ) V iterations of an O(V 2 ) update loop The constant is very small, so this is a fast O(V 3 )
• Slide 25
• Johnsons Algorithm Key ideas Reweight edge weights to eliminate negative weight edges AND preserve shortest paths Use Bellman-Ford and Dijkstras algorithms as subroutines Running time: O(V 2 log V + VE) Better than earlier algorithms for sparse graphs
• Slide 26
• Reweighting Original edge weight is w(u,v) New edge weight: w(u,v) = w(u,v) + h(u) h(v) h(v) is a function mapping vertices to real numbers Key observation: Let p be any path fro
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# 13709 (number)
13,709 (thirteen thousand seven hundred nine) is an odd five-digits prime number following 13708 and preceding 13710. In scientific notation, it is written as 1.3709 × 104. The sum of its digits is 20. It has a total of 1 prime factor and 2 positive divisors. There are 13,708 positive integers (up to 13709) that are relatively prime to 13709.
## Basic properties
• Is Prime? Yes
• Number parity Odd
• Number length 5
• Sum of Digits 20
• Digital Root 2
## Name
Short name 13 thousand 709 thirteen thousand seven hundred nine
## Notation
Scientific notation 1.3709 × 104 13.709 × 103
## Prime Factorization of 13709
Prime Factorization 13709
Prime number
Distinct Factors Total Factors Radical ω(n) 1 Total number of distinct prime factors Ω(n) 1 Total number of prime factors rad(n) 13709 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 9.52581 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 13,709 is 13709. Since it has a total of 1 prime factor, 13,709 is a prime number.
## Divisors of 13709
2 divisors
Even divisors 0 2 2 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 2 Total number of the positive divisors of n σ(n) 13710 Sum of all the positive divisors of n s(n) 1 Sum of the proper positive divisors of n A(n) 6855 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 117.085 Returns the nth root of the product of n divisors H(n) 1.99985 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 13,709 can be divided by 2 positive divisors (out of which 0 are even, and 2 are odd). The sum of these divisors (counting 13,709) is 13,710, the average is 6,855.
## Other Arithmetic Functions (n = 13709)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 13708 Total number of positive integers not greater than n that are coprime to n λ(n) 13708 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1623 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares
There are 13,708 positive integers (less than 13,709) that are coprime with 13,709. And there are approximately 1,623 prime numbers less than or equal to 13,709.
## Divisibility of 13709
m n mod m 2 3 4 5 6 7 8 9 1 2 1 4 5 3 5 2
13,709 is not divisible by any number less than or equal to 9.
• Arithmetic
• Prime
• Deficient
• Polite
• Prime Power
• Square Free
## Base conversion (13709)
Base System Value
2 Binary 11010110001101
3 Ternary 200210202
4 Quaternary 3112031
5 Quinary 414314
6 Senary 143245
8 Octal 32615
10 Decimal 13709
12 Duodecimal 7b25
20 Vigesimal 1e59
36 Base36 akt
## Basic calculations (n = 13709)
### Multiplication
n×y
n×2 27418 41127 54836 68545
### Division
n÷y
n÷2 6854.5 4569.67 3427.25 2741.8
### Exponentiation
ny
n2 187936681 2576423959829 35320196065295761 484204567859139587549
### Nth Root
y√n
2√n 117.085 23.9333 10.8206 6.72049
## 13709 as geometric shapes
### Circle
Diameter 27418 86136.2 5.9042e+08
### Sphere
Volume 1.07921e+13 2.36168e+09 86136.2
### Square
Length = n
Perimeter 54836 1.87937e+08 19387.5
### Cube
Length = n
Surface area 1.12762e+09 2.57642e+12 23744.7
### Equilateral Triangle
Length = n
Perimeter 41127 8.1379e+07 11872.3
### Triangular Pyramid
Length = n
Surface area 3.25516e+08 3.03634e+11 11193.4
## Cryptographic Hash Functions
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Skip navigation
George Jones put up a neat little puzzle on physicsforums. Following up on this taught me quite a bit about some of the mathematical subtleties of QM that physicists tend to gloss over.
“Suppose A is an observable, i.e., a self-adjoint operator, with real eigenvalue a and normalized eigenket $\left| a \right>$. In other words,
$A \left| a \right> = a \left| a \right>, \hspace{.5 in} \left< a | a \right> = 1.$
Suppose further that A and B are canonically conjugate observables, so
$\left[ A , B \right] = i \hbar I$,
where I is the identity operator. Compute, with respect to $\left| a \right>$, the matrix elements of this equation divided by $i \hbar$:
$\frac{1}{i \hbar} \left< a | \left[ A , B \right] | a \right>)= \left< a | I | a \right>$
$\frac{1}{i \hbar} \left( \left< a | AB | a \right> - \left \right) = \left< a | I | a \right>$
In the first term, let A act on the bra; in the second, let A act on the ket:
$\frac{1}{i \hbar} \left( a \left< a | B | a \right> - a \left \right)= $.
Thus,
0 = 1.
This is my favourite “proof” of the well-known equation 0 = 1.
What gives?”
When you’re done racking your brains, you can take a look here, here and on the arxiv for solutions to this and more such QM puzzlers. The last link is to a paper by F. Gieres called ‘Mathematical surprises and Dirac’s formalism in Quantum Mechanics’, which I found pretty fascinating. Apparently, when we’re taught as undergrads that state vectors in QM live in a Hilbert space, that’s one of those white lies that teachers tell.
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1. Linear transformation of vectors
I am in a matrix and power series class and I need to figure out how to answer the following type of question. I have read and reread the procedure the book uses and am still unable to figure this type of problem out,
I need to find all vectors in R^2 that satisfy T(v) = [-2 12 -20] or show that no such vector exists. The linear transformation T: R^2 ---> R^3 has matrix
A = [ 1 -3 1 ] T
[ 2 -8 8]
Could someone please show me the steps to attack this with? Thanks very much, Frostking
2. Originally Posted by Frostking
I am in a matrix and power series class and I need to figure out how to answer the following type of question. I have read and reread the procedure the book uses and am still unable to figure this type of problem out,
I need to find all vectors in R^2 that satisfy T(v) = [-2 12 -20] or show that no such vector exists. The linear transformation T: R^2 ---> R^3 has matrix
A = [ 1 -3 1 ] T
[ 2 -8 8]
Could someone please show me the steps to attack this with? Thanks very much, Frostking
This is a non-homogenoues linear system 3 x 2 (3 equations and 2 variables). Such a thing not always has a solution:
$\displaystyle \left(\begin{array}{cc}1&2\\ \! \! \! \! -3& \! \! \! \! -8\\1&8\end{array}\right)\left(\begin{array}{c}x\\y \end{array}\right)=\left(\begin{array}{c}-2\\\;\;12\\-20\end{array}\right) \Longleftrightarrow$ $\displaystyle \begin{array}{c}\;\;\;\;\;x+2y=\,\,\,\,\,-2\\-3x-8y=\,\,\,\,\,\,\,12\\\;\;\;\;\;\,x+8y=\;-20\end{array}$
If now you solve the first two equations as we used to do in 8th or 9th grade, we get the unique solution $\displaystyle (4,-3)$ . If you now plug in these two values in the 3rd equation you'll see you get an equality ==> there does exist a solution for the problem and it is unique (because it is unique for the first two eq's.)
Tonio
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[–]Applied Math 0 points1 point
sorry, this has been archived and can no longer be voted on
What about this way? We'll view the circle as an ellipse with equal length axes, and call it [; E_0 ;] Now squish the circle, so that the semi-minor axis (I'm visualizing this as the vertical axis, but squishing along a diagonal axis would give you your 'M' and offsetting the circle vertically would give you your 'b') is half the original radius, and the semi-major axis is twice the original radius, preserving the area of the circle. Call this [; E_1 ;]. We'll keep with this pattern, so that, letting [; r ;] be the radius of the original circle, [; E_n ;] is an elipse with semi-minor axis of length [; \frac{r}{2^n} ;] and semi-major axis of length [; r2^n ;]. In this way we keep stretching the circle out while preserving it's area.
Now let's take [; lim_{n\rightarrow \infty} E_n ;]. We find a sort of double-tailed Dirac Delta function with infinitessimal width, infinite length, and an area of [; \pi r^2 ;], which to my estimation is as much a line as anything else you could think of.
[–] 0 points1 point
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Do you want a "morph" kind of transform, like the Y axis in this animation?
[–] 0 points1 point
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To understand why this isn't as simple as it seems (because of the lack of a full homeomorphism between the two), try this lecture on Algebraic Topology.
Now, be warned: Professor Wildberger is rather famous for categorically denying the existence of infinite sets (and, with them, things like the Real numbers). However, this finitist "ultra-constructivism" has an advantage in pedagogy (in my opinion). He likes to show things very graphically and also uses projective geometry to create a compass and straightedge illustration of the group structure of the circle. So while I strongly disagree with his philosophical stance, it makes for a great way to learn algebraic topology and, in turn, why what you're asking for can't be done in a way that preserves structure.
[–] 11 points12 points
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I'm not sure if I totally understand your question, but you could use a stereographic projection to map (most of) the points of a circle to the points on a line. If you remove the north pole from your circle, this map is smooth and bijective, which is rather nice.
Alternatively if we write the equation for the circle in polar coordinates (r, θ), it's r = R, which technically is affine linear.
[–]Low-Dimensional Topology 8 points9 points
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Pick your favorite Möbius transformation z -> (az+b)/(cz+d), where you think of R2 as the complex plane and a,b,c,d are complex numbers such that ad-bc is nonzero. If the complex number -d/c lies on your circle, then this transformation will send the circle to a line. For example, if your circle is defined by |z|=r then the transformation z -> 1/(z-r) will suffice.
The reason for this is that Möbius transformations are really transformations of the Riemann sphere (a sphere composed of the complex plane plus a point at infinity) which send circles to circles. A line in the plane is the same thing as a circle through that point at infinity on the Riemann sphere, and the transformation z -> f(z) =(az+b)/(cz+d) will send any circle through -d/c to a circle through f(-d/c) = \infty, i.e. to a line.
[–] -1 points0 points
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Could you elaborate what you mean by "line" in this case? The full S1 cannot be mapped homeomorphically to a one-dimensional subvectorspace of R2 .
[–]Low-Dimensional Topology 5 points6 points
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The transformation is undefined at -d/c, so as a map from C to C it's really sending the complement of that point to the line. In the Riemann sphere the point -d/c would go to the point at infinity, sending the original S1 to the new circle formed by {line} u {infinity}.
I would also caution against using fancier language then necessary here (or in general), because it can either obfuscate what you're trying to say or make it wrong. A simple "R" or "the real line" would have sufficed instead of "a one-dimensional subvectorspace of R2 ". In fact, the subspace language makes it sound like you only want lines passing through the origin, when really this applies to any line at all in the plane, and for that matter the topology of the real line has nothing to do with how you might embed it in the plane. And if you were using that language to also imply that you want the image to be a straight line, then "homeomorphically" isn't the right choice of word because it applies just as well to, say, the graph of sin(x). This is why it's best to be concise, choose your definitions carefully, and even use active voice if you can: for example, "S1 is not homeomorphic to the real line."
[–] -1 points0 points
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I would also caution against using fancier language then necessary here (or in general), because it can either obfuscate what you're trying to say or make it wrong.
I intended to make myself more precise, as a "line" can be everything from a parametrisation of some compact intervall to the real line. Instead of going through all cases, I'd rather a fairly easy one. In any case, since you can identify any line parametrised over R with any one-dimensional vector space as a topological space, this discussion is really moot. As you seemed to know what you were talking about, I had hoped to make it more clear what I intended "line" to mean for the OP.
Homeomorphically is indeed the right choice of word in my view. It is the correct use both my statement and what I considered "making a circle into a line" could mean. Yes, the graph of sin(x) could also be a line, but since this is again homeomorphic to the real line, why should one bother to make it more complicated?
Of course, it is not the only one. But before one starts with isometries or anything more fancy, one should use more down-to-earth formulations, as you have told me yourself.
In this light, I'd like to make clear that I have no intentions to hide behind language, but I appreciate your input. I do not understand how this would apply to definitions or my wordings however, as I only use the very canonical ones everyone makes use of.
Now, I would like to make sure I understood what you stated, as I am not very familiar with Möbius transformations. The Möbius transformation sends an S1 to the compactification of R, which is, again S1?
And how is this less complicated than what I stated, or even a "line", as stated like this
If the complex number -d/c lies on your circle, then this transformation will send the circle to a line
by yourself?
[–]Low-Dimensional Topology 3 points4 points
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Let me address the math first: you're right about the image being S1 in the Riemann sphere. If you're just thinking about the Möbius transformation as acting on points in the usual complex plane than the image really is a line, because there's no point at infinity. (The Riemann sphere is a more natural setting -- in a complex analysis class or beyond you may see that the group formed by Möbius transformations is exactly its automorphism group, PGL(2,C) -- but we don't necessarily have to use it.) Also "the" compactification of R is misleading because S1 isn't the only compactification; in many other contexts it's natural to compactify it as a closed interval, by adding both a positive infinity and a negative infinity.
As for the wording, you need to consider your audience. The extra language about vector spaces could definitely confuse an OP who admits to having only taken courses up to differential equations, because they may not have ever seen linear algebra, much less homeomorphisms or any sort of topology. On the other hand, a lot of people learn about the complex plane in high school, so I hope (but can't guarantee) that my answer won't be lost on the OP.
If you had included the same language in a research paper I was refereeing anonymously, I would still ask you to change it in my report because it's less readable even to a reader who understands all of these terms than "the real line", which is precise and 100% unambiguous as far as I can tell. Why bring linear algebra into the picture when you're just talking about a particular topological space? If you only care about the homeomorphism type of that space, why mention an embedding of it into some other space? (Chekhov's gun applies to both of these questions.)
I don't mean to accuse you of having impure intentions or anything, and I don't claim to be a flawless writer myself, but when you spend lots of time reading other people's papers you really start to appreciate it when the author makes an effort to communicate things as clearly and as simply as possible. I think that a lot of mathematicians (and this is not a personal accusation directed at you) underestimate the importance of good writing skills in general.
[–] -1 points0 points
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Let me address the math first: you're right about the image being S1 in the Riemann sphere. If you're just thinking about the Möbius transformation as acting on points in the usual complex plane than the image really is a line, because there's no point at infinity. (The Riemann sphere is a more natural setting -- in a complex analysis class or beyond you may see that the group formed by Möbius transformations is exactly its automorphism group, PGL(2,C) -- but we don't necessarily have to use it.) Also "the" compactification of R is misleading because S1 isn't the only compactification; in many other contexts it's natural to compactify it as a closed interval, by adding both a positive infinity and a negative infinity.
Yes, you're right in all you say. I assumed I was talking to you, not the OP now, so I sloppily only called it "the compactification". Sorry about that.
I still need to know, however. I am not familiar with much complex analysis, but you can just about throw anything at me to explain it: is there anyway we can obtain a topological space to view the image of S1 under a Möbius transformation to be a line? That's what I read in your first post, and I am still confused.
I know I'm repeating myself here, but can you tell me what you mean by line, which space we are considering, and how the S1 is mapped to that?
As for the wording, you need to consider your audience. The extra language about vector spaces could definitely confuse an OP who admits to having only taken courses up to differential equations, because they may not have ever seen linear algebra, much less homeomorphisms or any sort of topology.
I might have overestimated what people do in differential equations courses then. How do you do ODEs or PDEs without knowing linear algebra and some metric spaces (of which topology is just the generalisation)?
Without this, you aren't even able to introduce the differential of maps between vector spaces of arbitary dimensions (or, if you like, just Rn and Rm ). How are you supposed to do differential equations without that?
I'm really confused now, as you can tell.
If you had included the same language in a research paper I was refereeing anonymously, I would still ask you to change it in my report because it's less readable even to a reader who understands all of these terms than "the real line", which is precise and 100% unambiguous as far as I can tell
And to me it isn't unambiguous, that's where our troubles start, I guess. If we first agree on calling it closed or open interval, then there's no problem. But just imagining a line .. what do you see? The real line, or maybe a closed interval?
As it wasn't clear what he meant by line, I didn't know if he knew that when I state "real line", I really mean it. It could just be "parts of the real line", or maybe he didn't really know the difference between taking different parts - I know, not a very common mistake, but I've seen it done once or twice already.
So I wanted to make sure this wouldn't occur again, and based my assumption on that you need linear algebra to do differential equations (which I now accept to be wrong, apparently).
I agree with you on the last part, but as you can see, I wasn't trying to make it more complicated. Quite to the contrary. My mistake then was that I was thinking too highly of the prerequisites of differential equations, or too much about how "real line" could be mistunderstood. How ironic.
[–]Low-Dimensional Topology 2 points3 points
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is there anyway we can obtain a topological space to view the image of S1 under a Möbius transformation to be a line?
If we view it as a transformation of the complex plane (no Riemann sphere, no point at infinity), then the answer is yes: the transformation z -> (az+b)/(cz+d) is simply undefined at z=-d/c, so if -d/c lies on the circle then the Möbius transformation when restricted to the circle is only defined on the complement of that point, and since S1\{*} is homeomorphic to the real line there's no problem. Otherwise, if you have a point at infinity then the answer is no, since the continuous image of a compact set (the circle) must be compact (not a line).
How do you do ODEs or PDEs without knowing linear algebra and some metric spaces (of which topology is just the generalisation)?
It's maybe not an ideal way to do it, but it does happen, especially in lower-level courses aimed at non-majors. See for example the basic ODE classes at MIT or at UT Austin, both highly ranked departments, where neither course has a linear algebra prerequisite. You just don't focus on things like systems of first-order equations (or you cover just enough basic material about eigenvalues and eigenvectors to solve them) or existence and uniqueness theorems, and you leave the more advanced material for later courses.
[–] 0 points1 point
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Thank you for your explanations. I think my problem here was that I'm learning in a different environment, where we begin with proof-based lectures from day one. I wasn't familiar with the American system.
[–] 4 points5 points
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You cannot. The "circle" you are thinking of is the n-sphere for n=1.
What you are trying to do is to obtain an homeomorphism between the S1 and a line (for example expressed as a one-dimensional subvectorspace of R2 ).
That is, however, impossible, due to the following: a property called connectedness is left untouched via homeomorphism. This means that if one of the two homeomorphic spaces is connected, the other one is too. In addition to this, if you take away one point away from each "space", they remain homeomorphic.
In this case, taking a point away from the S1 leaves it connected, but taking away one point from any line divides it into two connected components, while the line without a point itself is not connected. Ergo, the S1 and the line cannot be homeomorphic.
What others have told you is, canonically, the stereographic projection from the S1 without a point to the real line. As you can see, taking away two points of the S1 gives you two connected parts again .. just like with R. And indeed, this stereographic projection is a homeomorphism, even a smooth diffeomorphism!
[–] 2 points3 points
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I agree with the poster above that calling a line one dimensional subvectorspace is bizarre and not quite accurate, since if that line doesn't pass through the origin, it isn't even a vector space anymore, since it doesn't contain 0. Also, wouldn't it be more straightforward to say that S1 is compact and R is not? This avoids talking about removing a point.
[–] -1 points0 points
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We are talking about topological spaces. A "line" in Rn is nothing more than a subvectorspace translated. That's even a diffeomorphism .. where exactly is this confusing?* The OP had differential equations as prerequisite - how do you even do basic ODE, less alone PDE theory without knowing linear algebra?
You cannot use the compactness argument that easily for two reasons:
1. I'd have to tell you what compactness means in a topological sense. This is, in my view, more of a step than imagining taking two points away. If you only know about metric spaces, it might be a stretch to explain why it isn't always possible to have compactness by sequence or even Heine-Borel if you only dealt with R, but now you need that ugly definition about open covers.
2. I still wouldn't know if by line he meant a compact intervall or the real line. I cannot use the compactness argument if he meant a compact interval. Even if he meant an open intervall, what if the next question is about a compact one? I'd have to use the point-solution due to this natural extension of the question in any case. I assumed he meant an open interval now, but just in case I gave the idea for a proof that can be easily extended to a closed interval as well.
Nevertheless, yours is a good complaint. I thought of using compactness first as well, and then decided against it due to the above reasons.
*I now aknowledge that my assumptions about what you need for differential equations were wrong. I see how this can be confusing then.
[–]Mathematical Physics 3 points4 points
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Not exactly what you're looking for, but conformal mapping [see example 7 and pay attention to the colored dots] does a somewhat similar sort of thing. [Hint: it's helpful to already know complex analysis.]
To do exactly what what you mentioned, use polar coordinates.
[–] 0 points1 point
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Just for an idea of what your asking, describe a few circles and describe what lines you might want to get from them. Maybe a line through the center or tangent to the edge
[–] 0 points1 point
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Hmmm, I remember something about the W-plane where you can map lines into circles and vice versa, however I think it only works with complex numbers.
[–] 0 points1 point
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What does it mean? What do you want to accomplish? As nerkbot says, you can map all the points on a circle (or all but one) to points on a line, using a stereographic projection, or simply by giving the angle in radians between the point, the center of the circle, and a horizontal line passing through the center of the circle. There's no "mx+b" business here though, because a line in and of itself only has one coordinate; it's only when you place a line on a plane, with freedom to move it around and rotate it, that you need two.
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# Explanation for formula that counts number of numeric characters
1. ## Explanation for formula that counts number of numeric characters
The formula below was posted recently as a way to extract a numeric substring from somewhere in larger string.
When I first saw this formula, I had no idea how it "came up with the goods" and so I spent some time looking at
parts of it to see what they did.
Although I've made some progress, I still don't fully understand it, and I would be grateful if someone could
enlighten me. Heres the formula, and below Ive detailed how far Ive got:
=Iferror(lookup(9.99e+307,--mid(a1,min(find({1,2,3,4,5,6,7,8,9,0},a1&1234567890)),row(indirect("1:"&len(a1))))),0)
I hope my findings below are correct:
1. The formula is returning the first substring of contiguous numeric characters from the full string in Cell A1
2. 9.99e + 307 ensures that the search value won't be exceeded by the returned value.
3. The Mid function uses Min(Find( to get the character position of the first numeric in A1 (so MID then has its first 2 parameters)
So far so good.
4. Now... I cant work out where MID parameter 3 (the substring length) comes from. I suspect its the row function, but can't see how
as this isn't an array formula, and without CTRL+Shift+ Enter, I could only get Row to return 1 in my experiments (giving me only the first substring digit).
How does the formula return the correct number of numeric characters, (apparently) without counting them?
Thanks for any help
2. ## Re: Explanation for formula that counts number of numeric characters
That's what this part does, this is the third argument of the Mid function:
row(indirect("1:"&len(a1)))))
Len(a1) returns a number that is how many characters exist in A1
Indirect puts that number together with "1:" to get a range reference. So if there are 12 characters in A1, it would return Indirect("1:12")
Row() takes that range reference and converts it to an array of values, so Row(1:12) = {1,2,3,4,5,6,7,8,9,10,11,12}
Mid gets evaluated with each of those being the third argument. So it starts at the first numeric character found, then sees if length 1 is a number, then sees if length 2 is a number, then sees if length 3 is a number, and so on until it sees if length 12 is a number
Lookup is trying to find the largest number Excel can handle, 9.99e+307. If Lookup can't find its target, it returns the largest value that is still smaller than its target. Lookup is also a function that ignores error values in arrays, which makes it particularly suited to this task. So it will return the number of length x, where x was the largest number returned by the mid function.
"x 321 test x"
The first numeric value is 3 at location 3
Mid then returns this array:
{"3","32","321","321 ","321 t","321 te","321 tes","321 test","321 test ","321test x"}
The -- is called a double unary. It is used to return a value as itself. 0+ or 1* can accomplish the same thing. What this does is convert numbers stored as text to numbers. This is necessary because the Mid function can only have a text output, even if the output is numbers. If it tries to convert an actual text value to a number, the operation will fail and a #VALUE! error will be returned. So now the array looks like this:
{3,32,321,321,#VALUE!,#VALUE!,#VALUE!,#VALUE!,#VALUE!,#VALUE!}
Note that the quotation marks are gone and the numbers are now actual numbers instead of numbers stored as text. Any part in the array that had actual text is now an error. Now the Lookup function can return the largest value that is still below its target. It finds 321 to be that number so that is what is returned.
A useful tool that will also go through the steps of the formula is the Evaluate Formula tool. In Excel 2007+ select the cell with the formula, go to the Formulas tab, click on "Evaluate Formula"
3. ## Re: Explanation for formula that counts number of numeric characters
Hello Tigeravatar
Thanks for your response, which has cleared up all the points I didn't understand. Ill try the formula evaluate, which I wasn't aware of.
These formulae can be a bit difficult, as some of the processes can't be seen by breaking it down into smaller steps which is my
accustomed way of solving this type of problem. Your explanation was first class. Im very pleased to click your star!
regards
Hercules
4. ## Re: Explanation for formula that counts number of numeric characters
You're very welcome
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# How do you test for convergence or divergence?
• Oct 1st 2012, 04:41 AM
MathIsOhSoHard
How do you test for convergence or divergence?
Given:Sigma, n=1 going up to infinity.How would you test whether this is convergence or divergence?(n+3)/(n+2)My text book doesn't explain it to me properly and leaves me more confused.
• Oct 1st 2012, 06:09 AM
Plato
Re: How do you test for convergence or divergence?
Quote:
Originally Posted by MathIsOhSoHard
Given:Sigma, n=1 going up to infinity.How would you test whether this is convergence or divergence?(n+3)/(n+2)My text book doesn't explain it to me properly and leaves me more confused.
The series $\sum\limits_{n = 0}^\infty {a_n }$ converges only if $\left( {a_n } \right) \to 0$.
That means that if $\left( {a_n } \right)\not \to 0$ then the series diverges.
What about $\left( {\frac{n+3}{n+2} } \right)\to ?$
• Oct 1st 2012, 06:20 AM
MathIsOhSoHard
Re: How do you test for convergence or divergence?
My guess is that the fraction can be reduced to 3/2 which means that a_n is equal to 3/2.
Since it does not converge towards 0, the fraction is divergence.
But would it be correct to say the fraction is equal to 3/2 or would it be correct to say the fraction goes towards 3/2? Im kinda thinking that since the fraction does not contain any n value that goes towards infinity, then the correct term would be "equal" and not "towards"?
• Oct 1st 2012, 06:36 AM
Plato
Re: How do you test for convergence or divergence?
Quote:
Originally Posted by MathIsOhSoHard
My guess is that the fraction can be reduced to 3/2 which means that a_n is equal to 3/2.
Since it does not converge towards 0, the fraction is divergence.
But would it be correct to say the fraction is equal to 3/2 or would it be correct to say the fraction goes towards 3/2? Im kinda thinking that since the fraction does not contain any n value that goes towards infinity, then the correct term would be "equal" and not "towards"?
You are really confused on terminology.
Because the sequence $\left( {\frac{{{\rm{n + 3}}}}{{{\rm{n + 2}}}}} \right) \to 1$ the series $\sum\limits_{n = 1}^\infty {\left( {\frac{{{\rm{n + 3}}}}{{{\rm{n + 2}}}}} \right)}$ diverges.
• Oct 1st 2012, 06:45 AM
Deveno
Re: How do you test for convergence or divergence?
here is what we have:
$\frac{n+3}{n+2} = \frac{(n+2) + 1}{n+2} = 1 + \frac{1}{n+2} > 1$.
therefore:
$\sum_{n=1}^k a_n = \sum_{n=1}^k \frac{n+3}{n+2} > k$.
thus given ANY integer N, we can ALWAYS find an integer k such that: $\sum_{n=1}^k \frac{n+3}{n+2}> N$ (k = N will work nicely).
thus the series $\sum_{n=1}^\infty \frac{n+3}{n+2}$ diverges because the sequence of partial sums {Sk}:
$S_k = \sum_{n=1}^k \frac{n+3}{n+2}$ increases without bound.
• Oct 1st 2012, 09:03 AM
MathIsOhSoHard
Re: How do you test for convergence or divergence?
I'm a bit confused now because to me it seems like Plato is saying the sequence goes towards 1 but Deveno is saying it increases without bound.
I think there must be something I'm missing.
But we have the definition of the partial sums as:
$S_N=\frac{1+3}{1+2}+\frac{2+3}{2+2}+...+\frac{N+3} {N+2}=\sum_{n=1}^N \frac{n+3}{n+2}$
So the partial sums is the sum of the sequence up to $N$. Is that correct?
If we can determine that the sequences are convergent, then the series is convergent as well or vice versa (if the sequence is divergent then the series is divergent.)
The nth term test tells us that in series if $a_n$ does not go towards 0, then the series is divergent (so if $a_n$ goes towards 0 then the series is convergent).
Am I on the right track? So I need to figure out whether the sequence is convergent or not, and then I can use that information to conclude if the series is convergent or not.
• Oct 1st 2012, 09:21 AM
Plato
Re: How do you test for convergence or divergence?
Quote:
Originally Posted by MathIsOhSoHard
The nth term test tells us that in series if $a_n$ does not go towards 0, then the series is divergent (so if $a_n$ goes towards 0 then the series is convergent).
That is totally and completely false.
You are very confused on the vocabulary.
Series are about the convergence of partials sums.
Partials sums are finite sums of some sequence.
Now why is that false?
The sequence $\left( {\frac{1}{n}} \right) \to 0$ BUT $\sum\limits_{n = 1}^\infty {\left( {\frac{1}{n}} \right)}$ DIVERGES!
• Oct 1st 2012, 10:06 AM
MathIsOhSoHard
Re: How do you test for convergence or divergence?
If I remember correctly, $\frac{1}{n}$ is a harmonic function (not sure if that is the correct term but something "harmonic").
So the series for this harmonic function, starting with n=1, becomes:
$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}...$
We see in this series that it already starts at 1, and then it keeps getting added to it, so it surpasses 1.
Going by the nth term test, which says that if $a_n$ does not go towards 0, then the series is divergent.
This is the case here, since it is way over 1 and for every sequence more gets added to it making it go further away. So we can conclude by the nth term test that it is in fact divergent.
This is only for the series though!
Because the series is a sum where you add all the sequences together.
But for the sequence, we don't add all the previous together. Here $n\rightarrow\infty$ so the fraction becomes smaller and smaller as $n$ becomes larger and larger.
The difference between the sequence and the series that you mentioned is that the sequence is only one fraction where n becomes infinitely small and the series is a set of inifinite fractions that are all added together where n each time becomes 1 integer bigger.
I really hope some of it was correct lol
And you're right, the nth term test does not show us whether a series is convergent or not.
It can only be used to show us if a series is divergent (we need other types of tests to show the series is convergent).
• Oct 2nd 2012, 12:18 PM
Deveno
Re: How do you test for convergence or divergence?
yes. in fact, convergence of a series can be VERY hard to show. the reason is a convergent (or divergent) series can converge (or diverge) very slowly.
however we do NOT "add all the sequences together".
there are 3 separate things involved:
1. a sequence of TERMS $\{a_n\}$.
2. a sequence of "partial sums" (adding all the terms up to the k-th term} $\{S_k\},\ S_k = \sum_{n=1}^k a_n$.
3. the series itself: $\sum_{n=1}^\infty a_n = \lim_{k \to \infty} S_k$.
now, if the limit in number 3 exists, the partial sums need to get "closer and closer together": $\lim_{k \to \infty} S_k - S_{k-1} = 0$.
but what is $S_k - S_{k-1}$? it's just the k-th term, $a_k$.
so: IF a series is convergent (if the sequence of partial sums converges), the limit of the terms is 0.
however, just because the terms "go to 0" doesn't mean the series converges. the prime example of this is (as you said) the harmonic series:
$\sum_{n=1}^\infty \frac{1}{n}$.
it is clear that $\lim_{n \to \infty} \frac{1}{n} = 0$.
however, let's look at the " $2^k$-th" partial sum:
$\sum_{n = 1}^{2^k} \frac{1}{n}$
$= 1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right) + \cdots + \left(\frac{1}{2^{k-1}} + \cdots + \frac{1}{2^k}\right)$
$> \frac{1}{2} + \frac{1}{2} + \left(\frac{1}{4} + \frac{1}{4}\right) + \left(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}\right) + \cdots + \left(\frac{1}{2^k} + \cdots + \frac{1}{2^k}\right) = \frac{k+1}{2}$.
so, if we pick any positive integer N, and we pick k such that k > 4N, then:
$S_k > S_{4^N} = S_{2^{(2N)}} > \frac{2N + 1}{2} > \frac{2N}{2} = N$
which shows that $\lim_{k \to \infty} S_k$ does not exist.
• Oct 2nd 2012, 03:51 PM
MathIsOhSoHard
Re: How do you test for convergence or divergence?
I see, that does make sense, thanks!
But one thing that I'm still not sure about is the following example:
$\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$
This series is convergent and the sequence of partial sums is $S_k\to 1$
But how come this following series:
$\sum_{n=1}^{\infty}\frac{1}{2(n+2)}$
diverges?
What is the difference between the two of them and what convergence test would you use to prove the first series convergent and the second series divergent?
• Oct 2nd 2012, 09:31 PM
Deveno
Re: How do you test for convergence or divergence?
one way to tell if a series is convergent is: if it is "dominated" by another series, that is:
$0 \leq a_n \leq b_n$ for all n, and $\sum_{n=1}^\infty b_n$ converges. note that this only works if every term is non-negative.
by the same token, if some series dominates a divergent series, it must also be divergent.
now $\frac{1}{n(n+1)} < \frac{1}{n^2}$ so if:
$\sum_{n=1}^\infty \frac{1}{n^2}$ converges, so does your first series. it is a well-known result that the latter series does indeed converge to $\frac{\pi^2}{6}$.
but that doesn't tell us what the first series converges TO. however, since i "know the answer" i can prove what it converges to.
namely, i will prove by induction on m that:
$\sum_{n=1}^m \frac{1}{n(n+1)} = \frac{m}{m+1}$. this is clearly true for m = 1. assume it's true for m = k. then:
$\sum_{n=1}^{k+1} \frac{1}{n(n+1)} = \left(\sum_{n=1}^k \frac{1}{n(n+1)}\right) + \frac{1}{(k+1)(k+2)}$
$= \frac{k}{k+1} + \frac{1}{(k+1)(k+2)} = \frac{k(k+2)}{(k+1)(k+2)} + \frac{1}{(k+1)(k+2)}$
$= \frac{k^2 + 2k + 1}{(k+1)(k+2)} = \frac{k+1}{k+2}$, which is the assertion for m = k+1.
hence $\lim_{m \to \infty} S_m = \lim_{m \to \infty} \frac{m}{m+1} = \lim_{m \to \infty} 1 - \lim_{m \to \infty}\frac{1}{m+1} = 1 - 0 = 1$, so:
$\sum_{n=1}^\infty \frac{1}{n(n+1)} = 1$.
***************************
now for the second series:
$\frac{1}{2n+4} > \frac{1}{4n+4} = \left(\frac{1}{4}\right)\left(\frac{1}{n+1}\right)$.
if:
$\sum_{n=1}^\infty \frac{1}{2(n+2)}$ converged, it would follow that:
$\sum_{n=1}^\infty \frac{1}{n+1}$ converged as well. can you see why this can't be true?
• Oct 2nd 2012, 11:49 PM
MathIsOhSoHard
Re: How do you test for convergence or divergence?
I think I would use the comparison test here so we have:
$\frac{1}{2(n+2)}<\frac{1}{n+1}<\frac{1}{n}$
Since we know that $\frac{1}{n}$ is a harmonic series and it is divergent, then according to the comparison test, $\frac{1}{n+1}$ must also be divergent.
Dividing them both we get:
$\frac{\frac{1}{2(n+2)}}{\frac{1}{n+1}}=\frac{1}{2}$
Since $\frac{1}{2}>0$ we know for sure that if one is divergent, then the other must be divergent as well.
• Oct 3rd 2012, 11:39 AM
Deveno
Re: How do you test for convergence or divergence?
*sigh*
if the series $\sum_{n=1}^\infty \frac{1}{n}$ dominates $\sum_{n=1}^\infty \frac{1}{2(n+2)}$ that tells us NOTHING.
what we DO know is that the latter series dominates:
$\left(\frac{1}{4}\right)\left(\sum_{n=1}^\infty \frac{1}{n+1}\right)$
$=\left(\frac{1}{4}\right)\left(\sum_{n=2}^\infty \frac{1}{n}\right)$, which diverges.
basically:
less than a divergent series = ????
more than a convergent series = ????
less than a convergent series = convergent (for non-negative only!)
more than a divergent series = divergent
*********
as to "why", you can think about it this way: if we are adding terms that are reciprocals of polynomials in n, say p(n), linear polynomials don't grow "fast enough" so their reciprocals "shrink fast enough" to stop "adding meat to our infinite sum". but if deg(p) > 1, then the shrinkage is fast enough.
it can be very difficult to tell if 1/f(n) for an increasing function f shrinks fast enough for the series built on it to converge. there are two opposing forces at work:
1. we are adding more and more terms, so the partial sums keep increasing.
2. the terms are getting smaller and smaller, so the rate of increase is decreasing.
so checking for series convergence/divergence isn't an "automatic" process (like for example, differentiation) but more of an art form (like integration). the various "tests" for convergence/divergence only let us tell in some circumstances that some series converge or diverge. in fact, one of the BEST tests is the integral test, but evaluating an integral isn't always easy (some functions aren't integrable in "elementary terms").
why does the integral test work so well? because a series is actually a crude approximation for:
$\int_0^\infty \frac{1}{f(t)}\ dt$
and depending on whether you choose the intervals [n-1,n], or [n,n+1] a series is either a lower sum or an upper sum for said integral (one series has "one more term" than the other, but one term alone makes no difference as to convergence or divergence).
• Oct 3rd 2012, 01:15 PM
MathIsOhSoHard
Re: How do you test for convergence or divergence?
Quote:
Originally Posted by Deveno
*sigh*
if the series $\sum_{n=1}^\infty \frac{1}{n}$ dominates $\sum_{n=1}^\infty \frac{1}{2(n+2)}$ that tells us NOTHING.
what we DO know is that the latter series dominates:
$\left(\frac{1}{4}\right)\left(\sum_{n=1}^\infty \frac{1}{n+1}\right)$
$=\left(\frac{1}{4}\right)\left(\sum_{n=2}^\infty \frac{1}{n}\right)$, which diverges.
basically:
less than a divergent series = ????
more than a convergent series = ????
less than a convergent series = convergent (for non-negative only!)
more than a divergent series = divergent
I think I get it now. My mistake was that I used "less than a divergent series" which is inconclusive.
So in order to see if it is divergent, I need a sequence of term that is "more than a divergent series".
So I cannot use $\frac{1}{n}$.
So as you said earlier:
$\frac{1}{2n+4}>\frac{1}{4n+4}$
So we follow the rule that $0 which makes sure we only work with non-negatives. So if $\sum_{n=1}^{\infty}a_n$ is divergent, then $\sum_{n=1}^{\infty}b_n$ is divergent (more than a divergent series = divergent). If $\sum_{n=1}^{\infty}b_n$ is convergent, then $\sum_{n=1}^{\infty}a_n$ is convergent (less than a convergent series = convergent (for non-negative only!)).
These are the only two rules you can go by. Any other way (which I unfortunately did with $\frac{1}{n}$) is inconclusive and cannot be used.
So basically we have $b_n=\frac{1}{2n+4}$ and we need to find a divergent $a_n$ which is less than $b_n$.
So I'm trying to figure out why you use $\frac{1}{n+1}$.
You write that:
$\sum_{n=1}^{\infty}\frac{1}{n+1}=\sum_{n=2}^{\inft y}\frac{1}{n}$
Now to the right of the equal sign, you write that n=2.
You give the explanation to this here:
Quote:
and depending on whether you choose the intervals [n-1,n], or [n,n+1] a series is either a lower sum or an upper sum for said integral (one series has "one more term" than the other, but one term alone makes no difference as to convergence or divergence)
So basically what you do, is that you add one more term to the harmonic series $\frac{1}{n}$ which does not make any difference as to whether it converges or diverges.
I hope I understood that correctly.
EDIT: I don't know why a "y" appears on top of the last sum sign. It's supposed to be an infinity sign.
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# How do you work with negative integers?
Updated: 9/18/2023
Wiki User
13y ago
Multiply the integer by -1 this makes the number positive.
Wiki User
13y ago
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Q: How do you work with negative integers?
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Related questions
### Are all none negative integers are negative integers?
An integer is either positive (non-negative), or negative, or zero. There are no negative non-negative integers.
### What are non positive integers?
Non-positive integers are zero and the negative integers.
### What is the negative set of integers for 224?
There is no such thing as a negative set of integers. There can be a set of negative integers, but that is not the same thing. And even that does not make sense.There is no such thing as a negative set of integers. There can be a set of negative integers, but that is not the same thing. And even that does not make sense.There is no such thing as a negative set of integers. There can be a set of negative integers, but that is not the same thing. And even that does not make sense.There is no such thing as a negative set of integers. There can be a set of negative integers, but that is not the same thing. And even that does not make sense.
### Are negative one fourth and negative two fifths negative integers?
No. They are negative fractions, not integers.
### When the negative integers the positive integers and 0 are combined the integers are formed?
Negative integers, zero and the positive integers, together form the set of integers.
### How do you get a pair of negative integers that have the sum of five?
You cannot. The sum of negative integers will be negative.
### The Sum of 2 negative integers is always negative?
the sum of two negative integers is ALWAYS negative
### Why does negative integers less than positive integers?
the negative integers are below 0, for example -6.
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Home » News » GMAT » GMAT Tip: Consider the Possibilities and Make a Well-Reasoned Guess
# GMAT Tip: Consider the Possibilities and Make a Well-Reasoned Guess
In the GMAT Quantitative section, there are often problems that appear simple but turn into big time-wasters that detract from test takers’ ability to get to the tougher questions and/or finish the quantitative section of the exam. These time-wasters are a) often arithmetic questions b) also data sufficiency questions and c) seem very simple from the outset.
The GMAT is testing your ability to manage your time, and a test taker with excellent time management skills recognizes that when you are at 85% of the way in solving a problem, often an educated guess (and likely correct) guess is more valuable than allowing precious seconds count down from the count.
The below data sufficiency question perfectly illustrates what we are referring to:
If x and y are integers and 2 < x < y, does y = 16?
(1) The greatest common factor of x and y is 2.
(2) The lowest common multiple of x and y is 48.
Thinking back to our decision tree, Statement (1) should relatively easy to evaluate. If the GCF of x and y is 2 and 2 < x < y, then the sky’s is the limit with what the answer should be.
That should let us to our evaluation of Statement (2). The LCM of x and y is 48 – let’s think of what the possibilities might look like. Well, if x = 3 and y = 16, then indeed, y would have to be 16, right?
The problem is that if x = 24 and if y = 48, then the LCM would also be 48. y is not necessarily 16 – there are other combinations that would provide Statement (2) true.
Many test takers are then going to jump quickly select (E) as their answer choice, only to uncover later than they are incorrect because the correct answer is (C). In many ways, this question is testing your ability to understand factors, but there are other ways to get to the right answer.
When assessing Statement (2) we could also conceptually realize we need a “limiting factor” to the pairs of x and y that share a LCM of 48. When provided several parameters – 2 < x < y and the GCM of x and y is 2, and the LCM of x and y is 48, shouldn’t that be enough limiting factors to determine whether either y is (or isn’t) equal to 16?
Likely, we just don’t have sufficient time to think through all of the possibilities where that would prove true to give us the exact pair of what x and y is for this question. But, this is a data sufficiency, and not a problem-solving question. For Statement (1) the possibilities are too endless, and while not quite as numerous as with Statement (1), Statement (2) proves to have the same problem. But combined, it is more likely to prove there is a concrete possibility to sway yes versus no than the likelihood that there is not sufficient information.
When tackling tough time-wasting questions like this one, just think of what is more likely the right answer, and then move on to apply your time to another question – if you have come up with a bunch of possibilities already, then consider what that means to the problem and select your best guess.
The above GMAT Tip comes from Veritas Prep. Since its founding in 2002, Veritas Prep has helped more than 100,000 students prepare for the GMAT and offers the most highly rated GMAT Prep course in the industry.
Posted in: GMAT, GMAT - Quantitative, GMAT Tips, MBA Feature
## Lena Maratea
Lena Maratea is the Digital Marketing Manager at Clear Admit. She's a South Philadelphia native who graduated from Temple University’s Fox School of Business with a BBA in Marketing. She creates and curates essential digital content for the Clear Admit community.
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# This formula is on Wolfram Mathworld but I cannot find it anywhere else online?
$$\sum_{k=1}^{\infty}\ln\left[ \frac{(4k+1)^{1/(4k+1)^{n}}}{(4k-1)^{1/(4k-1)^{n}}} \right] = -\beta'(n)$$. Where $\beta$ is the Dirichlet Beta Function and $n$ is a positive integer.
I cannot find this cited anywhere nor values of the beta function derivative apart from at $-1,0,1$. How can I go about finding these things, I have searched googled and arxiv.
It is known that (uniformly and absolutely) $$\beta(n)=\sum^{\infty}_{k=1}\frac{\chi_4(k)}{k^n}\textrm{, }Re(n)>1.$$ Hence writing $1/k^n=e^{-n\log(k)}$, we have easily $$-\beta'(n)=\sum^{\infty}_{k=2}\frac{\chi_4(k)\log(k)}{k^n}.$$ But $$\chi_4(k)=\left\{ \begin{array}{cc} 0\textrm{ if }k\equiv 0 (mod)4\\ 1\textrm{ if }k\equiv 1 (mod)4\\ 0\textrm{ if }k\equiv 2 (mod)4\\ -1\textrm{ if }k\equiv 3 (mod)4 \end{array} \right\}.$$ Hence for $Re(n)>1$, we have $$-\beta'(n)=\sum^{\infty}_{k=1}\frac{\log(4k+1)}{(4k+1)^n}-\sum^{\infty}_{k=1}\frac{\log(4k-1)}{(4k-1)^n}=\sum^{\infty}_{k=1}\log\left(\frac{(4k+1)^{1/(4k+1)^{n}}}{(4k-1)^{1/(4k-1)^{n}}}\right).$$ QED
• I'd give the +50, but also asked about certain values and how they are computed Jul 1, 2018 at 14:03
• Here are some references: M. Waldschmidt, P. Moussa, J.-M. Luck, C. Itzykson (Eds.) "From Number Theory to Physics". Springer Verlag, Berlin, Heidelberg, New York, London, Paris, Tokyo, Hong Kong, Barcelona, Budapest. Printed (1992) in USA. Jul 1, 2018 at 15:08
• More: [2]: J.V. Armitage, W.F. Eberlein. 'Elliptic Functions'. Cambridge University Press. (2006)\\ [3]: Don. Zagier. 'Elliptic Modular Forms and Their Applications'. Available from [email protected] Jul 1, 2018 at 15:16
• [4]: J.M. Borwein, M.L. Glasser, R.C. McPhedran, J.G. Wan, I.J. Zucker. 'Lattice Sums Then and Now'. Cambridge University Press. New York, (2013). Jul 1, 2018 at 15:32
$\sum_{k=1}^{\infty}\ln\left[ \frac{(4k+1)^{1/(4k+1)^{n}}}{(4k-1)^{1/(4k-1)^{n}}} \right] = -\beta(n)$
I'll naively play with the left side and see what I can get.
The result is complicated expressions of dubious value, but here they are anyway.
$\begin{array}\\ t_k &=\ln\left[ \dfrac{(4k+1)^{1/(4k+1)^{n}}}{(4k-1)^{1/(4k-1)^{n}}} \right]\\ &=\ln((4k+1)^{1/(4k+1)^{n}})-\ln((4k-1)^{1/(4k-1)^{n}})\\ &=\dfrac1{(4k+1)^{n}}\ln(4k+1)-\dfrac1{(4k-1)^{n}}\ln(4k-1)\\ &=\dfrac{\ln(4k)}{(4k+1)^{n}}\ln(1+1/4k)-\dfrac{\ln(4k)}{(4k-1)^{n}}\ln(1-1/4k)\\ &=\dfrac{\ln(4k)}{(4k)^{n}}(1+1/(4k))^{-n}\ln(1+1/(4k))-\dfrac{\ln(4k)}{(4k)^{n}}(1-1/(4k))^{-n}\ln(1-1/4k)\\ &=\dfrac{\ln(4k)}{(4k)^{n}}\left((1+1/(4k))^{-n}\ln(1+1/(4k))-(1-1/(4k))^{-n}\ln(1-1/4k)\right)\\ &=\dfrac{\ln(4k)}{(4k)^{n}}\left(f(-1/(4k))-f(1/(4k))\right)\\ \text{where}\\ f(x) &=(1-x)^{-n}\ln(1-x)\\ &=-\sum_{i=0}^{\infty} \binom{n+i-1}{k}x^i\sum_{j=1}^{\infty}\dfrac{x^j}{j}\\ &=-x\sum_{i=0}^{\infty} \binom{n+i-1}{k}x^i\sum_{j=0}^{\infty}\dfrac{x^j}{j+1}\\ &=-x\sum_{m=0}^{\infty} x^m\sum_{i=0}^{m}\binom{n+i-1}{i}\dfrac{1}{m-i+1}\\ &=-x\sum_{m=0}^{\infty} x^m\sum_{i=0}^{m}g(n, m, i)\\ \text{so}\\ t_k &=\dfrac{\ln(4k)}{(4k)^{n}}\left(f(-1/(4k))-f(1/(4k))\right)\\ &=\dfrac{\ln(4k)}{(4k)^{n}}\left(-\dfrac{-1}{4k}\sum_{m=0}^{\infty} (-1)^m(4k)^{-m}\sum_{i=0}^{m}g(n, m, i)-\dfrac{1}{4k}\sum_{m=0}^{\infty} (4k)^{-m}\sum_{i=0}^{m}g(n, m, i)\right)\\ &=\dfrac{\ln(4k)}{(4k)^{n}}\left(\dfrac{1}{4k}\sum_{m=0}^{\infty} (-1)^m(4k)^{-m}\sum_{i=0}^{m}g(n, m, i)-\dfrac{1}{4k}\sum_{m=0}^{\infty} (4k)^{-m}\sum_{i=0}^{m}g(n, m, i)\right)\\ &=\dfrac{\ln(4k)}{(4k)^{n}}\left(\sum_{m=0}^{\infty}\dfrac{(4k)^{-m}}{4k}( (-1)^m-1)\sum_{i=0}^{m}g(n, m, i)\right)\\ &=\dfrac{\ln(4k)}{(4k)^{n}}\left(\sum_{m=0}^{\infty}(4k)^{-m-1}( (-1)^m-1)\sum_{i=0}^{m}g(n, m, i)\right)\\ &=-2\dfrac{\ln(4k)}{(4k)^{n}}\left(\sum_{m=0}^{\infty}(4k)^{-(2m+1)-1}\sum_{i=0}^{2m+1}g(n, 2m+1, i)\right)\\ &=-2\dfrac{\ln(4k)}{(4k)^{n}}\left(\sum_{m=0}^{\infty}(4k)^{-2m-2}\sum_{i=0}^{2m+1}g(n, 2m+1, i)\right)\\ \end{array}$
At this point, I don't know what to do. However, I don't want to waste all this algebra, so I'll sumit this and hope that it might help someone else.
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# Analysis — Chapter 1: continued — Real Variables part 9
9. Relations of magnitude between real numbers.
It is plain, that, now that we have extended our conception of number, we are bound to make corresponding extensions of our conceptions of equality, inequality, addition, multiplication, and so on. We have to show that these ideas can be applied to the new numbers, and that, when this extension of them is made, all the ordinary laws of algebra retain their validity, so that we can operate with real numbers in general in exactly the same way as with the rational numbers of Chapter 1, part 1 blog. To do all this systematically would occupy considerable space/time, and we shall be content to indicate summarily how a more systematic discussion would proceed.
We denote a real number by a Greek letter such as $\alpha$, $\beta$, $\gamma\ldots$; the rational numbers of its lower and upper classes by the corresponding English letters a, A; b, B; c, C; …We denote the classes themselves by (a), (A),…
If $\alpha$ and $\beta$ are two real numbers, there are three possibilities:
i) every $\alpha$ is a b and every A a B; in this case, (a) is identical with (b) and (A) with (B);
ii) every a in a b, but not all A’s are B’s; in this case (a) is a proper part of $(b)^{*}$, and (B) a proper part of (A);
iii) every A is a B, but not all a’s are b’s.
(These three cases may be indicated graphically on a number line).
In case (i) we write $\alpha=\beta$, in case (ii) $\alpha=\beta$, and in case (iii) $\alpha>\beta$. It is clear that, when $\alpha$ and $\beta$ are both rational, these definitions agree with the ideas of equality and inequality between rational numbers which we began by taking for granted; and that any positive number is greater than any negative number.
It will be convenient to define at this stage the negative $-\alpha$ of a positive number $\alpha$. If
$(\alpha)$, (A) are the classes, which consitute $\alpha$, we can define another section of the rational numbers by putting all numbers $-A$ in the lower class and all numbers $-\alpha$ in the upper. The real number thus defined, which is clearly negative, we denote by $-\alpha$. Similarly, we can define
$-\alpha$ when $\alpha$ is negative or zero; if $\alpha$ is negative, $-\alpha$ is positive, It is plain also that $-(-\alpha)=\alpha$. Of the two numbers $\alpha$ and $-\alpha$ one is always positive (unless $\alpha=0$). The one which is positive we denote by $|\alpha|$ and call the modulus of $\alpha$.
More later,
Nalin Pithwa
# Analysis — Chapter 1 — Real Variables — part 8
8. Real numbers. We have confined ourselves so far to certain sections of the positive rational numbers, which we have agreed provisionally to call “positive real numbers.” Before we frame our final definitions, we must alter our point of view a little. We shall consider sections, or divisions into two classes, not merely of the positive rational numbers, but of all rational numbers, including zero. We may then repeat all that we have said about sections of the positive rational numbers in part 6 and 7 merely omitting the word positive occasionally.
Definitions. A section of the rational numbers, in which both classes exist and the lower class has no greatest member, is called a real number, or simply a number.
A number which does not correspond to a rational number is called an irrational number.
If the real number does correspond to a rational number, we shall use the term “rational” as applying to the real number line.
The term “rational number” will, as a result of our definitions, be ambiguous, it may mean the rational number of part 1, or the, corresponding real number. If we say that $1/2 > 1/3$, we may be asserting either of the two different propositions, one a proposition of elementary arithmetic, the other a proposition concerning sections of the rational numbers. Ambiguities of this kind are common in mathematics, and are perfectly harmless, since the relations between different propositions are exactly the same whichever interpretation is attached to the propositions themselves. From $1/2>1/3$ and $1/3>1/4$ we can infer $1/2>1/4$; the inference is in no way affected by any doubt as to whether $1/2$, $1/3$ and $1/4$ are arithmetic fractions or real numbers. Sometimes, of course, the context in which (example) ‘$1/2$‘ occurs is sufficient to fix its interpretation. When we say (next blog part 9) that $1/2 < \sqrt{1/3}$we must mean by ‘$1/2$‘ the real number $1/2$.
The reader should observe, moreover, that no particular logical importance is to be attached to the precise form of definition of a ‘real number’ that we have adopted. We defined ‘a real number’ as being a section, that is, a pair of classes. We might equally well have defined it to being the lower, or the upper class; indeed it would be easy to define an infinity of classes of entities of each of which would possess the properties of the class of real numbers. What is essential in mathematics is that its symbols should be capable of some interpretation; generally they are capable of many, and then so far as mathematics is concerned, it does not matter which we adopt. Mr. Bertrand Russell has said that “mathematics is the science in which we do not know what we are talking about, and do not care what we say about it is true”, a remark which is expressed in the form of paradox but which in reality embodies a number of important truths. It would take too long to analyze the meaning of Mr Russell’s epigram in detail, but one at any rate of the implications is this, that the symbols of mathematics are capable of varying interpretations, and that we are in general at liberty to adopt whatever we prefer.
There are now three cases to distinguish. It may happen that all negative rational numbers belong to the lower class and zero and all positive rational numbers to the upper. We describe this section as the real number zero. Or, again it may happen that the lower class includes some positive numbers. Such a section we as a positive real number. Finally, it may happen that some negative numbers belong to the upper class. Such a section we describe as a negative real number.
Note: The difference between our presentation of a positive real number here and that or part 7 of the blogs amounts to the addition to the lower class of zero and all the negative rational numbers. An example of a negative real number is given by taking the property P of part 6 of the blogs to be $x+1<0$ and Q to be $x+1 \geq 0$/ This section plainly corresponds to the negative rational number $-1$. If we took P to be $x^{3}<-2$ and Q to be $x^{3}>-2$, we should obtain a negative real number which is not rational.
More later,
Nalin Pithwa
# Analysis — Chapter 1 Real Variables — part 7 — continued
Part 7. Irrational numbers (continued).
In the first two cases, we say that the section corresponds to a positive rational number a, which is l in the one case and r in the other. Conversely, it is clear that to any such number a corresponds a section which we shall denote by
$\alpha^{*}$. For we might take P and Q to be the properties expressed by
$x \leq a, x > a$
respectively, or by $x and $x \leq a$. In the first case, a would be the greatest number of L, and in the second case the least member of R. These are in fact just two sections corresponding to any positive rational number. In order to avoid ambiguity we select one of them; let us select that in which the number itself belongs to the upper class. In other words, let us agree that we will consider only sections in which the lower class L has no greatest number.
There being this correspondence between the positive rational numbers and the sections defined by means of them, it would be perfectly legitimate, for mathematical purposes, to replace the numbers by the sections, and to regard the symbols which occur in our formulae as standing for the sections instead of for the numbers. Thus, for example,
$\alpha > \alpha^{'}$ would mean the same as $a > a^{'}$. If $\alpha$ and $\alpha^{'}$ are
the sections which correspond to a and $a^{'}$.
But, when we have in this way substituted sections of rational numbers for the rational numbers themselves, we are almost forced to a generalization of our number system. For there are sections (such as that of blog on Chapter 1 — part 4) which do not correspond to any rational number. The aggregate of sections is a larger aggregate than that of the positive rational numbers; it includes sections corresponding to all these numbers, and more besides. It is this fact which we make the basis of our generalization of the idea of a number. We accordingly frame the following definitions, which will however be modified in the next blog, and must therefore be regarded as temporary and provisional.
A section of the positive rational numbers, in which both classes exist and the lower class has no greatest member, is called a positive real number.
A positive real number which does not correspond to a positive rational number is called a positive irrational
number.
More later,
Nalin Pithwa
# The Universal Appeal of Mathematics — Geetha S. Rao
I am reproducing an article, “The Universal Appeal of Mathematics — Geetha S. Rao” from “The Mathematics Student” , volume 83, Numbers 1 to 4, (2014), 01-04.
The purpose is just to share this beautiful article with the wider student community and math enthusiasts.
The Universal Appeal of Mathematics: Geetha S. Rao:
Mathematics is the Queen of all Sciences, the King of all Arts and the Master of all that is being surveyed. Such is the immaculate and immense potential of the all-pervasive, fascinating subject, that it transcends all geographical barriers, territorial domains and racial prejudices.
The four pillars that support the growth, development, flowering and fruition of this ever green subject are analytic thinking, logical reasoning, critical reviewing and decision thinking.
Every situation in real life can be modelled and simulated in mathematical language. So much so, every human must be empowered with at least a smattering of mathematical knowledge. Indeed, the field of Artificial Intelligence is one where these concepts are implemented and imparted to the digital computers of today.
From times immemorial, people know how to count and could trade using the barter system. Those who could join primary schools learnt the fundamental arithmetic and algebraic rules. Upon entry into high school and higher secondary classes, the acquaintance with the various branches of this exciting subject commences. It is at this point that effective communication skills of the teacher impact the comprehension and conceptual understanding of the students.
Unfortunately, if the teacher is unsure of the methods and rules involved, then begins a dislike of the subject by the students being taught. To prevent a carcinogenic spread of the dislike, the teacher ought to be suitably oriented and know precisely how to captivate the imagination of the students. If this is the case, the students enjoy learning process and even start loving the subject, making them eagely await Mathematics classes, with bated breath!
Acquiring necessary knowledge of algebraic operations, permutations and combinations, rudiments of probabilistic methods, persuasive ideas from differential and integral calculus and modern set theory will strengthen the bonds of mathematical wisdom.
From that stage, when one enters the portals of university education, general or technical, the opportunity to expand one’s horizon of mathematical initiation is stupendous. Besides, the effective use of Mathematics in Aeronautical, Agricultural, Biological, Chemical, Geographical and Physical Sciences, Engineering, Medicine, Meteorology, Robotics, Social Sciences and other branches of knowledge is indeed mind boggling.
Armed with this mathematical arsenal, the choice of a suitable career becomes very diverse. No two humans need to see eye to eye as far as such a choice is concerned, as the variety is staggering! So, it is crystal clear that studying Mathematics,at every level, is not only meaningful and worthwhile but absolutely essential.
A natural mathematical genius like Srinivasa Ramanujan was and continues to be an enigma and a Swayambhu, who could dream of extraordinary mathematical formulae, without any formal training.
A formally trained mathematician is capable of achieving laudable goals and imminent success in everything that he chooses to learn and if possible, discover for himself, the eternal truths of mathematics, provided he pursues the subject with imagination, passion, vigour and zeal.
Nothing can be so overwhelming as a long standing problem affording a unique solution, bu the creation of new tools, providing immense pleasure, a sense of reward and tremendous excitement in the voyage of discovery.
These flights of imagination and intuition form the core of research activities. With the advent of the computers, numerical algorithms gained in currency and greater precision, enabling the mathematical techniques to grow by leaps and bounds!
Until the enumeration of the Uncertainty Principle by Werner Heisenberg, in 1932, mathematics meant definite rules of certainty. One may venture to say that this is the origin of Fuzziness. Lotfi Zadeh wrote a seminal paper, entitled Fuzzy sets, Information and Control,
8, 1965, 328-353. He must be considered a remarkable pioneer who invented the subject of Fuzzy mathematics, which is the amalgam of mathematical rules and methods of probability put together to define domains of fuzziness.
Fuzzy means frayed, fluffy, blurred or indistinct. On a cold wintry day, haziness is seen around at dawn, and a person or an object at a distance, viewed through the mist, will appear hazy. This is a visual representation of fuzziness. The input variables in a fuzzy control systems are mapped into sets of membership functions known as fuzzy sets. The process of converting a crisp input value to a fuzzy value is called fuzzification.
A control system may also have various types of switches or on-off inputs along with its analog inputs, and such switch inputs will have a truth value equal to either 0 or 1.
Given mappings of input variables into membership functions and truth values, the micro controller makes decisions concerning what action should be taken, based on a set of rules. Fuzzy concepts are those that cannot be expressed as true or false, but rather as partially true!
Fuzzy logic is involved in approximating rather than precisely determining the value. Traditional control systems are based on mathematical models in which one or more differential equations that define the system’s response to the inputs will be used. In many cases, the mathematical model of the control process may not exist, or may be too expensive, in terms of computer processing power and memory, and a system based on empirical rules may be more effective.
Furthermore, fuzzy logic is more suited to low cost implementation based on inexpensive sensors, low resolution analog-to-digital converters and 4-bit or 8 bit microcontroller chips. Such systems can be easily upgraded by adding new rules/novel features to improve performance. In many cases, fuzzy control can be used to enhance the power of existing systems by adding an extra layer of intelligence to the current control system. In practice, there are several different ways to define a rule, but the most simple one employed is the max-min inference method, in which the output membership function is given the truth value generated by the underlying premise. It is important to note that rules involved in hardware are parallel, while in software they are sequential.
In 1985, interest in fuzzy systems was sparked by the Hitachi company in Japan, whose experts demonstrated the superiority of fuzzy control systems for trains. These ideas were quickly adopted and fuzzy systems were used to control accelerating, braking, and stoppage of electric trains, which led to the historic introduction, in 1987, of the bullet train, with a speed of 200 miles per hour, between Tokyo and Sendai.
During an international conference of fuzzy researchers in Tokyo, in 1987, T. Yamakawa explained the use of fuzzy control, through a set of simple dedicated fuzzy logic chips, in an inverted pendulum experiment. The Japanese soon became infatuated with fuzzy systems and implemented these methods in a wide range of astonishing commercial and industrial applications.
In 1988, the vacuum cleaners of Matsushita used micro controllers running fuzzy algorithms to interrogate dust sensors and adjust suction power accordingly. The Hitachi washing machines used fuzzy controllers to load-weight, fabric-mix and dirt sensors and automatically set the wash cycle for the optimum use of power, water and detergent.
The renowned Canon camera company developed an auto-focusing camera that used a charge coupled device to measure the clarity of the image in six regions in its field of view and use the information provided to determine if the image is in focus. It also tracks the rate of change of lens movement during focusing and controls its speed to prevent overshoot.
Work on fuzzy systems is also being done in USA, Europe, China and India. NASA in USA has studied fuzzy control for automated space docking, as simulation showed that a fuzzy control system can greatly reduce fuel consumption. Firms such as Boeing, General Motors, Allen-Bradley, Chrysler, Eaton and Whirlpool have used fuzzy logic to improve on automotive transmission, energy efficient electric meters, low power refrigerators, etc.
Researchers are concentrating on many applications of fuzzy control systems, have developed fuzzy systems and have integrated fuzzy logic, neural networks and adaptive genetic software systems, with the ultimate goal of building self-learning fuzzy control systems.
This, in my opinion, is sufficient reason to induce you to start learning mathematics!
Geetha S. Rao,
Ex Professor, Ramanujan Institute for Advanced Study in Mathematics, University of Madras,
Chepauk, Chennai 600005.
Email: [email protected]
**********************************************************************************************
More later,
Nalin Pithwa
# Analysis — Chapter 1 — Real Variables: part 6: Irrational numbers continued
6. Irrational numbers (continued).
In Part 4, we discussed a special mode of division of the positive rational numbers x into two classes, such that $x^{2}<2$ for the numbers of one class and $x^{2}>2$ for those of the others. Such a mode of division is called a section of the numbers in question. It is plain that we could equally well construct a section in which the numbers of the two classes were characterized by the inequalities
$x^{3}<2$ and $x^{3}>2$, or $x^{4}>7$ and $x^{4}<7$. Let us now attempt to state the principles of the construction of such “sections” of the positive rational numbers in quite general terms.
Suppose that P and Q stand for two properties which are mutually exclusive and one of which must be possessed by every positive rational number. Further, suppose that every such number which possesses P is less than any such number which possesses Q. Thus, P might be the property “$x^{2}<2$” and Q the property “$x^{2}>2$“. Then, we call the numbers which possess P the lower or left-hand class L and those which possess Q the upper or right hand class R. In general, both classes will exist; but, it may happen in special cases that one is non-existent and every number belongs to the other. This would obviously happen, for example, if P (or Q) were the property of being rational, or of being positive. For the present, however, we shall confine ourselves to cases in which both the classes do exist; and then it follows, as in Part 4, that we can find a member of L and a member of R, whose difference is as small as we please.
In the particular case, which we considered in Part 4, L had no greatest member and R no least. This question of the existence of greatest or least members of the classes is of the utmost importance. We observe first that it is impossible in any case that L should have a greatest member and R least. For, if l were the greatest member of L, and r the least of R, so that $l, then $(1/2)(l+r)$ would be a positive rational number lying between l and r, and so could neither belong to L nor to R, and this contradicts our assumption that every such number belongs to one class or to the other. This being so, there are but three possibilities, which are mutually exclusive. Either
(i) L has a greatest member l, or (ii) R has a least member, r, or (iii) L has no greatest member and R no least.
(In Part 4, there is an example of the last possibility.)
More later,
Nalin Pithwa
# Analysis — Chapter I — Real Variables — Part 5 — Irrational numbers continued
We have thus divided the positive rational numbers into two classes, L and R, such that (i) every member of R is greater than every member of L, and (ii) we can find a member of L and a member of R, whose difference is as small as we please, (iii) L has no greatest and R has not least member. Our common-sense notion of the attributes of a straight line, the requirements of our elementary geometry and our elementary algebra, alike demand the existence of a number x greater than all the members of L and less than all the members of R, and of a corresponding point P on $\Lambda$ such that P divides the points which correspond to members of L from those which correspond to members of R.
Let us suppose for a moment that there is such a number x and that it may be operated upon in accordance with laws of algebra, so that, for example, $x^{2}$ has a definite meaning. Then $x^{2}$ cannot either be less than or greater than 2. For suppose, for example, that $x^{2}$ is less than 2. Then, it follows from what precedes that we can find a positive rational number $\xi$ such that $\xi^{2}$ lies between $x^{2}$ and 2. That is to say, we can find a member of L greater than x; and this contradicts the supposition that x divides the members of L from those of R. Thus, $x^{2}$ cannot be less than 2, and similarly, it cannot be greater than 2. We are therefore driven to the conclusion that $x^{2}=2$, and that x is the number which in algebra we denote by $\sqrt{2}$. And, of course, this number $\sqrt{2}$ is not rational, for no rational number has its square equal to 2. It is the simplest example of what is called an irrational number.
But the preceding argument may be applied to equations other than $x^{2}=2$, almost word for word; for example, to
$x^{2}=N$, where N is an integer which is not a perfect square, or to
$latex$x^{3}=3\$ and $x^{2}=7$ and $x^{4}=23$,
or, as we shall see later on. to $x^{3}=3x+8$. We are thus led to believe for the existence of irrational numbers x and points P on $\Lambda$ such that x satisfies equations such as these, even when these lengths cannot (as $\sqrt{2}$ can) be constructed by means of elementary geometric methods.
The reader may now follow one or other of two alternative courses. He may, if he pleases, be content to assume that “irrational numbers” such as $\sqrt{2}$ and $\sqrt[5]{3}$ exist and are amenable to usual algebraic laws. If he does this, he will be able to avoid the more abstract discussions of the next few blogs.
If, on the other hand, he is not disposed to adopt so naive an attitude, he will be well advised to pay careful attention to the blogs which follow, in which these questions receive further consideration.
More later,
Nalin Pithwa
# What are worthwhile problems as per Richard Feynman, American, Physics Nobel Laureate
What are worthwhile problems as per Richard Feynman
The letter below is from Perfectly Reasonabe Deviations From The Beaten Track, a book of letters of Richard Feynman. It is one of the most moving letters that I have read. Tomonaga mentioned below shared the 1965 Nobel prize for physics along with Feynman and Schwinger.
A former student, who was also once a student of Tomonaga’s, wrote to extend his congratulations. Feynman responded, asking Mr. Mano what he was now doing. The response: “studying the Coherence theory with some applications to the propagation of electromagnetic waves through turbulent atmosphere… a humble and down-to-earth type of problem.”
Dear Koichi,
I was very happy to hear from you, and that you have such a position in the Research Laboratories. Unfortunately your letter made me unhappy for you seem to be truly sad. It seems that the influence of your teacher has been to give you a false idea of what are worthwhile problems. The worthwhile problems are the ones you can really solve or help solve, the ones you can really contribute something to. A problem is grand in science if it lies before us unsolved and we see some way for us to make some headway into it. I would advise you to take even simpler, or as you say, humbler, problems until you find some you can really solve easily, no matter how trivial. You will get the pleasure of success, and of helping your fellow man, even if it is only to answer a question in the mind of a colleague less able than you. You must not take away from yourself these pleasures because you have some erroneous idea of what is worthwhile.
You met me at the peak of my career when I seemed to you to be concerned with problems close to the gods. But at the same time I had another Ph.D. Student (Albert Hibbs) was on how it is that the winds build up waves blowing over water in the sea. I accepted him as a student because he came to me with the problem he wanted to solve. With you I made a mistake, I gave you the problem instead of letting you find your own; and left you with a wrong idea of what is interesting or pleasant or important to work on (namely those problems you see you may do something about). I am sorry, excuse me. I hope by this letter to correct it a little.
I have worked on innumerable problems that you would call humble, but which I enjoyed and felt very good about because I sometimes could partially succeed. For example, experiments on the coefficient of friction on highly polished surfaces, to try to learn something about how friction worked (failure). Or, how elastic properties of crystals depends on the forces between the atoms in them, or how to make electroplated metal stick to plastic objects (like radio knobs). Or, how neutrons diffuse out of Uranium. Or, the reflection of electromagnetic waves from films coating glass. The development of shock waves in explosions. The design of a neutron counter. Why some elements capture electrons from the L-orbits, but not the K-orbits. General theory of how to fold paper to make a certain type of child’s toy (called flexagons). The energy levels in the light nuclei. The theory of turbulence (I have spent several years on it without success). Plus all the “grander” problems of quantum theory.
No problem is too small or too trivial if we can really do something about it.
You say you are a nameless man. You are not to your wife and to your child. You will not long remain so to your immediate colleagues if you can answer their simple questions when they come into your office. You are not nameless to me. Do not remain nameless to yourself – it is too sad a way to be. now your place in the world and evaluate yourself fairly, not in terms of your naïve ideals of your own youth, nor in terms of what you erroneously imagine your teacher’s ideals are.
Best of luck and happiness.
Sincerely,
Richard P. Feynman.
An accomplished father giving heartfelt advice to a son struggling to find his way, a teacher who immediately feels from a few gestures what a pupil is going through and reaches out due to his love for his student and due to his own humility, a man who recognizes his greatness and his defects in equal measure
# Analysis — Chapter 1 — Real Variables — Part 4 Irrational numbers continued
Part 4. Irrational numbers (continued).
The result of our geometrical interpretation of the rational numbers is therefore to suggest the desirability of enlarging our conception of “number” by the introduction of further numbers of a new kind.
The same conclusion might have been reached without the use of geometrical language. One of the central problems of algebra is that of the solution of equations, such as
$x^{2}=1$, $x^{2}=2$.
The first equation has the two rational roots 1 and -1. But, if our conception of number is to be limited to the rational numbers, we can only say that the second equation has no roots; and the same is the case with such equations as $x^{3}=2$, $x^{4}=7$. These facts are plainly sufficient to make some generalization of our idea of number desirable, if it should prove to be possible.
Let us consider more closely the equation $x^{2}=2$.
We have already seen that there is no rational number x which satisfies this equation. The square of any rational number is either less than or greater than 2. We can therefore divide the rational numbers into two classes, one containing the numbers whose squares are less than 2, and the other those whose squares are greater than 2. We shall confine our attention to the positive rational numbers, and we shall call these two classes the class L, or the lower class, or the left-hand class, and the class R, or the upper class, or the right hand class. It is obvious that every member of R is greater than all the members of class R. Moreover, it is easy to convince ourselves that we can find a member of the class L whose square, though less than 2, differs from 2 by as little as possible, and a member of R whose square, though greater than 2, also differs from 2 by as little as we please. In fact, it we carry out the ordinary arithmetical process for the extraction of the square root of 2, we obtain a series of rational numbers, viz.,
1,1.4, 1.41, 1.414, 1.4142, $\ldots$
whose squares
1, 1.96, 1.9881, 1.999396, 1.99996164, $\ldots$
are all less than 2, but approach nearer and nearer to it, and by taking a sufficient number of the figures given by the process we can obtain as close an approximation as we want. And if we increase the last figure, in each of the approximations given above, by unity, we obtain a series of rational numbers
2, 1.5, 1.42, 1.415,1.413, $\ldots$
whose squares
4, 2.25, 2.0164, 2.002225, 2.00024449, $\ldots$
are all greater than 2, but approximate to 2 as closely as we please.
It follows also that there can be no largest member of L or smallest member of R. For if x is any member of L, then
$x^{2} < 2$. Suppose that $x^{2}=2-\delta$. Then we can find a member x, of L such that ${x_{1}}^{2}$ differs from 2 by less than $\delta$, and ${x_{1}}^{2}>x^{2}$ or $x_{1}>x$. Thus there are larger members of L than x; and, as x is any member of L, it follows that no member of L can be larger than all the rest. Hence, L has no largest member, and similarly, it has no smallest.
Note: A rigorous analysis of the above can be easily carried out. If you need help, please let me know and I will post it in the next blog.
More later,
Nalin Pithwa
# Analysis — Real Variables — Chapter 1 — Examples II
Examples II.
1) Show that no rational number can have its cube equal to 2.
Proof #1.
Let, if possible, $p/q, q \neq 0$, and p and q do not have any common factor and are integers. Then, if $(p/q)^{3}=2$, we have
$p^{3}=2q^{3}$. So, p contains a factor of 2. So, let $p=2k$. So, q contains a factor of 2. Hence, both p and q have a common factor have a common factor 2, contradictory to out assumption. Hence, the proof.
2) Prove generally that a rational function $p/q$ in its lowest terms cannot be the cube of a rational number unless p and q are both perfect cubes.
Proof #2.
Let, if possible, $p/q = (m/n)^{3}$ where m,n,p,q are integers, with n and q non-zero and p and q are in lowest terms. This implies that m and n have no common factor. Hence, $p=m^{3}, q=n^{3}$.
3) A more general proposition, which is due to Gauss and includes those which precede as particular cases, is the following: an algebraical equation
$z^{n}+p_{1}z^{n-1}+p_{2}z^{n-2}+ \ldots + p_{n}=0$ with integral coefficients, cannot have rational, but non-integral root.
Proof #3.
For suppose that, the equation has a root $a/b$, where a and b are integers without a common factor, and b is positive. Writing
$a/b$ for z, and multiplying by $b^{n-1}$, we obtain
$-(a^{n}/b)=p_{1}a^{n-1}+p_{2}a^{n-2}b+ \ldots + p_{n}b^{n-1}$,
a function in its lowest terms equal to an integer, which is absurd. Thus, $b=1$, and the root is a. It is evident that a must be a divisor of $p_{n}$.
4) Show that if $p_{n}=1$ and neither of
$1+p_{1}+p_{2}+p_{3}+\ldots$ and $1-p_{1}+p_{2}-p_{3}+\ldots$
is zero, then the equation cannot have a rational root.
Proof #4. Please try this and send me a solution.. I do not have a solution yet 🙂
5) Find the rational roots, if any, of $x^{4}-4x^{3}-8x^{2}+13x+10=0$.
Solution #5.
Use problem #3.
The roots can only be integral, and so find the roots by trial and error. It is clear that we can in this way determine the rational roots of any such equation.
More later,
Nalin Pithwa
# Analysis — Chapter I — part 3 — Real Variables — Irrational numbers
Part 3. Irrational numbers.
If the reader will mark off on the line all the points corresponding to the rational numbers whose denominators are 1,2,3, …in succession, he will readily convince himself that he can cover the line with rational points, as closely as he likes. We can state this more precisely as follows: If we take any segment BC on A, we can find as many rational points on it as we please on BC.
Suppose, for example, that BC falls within the segment $A_{1}A_{2}$. it is evident that if we choose a positive integer k such that
$k.BC>1$ Equation I
(The assumption that this is possible is equivalent to the assumption of what is known as the Axiom of Archimedes.)
and divide $A_{1}A_{2}$ into k equal parts, then at least one of the points of division (say P) must fall inside BC, without coinciding with either B or C. For if this were not so, BC would be entirely included in one of the k parts into which $A_{1}A_{2}$ has been divided, which contradicts the supposition I. But P obviously corresponds to a rational number whose denominator is k. Thus at least one rational point P lies between B and C. But, then we can find another such point Q between B and P, another between B and Q, and so on indefinitely; that is, as we asserted above, we can find as many as we please. We may express this by saying that BC includes infinitely many
rational points. (We will investigate the meaning of infinite more closely later).
From these considerations, the reader might be tempted to infer that an adequate view of the nature of the line could be obtained by imagining it to be formed simply by the rational points which lie on it. And, it is certainly the case that if we imagine the line to be made up of solely of the rational points, and all other points (if there are any such) to be eliminated, the figure would possess most of the properties which common sense attributes to the straight line, and would, to put the matter roughly, look and behave very much like a line.
A little further consideration, however, shows that this view would involve us in serious difficulties.
Let us look at the matter for a moment with the eye of common sense, and consider some of the properties which we may reasonably expect a straight line to possess if it is to satisfy the idea which we have formed of it in elementary geometry.
The straight line must be composed of points, and any segment of it by all the points which lie between its end points. With any such segment must be associated a certain entity called its length, which must be a quantity capable of numerical measurement in terms of any standard or unit length, and these lengths must be capable of combination with another, according to the ordinary rules of algebra, by means of addition or multiplication. Again, it must be possible to construct a line whose length is the sum or product of any two given lengths. If the length PQ along a given line is a, and the length QR, along the same straight line, is b, the length PR must be $a+b$.
Moreover, if the lengths OP and OQ, along one straight line, are 1 and a, and the length OR along another straight line is b, and if we determine the length OS by Euclid’s construction for a fourth proportional to the lines OP, OQ, OR, this length must be ab, the algebraic fourth proportional to 1, a and b. And, it is hardly necessary to remark that the sums and products thus defined must obey the ordinary laws of algebra; viz.,
$a+b=b+a$
$a+(b+c)=(a+b)+c$
$ab=ba$
$a(bc)=(ab)c$
$a(b+c)=ab+ac$
The lengths of our lines must also obey a number of obvious laws concerning inequalities as well as equalities: thus, if A, B, C are three points lying along A from left to right, we must have $AB, and so on. Moreover, it might be possible, on our fundamental line $\Lambda$ to find a point P such that $A_{0}P$ is equal to any segment whatever taken along $\Lambda$ or along any other straight line. All these properties of a line, and more, are involved in the presuppositions of our elementary geometry.
Now, it is very easy to see that the idea of a straight line as composed of a series of points, each corresponding to a rational number, cannot possibly satisfy all these requirements. There are various elementary geometrical constructions, for example, which purport to construct a length x such that $x^{2}=2$. For instance, we may construct an isosceles right angled triangle ABC such that $AB=AC=1$.. Then, if $BC=x$, $x^{2}=2$. Or we may determine the length x by means of Euclid’s construction for a mean proportional to a and 2, as indicated in the figure. Our requirements therefore involve the existence of a length measured by a number x, and a point P on $\Lambda$ such that $A_{0}P=x$, $x^{2}=2$.
But, it is easy to see that there is no rational number such that its square is 2. In fact, we may go further and say that there is no rational number whose square is $m/n$, where $m/n$ is say positive fraction in its lowest terms, unless m and n are both perfect squares.
For suppose, if possible, that $\frac {p^{2}}{q^{2}}=m/n$.
p having no common factor with q, and m no common factor with n. Thus, $np^{2}=mq^{2}$. Every factor of $q^{2}$ must divide $np^{2}$, and as p and q have no common factor, every factor of $q^{2}$ must divide n. Hence,
$n={\lambda}q^{2}$, where $\lambda$ is an integer. But, this involves $m={\lambda}p^{2}$: and as m and n have common factor, $\lambda$ must be unity. Thus, $m=p^{2}$ and $n=q^{2}$, as was to be proved. In particular, it follows by taking $n=1$, that an integer cannot be the square of a rational number, unless that rational number is itself integral.
it appears that our requirements involve the existence of a number x and a point P, not one of the rational points already constructed, such that $A_{0}P=x$ and $x^{2}=2$; and, (as the reader will remember from elementary algebra) we write $x = \sqrt {2}$.
Alternate proof.
The following alternate proof that no rational number can have its square equal to 2 is interesting.
Suppose, if possible, that $p/q$ is a positive fraction, in its lowest terms such that $(p/q)^{2}=2$. It is easy to see that this involves $(2q-p)^{2}=2(p-q)^{2}$, and so $\frac {2q-p}{p-q}$ is also another fraction having the same property. But, clearly,
$q and so $p-q. Hence, there is another fraction equal to $p/q$ and having a smaller denomination, which contradicts the assumption that $p/q$ is in its lowest terms.
In the next blog, we shall look at examples,
More later,
Nalin Pithwa
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# Ch 51: WEST Middle Grades Math: Statistics
If you need to review statistics for your WEST Middle Grades Math exam, these entertaining videos are as effective as they are enjoyable to watch. If you still have questions after viewing the lessons, ask for the instructors' support and guidance.
## WEST Middle Grades Math: Statistics - Chapter Summary
These appealing, easily-accessed videos describe the statistics-related details you're expected to know for the WEST Middle Grades Math certification assessment. As you navigate the chapter resources, you will examine:
• Measurements of average, such as mean, median and mode
• Average calculation methods
• Examples of quantitative data
• Different types of categorical variables
• Scales of measurement
• Rules for data ordering and ranking
• Methods for visual data demonstrations
• Distribution and spread within data sets
• Maximums, minimums, outliers, quartiles and interquartile range
Each of the quizzes in this chapter provides you with an opportunity to practice using the concepts and get an idea of what WEST statistics questions might look like. Refer to the video transcripts for additional review--they point out the most important statistics concepts from the lessons.
### WEST Middle Grades Math: Statistics Objectives
Statistics aren't glamorous, but WEST-certified math educators are expected to know them. These videos illustrate the concepts you need to know for the exam. Statistics, probability and discrete math questions will make up a full quarter of the assessment.
Another quarter is devoted to measurement and geometry, a third to algebra and functions, and the final seventeen percent to number sense and operations. There are 150 multiple-choice questions on the computer-administered test, and you'll be allotted four and a quarter hours to navigate them. Use the lesson quizzes to introduce yourself to the WEST format and determine how quickly you'll be able to work through the exam.
17 Lessons in Chapter 51: WEST Middle Grades Math: Statistics
Test your knowledge with a 30-question chapter practice test
Chapter Practice Exam
Test your knowledge of this chapter with a 30 question practice chapter exam.
Not Taken
Practice Final Exam
Test your knowledge of the entire course with a 50 question practice final exam.
Not Taken
### Earning College Credit
Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.
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# Does the James Bond Roulette Strategy work? (Full test with Results)
Does the James Bond Roulette Strategy work? We test this popular method and show you our results. Spoilers: we increased our bankroll by 30%. See how we did it.
The James Bond roulette strategy in practice
The James Bond Roulette strategy has been attracting a huge amount of attention in recent years. Despite 007’s edgy lifestyle and choice of career, his roulette method is surprisingly safe, with the added kick of an occasional inside win. Here we explore the strategy - and show you our results during 20+ game rounds.
What you will learn in this article:
• How to use the James Bond roulette method
• What our bankroll looked like after 20+ game rounds
• Whether the strategy is worth your time and chips
## How to use the James Bond Roulette Strategy
You start with a £200 bankroll (scale up or down depending on the size of your actual bankroll), and bet £14 on one column, £5 on another and £10 on the green slot. If none of your numbers come up, you simply double the amount you bet next time.
A great way to test out roulette strategies is via some real games, and that’s exactly what we did with some simulated roulette sessions at Pots of Gold, who have a number of European tables streamed in via Evolution Gaming’s facility in Latvia.
## James Bond Roulette Strategy Tested
So, what happened when we tried the James Bond Method? We planned 3 betting rounds of 10 spins each - a reasonable number for the purposes of illustration. Unfortunately, we couldn't keep to this number of betting rounds, as you will see later. We bet £14 on the top column (3,6,9, 12 15,18,21,24,27,30,33,36); £5 on the middle column (2,5,8,11,14,17,20,23,26,29,32,35) and £10 on the zero. Here are the outcomes:
#### Round 1
• The first spin landed on 3. At 2:1 that turned our £14 into £42, taking us up to £242 for a cost of which cost us £29, leaving us with a total bankroll of £213.
• Then a 9 was spun - another column win taking us to £226.
• Then 28 – our uncovered column. That cost us £29, taking us down to £197. Now it’s time to double up, with a total of £58 bet.
• Next came 14, taking us £30 up for a cost of £58, leaving us with £169.
• We now reduced our bet back to £29, and a 4 was spun, taking us down to £140.
• Then we doubled our bet to £58 again, and 5 was spun, taking us down to £112.
• Then we reduced our bet to £29 again and an 11 came up – taking us to £98.
• Then came a 32 appeared, taking us to £84.
• Another 3 took us to £97
Finally a 19 took us down to £68
#### Round 2
• An 11 took us to £86
• A 6 then took us to £99
• Next came a 1: This cost us £29, taking us to a round £70
• We then doubled up and spent £58. A 32 was spun, yielding £30 – so an overall loss of £28 taking us to £42
• We then dropped back to £29, and lost the lot when a 19 came up, taking us to £13.
Unable to double our bets, we wagered what we had left and were unable to continue using the James Bond method. We therefore quit with our £13 intact.
#### Round 3
This was our last chance for the day, and the numbers came up as follows:
• First to appear was a 12. This took our bankroll to £213
• Then came a 6, taking us to £226.
• Then a 33 was spun – now we were at £239.
• A 34 lost us £29, taking us to £210
• We then doubled up and spent £58 for a £30 win, taking us to £182
• Back at the £29 bet level a 1 came up, losing us £29, taking us to £153.
• We therefore had to double up again, and this time the zero came good: Great news, with a £58 bet yielding a return of £720. That takes us up by £662 to a total bankroll of £815.
• A 10 lost us £29 leaving us with £786
• We therefore doubled up, spending £58 and an 8 came up to take us down by £28 to £758
Finally, we ended up with a 36, taking us by £13 to arrive at £771.
As you can see, there will be many times in your James Bond-inspired roulette play sessions when things seem pretty bad. But the eventual, inevitable appearance of the magic zero will change everything.
We started out with three £200 bankrolls - £600 in total – and ended up with just £68, a complete disaster at £13 – and finally £771, taking the total bankroll to £852.
That’s not bad for a day at the tables, and show you just what sort of a ride you might expect.
## Should you use the James Bond Strategy?
These outcomes were taken directly from real games on real tables, and as you can see they ebb and flow as you’d expect.
But they show that decent returns are very achievable, so when trying this method don’t panic if things seem to be going out of hand.
After playing thousands of spins over a lifetime you’ll be seeing the ball land in green on plenty of occasions, and when it does whilst you’re using the James Bond strategy you can be sure of a huge upwards change in your bankroll.
To many, the idea of emulating 007 seems too good to be true, but the reality is that anyone can use his approach to roulette.
It won’t require you to perform any near-lethal stunts, take on highly-trained Soviet assassins or converse fluently in several oriental languages. You just follow the formula, and wait for the cash to start building up.
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# Probability Theory and Statistics
Your monthly electricity bill consists of a \$20 service charge, plus \$0.08 per kWh (kilowatt-hour)
used. Suppose the number of kWh used per month is W = 800 10X , where X ~ Binomial(40,0.5).
Calculate:
(a) E[C], the expected monthly cost of your electricity.
(b) The standard deviation of C.
(c) The probability P(C = E[C]).
Imagine that, before a measurement of some (discrete) random variable X you are asked to guess the
result. What should you choose? One strategy is to choose a number ? that is somehow ?closest? to
the most likely results of the measurement. Specifically, let the mean squared error be:
e=E[(X-?)2]
Explain the name and meaning of this quantity. Find the minimum of e as a function of ?, by
calculating the derivative de/d? and setting it equal to zero. What is the significance of your result?
Don't use plagiarized sources. Get Your Custom Essay on
Probability Theory and Statistics
Just from \$13/Page
Let X ~ Geometric(1/2). Derive a (simple!) formula for the CDF of X, and plot your result.
The CDF of the continuous random variable V is
?? 0 v<-5 FV(v)= c(v)2 -5=v<5 (a) Determine the value of the constant c required to make this CDF continuous. (b) What is P(V > 4)?
(c) What is fV (v)?
(d) CalculateE[V]andVar(V).
5. The Rayleigh random variable X has a parameter a > 0, and PDF
a2xe-a2x2/2 x=0
fX(x)= 0 otherwise
Prove that this PDF is normalized, and calculate the CDF of X.
hw05.pdf
Unformatted Attachment Preview
ENGR 3341
Probability Theory and Statistics
Homework #5, due 2/19
Prof. Gelb
Problems from textbook:
Section 3.3: problem 12, 13, 24
Section 4.1.4: problem 3
Section 4.4: problems 1, 2
Problems:
1. Your monthly electricity bill consists of a \$20 service charge, plus \$0.08 per kWh (kilowatt-hour)
used. Suppose the number of kWh used per month is W = 800 10X, where X ~ Binomial(40,0.5).
Calculate:
(a) E[C], the expected monthly cost of your electricity.
(b) The standard deviation of C.
(c) The probability P(C = E[C]).
2. Imagine that, before a measurement of some (discrete) random variable X you are asked to guess the
result. What should you choose? One strategy is to choose a number ? that is somehow ?closest? to
the most likely results of the measurement. Specifically, let the mean squared error be:
e = E[(X – ?)2 ]
Explain the name and meaning of this quantity. Find the minimum of e as a function of ?, by
calculating the derivative de/d? and setting it equal to zero. What is the significance of your result?
3. Let X ~ Geometric(1/2). Derive a (simple!) formula for the CDF of X, and plot your result.
4. The CDF of the continuous random variable V is
?
0
?
c(v 5)2
FV (v) =
?
1
v < -5 -5 = v < 5 5=v (a) Determine the value of the constant c required to make this CDF continuous. (b) What is P(V > 4)?
(c) What is fV (v)?
(d) Calculate E[V ] and Var(V ).
5. The Rayleigh random variable X has a parameter a > 0, and PDF
2 2
a2 xe-a x /2 x = 0
fX (x) =
0
otherwise
Prove that this PDF is normalized, and calculate the CDF of X.
1
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# Thread: Notate Coefficients of a Polynomial
1. ## Notate Coefficients of a Polynomial
Hi,
Not sure If this thread is in the right part of the forum.
Anyway, quite simply if one has a polynomial as below,
Can the coefficents of an arbitrary polynomial be notated as a series or set?
$\displaystyle (x+1)(x^4 + 2x^3 + 2x^2 + 2x + 1) \:=\:x^5 + 3x^4 + 4x^3 + 4x^2 + 3x + 1$
i.e. from this example i would need...
$\displaystyle s = [1, 3, 4, 4, 3, 1]$
If so how?
Thanks Alex
2. I dont think this would fall under the catagory of discrete mathematics.
I am not sure exactly what you want, but the first thing that came to my mind was this...
let $\displaystyle S=\{1,x,x^{2},...,x^{n-1},x^{n}\}$. This set spans $\displaystyle \mathbb{R}^{n}$ because any polynomial fucntion $\displaystyle p(x)=a_{0}+a_{1}x+...+a_{n-1}x^{n-1}+a_{n}x^{n}$ in $\displaystyle \mathbb{R}^{n}$ can be written as,
$\displaystyle p(x)=a_{0}(1)+a_{1}(x)+...+a_{n-1}(x^{n-1})+a_{n}(x^{n})$
So, in your example, your basis is the set $\displaystyle S=\{1,x,x^{2},x^{3},x^{4},x^{5}\}$ in $\displaystyle \mathbb{R}^{5}$. So any polynimial of degree 5 can be created by these vectors. Since you have coeficients $\displaystyle \{1,3,4,4,3,1\}$ we can write $\displaystyle p(x)=1(1)+3(x)+4(x^{2})+4(x^{3})+3(x^{4})+1(x^{5})$ which is your polynomial. Thus any polynomial or the form $\displaystyle p(x)=a_{0}+a_{1}x+...+a_{5}x^{5}$ can be created from the set S. The only thing that changes are your coeficients $\displaystyle (a_{0},a_{1},...,a_{5})$, so I guess you could just list them in a set.
Thats my stab at it.
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Polynomials
• May 15th 2010, 05:00 PM
VMM
Polynomials
What is the degree of the simplest polynomial with the zeroes:√2, 2√2 + i, and 2 + 2i ??
• May 15th 2010, 07:24 PM
NowIsForever
My guess would be 6 since the simplest polynomial that I can come up with that has all of these as roots is (x² - 2)(x² - 4√2 x + 9)(x² - 4x + 8).
• May 15th 2010, 07:29 PM
Bacterius
This means that $\displaystyle x - \sqrt{2}$, $\displaystyle x - (2 \sqrt{2} + \pi)$, $\displaystyle x - (2 + 2 \pi)$ are all factors of this polynomial. Therefore the simplest polynomial that has all these roots is :
$\displaystyle (x - \sqrt{2})(x - (2 \sqrt{2} + \pi))(x - (2 + 2 \pi)) = 0$
Or, equivalently :
$\displaystyle (x - \sqrt{2})(x - 2 \sqrt{2} - \pi)(x - 2 - 2 \pi) = 0$
And the degree of this polynomial is, trivially, 3 (can be checked by expanding).
• May 16th 2010, 06:52 AM
skeeter
Quote:
Originally Posted by VMM
What is the degree of the simplest polynomial with the zeroes:√2, 2√2 + i, and 2 + 2i ??
Roots or zeros of polynomials of degree greater than 2 - Topics in precalculus
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# If y is the midpoint of xz y is located at and z is located at
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mathematicsalgebra Physics
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# 傅里叶变换求解积分题2
Eufisky posted @ 2014年5月01日 04:21 in 数学分析 with tags 傅里叶变换 , 987 阅读
$\int_0^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx.}$
$\int_0^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx} = \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx} .$
\begin{align*}F\left( m \right) &= \int_{ - \infty }^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx} = \int_{ - \infty }^\infty {\frac{{\cos \left( {mx} \right)}}{{\left( {{x^2} - x + 1} \right)\left( {{x^2} + x + 1} \right)}}dx}\\&= \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\left( {x + 1} \right)\cos \left( {mx} \right)}}{{{x^2} + x + 1}}dx} - \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\left( {x - 1} \right)\cos \left( {mx} \right)}}{{{x^2} - x + 1}}dx}.\end{align*}
\begin{align*} \Rightarrow F'\left( m \right) &= - \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\left( {{x^2} + x} \right)\sin \left( {mx} \right)}}{{{x^2} + x + 1}}dx} + \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\left( {{x^2} - x} \right)sin\left( {mx} \right)}}{{{x^2} - x + 1}}dx} \\&= \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\sin \left( {mx} \right)}}{{{x^2} + x + 1}}dx} - \frac{1}{2}\int_{ - \infty }^\infty {\frac{{sin\left( {mx} \right)}}{{{x^2} - x + 1}}dx}\end{align*}
$I = \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\sin \left( {mx} \right)}}{{{x^2} + x + 1}}dx} - \frac{1}{2}\int_{ - \infty }^\infty {\frac{{sin\left( {mx} \right)}}{{{x^2} - x + 1}}dx} ,T = \frac{1}{2}\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} - \frac{1}{2}\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} .$
$I = {\mathop{\rm Im}\nolimits} T.$
$\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} - \int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx}$
$\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} .$
$\frac{{P\left( z \right)}}{{Q\left( z \right)}}{e^{i\left( {mz} \right)}} = \frac{{{e^{i\left( {mz} \right)}}}}{{{z^2} + z + 1}}$
$\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} = 2\pi i \cdot {\mathop{\rm Re}\nolimits} s\left( {\frac{{{e^{i\left( {mz} \right)}}}}{{{z^2} + z + 1}}, - \frac{1}{2} + \frac{{\sqrt 3 }}{2}i} \right).$
${\mathop{\rm Re}\nolimits} s\left( {\frac{{{e^{i\left( {mz} \right)}}}}{{{z^2} + z + 1}}, - \frac{1}{2} + \frac{{\sqrt 3 }}{2}i} \right) = \mathop {\lim }\limits_{z \to \alpha } \left( {z - \alpha } \right)\frac{{{e^{i\left( {mz} \right)}}}}{{\left( {z - \alpha } \right)\left( {z - \beta } \right)}} = \frac{{{e^{ - \frac{{\sqrt 3 }}{2}m - \frac{1}{2}im}}}}{{\sqrt 3 i}}.$
$\Rightarrow \int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} = \frac{{2\pi }}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m - \frac{1}{2}im}} = \frac{{2\pi }}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} - i\sin \frac{m}{2}} \right).$
$\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} = \frac{{2\pi }}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + i\sin \frac{m}{2}} \right).$
$\Rightarrow \int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} - \int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} = - \frac{{4\pi i}}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\sin \frac{m}{2}$
$F'\left( m \right) = I = {\mathop{\rm Im}\nolimits} T = {\mathop{\rm Im}\nolimits} \frac{1}{2}\left( {\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} - \int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} } \right) = - \frac{{2\pi }}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\sin \frac{m}{2}.$
$\Rightarrow F\left( m \right) = - \frac{{2\pi }}{{\sqrt 3 }} \cdot \left[ {\frac{{{e^{ - \frac{{\sqrt 3 }}{2}m}}}}{2}\left( { - \cos \frac{m}{2} - \sqrt 3 \sin \frac{m}{2}} \right)} \right] = \frac{\pi }{{\sqrt 3 }} \cdot {e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + \sqrt 3 \sin \frac{m}{2}} \right).$
\begin{align*} \Rightarrow \int_0^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx} &= \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx} = \frac{1}{2}F\left( m \right) \\&= \frac{\pi }{{2\sqrt 3 }} \cdot {e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + \sqrt 3 \sin \frac{m}{2}} \right) = \frac{\pi }{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\sin \left( {\frac{m}{2} + \frac{\pi }{6}} \right).\end{align*}
\begin{align*}{e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + \sqrt 3 \sin \frac{m}{2}} \right) &= \frac{2}{\pi }\int_0^\infty {\cos \left( {mx} \right)dx} \int_0^\infty {{e^{ - \frac{{\sqrt 3 }}{2}u}}\left( {\cos \frac{u}{2} + \sqrt 3 \sin \frac{u}{2}} \right)\cos \left( {ux} \right)du} \\&= \frac{{2\sqrt 3 }}{\pi }\int_0^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx}.\end{align*}
$\int_0^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx} = \frac{\pi }{{2\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + \sqrt 3 \sin \frac{m}{2}} \right){\rm{ = }}\frac{\pi }{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\sin \left( {\frac{m}{2} + \frac{\pi }{6}} \right).$
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https://blogfreely.net/waitergoose02/the-problem-on-morality-the-existence-of-bad-thing
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# The problem on Morality: The Existence of Bad thing
Today i want to first obtain the value designed for Sin(45), Cos(45) and Tan(45).
Let us have an isosceles right point of view triangle with base = height. In this article the viewpoint made by the hypotenuse with all the base is usually 45 college diplomas. By the pythogoreas theorm the square of this hypotenuse can be equal to the sum of this square of this base and the height. The square with the hypotenuse is usually thus sqrt(2) * bottom part or sqrt(2) * elevation.
Sin(45) can be hence height/length of hypotenuse = level / sqrt(2) * level = 1/ sqrt(2)
Cos (45) is identified as length of base / duration of height and as such it is base / sqrt(2) * foundation which is corresponding to 1/sqrt(2).
Tan(45) is hence Sin(45)/Cos(45) which is equal to 1 .
Let us obtain the expression to get Sin(60), Cosine(60) and Tan(60). Let us reflect on an equilateral triangle. Inside the equilateral triangular the three facets are add up to 60 degrees. Let us attract a verticle with respect between among the vertex on the opposite side. This will bisect the opposite side by really half mainly because perpendicular brand will also be a perpendicular bisector. Let us reflect on any one of the two triangles created with the verticle with respect bisector as your height. So that the length of the verticle with respect bisector is certainly nothing but sqrt( l ** l — l * * m /4) = l 1. sqrt(3)/2. By means of definition Sin(60) is consequently height of this triangle hcg diet plan hypotenuse, as a result Sin(60) can be calculated as l * sqrt(3/2) /l = sqrt(3)/2. Hence Cos(60) can be estimated as sqrt(1 – Sin(60) * Sin(60)) = sqrt(1 – ¾) = ½.
In Derivative Of sin2x is equal to twenty nine degrees. So Sin(30) = l/2 as well as l sama dengan ½ or perhaps 0. some. Using this Cos(30) can be determined as sqrt(1 – ¼) = sqrt(3)/2.
Let us get one stage further and derive ideals for Cos(15). Cosine(A + B) is termed as CosineACosineB – SinASinB hence when A sama dengan B in that case Cos(A plus B) = Cos2A as well as in other words is normally equal to Cos (A) 2. Cos(A) supports Sin(A) * Sin(A). Cos2A is comparable to sqrt(3)/2 is usually equal to Vergüenza * Cos A — Sin A good * Desprovisto A. Desprovisto A 5. Sin A fabulous can be written as one particular – Cosine A * Cos An important. So the appearance becomes only two Cosine A good * Cosine A supports 1 sama dengan sqrt(3)/2. Thus 2 Cos A 2. Cos A fabulous = (2 + sqrt(3))/2. Cos A fabulous * Cos A = (2 plus sqrt(3))/2. Consequently Cos 12-15 = Sqrt(2 + Sqrt(3))/2). Using this worth for Sin 15, Trouble 75, Cos 75, Trouble 7. 5 various. Sin a few, 75, Cos 3. 75 can be determined.
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https://sciencing.com/steps-learning-long-division-bases-other-10-10047617.html
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# Steps in Learning How to Do Long Division With Bases Other Than 10
••• Hemera Technologies/AbleStock.com/Getty Images
Print
Doing computations in a base other than ten can seem complicated, because you have always worked in base ten. Performing long division involves estimation, multiplication and subtraction, but the process is simplified by all the common math facts you have memorized since early elementary school. Since those math facts often do not apply in bases other than ten, you have to find ways to compensate for the disadvantage.
List the single-digit multiples of the divisor in the new base. As an example, here is a division problem in base seven. If you were dividing 1431 (base 7) by 23 (base 7), you would first list 23 x 1=23, 23 x 2=46, 23 x 3=102, 23 x 4=125, 23 x 5=151 and 23 x 6=204. Since you are working in base seven you do not need to multiply the divisor by more than 6. This eases the disadvantage of not knowing the multiplication facts in that base. If you were working with a different base, you would list other multiples
Choose the highest multiple that is not greater than the leading digits of the dividend. In the example, 125 would be the appropriate multiple, since 151 and 204 are both greater than 143. Write “4” above the dividend, since 23 (base 7) times 4 is 125 (base 7).
Subtract the appropriate multiple from the leading digits of the dividend. In the example, 143 (base 7) minus 125 (base 7) is 15 (base 7).
Bring down any trailing digits. In this example, bring down the "1" to make the temporary remainder 151 (base 7).
Repeat the steps until the remainder is less than the divisor. From the list of multiples, 23 x 5=151, so write “5” above the dividend to the right of the 4, and subtract 151 from 151, which leaves you with zero.
Write down any remainder greater than zero to the right of the answer, preceded by a capital “R.” In the example, the final remainder is zero, so there is no need to specify any remainder. The final answer to 1431 (base 7) divided by 23 (base 7) is 45 (base 7).
#### Tips
• When finding multiples and subtracting from the dividend, always remember that you are not working in base ten, so the usual multiplication facts may not apply. You can check your answer by converting the divisor, dividend, and answer to base ten. A calculator will probably not give the correct answer in the base you are using, unless it is capable of doing computations in bases other than ten. When working with bases larger than ten, remember that other symbols (such as the alphabet) will have to serve for digits for 11, 12, etc.
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http://fersch.de/beispiel?nr=RechtTrapez&nrform=GeoTrapFlaecheUh&datei=1003RechtTrapezGeoTrapFlaecheUh
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Geometrie-Viereck-Rechtwinkliges Trapez
• $A = \frac{a+c}{ 2}\cdot h$
1 2 3 4 5 6 7 8 9 10 11 12
$a = \frac{2\cdot A}{ h} - c$
1 2 3 4 5
$c = \frac{2\cdot A}{ h} - a$
1 2 3 4 5
$h = \frac{2\cdot A}{a+c}$
1 2 3 4 5
Beispiel Nr: 03
$\text{Gegeben:}\\\text{Grundlinie c} \qquad c \qquad [m] \\ \text{Grundlinie} \qquad a \qquad [m] \\ \text{Fläche} \qquad A \qquad [m^{2}] \\ \\ \text{Gesucht:} \\\text{Höhe} \qquad h \qquad [m] \\ \\ h = \frac{2\cdot A}{a+c}\\ \textbf{Gegeben:} \\ c=2\frac{2}{5}m \qquad a=2m \qquad A=20m^{2} \qquad \\ \\ \textbf{Rechnung:} \\ h = \frac{2\cdot A}{a+c} \\ c=2\frac{2}{5}m\\ a=2m\\ A=20m^{2}\\ h = \frac{2\cdot 20m^{2}}{2m+2\frac{2}{5}m}\\\\h=9\frac{1}{11}m \\\\\\ \small \begin{array}{|l|} \hline c=\\ \hline 2\frac{2}{5} m \\ \hline 24 dm \\ \hline 240 cm \\ \hline 2,4\cdot 10^{3} mm \\ \hline 2,4\cdot 10^{6} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline a=\\ \hline 2 m \\ \hline 20 dm \\ \hline 200 cm \\ \hline 2\cdot 10^{3} mm \\ \hline 2\cdot 10^{6} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline A=\\ \hline 20 m^2 \\ \hline 2\cdot 10^{3} dm^2 \\ \hline 2\cdot 10^{5} cm^2 \\ \hline 2\cdot 10^{7} mm^2 \\ \hline \frac{1}{5} a \\ \hline 0,002 ha \\ \hline \end{array} \small \begin{array}{|l|} \hline h=\\ \hline 9\frac{1}{11} m \\ \hline 90\frac{10}{11} dm \\ \hline 909\frac{1}{11} cm \\ \hline 9090\frac{10}{11} mm \\ \hline 9090909\frac{1}{11} \mu m \\ \hline \end{array}$
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https://www.gradesaver.com/textbooks/math/calculus/calculus-early-transcendentals-2nd-edition/chapter-6-applications-of-integration-6-10-hyperbolic-functions-6-10-exercises-page-503/50
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# Chapter 6 - Applications of Integration - 6.10 Hyperbolic Functions - 6.10 Exercises - Page 503: 50
$$f'\left( x \right) = \frac{{\left| x \right|}}{{x\sqrt {{x^2} + 4} }}$$
#### Work Step by Step
\eqalign{ & f\left( x \right) = {\operatorname{csch} ^{ - 1}}\left( {2/x} \right) \cr & {\text{find the derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left( {{{\operatorname{csch} }^{ - 1}}\left( {2/x} \right)} \right) \cr & {\text{use derivatives of the inverse hyperbolic functions}} \cr & f'\left( x \right) = - \frac{1}{{\left| {2/x} \right|\sqrt {1 + {{\left( {2/x} \right)}^2}} }}\frac{d}{{dx}}\left( {\frac{2}{x}} \right) \cr & f'\left( x \right) = - \frac{1}{{\left| {2/x} \right|\sqrt {1 + {{\left( {2/x} \right)}^2}} }}\left( { - \frac{2}{{{x^2}}}} \right) \cr & {\text{simplify}} \cr & f'\left( x \right) = \frac{{\left| x \right|}}{{2\sqrt {\frac{{{x^2} + 4}}{{{x^2}}}} }}\left( {\frac{2}{{{x^2}}}} \right) \cr & f'\left( x \right) = \frac{{\left| x \right|}}{{\frac{{\sqrt {{x^2} + 4} }}{x}}}\left( {\frac{1}{{{x^2}}}} \right) \cr & f'\left( x \right) = \frac{{\left| x \right|}}{{x\sqrt {{x^2} + 4} }} \cr}
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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http://demonstrations.wolfram.com/BlockOnAFrictionlessInclinedPlane/
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10176
# Block on a Frictionless Inclined Plane
The block speeds up as it slides down. How does the speed at the bottom depend on the steepness of the slope? Following Galileo, who did this type of experiment before Newton was born, you can investigate this using the slider and "trigger".
The Demonstration is a model showing the forces as "arrows" or "vectors". There are only two forces acting on the block: (1) the blue "normal" force, exerted by the plane on the block in a direction perpendicular or "normal" to the surface of the plane, and (2) the red gravitational force (, exerted downward by the Earth). These two forces "add" together (as vectors) to yield , which points down the plane.
Newton's second law states that the rate of change in the speed is directly proportional to As you change the angle, you can see that remains constant, while the normal force and change as required to keep equal to the vector sum of and the normal force.
### DETAILS
The key to analyzing this system successfully is to choose the correct orientation for the and axes. If you choose axes parallel and perpendicular to the plane, the analysis is (relatively) easy. You can use the conventional horizontal and vertical orientation, but be prepared for more difficult trigonometry!
Resolving into and components is the next step; the components (parallel and perpendicular to the plane) are also shown.
Physically, the block can only move parallel to the plane—assuming the plane is smooth and straight means the block is never going to jump off the surface of the plane and is confined to sliding along the surface. This means that the speed and net force in the direction, that is, perpendicular to the plane, must be zero.
Assuming that the plane is "frictionless" means that the plane does not exert any force on the block that is parallel to the surface. Therefore, any force exerted by the plane on the block must be perpendicular to the surface. This is the reason for the name "normal" force: it is always oriented "normal", or perpendicular, to the surface of the plane.
In order to have no net force on the block perpendicular to the plane, the "normal" force exerted by the plane and the component of must cancel exactly, leaving no net force on the block in the direction. As a result, the component of is the only remaining, or net, force. Thus, expected, the net force is pointing parallel to the plane, and in fact, down the plane.
Mathematically, the angle of the plane and the diagonal distance from the starting point to the bottom determines , the acceleration (or rate of change of the speed) from Newton's second law: or
This finds the time to reach the bottom from the relationship between distance and time for constant acceleration:
, or
.
This is the final speed at the bottom (or at any time ), assuming that the block starts from rest , from the definition of acceleration: , or , and, substituting for and from above,
.
This result for is much easier to get from the principle of conservation of energy (which holds because the motion is frictionless):
The kinetic energy at the bottom equals the gravitational field (potential) energy at the starting point, or, where is the vertical height of the block above the bottom of the plane before the block starts moving.
From the blue triangle, , so . Substituting this into the equation above and canceling yields
.
Solving, , which is the same as the result above.
### PERMANENT CITATION
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https://sisirch.com/qa/quick-answer-how-many-cups-is-3-oz-of-flour.html
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# Quick Answer: How Many Cups Is 3 Oz Of Flour?
## What does 3 oz of meat look like?
3 oz portion is similar in size to a deck of cards ▪ 1 oz of cooked meat is similar in size to 3 dice.
A 1-inch meatball is about one ounce.
4 oz of raw, lean meat is about 3 ounces after cooking.
3 oz of grilled fish is the size of a checkbook..
## How many tablespoons is 3 oz of flour?
10 7/8 tablespoons3 ounces flour equals 10 7/8 tablespoons.
## How do you measure 4 oz of flour?
Do you have a tablespoon? If so then 1oz of flour is approx 1 heaped tablespoon. 4oz of flour is one American “cup” which is 8 fluid oz.
## How many cups is 8 oz flour?
1 7/8 cupConvert 8 ounces or oz of flour to cups. 8 ounces flour equals 1 7/8 cup.
## How can I measure 1 cup of flour without a measuring cup?
Use an object as a reference point.A teaspoon is about the size of the tip of your finger.A tablespoon is about the size of an ice cube.1/4 cup is about the size of a large egg.1/2 cup is about the size of a tennis ball.A full cup is about the size of a baseball, an apple or a fist. X Research source
## How many cups does 4 oz equal?
0.50 cups4 oz equals 0.50 cups. 1 ounce is equivalent to 0.125 cups, and there are 0.50 cups in 4 ounces.
## How many cups is 4 oz of flour?
7/8 cupsConvert 4 ounces or oz of flour to cups. 4 ounces flour equals 7/8 cups.
## How do I measure 4 oz?
Fluid Ounces (oz) to Cups Conversion To convert fluid oz to cups, multiply the fluid oz value by 0.125 or divide by 8. For example, to calculate how many cups is 4 fl oz of water, multiply 4 by 0.125, that makes 0.5 cup is 4 fl oz.
## How can I measure 1/3 cup without a measuring cup?
Measurement Equivalents and Abbreviations3 teaspoons = 1 tablespoon.4 tablespoons = 1/4 cup.5 tablespoons + 1 teaspoon = 1/3 cup.8 tablespoons = 1/2 cup.1 cup = 1/2 pint.2 cups = 1 pint.4 cups (2 pints) = 1 quart.4 quarts = 1 gallon.More items…
## Is 1 oz the same as 1 fl oz?
How many oz in 1 fl oz? The answer is 1. We assume you are converting between ounce [US, liquid] and US fluid ounce. You can view more details on each measurement unit: oz or fl oz The SI derived unit for volume is the cubic meter.
## How can I measure 4 oz of flour without scales?
HOW DO I MEASURE FLOUR WITHOUT A SCALE?Use a spoon to fluff up the flour within the container.Use a spoon to scoop the flour into the measuring cup.Use a knife or other straight edged utensil to level the flour across the measuring cup.
## How many cups is 7 oz flour?
1 5/8 cupConvert 7 ounces or oz of flour to cups. 7 ounces flour equals 1 5/8 cup.
## How much is a tablespoon of flour in ounces?
One tablespoon of all purpose flour (APF) converted to ounce equals to 0.28 oz.
## What is 1 oz of flour in grams?
28.3495 gramsAn ounce of flour is 1/16 of a pound, and there are 28.3495 grams in one ounce. An ounce is a measure of flour weight. The ounce is a US customary and imperial unit of flour.
## How many ounces of flour equals a cup?
4 1/4 ouncesA cup of all-purpose flour weighs 4 1/4 ounces or 120 grams. This chart is a quick reference for volume, ounces, and grams equivalencies for common ingredients.
## How do you measure 3 oz?
equal to the end of your thumb, from the knuckle up. Three teaspoons equals 1 tablespoon. Choose lean poultry, fish, shellfish and beef. One palm size portion equals 3 oz.
## What is 4 oz of flour in tablespoons?
Flour Weight to Volume Conversion TableOuncesTablespoons (A.P. Flour)Tablespoons (Bread Flour)4 oz14 1/2 tbsp14 1/4 tbsp5 oz18 1/8 tbsp17 3/4 tbsp6 oz21 3/4 tbsp21 1/2 tbsp7 oz25 1/3 tbsp25 1/8 tbsp28 more rows
## How much flour is 4 oz?
Self-Rising Flour: 1 cup = 4 ounces = 113 grams.
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# Ncert solution class 11th maths pdf. Free Download NCERT Solutions Class 11 Maths PDF 2019-07-17
Ncert solution class 11th maths pdf Rating: 5,6/10 497 reviews
## NCERT Solutions for class 9 Maths In PDF [ Free Download ]*
To make it easier for them to learn, we are providing here solutions for all the students of 11th standard, such that, they can clarify their doubts for all types of questions. Your feedback is important and precious to us. Hence, by the principle of mathematical induction, statement P n is true for all natural numbers i. Pictorial representation of a function, domain, co-domain and range of a function. Definition of relation, pictorial diagrams, domain, co-domain and range of a relation. So all the students who are studying under this board have the best solution materials here.
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## NCERT Solutions Class 11 Maths Chapter 1 Sets Free PDF Download
If you want to prepare for more better performance in your examinations then these books can help you. After the shot 10 minutes break when you come back to continue the study, you will feel more powerful and energetic this time. Chapter 15: Statistics In this chapter, you will be studying important topics like Measures of dispersion, range, mean deviation, Mean deviation for ungrouped data, 1 Standard deviation and much more. This is a great material for students who are preparing for Class 11 exams. Standard equations and simple properties of parabola, ellipse and hyperbola.
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## NCERT solutions for class 11 Mathematics
You will also learn about the direction cosines, parallel and perpendicular ratios and so on. Here you will going to learn about the properties of probability. For a better understanding of this chapter, you should also see summary of Chapter 3 Trigonometric Functions , Maths, Class 11. Ans : Let the given statement be P n , i. Formulae for the special series sums. It is highly recommended that you practice thoroughly and learn properly all of the given solutions here.
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## NCERT Solutions for Class 11 Maths
Continuity and Differentiability In the chapter-5, new will learn about the continuity and differentiability with the help of simple algebraic solutions. You will understand the various factors vector algebra like: position vector, negative vector, geometrical interpretation and so on. Video — will be available soon. Standard equation of a circle. Hence, by the principle of mathematical induction, statement P n is true for all natural numbers i.
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After that you will be learning about binary operation, composite function, inverse function and so on. Otherwise you can also buy it easily online. Measuring angles in radians and in degrees and conversion from one measure to another. This solution is prepared by our hard work and highly knowledgable teachers. You may also have your own method of better and improved learning.
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## NCERT Solutions for Class 11 Maths
Additionally, to be a component of a set is likewise unclear term. Let P k be true for some positive integer k, i. These solutions are in accordance with 2019-2020 syllabus. Chapter 16: Probability Probability is the word we utilize computing the level of the conviction of occasions in perfect conditions. Real valued functions, domain and range of these functions, constant, identity, polynomial, rational, modulus, signum, exponential, logarithmic and greatest integer functions, with their graphs.
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It also supports state and central governments on educational policies. Sum, difference, product and quotient of functions. How much time you are spending daily on the studies. Here you will learn about all the vector algebra and the types. These materials are prepared by our expertise keeping on mind students understanding level. The difference of two sets A and B in this order is the set of elements which belong to A but not to B. This is the most recommended chapter to prepare for the best scoring in examination.
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## NCERT Solutions for Class 11 Maths (Updated for 2019
The curves like circles, oval, parabolas and hyperbolas are called conic segments or all the more normally conics. Just use the comment box below to leave any message. Toppr provides the last 10 years of question papers, free study materials, 1000+ hours of video lectures, mock tests, and live doubts solving. Hence, by the principle of mathematical induction, statement P n is true for all natural numbers i. After that you will learn about Bayes theorem.
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# Integral calculator
Free integral calculator - solve indefinite, definite and multiple integrals with all the steps. Type in any integral to get the solution, steps and grap Integral Calculator Calculus Add to Bookmarks How to use the Integral Calculator Type in the integral problem to solve To get started, type in a value of the integral problem and click «Submit» button. In a moment you will receive the calculation result. See a step-by-step solutio Matrix Inverse Calculator What are integrals? Integration is an important tool in calculus that can give an antiderivative or represent area under a curve. The indefinite integral of f (x) f ( x), denoted ∫ f (x)dx ∫ f ( x) d x , is defined to be the antiderivative of f (x) f ( x)
The integral calculator allows you to enter your problem and complete the integration to see the result. You can also get a better visual and understanding of the function and area under the curve using our graphing tool Free definite integral calculator - solve definite integrals with all the steps. Type in any integral to get the solution, free steps and grap
Integral Calculator An Integral Calculator is a free online tool that displays the antiderivative of the given function. BYJU'S online integral calculator tool makes the calculations faster, showing the integral value for the given function in a fraction of seconds. How to Use Online Integral Calculator Integral calculus calculator can be used to calculate improper integrals. This kind of integral is then solved by turning it into a problem of limits where c happens to approach infinity or negative infinity. Let's consider an example where one of the limits of integration is infinite and then solve it. ∫ 1 ∞ 1 x 2 d x a n d ∫ 1 ∞ 1 x d
### Integral Calculator - Symbola
• This calculator computes the definite and indefinite integrals (antiderivative) of a function with respect to a variable x. 1. For powers use ^. Example: x 1 2 = x^12 ; e x + 2 = e^ (x+2) 2. For square root use sqrt
• Integral calculator is a mathematical tool which makes it easy to evaluate the integrals. Online integral calculator provides a fast & reliable way to solve different integral queries. online integration calculator and its process is different from inverse derivative calculator as these two are the main concepts of calculus
• Using the online integral calculator is very easy, just enter the equation you need to solve. Alternatively, you can use the default button not to waste time. It is easy to find mistakes in your calculations when you can see every step of the process. Use the additional options on the calculator if you are not completely happy with the results
• How to use Integral Calculator 1 Step 1 Enter your integral problem in the input field. 2 Step 2 Press Enter on the keyboard or on the arrow to the right of the input field. 3 Step 3 In the pop-up window, select Find the Integral. You can also use the search. What is Integral in Mat
• Integral Calculator with Steps Enter the function. Use x as your variable. See Examples HELP Use the keypad given to enter functions. Use x as your variable. Click on SOLVE to process the function you entered. Here are a few examples of what you can enter. x 2 − 1 cos ( x) − 2 1 x Here is how you use the buttons GO TO HOME PAG
• integral-calculator. ar. Related Symbolab blog posts. Advanced Math Solutions - Integral Calculator, trigonometric substitution. In the previous posts we covered substitution, but standard substitution is not always enough. Integrals involving..
### Integral Calculator with Steps - 100% Fre
• imum or maximum value of intervals.. With this integral calculator, you can get step by step calculations of
• Integrals. Limits. Algebra Calculator. Trigonometry Calculator. Calculus Calculator. Matrix Calculator. Download. Topics Matrix Calculator. Type a math problem. Solve. Examples \int{ 1 }d x \int{ 3x }d x \int{ x^4 }d x \int{ 7x + 8 }d x \int{ \frac{1}{x} }d x \int{ cos(x) }d x. Quiz.
• This online calculator will calculate the integral of any function. When the integrand matches a known form, it applies fixed rules to solve the integral (e. g. partial fraction decomposition for rational functions, trigonometric substitution for integrands involving the square roots of a quadratic polynomial or integration by parts for products of certain functions)
• Definite Integral Calculator The definite integral calculator is a free online tool that displays the value of the integral function, when the lower and the upper limits are given. BYJU'S online definite integral calculator tool makes the calculations faster, where it shows the result of the integral function in a fraction of seconds
• An online integral calculator helps you to evaluate the integrals of the functions with respect to the variable involved and shows you the complete step-by-step calculations. When it comes to indefinite integral calculations, this antiderivative calculator allows you to solve indefinite integrals in no time
• The calculator decides which rule to apply and tries to solve the integral and find the antiderivative the same way a human would. Integration of constants and constant functions; Integration by Parts; Integration by Subsitution (u-substitution) Exponential and Logarithmic Functions; Trigonometric and Hyperbolic function
### Integral Calculator: Integrate with WolframAlph
1. Indefinite Integrals Calculator online with solution and steps. Detailed step by step solutions to your Indefinite Integrals problems online with our math solver and calculator. Solved exercises of Indefinite Integrals
2. What is an improper integral calculator? This is a tool used to evaluate improper Integral Calculator which works to provide the integrated value for the improper integral. The improper definite integral calculator is well manufactured to assist the users in computing complex improper integral functions in the blink of an eye
3. Evaluate an Integral Step 1: Enter an expression below to find the indefinite integral, or add bounds to solve for the definite integral. Make sure to specify the variable you wish to integrate with. Step 2: Click the blue arrow to compute the integral
4. definite integral calculator - Wolfram|Alpha. Area of a circle? Easy as pi (e). Unlock Step-by-Step. Natural Language. Math Input. NEW Use textbook math notation to enter your math. Try it
5. e whether the given function is convergent or divergent by using a convergent or divergent integral calculator. Before we start using this free calculator, let us discuss the basic concept of improper integral
6. آلة حاسبة لتكاملات محدودة. مشتقّات. مشتقّة أولى. مشتقّة ثانية. مشتقّة ثالثة. مشتقّة من رتبة أعلى. مشتقّة في نقطة. مشتقّة جزئيّة. مشتقّة دالّة ضمنيّة
7. Integral Calculator Get detailed solutions to your math problems with our Integral step-by-step calculator. Practice your math skills and learn step by step with our math solver. Check out all of our online calculators here
### Integral Calculator - Mathwa
Integral Calculator is used to integrate a function that could be in the form of either a definite or an indefinite integral. Integration is one of the most fundamental operations of calculus. It is the process of uniting infinitesimal data to find a whole The Reality of Mathematics. Mathematics is just a shorthand mental concept that simulates reality, or approximates reality or a possible reality or even an imaginary / impossible 'reality'. Mathematics is NOT reality itself. You can mathematically manipulate the alleged extra dimensions in String Theory but that doesn't mean of necessity tha Integral calculator. The integral calculator is able to calculate integrals online of the composition of common functions, using integral properties, the different mechanisms of integration and calculation online. The calculator precise calculation steps used to arrive at the result Integral Calculator computes an indefinite integral (anti-derivative) of a function with respect to a given variable using analytical integration. It also allows to draw graphs of the function and its integral. Please remember that the computed indefinite integral belongs to a class of functions F(x)+C, where C is an arbitrary constant
### Definite Integral Calculator - Symbola
Antiderivative Calculator. Integral calculator is an online tool that calculates the antiderivative of a function. It works as a definite integral calculator as well as an indefinite integral calculator and lets you solve the integral value in no time. Line integration calculator shows you all of the steps required to evaluate the integrals Integral Calculator. The integral calculator helps you compute antiderivatives and definite integrals. You can also easily calculate multiple integrals as well as use mathematical constants such as the π or Euler's number e . Traditional. Inline
Calculate a table of the integrals of the given function f(x) over the interval (a,b) using Trapezoid method. The integrand f(x) is assumed to be analytic and non-periodic. It is calculated by increasing the number of partitions to double from 2 to N It is very easy to solve integrals using calculator.here i use fx-991Ms calculator.It can solve proper integral only
Double Integral Calculator: Finding the definite integral for a function is similar as double integral. But in case of double integral, you need to perform the same operation twice. It is a bit difficult for anyone to get the double integral of an expression Evaluate an Integral Step 1: Enter an expression below to find the indefinite integral, or add bounds to solve for the definite integral. Make sure to specify the variable you wish to integrate with. Step 2: Click the blue arrow to compute the integral Integral Calculator is designed for students and teachers in Maths, engineering, phisycs and sciences in general. - Symbolic primitive, derivate and integral calculations. - System equations solver and matrix operations (Jordan form, eigenvalues, determinant, etc). - Plotting 2D and 3D functions. - Zeros and inflection points calculation
Definite Integral Calculator Free online Definite Integral Calculator tool computes the definite integral of a function over an interval using numerical integration. Give the variable range as the input and get the instant output in no time This integral calculator is a free online application that can illustrate the importance of either definite integrals or infinite integrals of all integral functions. As it shows the results in a fraction of seconds, it is a simpler way to measure integers. Only enter the integral type you want to add after installing this app from the play. Integral Calculator. An integral calculator is a mathematical instrument that makes evaluating integrals simple. Using an online integral calculator allows you to solve a variety of integral problems quickly and accurately by the help of math problem solver.Because these two are the core ideas of calculus, the online integration calculator and its methodology differ from the inverse derivative. Calculates the logarithmic integral li(x). Purpose of use Scientific activity. Bug report I notice that Logarithmic integral (chart) calculator is able to compute the li(x) function also for x higher than 1,000,000, but it is not possible to set the number of digits, whereas the logarithmic integral li(x) calculator goes on timeout for x higher than 1,000,000
The double integral calculator above computes the definite integral of your function with the x and y limits you provided. The region that it integrates over is a rectangle on the x-y plane. This 2-dimensional rectangle on the x-y plane extends upwards to the surface produced by f (x, y). The image below shows an example of a double integral. The Upside to Integral Calculator . Graph making with an equation or even given numbers isn't an effortless procedure. This calculator computes volumes for a few of the most usual straightforward shapes. Our Derivative Calculator tool supports all the most recent functions, computing and several other variables which are essential in 1 tool Here is a list of best free Integral Calculator Software to solve integrations. These integral calculator can be used to calculate and solve definite integrals and indefinite integrals.You can also use it to solve differential and integral equations. Other than calculating simple integration, these Integral calculators can also calculate and solve multiple integrals, like: double integration. Steps to use Convert Double Integral To Polar Coordinates Calculator:-. Follow the below steps to get output of Convert Double Integral To Polar Coordinates Calculator. Step 1: In the input field, enter the required values or functions. Step 2: For output, press the Submit or Solve button. Step 3: That's it Now your window will display. Improper Integral Calculator. What are improper integrals? Improper integrals are definite integrals that cover an unbounded area. One type of improper integrals are integrals where at least one of the endpoints is extended to infinity. For example, $$\int_{1}^{\infty}\frac{1}{x^2}dx$$ is an improper integral
Example: integral(fun,a,b,'ArrayValued',true) indicates that the integrand is an array-valued function. Waypoints — Integration waypoints vector. Integration waypoints, specified as the comma-separated pair consisting of 'Waypoints' and a vector of real or complex numbers. Use waypoints to indicate points in the integration interval that you. Math24.pro - this is a modern way of solving mathematics, including for comparing independent solutions with machine calculations. Using the service is convenient and understandable for every person who comes to the site for the first time. Immediately select the desired calculator, enter the necessary data for your task and click the. A double Integral Calculator is an online tool that helps to integrate a given function and obtain the value of the double integral. Double integrals can be used to find the volume under a surface and the average value of a function with two variables. To use the Double Integral Calculator, enter the values in the input boxes. How to Use Double. Integral Approximation Calculator. Integration, as we all know, is a method to calculate areas by adding consecutive slices together. It is one of the best ways to find the area of any curve drawn in between the axis. You might be thinking that such a large Integration is highly impossible to do
### Integral Calculator - Free online Calculato
1. The Integral Calculator is a free online integration calculator that has indefinite and definite integrals (antiderivatives) as well as how to integrate functions with different variables. Users can use it to compare the answers done manually
2. Integrals and Area Under the Curve. Integrals and Area Under the Curve. Log InorSign Up. Define your favorite function: 1. f x = x 2 − 1. 2. Compute the integral from a to b: 3. ∫ b a f t dt. 4. a = 0. 5. b = 2. 6.
3. An improper integral calculator is an online calculator that helps us calculate the integral with set limitations. A convergent or divergent integral calculator can also be used to rate if a given function is convergent or divergent. We can enter the function, upper, and lower limits in this calculator
4. The integral calculator is able to calculate integrals online of the composition of common functions, using integral properties, the different mechanisms of integration and calculation online. The calculator precise calculation steps used to arrive at the result. If the.
Integral Calculation. Integral calculus emerged when the scientists were searching how to compute the areas of surfaces, plane figures, solid body volumes, and addressed problems in hydrodynamics, statistics and other physics domains. An example of a problem that contributed to the development of integral calculus is finding the object's law. Double Integral Calculator. Our free handy Double Integral Calculator tool is aimed at giving the double integral of a function within fraction of seconds. The only one thing you need to do is just give your function and range for two variables as input and obtain the value as output immediately after hitting the calculate button Definite Integral Calculator. Author: Dr. Jack L. Jackson II. Topic: Definite Integral. Illustrating the Definite Integral. This app can be used to find and illustrate approximate values for any definite integral. The definite integral of a function over an interval [a, b] is the net signed area between the x-axis and the graph of the function. The calculator will help you calculate the double integral online. The double integral is a generalization of the notion of a definite integral to the two-dimensional case. The double integral of a function f (x, y) over a domain D is the limit of the integral sum lim S (d → 0), if it exists How to use the Derivative of the Integral Calculator. 1 Step 1. Enter your derivative problem in the input field. 2 Step 2. Press Enter on the keyboard or on the arrow to the right of the input field. 3 Step 3. In the pop-up window, select Find the Derivative of the Integral. You can also use the search
### Integral Calculator Best online Integration by parts
Steps to use Spherical Coordinates Integral Calculator:-. Follow the below steps to get output of Spherical Coordinates Integral Calculator. Step 1: In the input field, enter the required values or functions. Step 2: For output, press the Submit or Solve button. Step 3: That's it Now your window will display the Final Output of your Input Muito mais que uma calculadora de integral on-line. O Wolfram|Alpha é uma ótima ferramenta para calcular primitivas e integrais definidas, integrais duplas e triplas, e integrais impróprias. Também mostra gráficos, formas alternadas, e outras informações relevantes para melhorar a sua intuição matemática. Learn more about Surface integral of a vector field over a surface. Author: Juan Carlos Ponce Campuzano. Topic: Surface
### Online Integral Calculator - mathportal
1. In mathematics, an integral assigns numbers to functions in a way that describes displacement, area, volume, and other concepts that arise by combining infinitesimal data. The process of finding integrals is called integration.Along with differentiation, integration is a fundamental, essential operation of calculus, and serves as a tool to solve problems in mathematics and physics involving.
2. Free handy Triple Integral Calculator is here to calculate the triple integral of a function within a short span of time. Just provide the input function and bounds in the input section of the calculator and press on the calculate button to get the result immediately. Triple Integral Calculator: If you are struggling to figure out the triple.
3. Definite Integral: Enter a function for f(x) and use the sliders to choose the upper and lower limits of integration. Note that the definite integral only gives area if the function is above/on the x-axis for all x in the interval [a,b]
4. This calculus video tutorial explains how to calculate the definite integral of function. It provides a basic introduction into the concept of integration..
5. La calculadora de integrales está disponible en línea y de forma gratuita y le permite resolver problemas de integración definidos e indefinidos. Podrá encontrar respuestas, grafos, así como formas alternativas. Potenciada por Wolfram|Alpha
6. The Double Integral Calculator is a free tool that shows the resultant value of a definite double integral in the specified order. The online double integral calculator with steps from Protonstalk speeds up the calculation, displays result to you in few seconds and helps you in your work
### Integral Calculator The best Integration Calculato
• Indefinite Integral Calculator with steps. This is a very simple tool for Indefinite Integral Calculator. Follow the given process to use this tool. ☛ Step 1: Enter the complete equation/value in the input box i.e. across Provide Required Input Value: ☛ Step 2: Click Enter Solve Button for Final Output
• Double Integral Calculator - Free . Instructions to find a Double Integral (2 step process) 1) Solve the 1. Integral (Ex: Integrate x*y^3 with respect to x which yields 1/2*x^2*y^3) 2) Integrate the answer to 1) with respect to the other variable. (Ex: Integrate 1/2*x^2*y^3 with respect to y to get 1/2*x^2*1/4y^4
• A double integral is used in order to calculate the areas of regions, find the volumes of a given surface, or also the mean value of any given function in a plane region. How Do you Find The Integrals? Finding integrals is the inverse operation of finding the derivatives. A few integrals are remembered as formulas
• Finding the zero space (kernel) of the matrix online on our website will save you from routine decisions. We provide explanatory examples with step-by-step actions
• When int cannot compute the value of a definite integral, numerically approximate the integral by using vpa. syms x f = cos (x)/sqrt (1 + x^2); Fint = int (f,x, [0 10]) Fint =. Fvpa = vpa (Fint) Fvpa =. To approximate integrals directly, use vpaintegral instead of vpa
• The online definite and indefinite integral calculator is a tool with an advanced algorithm that can solve complex problems like integration. The integral calculator/solver is best because it provides solutions with steps and all possible methods for the integration
### ∫ Integral Calculator Online - with step
• This calculator for solving indefinite integrals is taken from Wolfram Alpha LLC. All rights belong to the owner! Indefinite integral. Finding an indefinite integral is a very common task in math and other technical sciences. Actually solution of the simplest physical problems seldom does without a few calculations of simple integrals
• Mit dem kostenlosen Online-Integralrechner können Sie bestimmte und unbestimmte Integrale lösen. Antworten, Graphen, alternative Darstellungen. Mit der Leistungskraft von Wolfram|Alpha
• Integral Calculator Level 2 v.1.0.0.0 Integral Calculator Level 2 1.0.0.0 is released to be a helpful and creative utility which can be found most useful by scientists, engineers, professors, and students. This calculator calculates definite integrals of differentiable functions.; Triple Integral Calculator Level 2 v.1.0.0.1 Triple Integral Calculator Level 2 1.0.0.1 is designed as a useful.
• Integral Calculator Derivative Calculator Algebra Calculator Standard Deviation Calculator Percentage Calculator Time and Date Calculator Time Card Calculator GPA Calculator Grade Calculator What is Limits? Calculus is known as one of the critical fields of study in Mathematics..
• Integral Calculator • With Steps! Above, enter the function to integrate. Variable of integration, integration bounds and more can be changed in Options. Click Go! to start the integral/antiderivative calculation. The result will be shown further below. How the Integral Calculator Work. www.integral-calculator.co
• Free Integral Approximation calculator - approximate the area of a curve using different approximation methods step-by-step This website uses cookies to ensure you get the best experience. By using this website, you agree to our Cookie Policy
• Amazon's Choice for integral calculator (CASIO) Scientific Calculator (FX-991ESPLUS) 4.7 out of 5 stars 8,800. $44.99$ 44. 99. Get it as soon as Fri, Apr 2. FREE Shipping by Amazon. Only 9 left in stock - order soon. FX-115ESPLUS Casio FX115ESPLUS Scientific Calculator Black 1-Pack 1
Improper integrals can't be computed employing a normal Riemann integral. for instance , the integral. (1) is an improper integral. Easy Steps to use Improper Integral Calculator. This is a very simple tool for Improper Integral Calculator. Follow the given process to use this tool Hi all! This video was made to help calculus students score well on their final. The steps to solving a Definite Integral with the calculator is simple:Press.. Double definite integrals online calculator. The double integral computed over some domain, such as domain S, shown in the picture: The double integral over specified domain is calculated by reducing it to the repeated integral: The repeated integral consists of two simple definite integrals, for each of which, based on analysis of domain S.
If you can parametrize the curve, you can always just throw the resulting (normal) integral into Wolfram Alpha, since it doesn't matter how ugly the parametrization makes things if you aren't doing it by hand. 1. r/math. Welcome to math. 1.7m This calculator for solving definite integrals is taken from Wolfram Alpha LLC. All rights belong to the owner! Definite integral. The online service at OnSolver.com allows you to find a definite integral solution online. The solution is performed automatically on the server and after a few seconds the result is given to the user
### Integral Calculator with Steps - Online & Free
1. Free Summation Calculator. The free tool below will allow you to calculate the summation of an expression. Just enter the expression to the right of the summation symbol (capital sigma, Σ) and then the appropriate ranges above and below the symbol, like the example provided
2. In this section we introduce the idea of a surface integral. With surface integrals we will be integrating over the surface of a solid. In other words, the variables will always be on the surface of the solid and will never come from inside the solid itself. Also, in this section we will be working with the first kind of surface integrals we'll be looking at in this chapter : surface.
3. Definite Integrals Calculator. Definite integrals calculator. Input a function, the integration variable and our math software will give you the value of the integral covering the selected interval (between the lower limit and the upper limit)
4. e if a particular location receives sufficient amounts of light for plants to grow well. Low light plants require between 5-10, medium light.
5. Double Integral Calculator - A Complete Overview. Double Integral Calculator: The indispensable double calculator over computes the definite indispensable. It include a feature with the x as well as y limitations you supplied. The area that it integrates above is a rectangular shape on the x-y plane. This 2-dimensional rectangle on the x-y.
6. A convolution is an integral that expresses the amount of overlap of one function g as it is shifted over another function f. It therefore blends one function with another. For example, in synthesis imaging, the measured dirty map is a convolution of the true CLEAN map with the dirty beam (the Fourier transform of the sampling distribution)
### Integral Calculator with Steps - Open Omni
• g to the details of this calculator. Inputs. The following are the inputs of the calculator. The function which need to be integrated three times. The order of integratio
• The calculation of the primitive is closely linked to the resolution of the integrals defined by the fundamental theorem of the integral calculation: in fact, the integral of a function is equal to the difference of the values of the primitive on the integration extremes
• To improve this 'Exponential integral Ei(x) (chart) Calculator', please fill in questionnaire. Age Under 20 years old 20 years old level 30 years old level 40 years old level 50 years old level 60 years old level or over Occupation Elementary school/ Junior high-school studen
• If any of the integration limits of a definite integral are floating-point numbers (e.g. 0.0, 1e5 or an expression that evaluates to a float, such as exp(-0.1)), then int computes the integral using numerical methods if possible (see evalf/int).Symbolic integration will be used if the limits are not floating-point numbers unless the numeric=true option is given
• Line integrals are a natural generalization of integration as first learned in single-variable calculus. Rather than an interval over which to integrate, line integrals generalize the boundaries to the two points that connect a curve which can be defined in two or more dimensions
• HELP. Use the keypad given to enter functions. Use x as your variable. Click on SOLVE to process the function you entered. Here are a few examples of what you can enter. Processes the function entered. Removes all text in the textfield. Deletes the last element before the cursor. Shows the alphabet
### آلة حاسبة لتكاملات - Symbola
Calculate a table of the integrals of the given function f(x) over the interval (a,b) using Trapezoid, Midpoint and Simpson's methods. The integrand f(x) is assumed to be analytic and non-periodic. It is calculated by increasing the number of partitions to double from 2 to N Advanced Math Solutions - Integral Calculator, inverse & hyperbolic trig functions In the previous post we covered common integrals ( click here ). There are a few more integrals worth mentioning before we continue with integration by parts; integrals involving inverse & hyperbolic trig functions 1. level 2. Goyomaster. Op · 1y. Website. To give an example, I have no clue how to say to wolfram alpha integrate (z 8/ (z8 - 1) )dz in the square of vertexes 1+i, -1+i. -1-i. 1-i. I have no idea how to give a curve to WolframAlpha to integrate in. Nor I know how to avoid literally giving it the curve and giving it an equivalent calculation. 1 Answer (1 of 16): Wolfram Alpha may suit you. Here is its link: Wolfram|Alpha: Making the world's knowledge computabl Integral Calculator Cesar Berard Education This is the first windows store app that solves definite integrals, even double integrals! The app includes trigonometric, logarithm and even more available expressions! Solve definite integrals and get a graph which shows the region to integrate
The Integral keyword modifies the method of computation and use of two-electron integrals and their derivatives. Options. Integration Grid Selection Option. Grid= grid-name. Specifies the integration grid to be used for numerical integrations SolveMyMath's Taylor Series Expansion Calculator. Input the function you want to expand in Taylor serie : Variable : Around the Point a = (default a = 0) Maximum Power of the Expansion You would type: fnInt (x^2,x,0,1) which would give the output. 0.3333333333. Double integrals with constant bounds: let's say you wanted to integrate the function z = x^2 + y^2 over the box 0 < x < 1 and 2 < y < 3. You could type: fnInt (fnInt (x^2 + y^2, x, 0, 1), y, 2, 3) However, depending on your calculator, you might not be able to nest. What I want to do in this video, you could have also used a typical graphing calculator to come up with the same result, and it would have actually evaluated the definite integral. So let's see how you do that. Now this, what I'm doing here, you could do this for a traditional, what if you're dealing with Cartesian coordinates, rectangular. Integrating using Samples¶. If the samples are equally-spaced and the number of samples available is $$2^{k}+1$$ for some integer $$k$$, then Romberg romb integration can be used to obtain high-precision estimates of the integral using the available samples. Romberg integration uses the trapezoid rule at step-sizes related by a power of two and then performs Richardson extrapolation on these.
49 Centroids by Integration Wednesday, November 7, 2012 Centroids from Functions ! If we can define the height of the loading diagram at any point x by the function q(x), then we can generalize out summations of areas by the quotient of the integrals y dx x i q(x) ( ) ( ) 0 0 L ii L i xq x dx x qx dx = ∫ ∫ 50 Centroids by Integratio Skills Comments on: Double Integral Calculator - A Complete Overvie
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# Test: Differential Equation (Competition Level) - 2
## 30 Questions MCQ Test Mathematics (Maths) Class 12 | Test: Differential Equation (Competition Level) - 2
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This mock test of Test: Differential Equation (Competition Level) - 2 for JEE helps you for every JEE entrance exam. This contains 30 Multiple Choice Questions for JEE Test: Differential Equation (Competition Level) - 2 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Differential Equation (Competition Level) - 2 quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Test: Differential Equation (Competition Level) - 2 exercise for a better result in the exam. You can find other Test: Differential Equation (Competition Level) - 2 extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1
### The degree of the differential equation satisfying
Solution:
putting x = sin A and y = sin B in the given relation, we get
cos ,4 + cos B = a(sin A - sin B)
⇒ A B = 2 cot-1 a
⇒ sin-1 x - sin-1 y = 2 cot-1 a Differentiating w.r.t. x, we get
Clearly, it is a differential equation of degree one
QUESTION: 2
Solution:
QUESTION: 3
### Differential equation of the family of circles touching the line y = 2 at (0, 2) is
Solution:
QUESTION: 4
The differential equation of all parabolas whose axis are parallel to the y-axis is
Solution:
The equation of a member of the family of parabolas having axis parallel toy-ax is is
y = Ax2 + Bx + C ......(1)
where A, B, and C are arbitrary constants
Differentiating equation (1) w.r.t. x, we .....(2)
which on again differentiating w.r.t. jc gives ......(3)
Differentiating (3) w.r.t. x, we get
QUESTION: 5
The differential equation of all circles which pass through the origin and whose centers lie on the y-axis is
Solution:
) be the centre on y-axis then its radius will be k as it passes through origin. Hence its equation is
QUESTION: 6
The differential equation whose general solution is given by, y = where c1, c2, c3, c4, c5 are arbitrary constants,
is
Solution:
QUESTION: 7
The equation of the curves through the point (1, 0) and whose slope is
Solution:
QUESTION: 8
The solution of the equation log(dy/dx) = ax + by is
Solution:
QUESTION: 9
Solution of differential equation dy – sin x sin ydx = 0 is
Solution:
QUESTION: 10
The solution of the equation
Solution:
Putting u = x - y, we get du/dx = 1 - dy/dx. The given equation can be written as 1 - du/dx = cos u
QUESTION: 11
Solution:
QUESTION: 12
The general solution of the differential equation
Solution:
QUESTION: 13
The solutions of (x + y + 1) dy = dx is
Solution:
Putting x + y 1 = u, we have du = dx + dy and the given equations reduces to u(du - dx) = dx
QUESTION: 14
The slope of the tangent at (x, y) to a curve passing through is given by then the equation of the curve is
Solution:
On integration, we get
This passes through (1,π/4), therefore 1 = log C.
QUESTION: 15
The solution of (x2 + xy)dy = (x2 + y2)dx is
Solution:
∴ equation reduces to
QUESTION: 16
The solution of (y + x + 5)dy = (y – x + 1) dx is
Solution:
The intersection of y – x + 1 = 0 and y + x + 5 = 0 is (-2, -3).
Put x = X - 2, y = Y - 3
The given equation reduces to
putting Y = vX, we get
QUESTION: 17
The slope of the tangent at (x, y) to a curve passing through a point (2, 1) is then the equation of the curve is
Solution:
∴ equation (1) transforms to
QUESTION: 18
The solution of satisfying y(1) = 1 is given by
Solution:
Rewriting the given equation as
Hence y2 = x(1 +x) - 1 which represents a system of hyperbola.
QUESTION: 19
Solution of the equation where
Solution:
The given differential equation can be written as which is linear differential equation of first order.
QUESTION: 20
A function y = f(x) satisfies (x + 1) f ' (x) - 2 (x2 + x)f(x) = 5, then f(x) is
Solution:
QUESTION: 21
The general solution of the equation
Solution:
QUESTION: 22
The solution of the differential equation
Solution:
Integrating, we get the solutions
QUESTION: 23
Which of the following is not the differential equation of family of curves whose tangent from an angle of π/4 with the hyperbola xy = c2
Solution:
QUESTION: 24
Tangent to a curve intercepts the y-axis at a point P. A line perpendicular to this tangent through P passes through another point (1, 0). The differential equation of the curve is
Solution:
The equations of the tangent at the point
The coor dinates of the point P are
The slope of the perpendiculer line
which is the rquried differential equation to the curve at y = f(x).
QUESTION: 25
The curve for which the normal at any point (x, y) and the line joining the origin to that point from an isosceles triangle with the x-axis as base is
Solution:
It is given that the triangle OPC is an isosceles triangle.
Therefore, OM= MG = sub-normal
On integration, we get x2 -y2 = C, which is a rectangular hyperbola
QUESTION: 26
The curve satisfying the equation and passing through the point (4, - 2 ) is
Solution:
It passes through the point (4, -2).
QUESTION: 27
A normal at P(x, y) on a curve meets the x-axis at Q and N is the foot of the ordinate at then the equation of curve given that it passes through the point (3,1) is
Solution:
Let the equation of the curve be y = f(x).
It is lincar differential equation.
QUESTION: 28
The equation of the curve passing through (2, 7/2) and having gradient is
Solution:
This passes through (2, 7/2),
Thus the equation of the curve is
QUESTION: 29
A normal at any point (x, y) to the curve y = f(x) cuts a triangle of unit area with the axis, the differential equation of the curve is
Solution:
Equation of normal at point p is
Y – y = (dx/dy)(X – x)
QUESTION: 30
The differential equation of all parabola each of which has a latus rectum 4a and whose axis parallel to the x-axis is
Solution:
Equations to the family of parabolas is (y - k)2 = 4a (x - h)
(substing y - k from equations (1))
Hence the order is 2 and the degree is 1.
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0
# What is the greatest common factor of -30x to the second power y to the third plus 15x to the second y to the second plus 60x to the fourth y to the third?
Updated: 11/2/2022
Wiki User
8y ago
The GCF is 15x^2y^2
Wiki User
8y ago
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Q: What is the greatest common factor of -30x to the second power y to the third plus 15x to the second y to the second plus 60x to the fourth y to the third?
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