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# What is the term”exclamation mark” in mathematics?
The answer is easy. Below are some approaches to inform as soon as an equation can be really actually a multiplication equation and also how to add and subtract using”exclamation marks”. For instance,”2x + 3″ imply”multiply two integers from about a few, making a value add up to 2 and 3″.
By way of example,”2x + 3″ are multiplication by write a paper online three. In addition, we can add the values of 2 and 3 collectively. To add the worth, we’ll use”e”that I” (or”E”). With”I” means”include the worth of one to the worth of 2″.
To add the worth , we can certainly do this similar to that:”x – y” means”multiply x by y, making a worth equal to zero”. For”x”y”, we’ll use”t” (or”TE”) for the subtraction and we’ll use”x y y” to solve the equation.
You might feel that you are not supposed to utilize”e” in addition as”I” implies”subtract” however it’s perhaps not so easy. https://www.masterpapers.com For instance, to express”2 – 3″ approaches subtract from three.
So, to add the values that we utilize”t”x” (or”TE”), which might be the numbers of the worth to be included. We will use”x” to subtract the price of a person in the price of both 2 plus that will provide us precisely exactly the outcome.
To multiply the worth , we certainly can certainly do this like this:”2x + y” necessarily imply”multiply two integers by y, creating a price add up to two plus one”. You may know this is really just a multiplication equation when we use”x ray” to subtract one from 2. Or, it can be”x – y” to subtract from 2. Be aware that you can publish the equation using a decimal position and parentheses.
Now, let us take an example. Let’s say that we would like to multiply the value of”nine” by”5″ and we all have”x nine”y https://www.eurachem.org/images/stories/Guides/pdf/recovery.pdf twenty-five”. Then we’ll use”x – y” to subtract the price of one in the worth of two.
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Our curriculum is spiral
Please note that our virtual Singapore Math Grade 3 curriculum is spiral and it provides for the review of the important concepts that students learned in Grade 2. Our online K-5 math curriculum is aligned with all standard Singapore Math textbook series and it includes all content that these series cover, from Kindergarten grade through 5th grade.
Our Singapore Math for 3rd Grade may introduce some topics one grade level earlier or postpone coverage of some topics until grade 4. In the few instances where 3rd grade level units don’t exactly align between our curriculum and textbooks, you will still be able to easily locate the corresponding unit in our program by referring to the table of contents one grade below or above.
Correspondence to 3A and 3B
For your reference, the following topics in our curriculum correspond to Singapore Math practice Grade 3 in levels 3A and 3B:
Singapore Math 3a
Multiplication and division, multiplication tables of 2, 5, and 10, multiplication tables of 3 and 4, multiplication tables of 6, 7, 8, and 9, solving problems involving multiplication and division, and mental math computation and estimation.
Singapore Math 3b
Understanding fractions, time, volume, mass, representing and interpreting data, area and perimeter, and attributes of two-dimensional shapes.
Student prior knowledge
Prior to starting third grade Singapore Math, students should already know how to relate three-digit numbers to place value, use place-value charts to form a number and compare three-digit numbers. The initial lessons in the Singapore Math 3rd Grade are both a review and an extension of content covered in the prior grade that include mental addition of 1-digit number to a 2-digit number and counting by 2s, 5s, and 10s.
3rd Grade Singapore Math Scope and Sequence
• Multiplication And Division
This unit covers understanding of multiplication and division. In this unit students will extend their knowledge of making equal groups to formalize their understanding of multiplication and division. The focus of this unit is on understanding multiplication and division using equal groups, not on memorizing facts. Students will learn how the multiplication symbol to represent addition of quantities in groups.
• Multiplication Tables Of 2, 5, And 10
This unit covers multiplying by 2 using skip-counting, multiplying by 2 using dot paper, multiplying by 5 using skip-counting, multiplying by 5 using dot paper, multiplying by 10 using skip-counting, dividing using related multiplication facts of 2, 5, or 10. Students will learn building multiplication tables of 2, 5, and 10 to formalize their understanding of multiplication and division for facts 2, 5, and 10. Students will learn to find their division facts by thinking of corresponding multiplication facts.
• Multiplication Tables Of 3 And 4
This unit covers multiplying by 3 using skip-counting, multiplying by 3 using dot paper, multiplying by 4 using skip-counting, multiplying by 4 using dot paper, dividing using related multiplication facts of 3 or 4. Students will learn building multiplication tables of 3 and 4 to formalize their understanding of multiplication and division for facts 3 and 14. Students will learn to find their division facts by thinking of corresponding multiplication facts.
• Multiplication Tables Of 6, 7, 8, And 9
This unit covers multiplication properties, multiplying by 6, multiplying by 7, multiplying by 8, multiplying by 9, dividing using related multiplication facts of 6, 7, 8, or 9. Students will learn building multiplication tables of 6, 7, 8, and 9 to formalize their understanding of multiplication and division for facts 6, 7, 8, and 9. Students will learn to find their division facts by thinking of corresponding multiplication facts.
• Solving Problems Involving Multiplication And Division
This unit covers solving one- and two-step word problems involving multiplication and division. Students will use a part-whole and comparison models to solve word problems involving multiplication and division.
• Mental Computation And Estimation
This unit covers learning mental math strategies to solve multiplication and division problems. Students will use place value to round whole numbers.
• Understanding Fractions
This unit covers parts and wholes, fractions and number lines, comparing unit fractions, equivalence of fractions, and comparing like fractions. Students will learn fractional notation that include the terms “numerator” and “denominator.” Students will understand that a common fraction is composed of unit fractions and they will learn to compare unit fractions.
• Time
This unit covers telling time, adding time, subtracting time, and time intervals. Students will review and practice to tell time to the minute, learn telling intervals of time in hours, convert units of time between hours, minutes, seconds, days and weeks.
• Volume
This unit covers understanding of volume, comparing volume, measuring and estimating volume, and word problems involving volume.
• Mass
This unit covers measuring mass in kilograms, comparing mass in kilograms, measuring mass in grams, comparing mass in grams, and word problems involving mass.
• Representing And Interpreting Data
This unit covers scaled picture graphs and scaled bar graphs, reading and interpreting bar graphs, and line plots. Students will learn to sort data into groups and categories and use numerical data to interpret bar graphs and line plots.
• Area And Perimeter
This unit covers understanding of area, measuring area using square centimeters and square inches, measuring area using square meters and square feet, area and perimeter, solving problems involving area and perimeter. Students will learn to find and measure area of figures in square units that include square centimeters, square inches, square meters and square feet.
• Attributes Of Two-Dimensional Shapes
This unit covers categories and attributes of shapes and partitioning shapes into equal areas. Students will understand that figures with different shapes can have the same area.
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# Height of the room
Given the floor area of a room as 24 feet by 48 feet and space diagonal of a room as 56 feet. Can you find the height of the room?
Correct result:
c = 16 ft
#### Solution:
We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!
Tips to related online calculators
Pythagorean theorem is the base for the right triangle calculator.
#### You need to know the following knowledge to solve this word math problem:
We encourage you to watch this tutorial video on this math problem:
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• Find diagonal
Find the length of the diagonal of a cuboid with length=20m width=25m height=150m
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# Equation solver with square root
This Equation solver with square root supplies step-by-step instructions for solving all math troubles. Our website will give you answers to homework.
## The Best Equation solver with square root
In this blog post, we will be discussing about Equation solver with square root. As any math student knows, calculus can be a difficult subject to grasp. The concepts are often complex and require a great deal of concentration to understand. Fortunately, there are now many calculus solvers available that can help to make the subject more manageable. These tools allow you to input an equation and see the steps involved in solving it. This can be a great way to learn how to solve problems on your own. In addition, calculus solvers with steps can also help you to check your work and ensure that you are getting the correct answer. With so many helpful features, it is no wonder that these tools are becoming increasingly popular among math students of all levels.
Word phrase math is a mathematical technique that uses words instead of symbols to represent numbers and operations. This approach can be particularly helpful for students who struggle with traditional math notation. By using words, students can more easily visualize the relationships between numbers and operations. As a result, word phrase math can provide a valuable tool for understanding complex mathematical concepts. Additionally, this technique can also be used to teach basic math skills to young children. By representing numbers and operations with familiar words, children can develop a strong foundation for future mathematics learning.
A rational function is any function which can be expressed as the quotient of two polynomials. In other words, it is a fraction whose numerator and denominator are both polynomials. The simplest example of a rational function is a linear function, which has the form f(x)=mx+b. More generally, a rational function can have any degree; that is, the highest power of x in the numerator and denominator can be any number. To solve a rational function, we must first determine its roots. A root is a value of x for which the numerator equals zero. Therefore, to solve a rational function, we set the numerator equal to zero and solve for x. Once we have determined the roots of the function, we can use them to find its asymptotes. An asymptote is a line which the graph of the function approaches but never crosses. A rational function can have horizontal, vertical, or slant asymptotes, depending on its roots. To find a horizontal asymptote, we take the limit of the function as x approaches infinity; that is, we let x get very large and see what happens to the value of the function. Similarly, to find a vertical asymptote, we take the limit of the function as x approaches zero. Finally, to find a slant asymptote, we take the limit of the function as x approaches one of its roots. Once we have determined all of these features of the graph, we can sketch it on a coordinate plane.
A binomial solver is a math tool that helps solve equations with two terms. This type of equation is also known as a quadratic equation. The solver will usually ask for the coefficients of the equation, which are the numbers in front of the x terms. It will also ask for the constants, which are the numbers not attached to an x. With this information, the solver can find the roots, or solutions, to the equation. The roots tell where the line intersects the x-axis on a graph. There are two roots because there are two values of x that make the equation true. To find these roots, the solver will use one of several methods, such as factoring or completing the square. Each method has its own set of steps, but all require some algebraic manipulation. The binomial solver can help take care of these steps so that you can focus on understanding the concept behind solving quadratic equations.
Basic mathematics is the study of mathematical operations and their properties. The focus of this branch of mathematics is on addition, subtraction, multiplication, and division. These operations are the foundation for all other types of math, including algebra, geometry, and trigonometry. In addition to studying how these operations work, students also learn how to solve equations and how to use basic concepts of geometry and trigonometry. Basic mathematics is an essential part of every student's education, and it provides a strong foundation for further study in math.
## Help with math
It has helped me more than my math teacher, of course there are a few problems here and there like not scanning the exercises correctly and it never scans "±" this sign but apart from that I really love the app. Although it would be nice to not have to pay for explanations, maybe other features but explanation is an important part.
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App that solves math problems with a picture Math solution calculator Free online math websites Math probability solver Geometry tutor online
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# Period Of Function
by -2 views
Period of sinusoidal functions from equation. You can figure this out without looking at a graph by dividing with the frequency which in this case is 2.
Graphing Trigonometric Functions Graphing Trigonometric Functions Neon Signs
### Periodic functions are used throughout science to describe oscillations waves and other phenomena that exhibit periodicity.
Period of function. This is why this function family is also called the periodic function family. Tap for more steps. Therefore in the case of the basic cosine function f x.
When this occurs we call the horizontal shift the period of the function. If a function has a repeating pattern like sine or cosine it is called a periodic function. Amplitude a Let b be a real number.
You might immediately guess that there is a connection here to finding points on a circle. F x k f x. Horizontal stretch is measured for sinusoidal functions as their periods.
For basic sine and cosine functions the period is 2 π. Any function that is not periodic is called aperiodic. The period is defined as the length of one wave of the function.
The period is the length on the x axis in one cycle. However the amplitude does not refer to the highest point on the graph or the distance from the highest point to the x axis. Midline of sinusoidal functions from equation.
The x-value results in a unique output eg. The period of a sinusoid is the length of a complete cycle. By using this website you agree to our Cookie Policy.
Periodic Functions A periodic function occurs when a specific horizontal shift P results in the original function. More formally we say that this type of function has a positive constant k where any input x. The period is defined as the length of a functions cycle.
Period of a Function The time interval between two waves is known as a Period whereas a function that repeats its values at regular intervals or periods is known as a Periodic Function. According to periodic function definition the period of a function is represented like f x f x p p is equal to the real number and this is the period of the given function f x. In other words a periodic function is a function that repeats its values after every particular interval.
Trig functions are cyclical and when you graph them youll see the ups and downs of the graph and youll see that these ups and downs. Where f x P f x for all values of x. Replace with in the formula for period.
Amplitude and Period of Sine and Cosine Functions The amplitude of y a sin x and y a cos x represents half the distance between the maximum and minimum values of the function. This is the currently selected item. Any part of the graph that shows this pattern over one period is called a cycle.
Notice that in the graph of the sine function shown that f x sin x has period. Period of sinusoidal functions from graph. The distance between and is.
Midline amplitude and period review. The absolute value is the distance between a number and zero. The period of a periodic function is the interval of x -values on which the cycle of the graph thats repeated in both directions lies.
In this case one full wave is 180 degrees or radians. So the period of or is. You are partially correct.
Khan Academy is a 501c3 nonprofit organization. The period of a periodic function is the interval of x -values on which one copy of the repeated pattern occurs. This is the currently selected item.
A periodic function repeats its values at set intervals called periods. Period can be defined as the time interval between the two occurrences of the wave. A periodic function is a function that repeats its values at regular intervals for example the trigonometric functions which repeat at intervals of 2π radians.
The period of the function can be calculated using. Our mission is to provide a free world-class education to anyone anywhere. Grade 12 trigonometry problems and questions on how to find the period of trigonometric functions given its graph or formula are presented along with detailed solutions.
So how to find the period of a function actually. Free function periodicity calculator – find periodicity of periodic functions step-by-step This website uses cookies to ensure you get the best experience. In the problems below we will use the formula for the period P of trigonometric functions of the form y a sin bx c d or y a cos bx c d and which is given by.
The period is the length of the smallest interval that contains exactly one copy of the repeating pattern. A function is just a type of equation where every input eg.
Amplitude Period Phase Shift Vertical Translation And Range Of Y 3 Math Videos Translation Period
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Students Will Match 10 Graphs To 10 Sine Or Cosine Equations By Finding The Amplitude And Period Of Each Function Students Will Then Graphing Sines Equations
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READ: Which One Of The Following Is An Example Of A Period Cost?
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https://fiveable.me/ap-stats/unit-6/justifying-claim-based-confidence-interval-difference-population-proportions/study-guide/m6d2wvAbSyMqmNwEyt43?
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Unit 6
6.9 Justifying a Claim Based on a Confidence Interval for a Difference of Population Proportions
2 min readjune 5, 2020
Josh Argo
AP Statistics📊
Bookmarked 4.3k • 246 resources
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Interpretation
When interpreting a confidence interval for the difference in two population proportions, we need to show that the bounds given by the confidence interval give us an estimate of what the difference in our two population proportion are within. Our interpretation and interval is also strongly based on what our confidence level is. The standard confidence level is 95%.
When completing a confidence interval on the AP Statistics Exam, you will generally be graded on if you included the following three components:
👉 Including confidence level (given in problem)
👉 Including that our interval is making inference about the difference in population proportions, not sample proportions.
👉 Including context of the problem.
If you do these three things and get the correct answer, you will be on your way to 💯.
Testing a Claim
When we are testing a claim using a confidence interval, we want to see if 0 is included in our interval. If 0 is included in our interval, it is quite possible that there is no difference in the two population proportions we are testing. If 0 is not included in our interval, we have reason to suspect that the two population proportions are in fact different.
Image Courtesy of cbssports
Example
Recall from Unit 6.8 we constructed a confidence interval for the difference in proportions for shots made for Michael Jordan and Lebron James. We got the following output from our calculator:
A correct way to interpret this would be:
"We are 95% confident that the true difference in the population proportions for shots made between Michael Jordan and Lebron James is between (0.063, 0.133). Since 0 is not included in our interval, we have reasonable evidence that the two population proportions are actually different"
So it appears that MJ is better than Lebron. ¯\_(ツ)_/¯
However, some things to note here is that we took a sample from both of their first seasons. If we refer back to basketball-reference.com and compare the two players 10th seasons in the league, we might find a VERY different result. Another possible confounding variable is the teammates these two players have been surrounded by throughout their career.
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# How to solve differential equations
Differential equations make use of mathematical operations with derivatives. Solving differential equations is a very important, but also hard concept in calculus. There exist more methods of solving this kind of exercises.
Firstly, we will have a look on how first order differential equation can be solved. Below, you will find a series of methods of solving this kind of differential equations. The first one is the separation method of variables. Practically, as the name says itself, you have to separate the variables, obtaining an equality of two functions with different variables. Then you integrate the expression and you obtain an equality of two integrals. Then you will solve the integrals and find the solution.
For the homogeneous differential equations, we use the substitution method and we reduce the equation to the variable separable. Having an exercise in which you have to solve the differential equation, you firstly have to figure out what kind of differential equation is the equation, so you know what method it's better to use.
Another method to solve differential equation is the exact form method. You know your equation is in an exact form if it has the following form: M(x,y) dx + N(x,y) dy = 0, where M and N are the functions of x and y in such a way that:
A differential equation is called linear as long as the dependent variable and its derivatives occur in the first degree and are not multiplied together.
f to fn are the functions of x.
In order to figure out how to solve differential equation, you firstly have to determine the order of the differential equation. For example, for the second order differential equation there is a more special method of finding the solution: divide the second order differential equation in 2 parts: Q(x)=0 and Q(x) is a function of x. For both members calculate the auxiliary equation and find the complementary function. Next, if Q(x) is a part of the equation, find the particular integral of the equation. In the end, sum up the complementary function with the particular integral.
## Solving differential equations video lesson
Ian Roberts Engineer San Francisco, USA "If you're at school or you just deal with mathematics, you need to use Studygeek.org. This thing is really helpful." Lisa Jordan Math Teacher New-York, USA "I will recommend Studygeek to students, who have some chalenges in mathematics. This Site has bunch of great lessons and examples. " John Maloney Student, Designer Philadelphia, USA " I'm a geek, and I love this website. It really helped me during my math classes. Check it out) " Steve Karpesky Bookkeeper Vancuver, Canada "I use Studygeek.org a lot on a daily basis, helping my son with his geometry classes. Also, it has very cool math solver, which makes study process pretty fun"
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Refraction & Reflection of Plane Waves using Huygens Principle
# Refraction & Reflection of Plane Waves using Huygens Principle | Physics Class 12 - NEET PDF Download
As we know that when light falls on an object, it bends and move through the material, this is what refraction is. Also when the light bounces off the medium it is called a reflection. Let us know study reflection and refraction of waves by Huygen’s principle.
Reflection using Huygens Principle
We can see a ray of light is incident on this surface and another ray which is parallel to this ray is also incident on this surface. Plane AB is incident at an angle ‘ i ‘ on the reflecting surface MN. As these rays are incident from the surface, so we call it incident ray. If we draw a perpendicular from point ‘A’ to this ray of light, Point A, and point B will have a line joining them and this is called as wavefront and this wavefront is incident on the surface.
These incident wavefront is carrying two points, point A and point B, so we can say that from point B to point C light is travelling a distance. If ‘ v ‘ represents the speed of the wave in the medium and if ‘ r ‘ represents the time taken by the wavefront from the point B to C then the distance
BC = vr
In order the construct the reflected wavefront we draw a sphere of radius vr from the point A. Let CE represent the tangent plane drawn from the point C to this sphere. So,
AE = BC = vr
If we now consider the triangles EAC and BAC we will find that they are congruent and therefore, the angles ‘ i ‘ and ‘r ‘ would be equal. This is the law of reflection
### Refraction using Huygen’s principle
We know that when a light travels from one transparent medium to another transparent medium its path changes. So the laws of refraction state that the angle of incidence is the angle between the incident ray and the normal and the angle of refraction is the angle between the refracted ray and the normal.
The incident ray, reflected ray and the normal, to the interface of any two given mediums all lie in the same plane. We also know that the ratio of the sine of the angle of incidence and sine of the angle of refraction is constant.
A plane wave AB is incident at an angle i on the surface PP' separating medium 1 and medium 2. The plane wave undergoes refraction and CE represents the refracted
wavefront. the figure corresponds to v2 < v1 so that the refracted waves bends towards the normal.
We can see a ray of light is incident on this surface and another ray which is parallel to this ray is also incident on this surface. As these rays are incident from the surface, so we call it incident ray.
Let PP’ represent the medium 1 and medium 2. The speed of the light in this medium is represented by v1 and v2. If we draw a perpendicular from point ‘A’ to this ray of light, Point A, and point B will have a line joining them and this is called as wavefront and this wavefront is incident on the surface.
If ‘ r ‘ represents the time taken by the wavefront from the point B to C then the distance,
BC = v1r
So to determine the shape of the refracted wavefront, we draw a sphere of radius v2r from the point A in the second medium. Let CE represent a tangent plane drawn from the point C on to the sphere. Then, AE = v2r, and CE would represent the refracted wavefront. If we now consider the triangles ABC and AEC, we readily obtain
where’ i ‘ and ‘ r ‘ are the angles of incidence and refraction, respectively. Substituting the values of v1 and v2 in terms of we get the Snell’s Law,
n1 sin i = n2 sin r
The document Refraction & Reflection of Plane Waves using Huygens Principle | Physics Class 12 - NEET is a part of the NEET Course Physics Class 12.
All you need of NEET at this link: NEET
## Physics Class 12
105 videos|425 docs|114 tests
## FAQs on Refraction & Reflection of Plane Waves using Huygens Principle - Physics Class 12 - NEET
1. What is refraction and reflection of plane waves?
Ans. Refraction is the bending of a wave as it passes from one medium to another, while reflection is the bouncing back of a wave after striking a surface. In the context of plane waves, refraction and reflection refer to the behavior of the wavefronts as they encounter different mediums or surfaces.
2. How does Huygens Principle explain refraction and reflection of plane waves?
Ans. According to Huygens Principle, every point on a wavefront acts as a source of secondary spherical wavelets. These secondary wavelets combine to form the new wavefront. In the case of refraction, the change in speed of the wavefront as it enters a new medium causes the wavefront to change direction. In reflection, the wavefront encounters a surface and the secondary wavelets bounce back, resulting in the reflection of the wave.
3. Can you explain how refraction and reflection affect the direction of plane waves?
Ans. Refraction causes a change in the direction of plane waves when they pass from one medium to another. This change occurs due to the change in speed of the wavefront in different mediums. The angle at which the wavefront enters the new medium, known as the angle of incidence, determines the angle at which it is refracted, known as the angle of refraction. Reflection, on the other hand, causes the wavefront to bounce back in the opposite direction when it encounters a surface.
4. What factors determine the extent of refraction and reflection in plane waves?
Ans. The extent of refraction depends on the change in speed and the angle of incidence of the wavefront. The greater the difference in speed between the two mediums, the greater the change in direction of the wavefront. Additionally, the angle of incidence plays a role in determining the angle of refraction. The extent of reflection depends on the nature of the surface encountered by the wavefront, as well as the angle of incidence. Smooth surfaces tend to have more reflection compared to rough surfaces.
5. How are refraction and reflection of plane waves useful in practical applications?
Ans. Refraction and reflection of plane waves have numerous practical applications. For example, in optics, lenses are designed based on the principles of refraction to focus or diverge light. In telecommunications, the reflection of radio waves off the ionosphere allows for long-distance communication. Refraction is also used in the field of medicine for techniques such as ultrasound imaging and laser eye surgery.
## Physics Class 12
105 videos|425 docs|114 tests
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# Often asked: What Is A Space Figure?
A space figure or three-dimensional figure is a figure that has depth in addition to width and height. Some common simple space figures include cubes, spheres, cylinders, prisms, cones, and pyramids. A space figure having all flat faces is called a polyhedron.
## What is the difference between a space figure and a solid figure?
A plane figure is two-dimensional, and a solid figure is three-dimensional. The difference between plane and solid figures is in their dimensions. The same shape takes on extra dimension by adding additional points and lines to give the shape height, width and depth.
## What are the example of spatial figures?
In mathematics, spatial figures are defined simply as three-dimensional objects. For example, a basketball or a cardboard box are spatial figures that we have likely encountered in our lives.
## How are space figures useful?
Three-dimensional geometry, or space geometry, is used to describe the buildings we live and work in, the tools we work with, and the objects we create. They made important discoveries and consequently they got to name the objects they discovered. That’s why geometric figures usually have Greek names!
## What is the space figure of volleyball?
Dimensions. The playing court is 18m long and 9m wide and is surrounded by a free zone 3m wide on all sides. The space above the playing area is known as the free playing space and is a minimum of 7m high from the playing surface.
## Is a figure in space or space figure?
Space figures are figures whose points do not all lie in the same plane. In this unit, we’ll study the polyhedron, the cylinder, the cone, and the sphere. Polyhedrons are space figures with flat surfaces, called faces, which are made of polygons. Prisms and pyramids are examples of polyhedrons.
You might be interested: What large body of water borders mexico on the east
## Can prisms have circular bases?
A triangular prism has a triangle as its base, a rectangular prism has a rectangle as its base, and a cube is a rectangular prism with all its sides of equal length. A cylinder is another type of right prism which has a circle as its base.
## What is a solid figure?
Solid figures are basically three-dimensional objects, which means that they have length, height and width. Because the solid figures have three dimensions, they have depth and take up space in our universe. Solid figures are identified according to the features that are unique to each type of solid.
## What do you call a space figure that has two circular bases and a curved surface?
A cylinder is similar to a prism, but its two bases are circles, not polygons. Also, the sides of a cylinder are curved, not flat.
## What jobs do you need geometry for?
Jobs that use geometry
• Animator.
• Mathematics teacher.
• Fashion designer.
• Plumber.
• Game developer.
• Interior designer.
• Surveyor.
## Can we live without geometry?
Without Geometry things would have been very challenging in Day to day life as well in various technological fields. Lines, Angles, Shapes, 2d & 3d designs plays a vital role in designing of home and commercial infra, mechanical and engineering design. This is feasible only because of Geometry.
## How do you explain geometry to a child?
Geometry is a kind of mathematics that deals with shapes and figures. Geometry explains how to build or draw shapes, measure them, and compare them. People use geometry in many kinds of work, from building houses and bridges to planning space travel.
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## How far apart should volleyball posts be?
Posts should be placed 1m (3′-4”) from each side line, 36′-8” from each other. A recommended free or clearance zone of at least 10 ft is recommended. For the most versatile facility, it is recommended to install poles 36′-8” from each other to allow for both competition and recreational play.
## Who created volleyball?
Volleyball has come a long way from the dusty-old YMCA gymnasium of Holyoke, Massachusetts, USA, where the visionary William G. Morgan invented the sport back in 1895.
## Why does a volleyball court need a free zone?
The volleyball court is surrounded by a free zone. The free zone is the area outside the court that players may enter to make a play on the ball. The free zone should be at least 3 meters wide from the court. If any part of the ball crosses the net directly above or outside the antenna, the ball is out of play.
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# Mechanics Question
A lift of mass Mkg is being raised by a vertical cable attached to the top of the lift. A person of mass mkg stands on the floor inside the lift, as shown in Figure 1. The lift ascends vertically with constant acceleration 1.4 m s. The tension in the cable is 2800 N and the person experiences a constant normal reaction of magnitude 560 N from the floor of the lift. The cable is modelled as being light and inextensible, the person is modelled as a particle and air resistance is negligible.By writing an equation of motion for the person only, and an equation of motion for the lift only, find the value of M and the value of m.
So you need to apply Newton's second law, resultant force =mass x acceleration, on the person and lift separately.
You should get 2 equations to solve simultaneously.
Original post by GhostWalker123
A lift of mass Mkg is being raised by a vertical cable attached to the top of the lift. A person of mass mkg stands on the floor inside the lift, as shown in Figure 1. The lift ascends vertically with constant acceleration 1.4 m s. The tension in the cable is 2800 N and the person experiences a constant normal reaction of magnitude 560 N from the floor of the lift. The cable is modelled as being light and inextensible, the person is modelled as a particle and air resistance is negligible.By writing an equation of motion for the person only, and an equation of motion for the lift only, find the value of M and the value of m.
It would first be helpful to draw a diagram when attempting any mechanical problem such as this.
Motion of lift:
F=Ma
2800-Mg=M(1.4)
g=-9,8
Rearrange for M: M=250kg
Motion of person:
F=ma
560-mg=m(1.4)
g=-9,8
Rearrange for m:
Rearrange for m: m=50kg
Original post by Drogonmeister
It would first be helpful to draw a diagram when attempting any mechanical problem such as this.
Motion of lift:
F=Ma
2800-Mg=M(1.4)
g=-9,8
Rearrange for M: M=250kg
Motion of person:
F=ma
560-mg=m(1.4)
g=-9,8
Rearrange for m:
Rearrange for m: m=50kg
This is wrong for the motion of the lift u have to consider newtons 3rd law and hence there should be a downwards reaction force of 560 acting on the lift from the man. Even tho we ignore the man he still exerts a force on the floor of the lift and this force will be the same as that ofbthe lift exerting 560 to the man. The reason they dont cancel is because 1 force only acts of lift and the other force only acts on man. So ur calculations shld be 2800-560-Mg=M(1.4) and so on
Original post by Drogonmeister
It would first be helpful to draw a diagram when attempting any mechanical problem such as this.
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1. ## havent a clue!
havent a clue!
6x2+x-15=0
2. Originally Posted by jenko
havent a clue!
6x2+x-15=0
Let's try this using the method I posted in your other thread.
Mulitply 6 and -15: -90
Now list all pairs of factors of -90:
1, -90
2, -45
3, -30
6, -15
9, -10
10, -9
15, -6
30, -3
45, -2
90, -1
Now which of these pairs add to 1? The 10 and -9, of course. So 1 = 10 - 9:
$6x^2 + x - 15 = 0$
$6x^2 + (10 - 9)x - 15 = 0$
$6x^2 + 10x - 9x - 15 = 0$
$(6x^2 + 10x) + (-9x - 15) = 0$
$2x(3x + 5) - 3(3x + 5) = 0$
$(2x - 3)(3x + 5) = 0$
So set each factor equal to 0:
$2x - 3 = 0 \implies x = \frac{3}{2}$
or
$3x + 5 = 0 \implies x = -\frac{5}{3}$
-Dan
3. BTW, after a lot of practice of factorials, you will be able to look at them for a couple seconds and know what the constants will be.
I understand your plight. When I was a kid learning about factoring, my teacher didn't even show me about the additives and multiples that make up the constants right away. Needless to say, I taught it to myself.
4. i know the basics of mathematics e.g multiples etc it jus some of the terminologies that are used which i have never heared of or are unsure of ite been a while since i have done this last time i done then wen i was in high school two years ago and ever since that day i forgot it all as i thought i would never use it on it later life which i realy regret otherwise i wouldnt be here now or better still would be here with even more harder questions for you guys to help me with!
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## How to calculate the interest you can earn
Here’s a quick explanation on how to calculate the interest that you can earn on your savings (or pay on your debt).
Let’s use an example of R1,500 (don’t worry about the currency – the maths is the same) and you deposit it in your bank account that earns 3.5% interest.
First thing to note is that the interest rate is always given as the annual rate (unless very specifically detailed otherwise). If you use a calculator (or a spreadsheet such as Excel) and do the following : 1500 x 3.5% the answer you get is 52.50.
Thus R1,500 invested for 1 year at 3.5% interest will earn R52.50. That’s easy!
To calculate what the monthly interest is, simply divide 52.50 by 12 and you will get R4.38 (rounded up) interest for 1 month. You could be more accurate and divide 52.5 by 365.4 (number of days in the year) and then multiply by the exact number of days in the month (e.g. 30).
E.g. 52.5 divided 365.4 multiplied by 30 = R4.31 (a very slight difference to the first answer we received)
So, if you invest R1,500 for 1 year at an interest rate of 3.5%, the interest at the end of the year would be R52.50. However, banks and all financial institutions calculate interest on a daily basis and pay it out monthly. If you withdrew each months interest (and leave the original amount of money in the account) then you would effectively earn “simple” interest – just as we calculated above and you would have R52.50 (assuming no fees or charges)
However, if you leave the interest that you earn each month in your account, then the bank would calculate the next months interest on your original amount plus whatever extra interest is already in your account. So each month you would earn slightly more. This is what is called “compound” interest.
In this case the difference is not great, but the larger the starting amount is, and the longer time you keep reinvesting the interest, the larger the difference is! In fact, after a few years, compound interest can be really large. Remember of course that we are assuming that no fees or charges apply.
This example may seem simple, but there is no need to complicate things!
## Find Extra Money
We’d all love some extra money I’m sure, so here are some steps to literally find it.
Step 1 : Look at your bank statement
Look at your bank statement as well as your credit card statement for the past month and make a list of all automatic payments that are deducted (eg internet bill, mobile phone, insurance, Netflix, etc)
Step 2 : Analyze all payments
One-by-one look at each automatic payment and ask:
• Do I need this service / subscription or whatever it is?
• Do I need all the features I pay for? Can I downgrade at all or pay for less?
• If it is a store card, do you need the extra’s such as insurance, magazine, etc?
• If it is a debt repayment, how much interest am I paying? What must I do to pay it off?
Step 3 : Insurance
Pay special attention to your insurance payment. Look at your latest insurance policy or contact the company for it. Are the values correct? Now, get new quotes and see if you can find a cheaper option.
Look at optional features you may have on any insurance, store card or debt policies. Perhaps you have the same benefits offered by different policies; in that case you can cancel some.
Step 4 : Look for wasted money
Look at your monthly expenses and take special note of the following:
• Eating Out / Take-Aways
• Groceries
• Clothing
• Entertaining at home
• Alcohol bought
• Luxuries
• Telephone
• Electricity
• Internet
• Make-up
Can you spend less on any of the above? Try for just 1 month to spend less than the previous month and you will see that you really can survive quite easily!
Important note is that what you save today should not be spent tomorrow, don’t feel tempted to spend your savings on other things. See when saving is not really saving.
Step 5 : Be strict with yourself
Make a list of every area where you feel you can save a little and be strict with yourself over the next month to actually do this!
## When saving ins’t saving
I am constantly looking at how and where I spend my money and deciding whether what I am doing is worth it or not. There needs to be balance between enjoying live but at the same time living well within ones means and achieving ones financial goals. All it takes to be conscious about what you spend. I have come across times though when I think I am saving money when in fact I am not.
### Hosting people vs Going out
The first instance is the fallacy that entertaining at home is cheaper than going out. This can be true, but not necessarily all the time. It is great to host friends and family and one should enjoy every moment of it; life is too short not too. I am definitely not implying that one should not host people in your home, but the point of this post is about saving money. If you specifically decide to host people at your home instead of going out (in order to save) then you need to ensure that you do actually spend less money.
Think of the average restaurant bill that you would be paying if you went out, and now look at what you are about to spend in order to host your friends. Flowers, candles, snacks, wine, groceries, dessert, etc… It is easy to go overboard and spend more than you would have. (Especially if you’re married to a chef, which is the case for me.) The point is not to be stingy though. If you wish to save money then calculate the costs of going versus what you would like to spend when entertaining and chose the cheaper option.
### A picnic vs lunch out
As with the first example, if you decide to go on a picnic in order to save money (rather than an expensive restaurant), then be sure to do some calculations. It’s funny how in our minds we convince ourselves that because we’re taking the cheaper option, we can now spend more. A bottle of bubbly, expensive cheeses, some tapas, etc… It’s really easy to spend more on the picnic than in a restaurant.
###### The point of these examples is that if you choose to do something in order to save some money, keep that in mind and be sure to actually save! Don’t let yourself be tempted to buy the more expensive items to compensate for the cheaper option. You will actually end up spending more.
Now, take the money you calculated you would save and immediately transfer it into your savings account! Do it now before you find something else to spend it on!
## Calculate the Future Value of an investment
In this article we looked at the effect of extra payments into ones home loan (mortgage bond). That’s all good and well, but what if you don’t have a loan? Or perhaps you would rather invest in a Unit Trust or Interest Bearing bank account.
Microsoft Excel has a simple formula called “FV” which will help you to calculate the future value of your investment. This does not account for changing interest rates or complicated scenarios, but it will give you a simple tool to give you an idea of how your money can grow.
To start with a simple example, let’s save 600 each month for 10 years (120 months) at an interest of 6%. The answer as you see is 98,327.61.
That’s pretty easy. Let’s quickly look at the input parameters:
Rate: This is the interest rate per period. Generally speaking, financial institutions will quote an annual interest rate. In this example it is 6%. To get the monthly interest rate we simply use 6%/12. If we change the example an decide to make quarterly payments we would need to use 6%/4.
Nper: This is the number of periods (in total) for which we will make a payment. If we wish to pay monthly, this is the number of months.
Pmt: This is the amount we wish to pay each period. This figure cannot change over the lifetime of the investment. But, see further down how we can circumvent this potential problem. Excel expects the payment to be a negative figure as payments out of your account would “minus” from your account. You will see that if you use a positive figure, your answer will show negative. The figure will be the same though. (If that is confusing, try it yourself)
Pv: The present value. If you are adding money to an existing bank account, use Pv as the present value in the account. If this to calculate a new investment leave it blank or type a zero in the box.
Type: This is to specify whether the payment is being made at the beginning of each period, or at the end. This will affect the interest that you earn. By default Excel will assume you are paying at the end of each period.
So, in the example above we assumed a constant 600 payment each month for 10 years. But what if you decide to pay 10% more each year? For that we will need to spice things up a bit. See how I have calculated the Future Value for 12 months at a time. I then use the answer as my Present Value for the following year. The monthly payment is simply increased by 10% in each row.
## How much is your coffee costing you?
I love coffee! The very thought of drinking less makes me quiver with fear. But, sometimes one must take the emotion out of decisions and just look at cold, hard facts.
In a previous article on being more conscious with cash, I discussed how I buy 2 coffees each day, along with a breakfast and lunch. I’m not good at planning meals to bring to work but I have decided to spend slightly less. The savings I calculated was only R150 per week, but I’ve decided to look at how this could affect my home loan (mortgage bond) if I pay R600 extra per month.
These figures may not make sense to you if you use a different currency and your countries interest rates may be significantly different. I’ll post something soon about how you can do these calculations yourself.
The calculations can get messy so take note of the following assumptions:
Home Loan Value: R1,500,000
Interest Rate: 11% (annual)
Total Loan period: 20 years
For this example I will assume 2 years of the loan are already complete, and that up to now no additional payments have been made. Also, the R600 p.m. will remain constant (with other words I won’t save more in years to come)
This may all sound complicated, but the end result is simple to see. If I pay R600 p.m. extra into my bond until my bond is paid off I will:
• Save a total of R236,000 (rounded to nearest thousand)
• Pay the loan off 22 months earlier
That’s pretty amazing don’t you think? Considering that I am making this extra saving by cutting down on my weekly coffee/food expenses at my office. In fact, it was quite easy to find the extra R150 per week. Imagine if I actually analyze my expenses properly and relook at my insurance policies, health care, mobile phone contract, bank charges, etc. Imagine the savings I can make then!
What if you don’t have a home loan?
That’s really not a problem. If you invest this money in either a unit trust or interest bearing account for the next 15 – 20 years you will have quite a large lump sum. We’ll look at how to use the Future Value formula in Excel to calculate this.
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## Want to keep learning?
This content is taken from the UNSW Sydney's online course, Through Engineers' Eyes: Engineering Mechanics by Experiment, Analysis and Design. Join the course to learn more.
4.7
## UNSW Sydney
Skip to 0 minutes and 9 seconds We learned earlier that a plane might crash if its centre of gravity is outside permissible limits. How can you check it? You can use the method of composite bodies.
Skip to 0 minutes and 24 seconds This table shows how to find the CG, that is the centre of gravity, of an empty aircraft by weighing the load on each wheel. The location of the CG is obtained by dividing the total moment by the total weight. More on this later. But, how do you check the CG position for each flight when the numbers of passengers and the amount of baggage and fuel might vary? There is a standard calculation and, to implement it, we can use a table like this. There are several sets of calculations here. One with just the pilot, one fully loaded, and others with various amounts of fuel. It must be OK with all of these scenarios if the aircraft is going to be safe.
Skip to 1 minute and 14 seconds The table assumes that we know the weight and CG of each component. Together, they make up our composite body. We’ve already got the weight and the CG of the bare aircraft by weight. We can now add the effect of having a pilot, a passenger, baggage, and fuel. These are the components of our composite body. We did an experiment with composite bodies. Our one was made up of squares, triangles, and circles. We can find the weight and the CG of each of these components, then combine them. But we’re going to start with a simpler example. Suppose we had two spheres connected by a weightless bar. One sphere is twice the mass, hence twice the weight of the other.
Skip to 2 minutes and 11 seconds Here are two free-body diagrams. One shows the two weights. The other shows a combined weight at the centre of gravity. Notice that the CG is not in the middle. It is nearer the larger mass. But how much nearer? We could find the position experimentally by supporting it and finding the point of balance. But that is sometimes impractical. In that case, we can calculate the position. Here’s how. These two representations must be equivalent, which means that they must generate the same force in any direction, and must generate the same moment about any axis. We can express this mathematically. It’s related to equilibrium, so you won’t be surprised to learn that you can check for equivalence by these equations.
Skip to 3 minutes and 9 seconds Firstly, sum of the forces in the y direction on diagram one equals sum of the forces in the y direction on diagram two. This gives us the total weight. Next, we can use the fact that sum of the moments about our point on diagram one must equal sum of the moments about the same point on diagram two. This locates the centre of gravity along the bar, pause the video, and uses equations to find the expressions for w, the total weight, and x, its position along the bar. You could take moments anywhere, so long as the point is the same on both diagrams. Moments about the centre of the left-hand sphere work well here.
Skip to 4 minutes and 11 seconds It’s easy to find the total weight. We get W equals W1 plus W2. It’s slightly more complicated to find the x-coordinate of the centre of gravity. Here’s what you do. Take sum of the moments about the left-hand sphere on diagram one, and put it equal to sum of the moments about the same point on diagram two. If you would like to follow the development of this, pause the video and look at each of the lines of the explanation. You can see that the final result is x equals (2/3)L, which is 2/3 of the distance between the two spheres. All this can be summed up in three equations.
Skip to 5 minutes and 1 second Capital X, capital Y, and capital Z are coordinates of the centroid of the complete body. XC, YC, and ZC are coordinates of the centroid of each component. W represents the weight of each component. For more complicated objects, like an aeroplane, you can use a table to keep track of the calculations. The summations in the equation are easily found from the table. To get the location of the centroid, you just divide one total by the other, which is what you will do next. It will be described in the design task. Now you’ve got the method sorted out. You can use it to predict where to add weight to a paper aeroplane to make it fly.
# Analysis: Centres of gravity of composite bodies
If you know the location of the centre of gravity of each component of an object you can find the overall centre of gravity of the whole thing.
You do it by taking moments.
We’ll use the method on composite objects where the components are rectangles, circles or triangles. We know the location of the centre of gravity for all of these.
If you know calculus you can apply this method to a wide range of geometric shapes. For example you can prove the standard result for a triangle. But that’s for another time.
For now we’ll stick with simple shapes and show you how to keep track of your working by using a table. A table suits a spreadsheet perfectly.
### Talking points
• What do you think are the benefits of a table when using the method of composite bodies?
• Under what the circumstances, if any, would you not use a table?
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Simultaneous equations can be thought of as being two equations in two unknowns, say x and y. Note that the word simultaneous means ‘at the same time’. It follows that for the values of x and y found both equations must be true at the same time. Sometimes it is easy to inspect the equations and guess the answers. However, when one of the equations is quadratic this becomes less likely. The answers could be surds, in which case, this is very difficult to guess.
Note that a question may ask you to solve simultaneous equations explicitly. In others, it will be implied and you must deduce that it is simultaneous equations to solve. For example, you could be asked to find out which points two curves have in common. See Example 2 below.
## Methods for solving Simultaneous Equations
There are three methods for solving simultaneous equations:
1. Elimination – this is where you multiply both equations through by different coefficient in order to eliminate one of the unknowns. This page will focus on substitution since it works for more complicated simultaneous equations. For example, when one of the equations is a quadratic. Click here to see an example using elimination.
2. Substitution – one of the equations can be quadratic, in which case, substitution is the method to use. You will need to know how to solve quadratics. By making x or y the subject of one of the equations, it can be substituted into the other. See the Worked Example and Example 1 below.
3. Graphical method – the solution of simultaneous equations can be interpreted as the intersection of their graphs. This plot shows the graphs of $y=2x-3$ in red and $4x+5y=6$ in blue. Their intersection lies on the x-axis and has coordinates (1.5,0). This is the solution when solving simultaneously. Also see Example 2 below.
## Simultaneous Equations Worked Example
Solve the simultaneous equations
$x^2+y^2=10$
and
$x+2y=5$.
This example requires solution via substitution, i.e. make either x or y the subject of one equation and insert it into the other. The obvious choice would be to make x the subject of the second equation – it is the quickest, least complicated choice. The second equation tells us that $x=5-2y$. We can insert this into the first equation: $(5-2y)^2+y^2=10$. By multiplying out the brackets and simplifying we see that this is a quadratic equation in y:
$(5-2y)^2+y^2=10$
Write out the brackets: $(5-2y)(5-2y)+y^2=10$
Expand the brackets: $25-10y-10y+4y^2+y^2=10$
Simplify: $5y^2-20y+15=0$
Divide both by sides by 5: $y^2-4y+3=0$
Factorise: $(y-3)(y-1)=0$
This tells us that y has to be either 3 or 1. If $y=3$, then $x=5-2\times 3=-1$ (from the second equation rearranged) and if $y=1$ then $x=5-2\times 1=3$.
We obtain the solutions $(x_1,y_1)=(-1,3)$ and $(x_2,y_2)=(3,1)$.
### Example 1
Solve the simultaneous equations:
$y=x-4$
$2x^2-xy=8$
### Example 2
Sketch the graphs of $x^2+y^2=10$ and $x+2y=5$ on the same plot. Determine the coordinates of the intersection points.
Click here to find Questions by Topic all scroll down to all past SIMULTANEOUS EQUATIONS questions to practice some more questions.
Are you ready to test your Pure Maths knowledge? If so, visit our Practice Papers page and take StudyWell’s own Pure Maths tests. Alternatively, try the
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APPLICATIONS OF ANTIDERIVATIVES; DIFFERENTIAL EQUATIONS - Antidifferentiation - Calculus AB and Calculus BC
CHAPTER 5 Antidifferentiation
E. APPLICATIONS OF ANTIDERIVATIVES; DIFFERENTIAL EQUATIONS
The following examples show how we use given conditions to determine constants of integration.
EXAMPLE 48
Find f (x) if f (x) = 3x2 and f (1) = 6.
SOLUTION:
Since f (1) = 6, 13 + C must equal 6; so C must equal 6 − 1 or 5, and f (x) = x3 + 5.
EXAMPLE 49
Find a curve whose slope at each point (x, y) equals the reciprocal of the x-value if the curve contains the point (e, −3).
SOLUTION: We are given that and that y = −3 when x = e. This equation is also solved by integration. Since
Thus, y = ln x + C. We now use the given condition, by substituting the point (e, −3), to determine C. Since −3 = ln e + C, we have −3 = 1 + C, and C = −4. Then, the solution of the given equation subject to the given condition is
y = ln x − 4.
DIFFERENTIAL EQUATIONS: MOTION PROBLEMS.
An equation involving a derivative is called a differential equation. In Examples 48 and 49, we solved two simple differential equations. In each one we were given the derivative of a function and the value of the function at a particular point. The problem of finding the function is called aninitial-value problem and the given condition is called the initial condition.
In Examples 50 and 51, we use the velocity (or the acceleration) of a particle moving on a line to find the position of the particle. Note especially how the initial conditions are used to evaluate constants of integration.
EXAMPLE 50
The velocity of a particle moving along a line is given by v(t) = 4t3 − 3t2 at time t. If the particle is initially at x = 3 on the line, find its position when t = 2.
SOLUTION: Since
Since x(0) = 04 − 03 + C = 3, we see that C = 3, and that the position function is x(t) = t4 t3 + 3. When t = 2, we see that
x(2) = 24 − 23 + 3 = 16 − 8 + 3 = 11.
EXAMPLE 51
Suppose that a(t), the acceleration of a particle at time t, is given by a(t) = 4t − 3, that v(1) = 6, and that f (2) = 5, where f (t) is the position function.
(a) Find v(t) and f (t).
(b) Find the position of the particle when t = 1.
SOLUTIONS:
Using v(1) = 6, we get 6 = 2(1)2 − 3(1) + C1, and C1 = 7, from which it follows that v(t) = 2t2 − 3t + 7. Since
Using f (2) = 5, we get + 14 + C2, so Thus,
For more examples of motion along a line, see Chapter 8, Further Applications of Integration, and Chapter 9, Differential Equations.
Chapter Summary
In this chapter, we have reviewed basic skills for finding indefinite integrals. We’ve looked at the antiderivative formulas for all of the basic functions and reviewed techniques for finding antiderivatives of other functions.
We’ve also reviewed the more advanced techniques of integration by partial fractions and integration by parts, both topics only for the BC Calculus course.
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Three common formats for numbers room fractions, decimals, and percents.
You are watching: How do you write 0.6 as a fraction
Percents are regularly used to connect a family member amount. You have probably seen them supplied for discounts, where the percent the discount can apply to different prices. Percents are likewise used when stating taxes and also interest rates on savings and also loans.
A percent is a proportion of a number come 100. Every cent way “per 100,” or “how numerous out that 100.” You usage the price % ~ a number to show percent.
Notice the 12 that the 100 squares in the grid below have to be shaded green. This to represent 12 percent (12 per 100).
12% = 12 percent = 12 parts out the 100 =
How countless of the squares in the grid over are unshaded? due to the fact that 12 space shaded and there room a full of 100 squares, 88 space unshaded. The unshaded section of the totality grid is 88 components out of 100, or 88% of the grid. Notice that the shaded and also unshaded portions together make 100% of the network (100 out of 100 squares).
Example Problem What percent of the network is shaded? The net is separated into 100 smaller sized squares, v 10 squares in each row. 23 squares out of 100 squares room shaded. Answer 23% of the network is shaded.
Example Problem What percent of the big square is shaded? The net is divided into 10 rectangles. For percents, you need to look at 100 equal-sized parts of the whole. You can divide each of the 10 rectangles right into 10 pieces, providing 100 parts. 30 tiny squares out of 100 space shaded. Answer 30% of the huge square is shaded.
What percent that this network is shaded?
A) 3%
B) 11%
C) 38%
D) 62%
A) 3%
Incorrect. Three complete columns that 10 squares space shaded, plus an additional 8 squares indigenous the next column. So, there space 30 + 8, or 38, squares shaded the end of the 100 squares in the large square. The exactly answer is 38%.
B) 11%
Incorrect. Three complete columns of 10 squares room shaded, plus another 8 squares native the following column. So, there space 30 + 8, or 38, squares shaded the end of the 100 squares in the big square. The correct answer is 38%.
C) 38%
Correct. Three complete columns that 10 squares room shaded, plus an additional 8 squares from the next column. So, there space 30 + 8, or 38, squares shaded the end of the 100 squares in the large square. This means 38% that the huge square is shaded.
D) 62%
Incorrect. There room 62 tiny unshaded squares out of the 100 in the large square, therefore the percent of the large square the is unshaded is 62%. However, the inquiry asked what percent is shaded. There room 38 shaded squares that the 100 squares in the huge square, therefore the exactly answer is 38%.
Rewriting Percents, Decimals, and also Fractions
It is often helpful to readjust the layout of a number. Because that example, girlfriend may find it less complicated to add decimals 보다 to include fractions. If you have the right to write the fractions together decimals, girlfriend can add them as decimals. Then you can rewrite your decimal amount as a fraction, if necessary.
Percents can be composed as fractions and also decimals in very few steps.
Example Problem Write 25% as a simplified portion and as a decimal. Write together a fraction. 25% = Since % method “out the 100,” 25% way 25 the end of 100. You create this together a fraction, utilizing 100 together the denominator. Simplify the portion by separating the numerator and denominator through the usual factor 25. Write together a decimal. 25% = = 0.25 You can also just move the decimal suggest in the whole number 25 two locations to the left to acquire 0.25. Answer 25% = = 0.25
Notice in the diagram listed below that 25% that a network is additionally of the grid, as you found in the example.
Notice that in the vault example, rewriting a percent together a decimal takes just a shift of the decimal point. You have the right to use fractions to recognize why this is the case. Any type of percentage x can be represented as the portion , and any fraction can be created as a decimal by moving the decimal suggest in x two places to the left. For example, 81% can be composed as
, and dividing 81 by 100 results in 0.81. People often skip end the intermediary portion step and just convert a percent come a decimal by relocating the decimal suggest two places to the left.
In the same way, rewriting a decimal together a percent (or as a fraction) requires few steps.
Example Problem Write 0.6 together a percent and also as a streamlined fraction. Write as a percent. 0.6 = 0.60 = 60% Write 0.6 as 0.60, i beg your pardon is 60 hundredths. 60 hundredths is 60 percent. You can additionally move the decimal point two places to the appropriate to discover the percent equivalent. Write as a fraction. 0.6 = To compose 0.6 together a fraction, you read the decimal, 6 tenths, and write 6 tenths in fraction form. Simplify the portion by splitting the numerator and also denominator by 2, a usual factor. Answer 0.6 = 60% =
In this example, the percent is not a entirety number. You can handle this in the exact same way, yet it’s usually less complicated to convert the percent come a decimal and also then transform the decimal to a fraction.
Example Problem Write 5.6% as a decimal and also as a streamlined fraction. Write as a decimal. 5.6% = 0.056 Move the decimal point two locations to the left. In this case, insert a 0 in front of the 5 (05.6) in bespeak to be able to move the decimal come the left two places. Write as a fraction. 0.056 = Write the portion as friend would check out the decimal. The last digit is in the thousandths place, therefore the denominator is 1,000. Simplify the portion by separating the numerator and also denominator through 8, a usual factor. Answer 5.6% = = 0.056
Write 0.645 together a percent and as a simplified fraction. A) 64.5% and B) 0.645% and also C) 645% and also D) 64.5% and also Show/Hide Answer A) 64.5% and Correct. 0.645 = 64.5% = . B) 0.645% and also Incorrect. 0.645 = 64.5%, not 0.645%. Psychic that when you convert a decimal to a percent you have to move the decimal suggest two locations to the right. The correct answer is 64.5% and . C) 645% and Incorrect. 0.645 = 64.5%, not 645%. Remember that when you convert a decimal to a percent you need to move the decimal point two areas to the right. The correct answer is 64.5% and . D) 64.5% and also Incorrect. To create 0.645 as a percent, move the decimal ar two locations to the right: 64.5%. To create 0.645 together a fraction, usage 645 as the numerator. The place value that the critical digit (the 5) is thousandths, for this reason the denominator is 1,000. The fraction is . The greatest common factor the 645 and also 1,000 is 5, therefore you deserve to divide the numerator and denominator by 5 to gain . The exactly answer is 64.5% and also .
In stimulate to create a portion as a decimal or a percent, you have the right to write the fraction as an equivalent fraction with a denominator that 10 (or any kind of other strength of 10 such as 100 or 1,000), which deserve to be then converted to a decimal and then a percent.
Example Problem Write as a decimal and as a percent. Write together a decimal. Find one equivalent portion with 10, 100, 1,000, or various other power that 10 in the denominator. Due to the fact that 100 is a lot of of 4, you deserve to multiply 4 through 25 to obtain 100. Multiply both the numerator and also the denominator by 25. = 0.75 Write the fraction as a decimal through the 5 in the percentage percent place. Write as a percent. 0.75 = 75% To create the decimal together a percent, move the decimal suggest two locations to the right. Answer = 0.75 = 75%
If that is complicated to find an equivalent portion with a denominator of 10, 100, 1,000, and so on, friend can constantly divide the molecule by the denominator to discover the decimal equivalent.
Example Problem Write as a decimal and also as a percent. Write as a decimal. Divide the molecule by the denominator. 3 ÷ 8 = 0.375. Write as a percent. 0.375 = 37.5% To create the decimal together a percent, relocate the decimal allude two areas to the right. Answer = 0.375 = 37.5%
Write as a decimal and also as a percent.
A) 80.0 and 0.8%
B) 0.4 and 4%
C) 0.8 and also 80%
D) 0.8 and also 8%
A) 80.0 and also 0.8%
Incorrect. An alert that 10 is a multiple of 5, so you have the right to rewrite using 10 together the denominator. Main point the numerator and also denominator by 2 to gain . The indistinguishable decimal is 0.8. You deserve to write this together a percent by moving the decimal suggest two places to the right. Since 0.8 has only one location to the right, encompass 0 in the percentage percent place: 0.8 = 0.80 = 80%. The correct answer is 0.8 and also 80%.
B) 0.4 and also 4%
Incorrect. To uncover a decimal indistinguishable for , an initial convert the portion to tenths. Multiply the numerator and denominator by 2 to acquire . The tantamount decimal is 0.8. So, and 0.4 space not indistinguishable quantities. The correct answer is 0.8 and 80%.
C) 0.8 and also 80%
Correct. The price is = 0.8 = 80%.
D) 0.8 and also 8%
Incorrect. It is true the = 0.8, however this does not equal 8%. To create 0.8 as a percent, relocate the decimal allude two locations to the right: 0.8 = 0.80 = 80%. The correct answer is 0.8 and also 80%.
Mixed Numbers
All the previous examples involve fractions and also decimals less than 1, so all of the percents you have actually seen so far have been much less than 100%.
Percents greater than 100% are feasible as well. Percents more than 100% are used to describe instances where there is more than one entirety (fractions and decimals higher than 1 are offered for the same reason).
In the diagram below, 115% is shaded. Every grid is taken into consideration a whole, and also you require two grids for 115%.
Expressed as a decimal, the percent 115% is 1.15; together a fraction, that is
, or
. Notice that you deserve to still convert amongst percents, fractions, and also decimals once the quantity is higher than one whole.
Numbers better than one that incorporate a fractional component can be composed as the sum of a totality number and also the fractional part. For instance, the mixed number is the amount of the entirety number 3 and the portion . = 3 + .
Example Problem Write as a decimal and as a percent. Write the mixed fraction as 2 wholes to add the spring part. Write together a decimal. Write the fractional component as a decimal by splitting the numerator by the denominator. 7 ÷ 8 = 0.875. Add 2 come the decimal. Write together a percent. 2.875 = 287.5% Now you can move the decimal allude two areas to the right to create the decimal as a percent. Answer = 2.875 = 287.5%
Note that a totality number can be created as a percent. 100% means one whole; so 2 wholes would certainly be 200%.
Example Problem Write 375% as a decimal and also as a streamlined fraction. Write together a decimal. 375% = 3.75 Move the decimal allude two areas to the left. Keep in mind that over there is a entirety number together with the decimal together the percent is more than 100%. Write as a fraction. 3.75 = 3 + 0.75 Write the decimal as a amount of the totality number and also the fractional part. 0.75 = Write the decimal part as a fraction. Simplify the portion by splitting the numerator and also denominator by a usual factor that 25. 3 + = Add the totality number component to the fraction. Answer 375% = 3.75=
Write 4.12 as a percent and also as a simplified fraction. A) 0.0412% and B) 412% and also C) 412% and also D) 4.12% and also Show/Hide Answer A) 0.0412% and Incorrect. To convert 4.12 to a percent, move the decimal suggest two locations to the right, not the left. The exactly answer is 412% and also . B) 412% and also Correct. 4.12 amounts to 412%, and also the simplified kind of is . C) 412% and Incorrect. 4.12 does equal 412%, yet it is also equivalent to , no . The correct answer is 412% and also . D) 4.12% and also Incorrect. To transform 4.12 come a percent, relocate the decimal allude two places to the right. The exactly answer is 412% and .See more: How To Build An Indoor Pitching Mound Plans: Step By Step Instructions Summary Percents space a common means to stand for fractional amounts, simply as decimals and fractions are. Any number that deserve to be composed as a decimal, fraction, or percent can also be written utilizing the various other two representations. .tags a { color: #fff; background: #909295; padding: 3px 10px; border-radius: 10px; font-size: 13px; line-height: 30px; white-space: nowrap; } .tags a:hover { background: #818182; } Home Contact - Advertising Copyright © 2022 dearteassociazione.org #footer {font-size: 14px;background: #ffffff;padding: 10px;text-align: center;} #footer a {color: #2c2b2b;margin-right: 10px;}
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# Difference Between Algorithm and Pseudocode
By BYJU'S Exam Prep
Updated on: September 25th, 2023
Difference Between Algorithm and Pseudocode: In a programming language, both algorithm and pseudocode play an important role. Where an algorithm is considered the foundation of the programming language pseudocode is used to make the programming language more human-friendly. The major difference between algorithm and pseudocode is that pseudocode is a method of writing an algorithm and an algorithm is a step-by-step description of the procedure of a task.
Here, we will first read what is algorithm and pseudocode in brief then we will discuss the difference between algorithm and pseudocode on various factors.
Table of content
## Difference Between Algorithm and Pseudocode
Although there are various similarities between algorithm and pseudocode, there are a few differences between the two which are explained in the table provided below:
### Key Differences Between Algorithm and Pseudocode
Algorithm Pseudocode It is a step-by-step description of the solution. It is an easy way of writing algorithms for users to understand. It is always a real algorithm and not fake codes. These are fake codes. They are a sequence of solutions to a problem. They are representations of algorithms. It is a systematically written code. These are simpler ways of writing codes. They are an unambiguous way of writing codes. They are a method of describing codes written in an algorithm. They can be considered pseudocode. They can not be considered algorithms There are no rules to writing algorithms. Certain rules to writing pseudocode are there.
The Difference Between Algorithm, Pseudocode, and Program to know more about these topics.
## What is an Algorithm?
In the programming language, algorithms are a procedure to solve a given problem with step by step description of the solution. The steps are carried out in a finite amount of time. The problems of complex nature can be solved by a simple step-by-step description of an algorithm.
The algorithm will have a well-defined set of steps. Problems are solved with a specific solution. Natural languages, flow charts, etc can be used to represent an algorithm. Candidates can check out Prim’s Algorithm to know more about Algorithm.
## What is a Pseudocode?
Pseudocode is also known as fake codes. It is used to give a simple human-friendly description of the steps used in an algorithm. It is an informal description. It is often used to summarise the steps or flow of the algorithm but it does not specify the detail of the algorithm. It is written by the system designers so that aligned codes and requirements can be understood by the programmers.
Pseudocode is used to plan an algorithm. They are not used in complex programming languages. As we have seen the algorithm and pseudocode, let us now see the major differences between the two in the next section.
Check out some important topics related to the difference between Algorithm and Pseudocode:
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# Into Math Grade 2 Module 12 Lesson 3 Answer Key Represent and Record Two-Digit Addition
We included HMH Into Math Grade 2 Answer Key PDF Module 12 Lesson 3 Represent and Record Two-Digit Addition to make students experts in learning maths.
## HMH Into Math Grade 2 Module 12 Lesson 3 Answer Key Represent and Record Two-Digit Addition
I Can represent and record two-digit addition with and without regrouping.
How can you represent Brianna’s cat and dog books? How many books about cats or dogs does she have?
Brianna has _________ cat or dog books.
Read the following: Brianna has 12 books about cats. She has 11 books about dogs. How many books about cats or dogs does she have?
Given that,
The total number of books about cats near Brianna is 12
The total number of books about dogs near Brianna is 11
Therefore 12 + 11 = 23
There are 23 books she has.
Build Understanding
Question 1.
Kurt has 57¢. His friend gives him 35¢. How much money does Kurt hove now?
A. How can you use tools to show the two addends for this problem? Draw to show what you did.
Given that
Kurt has money = 57 cents.
Her friend given = 35 cents.
The total money near Kurt = 57 + 35 = 92
Kurt has 92 cents.
B. Are there 10 ones to regroup?
Yes, there are 10 ones to regroup.
Adding 57 + 35 in this case you need to regroup the numbers.
when you add the ones place digits 7 + 5, you get 12 which means 1 ten and 2 ones.
Know to regroup the tens into the tens place and leave the ones. Then 57 + 35 = 92.
C. Regroup 10 ones as 1 ten. Write a 1 in the tens column to show the regrouped ten.
D. How many ones are left after regrouping? Write the number of ones left over in the ones place.
After regrouping the number of ones left over in the one place is 2.
E. How many tens are there in all? Write the number of tens ¡n the tens place.
The number of tens in the tens place is 9.
F. How much money does Kurt have now?
________ ¢
Given that
Kurt has money = 57 cents.
Her friend given = 35 cents.
The total money near Kurt = 57 + 35 = 92
Kurt has 92 cents.
Question 2.
Mateo and his friends make a list of two-digit numbers. He chooses two of the numbers to add.
A. How can you draw quick pictures to help you find the sum of 26 and 46?
B. How can you add the ones? Regroup if you need to. Show your work in the chart.
26 + 46 = 72
Adding 26 + 46 in this case you need to regroup the numbers.
when you add the ones place digits 6 + 6, you get 12 which means 1 ten and 2 ones.
Know to regroup the tens into the tens place and leave the ones. Then 26 + 46 = 72.
C. How can you odd the tens? Show your work in the chart.
D. What is the sum?
26 + 46 = 72
Adding 26 with 46 then we get 72.
Turn and Talk Are there two numbers from that Mateo could add without regrouping?
52, 11, 25 and 74
Any two numbers can add without regrouping. Because the addition of one’s place digit is less than the 10.
Step It Out
Question 1.
Add 47 and 37.
A. Find How many ones in all. Regroup if you need to. Write a I in the tens column to show the regrouped ten.
Adding 47 + 37 in this case you need to regroup the numbers.
when you add the ones place digits 7 + 7, you get 14 which means 1 ten and 4 ones.
Know to regroup the tens into the tens place and leave the ones. Then 47 + 37 = 84.
B. Write the number of ones left over in the ones place.
Number of ones left over in the ones place is 4.
C. Write the number of tens in the tens place.
Number of tens in the tens place is 8.
D. Write the sum.
47 + 37 = 84
Adding 47 with 37 then we get 84.
Check Understanding
Question 1.
There are 65 apples on a tree. There are 28 apples on another tree. How many apples are on the trees? Draw to show the addition.
________ apples
Given that,
The total number of apples on the tree = 65
The total number of apple on the another tree = 28
The total number of apples = 65 + 28 = 93.
Question 2.
Attend to Precision Mrs. Meyers plants 34 flowers. Mrs. Owens plants 42 flowers. How many flowers do they plant? Draw to show the addition.
_________ flowers
Given that,
Mrs. Meyers plants 34 flowers.
Mrs. Owens plants 42 flowers.
The total number of flowers = 34 + 42 = 76.
Question 3.
Reason Did you need to regroup 10 ones as 1 ten in Problem 2? Explain.
No need to regroup 10 ones as 1 ten. Because the addition of one’s place digits is less than 10.
So, there is no need to regroup.
Question 4.
Open Ended Rewrite Problem 2 with different numbers so that you need to regroup when you odd. Then solve.
Mrs. Meyers plants 36 flowers. Mrs. Owens plants 45 flowers. How many flowers do they plant? Draw to show the addition.
36 + 45 = 81
Adding 36 + 45 in this case you need to regroup the numbers.
when you add the ones place digits 6 + 5, you get 11 which means 1 ten and 1 ones.
Know to regroup the tens into the tens place and leave the ones.
Question 5.
Use Structure There are 25 big dogs and 19 small dogs at the dog park. How many dogs are at the park?
_________ dogs
Given that,
The total number of big dogs = 25.
The total number of small dogs = 19.
The total number of dogs = 25 + 19 = 44.
Question 6.
Add 16 and 23.
16 + 23 = 39
There is no need of regrouping.
Question 7.
Add 44 + 49
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# Parabolas math problem
1. Mar 29, 2005
### vitaly
I'm having difficulty with this question. All help is appreciated.
*The cross section of television antenna dish is a parabola and the receiver is located at the focus.
A. If the receiver is located 5 feet above the vertex, assume the vertex is the origin, find an equation for the cross section of the dish.
Okay, I know the vertex is 0,0. The focus is 0, 5. The equation is x^2=4ay.
I don't know where to go from there, or what equation is needed to find the cross section.
2. Mar 29, 2005
### vitaly
Actually, I figured it out. x^2 = 4ay, and a must equal 5 because the focus is (0,5).
That means teh equation is x^2 = 4(5)y or x^2 = 20y.
What I can't figure out is part B:
If the dish is 10 feet wide, how deep is it?
I have never had a question like this before. How do you know how "deep" a dish is?
3. Mar 29, 2005
### Kamataat
So the equation of the parabola is $y=x^2/20$. If it's 10 feet wide and centered at the origin, then it's cross section is between -5 and 5 on the x-axis. So, to find the depth, you need to calculate "y" for x=5... that is, if I understand the question correctly.
- Kamataat
4. Mar 29, 2005
### vitaly
Thank you for the help. I think that's right. Solving for y, it would be 1.25, which is the answer. I just didn't know how to come to it and show my work. Thanks again.
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# How do you determine if a set is open or closed examples?
## How do you determine if a set is open or closed examples?
Definition 5.1.1: Open and Closed Sets A set U R is called open, if for each x U there exists an > 0 such that the interval ( x – , x + ) is contained in U. Such an interval is often called an – neighborhood of x, or simply a neighborhood of x. A set F is called closed if the complement of F, R \ F, is open.
## Which functions are not continuous?
The function value and the limit aren’t the same and so the function is not continuous at this point. This kind of discontinuity in a graph is called a jump discontinuity.
## What is a discontinuity in a graph?
Discontinuities can be classified as jump, infinite, removable, endpoint, or mixed. Removable discontinuities are characterized by the fact that the limit exists. Removable discontinuities can be “fixed” by re-defining the function. Jump Discontinuities: both one-sided limits exist, but have different values.
## How do you know a function is closed?
A domain (denoted by region R) is said to be closed if the region R contains all boundary points. If the region R does not contain any boundary points, then the Domain is said to be open. If the region R contains some but not all of the boundary points, then the Domain is said to be both open and closed.
## How do you know if a graph is discontinuous?
If the function factors and the bottom term cancels, the discontinuity at the x-value for which the denominator was zero is removable, so the graph has a hole in it. After canceling, it leaves you with x – 7. Therefore x + 3 = 0 (or x = –3) is a removable discontinuity — the graph has a hole, like you see in Figure a.
## Is a jump discontinuity removable?
Then there are two types of non-removable discontinuities: jump or infinite discontinuities. Removable discontinuities are also known as holes. Jump discontinuities occur when a function has two ends that don’t meet, even if the hole is filled in at one of the ends.
## What makes a graph discontinuous?
A discontinuous function is the opposite. It is a function that is not a continuous curve, meaning that it has points that are isolated from each other on a graph. When you put your pencil down to draw a discontinuous function, you must lift your pencil up at least one point before it is complete.
## Where is the epigraph found?
An epigraph is a quote, paragraph, or short excerpt typically found at the beginning of a book. It usually serves as a preface or introduction to your story before any character makes an appearance or the action begins.
## What are the 3 conditions of continuity?
Key Concepts. For a function to be continuous at a point, it must be defined at that point, its limit must exist at the point, and the value of the function at that point must equal the value of the limit at that point.
## What is discontinuity in Earth?
Earth’s interior is made of different kinds of materials. Unique layers are there according to their characteristics inside the earth. All those layers are separated from each other through a transition zone. These transition zones are called discontinuities.
## What does a continuous graph look like?
Continuous graphs are graphs that appear as one smooth curve, with no holes or gaps. Intuitively, continuous graphs are those that can be drawn without lifting a pencil. Sometimes discrete graphs will show a pattern that seems to come from a continuous graph.
## Where is a function discontinuous on a graph?
We say the function is discontinuous when x = 0 and x = 1. There are 3 asymptotes (lines the curve gets closer to, but doesn’t touch) for this function. They are the x-axis, the y-axis and the vertical line x=1 (denoted by a dashed line in the graph above).
## What is an essential discontinuity?
Any discontinuity that is not removable. That is, a place where a graph is not connected and cannot be made connected simply by filling in a single point. Step discontinuities and vertical asymptotes are two types of essential discontinuities.
## What is irony sentence?
Definition of Irony. a state of affairs that is contrary to what is expected and is therefore amusing. Examples of Irony in a sentence. 1. The irony of the situation is that Frank wanted to scare his little sister, but she ended up scaring him instead.
## Do discontinuous functions have limits?
3 Answers. No, a function can be discontinuous and have a limit. The limit is precisely the continuation that can make it continuous. Let f(x)=1 for x=0,f(x)=0 for x≠0.
## What are the 3 types of discontinuity?
Continuity and Discontinuity of Functions Functions that can be drawn without lifting up your pencil are called continuous functions. You will define continuous in a more mathematically rigorous way after you study limits. There are three types of discontinuities: Removable, Jump and Infinite.
## What is an epigraph in an essay?
A quote used to introduce an article, paper, or chapter is called an epigraph. It often serves as a summary or counterpoint to the passage that follows, although it may simply set the stage for it.
## How do you find a closed form expression?
A closed form is an expression that can be computed by applying a fixed number of familiar operations to the arguments. For example, the expression 2 + 4 + … + 2n is not a closed form, but the expression n(n+1) is a closed form. ” = a1 +L+an .
## How do you use an epigraph in a sentence?
Epigraph in a Sentence ?
1. One of the explorer’s quotes was used as an epigraph on the school building named after him.
2. Before the headstone is finished, it will be etched with an epigraph befitting a former president of our nation.
3. We asked one of the islanders to translate the statue’s epigraph for us.
## How do you write an epigraph in an essay?
Write your epigraph one double space beneath your title. Indent 2 inches on both sides of the epigraph, so it’s 1 inch further from the standard margin. Use single spacing for the epigraph, and center the text on the page. Put quotation marks around the text.
## How do you know if a function is continuous or discontinuous?
A function being continuous at a point means that the two-sided limit at that point exists and is equal to the function’s value. Point/removable discontinuity is when the two-sided limit exists, but isn’t equal to the function’s value….
1. f(c) is defined.
2. lim f(x) exists.
3. They are equal.
## What is a closed equation?
An equation is said to be a closed-form solution if it solves a given problem in terms of functions and mathematical operations from a given generally-accepted set. For example, an infinite sum would generally not be considered closed-form.
## What does it mean for a function to be closed?
In mathematics, a function is said to be closed if for each , the sublevel set. is a closed set. Equivalently, if the epigraph defined by is closed, then the function. is closed. This definition is valid for any function, but most used for convex functions.
## Why are epigraphs used?
Epigraphs serve to give readers some idea of the themes and subjects that will appear later in your work, while also establishing context for your story.
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Are you struggling with Solving quadratic equations calculator? In this post, we will show you how to do it step-by-step. So let's get started!
We will also give you a few tips on how to choose the right app for Solving quadratic equations calculator. There are a number of websites that allow users to input a math problem and receive step-by-step solutions. This can be a helpful resource for students who are struggling to understand how to solve a particular type of problem. It can also be a good way for students to check their work, as they can compare their own solutions to the ones provided online.
Factoring algebra is a process of breaking down an algebraic expression into smaller parts that can be more easily solved. Factoring is a useful tool for simplifying equations and solving systems of equations. There are a variety of methods that can be used to factor algebraic expressions, and the best method to use depends on the specific equation being considered. In general, however, the goal is to identify common factors in the equation and then to cancel or factor out those common factors. Factoring is a fundamental skill in algebra, and it can be used to solve a wide variety of problems. With practice, it can be mastered by anyone who is willing to put in the effort.
To solve for the hypotenuse of a right angled triangle, you can use the Pythagorean Theorem. This theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides. So, in order to solve for the hypotenuse, you would need to square the other two sides and then add them together. Afterwards, you would need to take the square root of the result in order to find the value of the hypoten
Range is a psychological term that refers to the discrepancy between how much we feel like eating and when we actually eat. There are two main reasons why people may be range deprived: 1) they eat too little, or 2) they eat too much. Eating too little can lead to range deprivation because you’re not eating enough food to properly fuel your body. This can lead to cravings, overeating and weight gain. Eating too much can lead to range deprivation because you’re eating more food than your body needs, which can cause weight gain as well as health problems such as high blood pressure and heart disease. To solve range, you must first identify the source of your problem. For example, if you’re only eating 200 calories at dinner but feeling hungry, it may be because you’re not eating enough throughout the day. You can then adjust your caloric intake accordingly so that you’re eating enough for the day but not too much for the night.
## We cover all types of math problems
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# 6.06 Sales tax and tip
Lesson
Two ways that percentages are commonly used in the U.S. are when calculating the amount of sales tax we will need to pay for purchasing an item and when calculating the proper tip to leave the waitstaff at a restaurant. Interestingly, we will find that in different parts of the United States the amount of tax that we will pay for the purchase of goods or services can vary. In addition, while tipping for good service is a common practice in the U.S., there are other countries that do not engage in the practice of tipping at all.
### Tips
This image below, from mint.com, displays the tipping customs of many countries around the world. You can see that tips vary from $0$0 to $20%$20%
Tipping is an amount of money left for the staff, in addition to paying the bill, as a sign that we appreciate good service. Tips are common in the service industry, but in other sectors like government receiving a tip can be considered illegal. So, it is important to know the customary amount to tip for different services and who we should not offer a tip to.
#### Worked examples
##### Question 1
David is paying for a meal with lots of friends. They received great service, so he is giving a $20%$20% tip. The meal came to $\$182.30$$182.30. How much will he leave as a tip? Think: I need to work out 20%20% of the total meal charge. 20%20% as a fraction is \frac{20}{100}20100. Do: 20%20% of \182.30$$182.30
$20%$20% of $\$182.30$$182.30 == \frac{20}{100}\times\182.3020100×182.30 20%20% is \frac{20}{100}20100 and of in mathematics means multiplication. == \frac{20\times182.30}{100}20×182.30100 == \36.46$$36.46
### Outcomes
#### 7.RP.3
Use proportional relationships to solve multi-step ratio, rate, and percent problems. Examples: simple interest, tax, price increases and discounts, gratuities and commissions, fees, percent increase and decrease, percent error.
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### Smart Word Wrap
Recursion? Oh, we all love it!
This post will try to illustrate how useful the recursion is and how complicated it can get sometimes.
We, programmers, use recursion frequently to create small and elegant programs which can do complicated things.
Just think about how would you split the following sequence "aaa bb cc ddddd" into individual rows with a maximum width of 6 characters per row.
Well, one solution is to go from the left side of the string and just take words, and when the width limit is reached, append a newline. This is called the 'greedy word wrap algorithm'. It is very fast, indeed, but the output is not always pretty.
Here enters another algorithm into play. We call it the 'smart word wrap algorithm'. Don't be fooled by the name. It's not smart at all! It can be, but not this one. For now it just tries in its head all the possible combinations, does some math and returns the prettiest result possible. It is much slower than the first algorithm, especially on very large strings with a great width value because it has to create and remember more combinations.
How many combinations do we have for the above sequence? Well, there are three of them:
1. ["aaa", "bb", "cc", "ddddd"]
2. ["aaa", "bb cc", "ddddd"]
3. ["aaa bb", "cc", "ddddd"]
How do we know which one is the best? We can find this by summing the squared number of remaining spaces on each line.
The combinations are represented as arrays. Each element of the array is a line. For example, the first array has four lines. Let's do the math to see what it says. Remember, the maximum width == 6;
1. "aaa" -> (6-3)**2 = 9
"bb" -> (6-2)**2 = 16
"cc" -> (6-2)**2 = 16
"ddddd" -> (6-5)**2 = 1
-------------------------
TOTAL: 42
2. "aaa" -> (6-3)**2 = 9
"bb cc" -> (6-5)**2 = 1
"ddddd" -> (6-5)**2 = 1
-------------------------
TOTAL: 11
3. "aaa bb" -> (6-6)**2 = 0
"cc" -> (6-2)**2 = 16
"ddddd" -> (6-5)**2 = 1
-------------------------
TOTAL: 17
Clearly, we can see that the first combination is the worst of all. A lower sum indicates more uniformity near the edges. The best solution is the second one, which has a sum of 11, and is the result of a smart word wrap algorithm. The third combination is achieved by a greedy word wrap algorithm.
Here are the steps of a smart word wrap algorithm with combinations:
1. Split the string into words
2. Create all the possible paths
3. Normalize the paths
4. Create all the possible combinations
5. Normalize the combinations
6. Find the best result
In phase 1, the words will look like this:
("aaa", "bb", "cc", "ddddd")
In phase 2, we will have an array with two sub-arrays, which contains other sub-arrays:
(
["aaa", ["bb", ["cc", ["ddddd"]]], ["bb", "cc", ["ddddd"]]],
["aaa", "bb", ["cc", ["ddddd"]]],
)
The phase 3 represents a small transformation of the paths from the phase 2:
(
[
{ "aaa" => [{ "bb" => [{ "cc" => "ddddd" }] }] },
{ "aaa" => [{ "bb cc" => "ddddd" }] },
],
[{ "aaa bb" => [{ "cc" => "ddddd" }] }],
)
In phase 4, we need to create the combinations using the paths from the phase 3:
(
[[[["aaa", "bb", "cc", "ddddd"]]]],
[[["aaa", "bb cc", "ddddd"]]],
[[["aaa bb", "cc", "ddddd"]]],
)
In phase 5, the combinations have to be arrays of strings, so we need to normalize them:
(
["aaa", "bb", "cc", "ddddd"],
["aaa", "bb cc", "ddddd"],
["aaa bb", "cc", "ddddd"],
)
Finally, in phase 6, after some calculations, the best result pops up:
("aaa", "bb cc", "ddddd")
As shown in the above phases (or steps), the algorithm does many useless transformations before it gets to the best result. The transformations take time, but they are beautiful. :)
Here is an example for a random text with MAX_WIDTH=20:
*** SMART WRAP
-----------------------------------------------------------------
As shown in the |
above phases |
(or steps), the |
algorithm does |
many useless |
transformations |
-----------------------------------------------------------------
*** GREEDY WRAP (Text::Wrap)
-----------------------------------------------------------------
As shown in the |
above phases (or |
steps), the |
algorithm does many |
useless |
transformations |
-----------------------------------------------------------------
An implementation of the algorithm described in this post can be found by clicking on one of the following links:
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# Painting a Wall
Alignments to Content Standards: 5.NF.B
Nicolas is helping to paint a wall at a park near his house as part of a community service project. He had painted half of the wall yellow when the park director walked by and said,
This wall is supposed to be painted red.
Nicolas immediately started painting over the yellow portion of the wall. By the end of the day, he had repainted $\frac56$ of the yellow portion red.
What fraction of the entire wall is painted red at the end of the day?
## IM Commentary
The purpose of this task is for students to find the answer to a question in context that can be represented by fraction multiplication. This task is appropriate for either instruction or assessment depending on how it is used and where students are in their understanding of fraction multiplication. If used in instruction, it can provide a lead-in to the meaning of fraction multiplication. If used for assessment, it can help teachers see whether students readily see that this is can be solved by multiplying $\frac56\times \frac12$ or not, which can help diagnose their comfort level with the meaning of fraction multiplication.
The teacher might need to emphasize that the task is asking for what portion of the total wall is red, it is not asking what portion of the yellow has been repainted.
## Solutions
Solution: Solution 1
In order to see what fraction of the wall is red we need to find out what $\frac56$ of $\frac12$ is. To do this we can multiply the fractions together like so:
$\frac56 \times \frac12 = \frac{5 \times 1}{6 \times 2} = \frac{5}{12}$
So we can see that $\frac{5}{12}$ of the wall is red.
Solution: Solution 2
The solution can also be represented with pictures. Here we see the wall right before the park director walks by:
And now we can break up the yellow portion into 6 equally sized parts:
Now we can show what the wall looked like at the end of the day by shading 5 out of those 6 parts red.
And finally, we can see that if we had broken up the wall into 12 equally sized pieces from the beginning, that finding the fraction of the wall that is red would be just a matter of counting the number of red pieces and comparing them to the total.
And so, since 5 pieces of the total 12 are red, we can see that $\frac{5}{12}$ of the wall is red at the end of the day.
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# Plot
The land is in the shape of a square with a dimension of 22 meters. How much will we pay for the fence around the entire plot?
Result
x = 1760 Eur
#### Solution:
$a = 22 \ m \ \\ o = 4 \cdot \ a = 4 \cdot \ 22 = 88 \ m \ \\ \ \\ x = 20 \cdot \ o = 20 \cdot \ 88 = 1760 = 1760 \ \text{ Eur }$
Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!
Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...):
Be the first to comment!
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Find the Missing Number in the Given Table
Look at the table given in the riddle and find the connection between the numbers. Once you crack the logic, find the value of the missing number in the table. 432 975 543 234 579 345 123 ? 345 So were you able to solve the riddle? Leave your answers in the comment section below.
Fun With Maths Riddle: Find The Result For the Given Equations
Let’s test your skills in maths with this Riddle. Find the result for the given 2 equations. 100 + 101 x 102 – 100 ______________________ 10 4! x √42 4 √4 4 So were you able to solve the riddle? Leave your answers in the comment section below. You can check if your answer
Numerical Riddle: Find the Missing Number in the Riddle
Look at the table given in the riddle and find the connection between the numbers. Once you establish a logic connecting the number, you can easily find the missing number in the riddle. 16 100 49 ? 25 144 64 81 36 So were you able to solve the riddle? Leave your answers in the
Number Riddles: Find the Missing Number in the Last Triangle
Look at the numbers in the triangle and find how they are connected to each other. Once you get the logic behind the numbers you will be able to find the missing number in the last triangle. So were you able to solve the riddle? Leave your answers in the comment section below. You can check
Find the Value of the Missing Number in the Given Table
In this number riddle you will see a table with some numbers which are connected to each other in some way. Find the logic behind the numbers and then find the value of the missing number. Find the value of ? in the table below; 6 3 21 7 3 24 8 4 36 9
Fun Riddles: What Goes Through A Door But Never Goes In Or Comes Out?
Solve this riddle and leave your answers in the comment section below. What goes through a door but never goes in or comes out? So were you able to solve the riddle? Then you have all the bragging rights. If you get the correct answer, please share it with your friends and family on WhatsApp,
What Comes Next In The Sequence: 1, 11, 21, 1211, 111221, 312211, ?, ?
Look at the given set of numbers in the riddle and find the connection between then. Then you can find out what Comes Next In The Sequence: 1, 11, 21, 1211, 111221, 312211, ?, ? So were you able to solve the riddle? Leave your answers in the comment section below. If you get the
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# How Many Weeks in a Month
Currently, we have a seven-day week because of the Babylonians, who lived in what is now called Iraq. They chose the number seven because they were astronomers and professional observers who had studied all seven planets: the Sun, Moon, Mercury, Venus, Mars, Jupiter, and Saturn. As a result, they valued this number greatly. Get a better understanding of how to complete your assignments on time by looking at how many weeks are in a month.
## How many weeks are there in a month?
Have you ever considered how the calendar operates and assists us in timing our daily activities? With the exception of leap years, which contain 366 days instead of 365, the Gregorian calendar is the one that is most often used worldwide. These days are split into 12 months, each of which has either 31 or 30 days, or 28 days (in February), allowing us to calculate the number of weeks in a month. The days are compressed into seven-day weeks. Therefore, a year typically equals 52 weeks plus one extra day.
## Weeks in a Month
Since every month on the calendar includes at least 28 days, it has 4 complete weeks every month. One week is equal to seven days. However, some months have some additional days, but they do not add up to form a new week, so such days are not included in the week.
For instance, the month of August contains 31 days (one week is equal to seven days). Therefore, 31/7 is 4 weeks plus 3 days. This displays 4 whole weeks plus 3 additional days.
## Weeks in a Month: Table Illustration
To count the number of weeks in each of the 12 months of the year, we must count the days in a month and divide that number by 7 to determine how many weeks there are in a month (one week equals seven days). Let's examine the chart that details the precise number of weeks and days in each month of the year along with the remaining additional days.
31 January 4 3
28 (29 in a leap year) February 4 0 (regular year), 1 (leap year)
31 March 4 3
30 April 4 2
31 May 4 3
30 June 4 2
31 July 4 3
31 August 4 3
30 September 4 2
31 October 4 3
30 November 4 2
31 December 4 3
## What indicators are there indicating that a year is a leap year?
We multiply the year digits (value) by 4 to determine if the year is a leap year. When a number divides evenly by four, a leap year is indicated. For example, since 2020 is exactly divisible by 4, the year 2020 is a leap year, in which February has a total of 29 days.
## Century Year is an exception!
The previously mentioned test is valid for all regular years, but to establish whether a century year is a leap year, it must be divided by 400 instead of 4. Examples of such century years are 300, 700, 1900, 2000, and many more.
2000, for instance, is a leap year since it can be completely divided by 400. A century year, for example, 1900, is divisible by four, but it is still not a leap year because it is not divisible by four hundred. We need to remember that for every centenary year to be a leap year, it must be exactly divisible by 400, not 4.
## Essential Facts to Remember
Why is there a leap year, in which there is a month of February with an extra day every four years?
• The Earth requires 365 and a quarter days to complete one orbit around the sun.
• The quarter is excluded when a regular year is calculated as 365 days alone.
• These quarter days are added every four years by adding four quarters together (1/4+1/4+1/4+1/4=1) to make a full extra day.
• For the leap year, this extra day is added. As a result, a leap year has 366 (365 + 1 =366) days.
## Summary
In short, there is no firm agreement about the number of weeks in a month. It is a popular misconception that there are four weeks in a month because this is simplistic to prove. The Gregorian calendar in common use today is composed of months that range in length from 28 to 31 days. As a result, the number of weeks in a month may change because it has extra days.
Let's go into some specifics so that we can better understand this idea. Given that there are seven days in a week and there can be at most 31 in a month, when divided by 7, the result is 4.43 weeks. If we round this figure, every month will have four full weeks. However, these are actually exceptions. Still, we stand with this theory.
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# Length of Pole | 29 Sep 2009
Discussion in '\$1 Daily Competition' started by shabbir, Sep 29, 2009.
1. ### shabbirAdministratorStaff Member
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Pole is in a lake. Half of the pole is under the ground, One-third of it is covered by water. 8 ft is out of the water.
What could be the possible length of the pole?
2. ### sameer_havakajokaNew Member
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48
3. ### sameer_havakajokaNew Member
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48ft is correct
Solution:
Fraction of pole in the ground = 1/2
Fraction of pole covered by water = 1/3
Fraction of pole in the ground and covered by water = 1/2 + 1/3 = (3 + 2)/6 = 5/6
Fraction of pole out of water = 1 - 5/6 = 1/6
Thus, one-sixth of the pole (out of water) is 8 ft.
So, total length of pole = 48 ft.
It may be noted that:
Length of pole in the ground = 48/2 = 24 ft.
Length of pole covered by water = 48/3 = 16 ft.
Length of pole out of water = 8 ft.
The problem may also be solved by setting up the following equation: x/2 + x/3 + 8 = x
where x denotes the total length of the pole in ft.
The equation may be solved as shown below.
5x/6 + 8 = x
8 = x - 5x/6 = x/6
x/6 = 8 or x = 48.
4. ### shabbirAdministratorStaff Member
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# Fun Math Games for Kids at Age 6
## Math Concepts for 6-Year-Olds
In first grade, your 6-year-old will be learning strategies for adding and subtracting within 20. To be successful, it will also be important for her to understand place value and how to correctly group items into tens and ones. Other skills your child will develop at the first grade level are measuring the length of items and composing geometric shapes. Reinforce these concepts at home using fun review games.
### Even or Odd?
Before beginning this game, remove all the face cards from a deck of cards. Turn the cards face down on the table. Players will take turns flipping over two cards. If the sum of the two cards is even, the player will keep both cards.
If the sum of the two cards is odd, the cards will be returned to play. The player with the most cards at the end of the game wins! For more advanced players, try turning over three cards for each round of play.
### Show Me This!
For this activity, use paper clips to have your child practice grouping tens and ones. Use large paper clips to represent tens, and smaller paper clips to represent ones. For example, if you ask your child to group the number 24, he would use two large paper clips (representing the 20) and four small paper clips (representing the four). As an extension to this activity, have your child model number sentences using paper clips.
### Measure It
Before beginning this activity, ask your child to predict the length of certain areas of your house. After making predictions, have your child measure the areas using a ruler or a yardstick. For instance, have your child predict the length of her bedroom and then measure the bedroom using a ruler.
Keep in mind that your child may not understand that a ruler is one foot in length; she may only be able to tell you the number of rulers it takes to get from one side of the room to the other. It is also important that you help your child count the number of times she 'flips' the ruler when measuring.
### Piece It Together
To prepare for this activity, cut out different shapes in a variety of sizes from construction paper. Have your child explore making different shapes or symbols using the pieces of construction paper. For example, your child could make a square by piecing together two triangles. Or you could challenge your child to create an object, such as an arrow, using his choice of shapes.
Did you find this useful? If so, please let others know!
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• 5 Free and Fun Math Games for Kids
Looking for a way to get your child engaged with math? There are many free, fun math games online that explore basic concepts such as addition, subtraction, multiplication and division, as well as more advanced games that offer practice with decimals and fractions. Read on to discover five of our favorite educational - and fun!...
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https://www.enotes.com/homework-help/two-mile-pier-one-mile-pier-extend-545534
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# A two mile pier and a one mile pier extend perpendicularly into the ocean with four miles of shore between the two piers. A swimmer wishes to swim from the end of the longer pier to the end of the...
A two mile pier and a one mile pier extend perpendicularly into the ocean with four miles of shore between the two piers. A swimmer wishes to swim from the end of the longer pier to the end of the shorter pier with one rest stop on the beach. Assuming that the swimmer recalls that, in any right triangle, the hypotenuse2 = leg2 + leg2, find the shortest possible swim.
http://math.kendallhunt.com/documents/ALookInside/ThoughtProvokers/ThoughtProvokers_pp_44-45.pdf
embizze | High School Teacher | (Level 2) Educator Emeritus
Posted on
The swimmer begins on the end of the 2 mile pier, swims to the beach and then out to the end of the 1 mile pier, and we are asked to find the shortest possible swim.
The easiest solution is to reflect the 1 mile pier across the line formed by the beach. (See attachment.) Now draw a line from the end of the 2 mile pier to the reflected end of the 1 mile pier. Extending the 2 mile pier 1 mile "below" the beach and connecting this point to the reflection of the 1 mile pier creates a right triangle whose sides are 3 miles and 4 miles long. The hypotenuse of this triangle is 5 miles long.
If d1 is the distance from the 2 mile pier to the shore, and d2 is the distance from the shore to the 1 mile pier, and if we assume that the point on the beach is where the line from the 2 mile pier to the reflected 1 mile pier intersects the beach, then d1+d2=5 miles.
Claim: 5 miles is the shortest distance. Let A be the end of the 2 mile pier, B the end of the 1 mile pier, B' the reflected end, and X the intersection of AB' and the shore. Suppose Y is a point on the shore between the piers and Y is not X. Then the total path is AY+YB. But by construction the total path using X is AX+XB=AX+XB'. Now consider triangle AYB'. By the triangle inequality theorem, AY+YB'>AB'=AX+XB'. Thus AY+YB>AX+XB for any choice of Y not equal to X.
----------------------------------------------------------------------------------------
The shortest swimming distance is 5 miles.
----------------------------------------------------------------------------------------
You can use calculus to minimize the sum of the distance functions, or algebra to find the minimum, but the geometric argument is easier.
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https://skydivehayabusa.com/kitesurfing/best-answer-is-a-rhombus-always-a-kite.html
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# Best answer: Is a rhombus always a kite?
Contents
## Why is a rhombus always a kite?
With a hierarchical classification, a rhombus (a quadrilateral with four sides of the same length) is considered to be a special case of a kite, because it is possible to partition its edges into two adjacent pairs of equal length, and a square is a special case of a rhombus that has equal right angles, and thus is …
## Is a rhombus always sometimes or never a kite?
A square is a rhombus is a kite is a quadrilateral. A kite is not always a rhombus.
## What is a rhombus that is not a kite?
A kite is a convex quadrilateral with two pairs of adjacent equal sides. A rhombus has two pairs of adjacent equal sides too, but all four sides are the same length. This answer assumes a positive definition of “kite” that does not deliberately exclude such special cases as rhombuses or squares.
## What makes a kite a kite?
A Kite is a flat shape with straight sides. It has two pairs of equal-length adjacent (next to each other) sides. It often looks like. a kite! Two pairs of sides.
## How is a kite different to a rhombus?
The main difference between a kite and a rhombus is that a rhombus has all equal sides whereas a kite has two pairs of adjacent equal sides.
INTERESTING: Frequent question: What type of bird is a red kite?
## Is a kite sometimes always or never a parallelogram?
Explanation: A kite is a quadrilateral with two disjoint pairs (no side is in both pairs) of equal-length, adjacent (sharing a vertex) sides. A parallelogram also has two pairs of equal-length sides, however they must be opposite, as opposed to adjacent.
## What shape is a kite?
A kite is a four-sided shape with straight sides that has two pairs of sides. Each pair of adjacent sides are equal in length. A square is also considered a kite.
## Which is not a rhombus?
One of the two characteristics that make a rhombus unique is that its four sides are equal in length, or congruent. … If you have a quadrilateral with only one pair of parallel sides, you definitely do not have a rhombus (because two of its sides cannot be the same length). You have a trapezoid.
## How do you prove a rhombus is a kite?
Here are the two methods:
1. If two disjoint pairs of consecutive sides of a quadrilateral are congruent, then it’s a kite (reverse of the kite definition).
2. If one of the diagonals of a quadrilateral is the perpendicular bisector of the other, then it’s a kite (converse of a property).
## What forces act on a kite?
Just like rockets, jets, or birds, all kites experience a combination of forces as they fly. The main forces that determine whether or not a kite is able to fly are weight, lift, tension, and drag.
## Can a rhombus and kite be congruent?
Sometimes a kite can be a rhombus (four congruent sides), a dart, or even a square (four congruent sides and four congruent interior angles). Some kites are rhombi, darts, and squares. Not every rhombus or square is a kite.
## Can you fly a kite without wind?
Before you can fly your kite, you need wind. … Others are especially made to fly in light wind. But most kites are made to fly in average winds of between four and ten miles per hour. If you can feel the wind on your face, there is probably enough to fly.
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Boolean Algebra and Reduction Techniques
# Boolean Algebra Laws and Rules
There are three laws of Boolean Algebra that are the same as ordinary algebra.
1. The Commutative Law
addition A + B = B + A (In terms of the result, the order in which variables are ORed makes no difference.)
multiplication AB = BA (In terms of the result, the order in which variables are ANDed makes no difference.)
2. The Associative Law
addition A + (B + C) = (A + B) + C (When ORing more than two variables, the result is the same regardless of the grouping of the variables.)
multiplication A(BC) = (AB)C (When ANDing more than two variables, the result is the same regardless of the grouping of the variables.)
3. The Distributive Law - The distributive law is the factoring law. A common variable can be factored from an expression just as in ordinary algebra.
A(B + C) = AB + AC
(A + B)(C + D) = AC + AD + BC + BD Remeber FOIL(First, Outer, Inner, Last)?
# Ten Basic Rules of Boolean Algebra
1. Anything ANDed with a 0 is equal to 0. A * 0 = 0
2. Anything ANDed with a 1 is equal to itself. A * 1 = A
3. Anything ORed with a 0 is equal to itself. A + 0 = A
4. Anything ORed with a 1 is equal to 1. A + 1 = 1
5. Anything ANDed with itself is equal to itself. A * A = A
6. Anything ORed with itself is equal to itself. A + A = A
7. Anything ANDed with its own complement equals 0.
8. Anything ORed with its own complement equals 1.
9. Anything complemented twice is equal to the original.
10. The two variable rule.
# Simplification of Combinational Logic Circuits Using Boolean Algebra
• Complex combinational logic circuits must be reduced without changing the function of the circuit.
• Reduction of a logic circuit means the same logic function with fewer gates and/or inputs.
• The first step to reducing a logic circuit is to write the Boolean Equation for the logic function.
• The next step is to apply as many rules and laws as possible in order to decrease the number of terms and variables in the expression.
• To apply the rules of Boolean Algebra it is often helpful to first remove any parentheses or brackets.
• After removal of the parentheses, common terms or factors may be removed leaving terms that can be reduced by the rules of Boolean Algebra.
• The final step is to draw the logic diagram for the reduced Boolean Expression.
# Some Examples of Simplification
Perform FOIL (Firt - Outer - Inner - Last)
AA = A (Anything ANDed with itself is itself)
Find a like term (A) and pull it out. (There is an A in A, AC, and AB). Make sure you leave the BC alone at the end.
Anything ORed with a 1 is a 1 (1+C+B=1).
Anthing ANDed with a 1 is itself (A1=A)
# Some Examples of Simplification (cont.)
Find like term (B) and pull it out.
Anything ORed with its own complement equals 1.
Anything ANDed with 1 is itself.
# Some Examples of Simplification (cont.)
Find like term and pull them out. Make sure you leave the one.
Anything ORed with a 1 is 1.
Anything ANDed with a 1 is itself
# Some Examples of Simplification (cont.)
Find like terms and pull them out.
Anything ORed with its own complement equals 1.
Anything ANDed with 1 equals itself.
NOTE: I will workout many examples in the video.
# DeMorgan's Theorem
• De Morgan's theorem allows large bars in a Boolean Expression to be broken up into smaller bars over individual variables.
• De Morgan's theorem says that a large bar over several variables can be broken between the variables if the sign between the variables is changed.
• De Morgan's theorem can be used to prove that a NAND gate is equal to an OR gate with inverted inputs.
• De Morgan's theorem can be used to prove that a NOR gate is equal to an AND gate with inverted inputs.
• In order to reduce expressions with large bars, the bars must first be broken up. This means that in some cases, the first step in reducing an expression is to use De Morgan's theorem.
• It is highly recommended to place parentheses around terms where lines have been broken.
For example:
# DeMorgan (cont.)
Applying DeMorgan's theorem and the distribution law:
# Bubble Pushing
• Bubble pushing is a technique to apply De Morgan's theorem directly to the logic diagram.
1. Change the logic gate (AND to OR and OR to AND).
2. Add bubbles to the inputs and outputs where there were none, and remove the original bubbles.
• Logic gates can be De Morganized so that bubbles appear on inputs or outputs in order to satisfy signal conditions rather than specific logic functions. An active-low signal should be connected to a bubble on the input of a logic gate.
# The Universal Capability of NAND and NOR Gates
• NAND and NOR gates are universal logic gates.
• The AND, Or, Nor and Inverter functions can all be performed using only NAND gates.
• The AND, OR, NAND and Inverter functions can all be performed using only NOR gates.
• An inverter can be made from a NAND or a NOR by connecting all inputs of the gate together.
• If the output of a NAND gate is inverted, it becomes an AND function.
• If the output of a NOR gate is inverted, it becomes an OR function.
• If the inputs to a NAND gate are inverted, the gate becomes an OR function.
• If the inputs to a NOR gate are inverted, the gate becomes an AND function.
• When NAND gates are used to make the OR function and the output is inverted, the function becomes NOR.
• When NOR gates are used to jake the AND function and the output is inverted, the function becomes NAND.
# AND-OR-Invert Gates for Implementing Sum-of-Products Expressions
• Most Boolean reductions result in a Product-of-Sums (POS) expression or a Sum-of-Products (SOP) expression.
• The Sum-of-Products means the variables are ANDed to form a term and the terms are ORed. X = AB + CD.
• The Product-of-Sums means the variables are ORed to form a term and the terms are ANDed. X = (A + B)(C + D)
• AND-OR-Inverter gate combinations (AOI) are available in standard ICs and can be used to implement SOP expressions.
• The 74LS54 is a commonly used AOI.
• Programmable Logic Devices (PLDs) are available for larger and more complex functions than can be accomplished with an AOI.
# Karnaugh Mapping
• Karnaugh mapping is a graphic technique for reducing a Sum-of-Products (SOP) expression to its minimum form.
• Two, three and four variable k-maps will have 4, 8 and 16 cells respectively.
• Each cell of the k-map corresponds to a particular combination of the input variable and between adjacent cells only one variable is allowed to change.
• Use the following steps to reduce an expression using a k-map.
1. Use the rules of Boolean Algebra to change the expression to a SOP expression.
2. Mark each term of the SOP expression in the correct cell of the k-map. (kind of like the game Battleship)
3. Circle adjacent cells in groups of 2, 4 or 8 making the circles as large as possible. (NO DIAGONALS!)
4. Write a term for each circle in a final SOP expression. The variables in a term are the ones that remain constant across a circle.
• The cells of a k-map are continuous left-to-right and top-to-bottom. The wraparound feature can be used to draw the circles as large as possible.
• When a variable does not appear in the original equation, the equation must be plotted so that all combinations of the missing variable(s) are covered.
This is a very visual problem so watch the video for examples on how to complete and solve Karnaugh Maps!
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Resources tagged with: Digit cards
Filter by: Content type:
Age range:
Challenge level:
There are 12 results
Broad Topics > Physical and Digital Manipulatives > Digit cards
The Thousands Game
Age 7 to 11 Challenge Level:
Each child in Class 3 took four numbers out of the bag. Who had made the highest even number?
Song Book
Age 7 to 11 Challenge Level:
A school song book contains 700 songs. The numbers of the songs are displayed by combining special small single-digit cards. What is the minimum number of small cards that is needed?
Magic Circles
Age 7 to 11 Challenge Level:
Put the numbers 1, 2, 3, 4, 5, 6 into the squares so that the numbers on each circle add up to the same amount. Can you find the rule for giving another set of six numbers?
Twenty Divided Into Six
Age 7 to 11 Challenge Level:
Katie had a pack of 20 cards numbered from 1 to 20. She arranged the cards into 6 unequal piles where each pile added to the same total. What was the total and how could this be done?
All the Digits
Age 7 to 11 Challenge Level:
This multiplication uses each of the digits 0 - 9 once and once only. Using the information given, can you replace the stars in the calculation with figures?
Reach 100
Age 7 to 14 Challenge Level:
Choose four different digits from 1-9 and put one in each box so that the resulting four two-digit numbers add to a total of 100.
Magic Vs
Age 7 to 11 Challenge Level:
Can you put the numbers 1-5 in the V shape so that both 'arms' have the same total?
Sealed Solution
Age 7 to 11 Challenge Level:
Ten cards are put into five envelopes so that there are two cards in each envelope. The sum of the numbers inside it is written on each envelope. What numbers could be inside the envelopes?
Finding Fifteen
Age 7 to 11 Challenge Level:
Tim had nine cards each with a different number from 1 to 9 on it. How could he have put them into three piles so that the total in each pile was 15?
Four-digit Targets
Age 7 to 11 Challenge Level:
You have two sets of the digits 0 – 9. Can you arrange these in the five boxes to make four-digit numbers as close to the target numbers as possible?
Penta Primes
Age 7 to 11 Challenge Level:
Using all ten cards from 0 to 9, rearrange them to make five prime numbers. Can you find any other ways of doing it?
Fifteen Cards
Age 7 to 11 Challenge Level:
Can you use the information to find out which cards I have used?
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# 3.2 Forces and Potential Energy
Riya Patel
Riya Patel
61 resources
See Units
## Hooke's Law
Robert Hooke came up with an equation to describe an ideal "linear" spring acting in a system.
The equation for Hooke's Law is as follows:
Where Fs is the spring force, k is the spring constant, and Δ x is the displacement of the spring from its equilibrium position. Keep in mind spring force is a restoring force!
⚠️Hang on...what's a spring constant?
A spring constant is a number used to describe the properties of a spring, primarily its stiffness. Essentially, the easier a spring is to stretch, the smaller the spring constant is (start thinking about how this relates back to Newton's Third Law).
Many times, students will be asked to graph this relationship in order to find the spring constant k. Here's an example of a graph from a lab:
As you may be able to tell, the spring constant of the spring used in the lab should be the slope of the graph. (You can also see why this law describes "linear" springs)
Now let's connect this back to energy! The elastic potential energy of a spring can be defined as:
## Conservative Forces
A conservative force is a force where the total work done on an object is solely dependent on the final and initial positions of the object. Dissipative forces are the opposite of conservative forces, and the ones typically seen are friction or external applied forces.
Fast facts of conservative forces:
• Independent of the path taken
• Total work on a closed path is zero
Examples of conservative forces:
• Gravitational force
• Spring force
Work done by conservative forces is also equal to the negative change in potential energy (U).
It can also be written as:
Where F is a conservative force and a and b are typically the the initial and final radius.
The differential version of this equation is:
We can do some fun things with this version of the equation, especially with graphs. The most important thing to note is that force is the negative slope of a potential energy versus position graph. AP loves to make students analyze energy graphs. Let's take a look at some examples!
Taken from LibreTexts
From your expert calculus knowledge, you should be able to see that equilibrium is wherever the slope is zero, meaning there is no net force. When analyzing these graphs, you should attempt to determine the total mechanical energy and draw a horizontal line for it. Occasionally you will find graphs that contain a section known as Potential Energy Wells which are typically caused by oscillations. You can spot a potential energy well at a local minimum!
## Gravitational Potential Energy
The gravitational potential energy of a system with an object very near/on the Earth in a uniform gravitational field is:
Where delta U is the change in potential energy, m is mass, g is acceleration due to gravity, and delta h is the change in height.
Here's the derivation for the equation for Gravitational Potential Energy (for large masses at a distance:
Where W is work, F(r) is a function for the force and r is the radius/distance.
Plug in the formula for Newton's Law of Universal Gravitation as a function of F(r). Where m1 and m2 are the masses in the system and G is the gravitational constant.
Take the integral evaluated from the initial radius to the final radius.
Gravity is a conservative force, so:
To make equations work nicely, we usually state that Ro(initial r) is set at infinity and that the initial potential energy is 0. So it simplifies the above to be:
Which should be the formula you see on your formula chart!
Practice Questions
1. A 5.00 × 10^5-kg subway train is brought to a stop from a speed of 0.500 m/s in 0.400 m by a large spring bumper at the end of its track. What is the force constant k of the spring? (Taken from Lumen Learning)
Energy is not conserved because there is acceleration from a force, therefore we can tackle this problem with work!
So we know that W = Fd, and we know our d, so let's try to find the force.
F=ma
We don't know acceleration! But we know our displacement, our initial velocity, and our final velocity...so we can recall an equation from unit 1.
Then we know that W = F*d so:
Another relationship we know about work is:
Spring energy is elastic potential energy so we can plug that formula into the equation.
2. Suppose a 350-g kookaburra (a large kingfisher bird) picks up a 75-g snake and raises it 2.5 m from the ground to a branch. (a) How much work did the bird do on the snake?
(b) How much work did it do to raise its own center of mass to the branch? (Taken from Lumen Learning)
(a)
(b)
3.
Taken from CollegeBoard
Make sure to use the variables they want you to use and place bounds if they exist in the problem.
4.
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# 11 Times Tables Worksheets
• ###### Wheel of 11s
The multiplication table of 11 has some unique characteristics that make it fascinating and slightly different from other times tables. For instance, when multiplying 11 by any single-digit number, the result is that number repeated twice. So, 11 x 3 is 33, 11 x 5 is 55, and so forth.
However, when you multiply 11 by a two-digit number, the trick is slightly different. That’s where the 11 Times Tables Worksheets come in handy. They allow you to repeatedly practice these problems, understand the pattern, and improve your speed and accuracy.
When you look at the worksheet, you will see multiple multiplication problems. These problems are usually arranged in a structured way. For example, the first few problems might involve multiplying 11 by single-digit numbers. As you move further down the sheet, you might find problems where you have to multiply 11 by two-digit numbers.
You may be wondering how this worksheet will help you, given that calculators are so readily available and easy to use. However, understanding the multiplication table and being able to do it mentally gives you a strong foundation in mathematics. It helps you quickly solve problems and gain confidence in math. For instance, when you have to solve complex equations later in algebra or geometry, this fundamental knowledge will help you simplify and solve problems quickly.
Using these worksheets is pretty straightforward. You usually start by solving the problems from the top of the worksheet, then gradually make your way down. If you get stuck, don’t worry. Remember, practice makes perfect. The more you work with these problems, the better you will understand the patterns and the quicker you will be able to solve them.
What’s more, these worksheets all have an answer key at the end. This means you can check your work once you’re done. This immediate feedback can help you understand your mistakes and learn from them. If you get a problem wrong, don’t get discouraged. Instead, try to figure out what went wrong and how to avoid making the same mistake in the future.
### What Is the 11 Times Table Trick?
There is a neat trick you can use to quickly find products when a multiplier of 11 is involved. Please note that is only works under certain conditions. here’s a trick for multiplying a two-digit number by 11:
Step 1) Separate the Digits (Multiplicand) – If you’re multiplying 11 by a two-digit number, say 35, you would first separate the two digits (3 and 5).
Step 2) Add the Two Digits Together – Next, you would add those two numbers together. In this case, 3 + 5 equals 8.
Step 3) Insert the Sum – Insert this sum between the two digits you started with. So you now have 385.
Therefore, 35 times 11 equals 385.
But remember, there’s a catch when the sum of two digits is more than 9. In that case, you should carry over the value to the next digit.
For example, let’s calculate 11 times 78:
1) Separate the digits, 7 and 8.
2) Add those two numbers together. 7 + 8 equals 15.
3) Here, since 15 is a two-digit number, we only place the unit digit (5) in the middle and carry over the tens digit (1) to the 7, which then becomes 8.
So, the final result is 858.
Therefore, 78 times 11 equals 858.
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offers hundreds of practice questions and video explanations. Go there now.
# Don’t Get “Worked” Up!
See if you can crack this problem in less than 2 minutes.
Mike paints a fence in 9 hours
Marty can paint the same fence in 5 hours
Column A
The time it takes Mike and Marty,
working at a constant rate, to
paint the fence
Column B
3 hours
Work rates are one of my favorite problems to teach. Students usually have a formula in their heads that they vaguely remember. Even if they know the formula, they take awhile to put the numbers in the correct places. Assuming they don’t make a mistake, the problem can take them 2 minutes to finish.
What if I could show you a way to finish the problem in less than 15 seconds?
And that’s with no messy formulas.
Okay first things first let’s conceptually work through the problem.
Ask yourself, how much of the job does each person finish in one hour. With Mike, he finishes 1/9 of the job in one hour, because it takes him 9 hours to finish the entire job. With Marty he finishes 1/5 of the job in one hour. Add those two rates together, 1/9 + 1/5 = 14/45 and then Flip It! and you get 45/14. That is greater than 3, so the answer is (A).
Not bad. No cumbersome x and y, or Work Rate 1 and Work Rate 2, Total Work Rate, as many books on the market show you.
But imagine an even faster way. Ready?
All you have to do is multiple the hourly rate to find the numerator and add the rates to find the denominator.
Or more succinctly put, multiply the top; add the bottom.
9 x 5 = 45, 9 + 5 = 14. 45/14.
It’s that easy.
Let’s try two new numbers.
Mike = 15 hrs, John 5 hrs.
Now here’s all you have to do: multiply the top; add the bottom. In other words, multiply the time it takes Mike to do the job by the time it takes John to do the job. Then divide that by the sum of the time it takes Mike to do the job and the time it takes John to do the job.
(15 x 5)/(15 + 5) = 75/20 = 3 ¾ hrs.
Because it’s so easy try the next numbers:
7 hrs and 4hrs, Combined work rate: (Don’t look below till you’ve solved it)
Ans: 28/11 hrs.
I told you—no need to get “worked” up!
By the way, students who use Magoosh GRE improve their scores by an average of 8 points on the new scale (150 points on the old scale.) Click here to learn more.
### 32 Responses to Don’t Get “Worked” Up!
1. rajeev October 29, 2014 at 10:24 pm #
This trick can also be use to add fractions by skipping LCM part . I developed this method in school and have saved lots of time since then especially if fractions have 1 in the numerator
like if you have to add 1/6+ 1/8 then, normally you would take 24 as the lcm and multiply the numerators with appropriate numbers and then add and so on
from thr trick its [8+6]/[8*6]= 14/48= 7/24
• kathy May 23, 2016 at 6:23 am #
Hey, what trick trick do you have if you need add fractions like 7/9 and 8/9
• Magoosh Test Prep Expert May 23, 2016 at 11:19 am #
This may be an instance where the basic operation is so easy that you don’t need a trick. With the denominators being the same, all you really need to do is add the numerators. 7 + 8 = 15, so 7/9 + 8/9 = 15/9. Then you can simplify as needed, creating either a whole fraction (5/3) or a mixed numeral (1 and 2/3).
2. nicky October 7, 2014 at 9:24 am #
Hey Chris
That’s awesome trick
You are genius buddy!!
• Chris Lele October 13, 2014 at 11:45 am #
You are welcome 🙂
3. Karan September 6, 2014 at 3:50 am #
Hey Chris,
While no doubt that this trick is AWESOME and a real time saver… the wording in the blog is misleading when you say, “All you have to do is multiple the hourly rate to find the numerator and add the rates to find the denominator.”
The correct way to phrase this in a formulaic manner would be:
Let Ta be the time A takes to finish the job while working alone,
Let Tb be the time B takes to finish the job while working alone and
Let Tab be the time take to finish the job when both A and B work together.
The correct formula is:
Tab = (Ta x Tb) / (Ta + Tb)
• Chris Lele September 8, 2014 at 2:02 pm #
Yes, thanks for catching that! I meant to say “multiply the time A takes to finish the job…” Not “hourly rate”.
I’ll correct that in the blog 🙂
• Samy October 10, 2014 at 5:41 pm #
Hi Chris,
Just going back to what Karan pointed out or mentioned, and you also highlighted, you may want to correct the wording in this blog post above meaning: Multiply the time A takes to finish the job by time B takes to finish the job, and divide product by sum of time A takes to finish the job and B takes to finish the job.
This really is brilliant and saves a lot of time.
Cheers,
Samy
• Chris Lele October 13, 2014 at 11:46 am #
Great! Thanks for the feedback. Made some changes 🙂
4. Aamir August 21, 2014 at 8:51 am #
This is awesome trick !!! Thank you Chris 🙂
• Chris Lele August 22, 2014 at 10:12 am #
You are welcome!
5. Padmaja June 17, 2014 at 7:19 pm #
Chris,
You are really really great! thanks for posting such easy tricks 🙂
• Chris Lele June 18, 2014 at 11:35 am #
You are welcome!
6. shanna December 17, 2013 at 3:58 pm #
This is one of the best methods/ ideas i have seen in the entire package. Saves me so much time doing these problems and anxiety. Thank you! I wish you could give tips like this for all the lessons instead of remembering formula after formula.
• Chris Lele December 18, 2013 at 3:55 pm #
Great, I’m happy the trick made life easier :).
Sadly, there aren’t too many concepts that can be broken down this easily. I’ve a shortcut for combinations/permutations, some for rates and weighted averages, but otherwise there aren’t too many.
7. Jessica October 31, 2013 at 2:53 am #
Chris,
You are the math teacher I wish I had in high school!! I am an old lady (45) going back to grad school after having kids, etc. It has been forever since I took math. You are making it possible for me to do this type of thinking again!!! Thank you so much
• Chris Lele November 4, 2013 at 9:18 am #
You are welcome! Thanks for the kudos and good luck :).
8. abcStudent September 9, 2013 at 7:10 am #
OMG! Cant believe you made it that easy……..not even 15 seconds, it just takes less than 5 seconds to solve the answer
9. siddharth mehra June 6, 2013 at 3:36 am #
awesome chris!!!! one question i am subscribed with magoosh material …… I want to know when and where can we use this flip technique!!! and in which kind of rate problems can we use the techniwue suggested by you as its a huge time saver and provides better understnding!!1
• Chris Lele September 4, 2013 at 1:10 pm #
Hi Siddarth,
The “flip method” can only be used for work rate problems that give two differing rates. You might see one of these questions per GRE test. So definitely great as a time-saver, but limited in the type of problem you can use it on.
Hope that helps!
10. Nitish April 27, 2013 at 7:15 am #
Thnxx Chris 😀 !!
11. annu March 2, 2013 at 8:30 pm #
Hi Chris,
I’m big fan of yours..the way you make things possible is tremendous I don’t even have the words..I’m preparing for gre and I follow everything you write with care…I don’t know how well I’ll do but you’ll always be my hero…
Thanks a lot for your work and thanks to the team behind magoosh
• Chris Lele March 5, 2013 at 3:01 pm #
Wow, thanks for the kind words :). I’m so happy I am been helpful thus far. Good luck with your test and let me know whenever you have any questions :).
12. Z August 12, 2012 at 1:38 pm #
I usually never comment on these types of websites. But this lesson absolutely blew my mind. Thank you very much; all of your advice is great.
• Chris August 13, 2012 at 1:25 pm #
You are welcome!
• David September 3, 2013 at 9:17 pm #
Same here Chris! Blew my mind… this literally takes 10 seconds to answer a problem using this technique.
13. Tayyaba January 26, 2012 at 6:44 am #
That was really great!
• Chris January 26, 2012 at 2:33 pm #
I am happy that helped!
14. Julia September 26, 2011 at 5:59 am #
You are my hero. Wow. Thank you!
• Chris September 26, 2011 at 4:09 pm #
Glad I could help! Thanks so much!!
15. Erika July 6, 2011 at 12:09 am #
Wow! Thanks Chris! I just can’t wait for the new GRE material to come out tomorrow!
• Chris July 6, 2011 at 9:33 am #
Yep, Magoosh’s new GRE product is here! Also feel free to recommend any possible blog topics if there is a type of question/concept – math or verbal – that you find especially tricky while going through the new questions. Good luck!
Magoosh blog comment policy: To create the best experience for our readers, we will only approve comments that are relevant to the article, general enough to be helpful to other students, concise, and well-written! 😄 Due to the high volume of comments across all of our blogs, we cannot promise that all comments will receive responses from our instructors.
We highly encourage students to help each other out and respond to other students' comments if you can!
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# Break a stick in two random points, can you then build a triangle?
Here is an interesting problem that involves probability. Can you guess the answer?
Consider a stick of length 1. Select two points uniformly at random on the stick and break the stick at those points. What is the probability that the three segments form a triangle?
Where does this problem come from? After a short research, this problem appeared in one of the Senate-House Examinations (or later known as Mathematical Tripos) back in 1850 in Cambridge. From that point on, it became one of the many classical problems in continuous random variables. For instance, Poincaré included this problem in his Calcul des Probabilités (1986). Although not so recent, this paper by Dr. Goodman brings a great review on the multiple solutions and the history behind this problem. What makes this problem interesting to me is not the actual value of the probability of forming a triangle, but how the to get there.
The first challenge is to acknowledge that is is not always possible to make a triangle out of 3 segments. Asking friends, I’ve got the answers 1, 1/2 and 0 (in order of frequency). One of the reasons why I am writing this blog entry is because both of these solutions are very different from the canonical solution I find over the web.
I came across this problem during the 3rd QCBio Retreat, during the lunch break. I immediatelly started to give a shot solving it, but got stuck. While I was doing that, another guy in our table came up with a quite interesting solution. In the afternoon I was able to close my calculations.
Let me thank Brenno Barbosa for helping me cut a step in my formal solution.
And since we’re talking about triangles…
## Why is it not always possible to form a triangle?
The first challenge is to recognize that you can’t always build a triangle with the three resulting segments regardless of where you break the sick.
A simple counterexample is when you break the stick in two very close to one of the ends. As a result, two of the segments are too short, and the third one is very long. In this case, you can’t connect all three vertices.
The actual constraint to form a triangle is the triangle inequality. Let $x$, $y$, and $z$ be length of each of the three vertices of the triangle. Then, the triangle inequality can be written as
Here is a way to visualize this inequality:
To connect this with the problem at hand, assume that $x + y + z = 1$ and that the two randomly chosen points were $x$ and $x+y$. However, in this case, we also must satisfy
and
Any solution $(x,y,z)$ that does not satisfy all of the equations above represents a possibility of breaking the stick and not being able to build a triangle with the resulting three segments.
## A clever and direct solution: graphically enumerating the solutions
Let’s assume points $A$ and $B$ are the two points selected. The three inequalities from the previous section define a region of the plane $A$ and $B$ in which solutions are admissible. If we can somehow calculate the area of that region and normalize it, that solves the problem.
Here is how. Assume $A < B$, i.e., $A = x$ and $B = y + x$. From the first inequality, $A < 1/2$ must be satisfied. If $A$ is larger than half of the stick, then no point $B$ satisfies the triangle inequality.
Let’s draw the $A \times B$ plane: on the x-axis, all possible values of $A$; and the y-axis, all values of $B$. So, only the green shaded area is valid:
Because of the last inequality, if $B$ is also lower than $1/2$, then the third segment ($z$) is larger than the sum of the first two segments. This fact restricts the region of admissible regions further (on the green area):
The last condition is that the middle segment, between points $A$ and $B$, is not larger than the other two. To ensure that, $B < A + \tfrac{1}{2}$. This last inequality defined a straight line in the $A \times B$ plane:
Thus, the area is $1/4$ of the the admissible area, which is the probability of forming a triangle with the resulting broken segments.
I think this is a pretty creative and powerful solution, and to me . The only problem is when the problem gets harder, it becomes harder to come up with solutions like this one. So, next we will address this same problem but using elementary probability theory to come up with a purely anaytical solution.
## Quick Python script to numerically test the answer
Why not estimate this probability using a simple simulation? We can do that with a few simple lines of code. Here is an example.
With 10000 samples, the result is:
## Why bother with probability if there is a faster solution…?
This is always a great question: why do we bother (and why should you) with a more formal solution to this problem, if the one above does the job?
Problems like the Broken Stick can be extended very easily. Dr. Hildebrand, for instance, has wrote these interesting notes - probably for his calculus class - where he discusses this problem and proposes other possible related problems. In some cases, he studies the solution numerically.
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# Related Rate Problems II
If you look in most textbooks for related rate questions you will find pretty much the same related rate problems: ladders sliding down walls, people walking away from lamppost, water running into or out of cone shaped tanks, etc.
Here are two somewhat different related problems you may like.
1. A girl starts riding down an escalator at the same time a boy starts riding up a parallel escalator. The escalators are 5 meters apart, 30 meters long, and they both move at the rate of 1 m/sec.
1. How fast is the distance between the kids changing 10 seconds later? (Answer: -1.789 m/sec.)
2. Variation: Suppose that the girl is moving at 1 m/sec and the boy at 3 m/sec. The boy reaches the end of the escalator first and stands there. Now how fast is the distance between them changing 8 seconds after they start? (1.486 m/sec)
3. Variation: With the situation given in part b., write and graph a piece-wise defined function that gives the distance between the kids. Find where this graph is continuous but not differentiable. Why does this happen? (Answer: the boy reaches the top in 10 seconds and stands still. Up to this point the distance between them is represented by a hyperbola; here it now becomes a different parabola with an abrupt change in the graph. At this point, the function is not differentiable. This second parabola appears almost linear from the 10 second point on.
1. A 60-foot-long rope is attached to a pulley 36 feet above the ground. A lantern his attached to one end of the rope and a man holds the other end on the top of his head 6 feet off the ground. He walks away at the rate of 5 feet/sec.
1. Find the rate at which the distance between the pulley and the top of the man’s head is changing when he is 40 feet from the point directly under the pulley? (Answer: 4 ft./sec.)
2. Find the rate of change of the length of the man’s shadow when he is 40 feet from a point directly under the pulley? (Answer: -0.90 ft./sec.)
3. (Extension – extreme value problem, rather difficult) When the man starts walking the lantern is at the height of his head and his shadow is infinitely long. When is the tip of the man’s shadow closest to the point directly under the pulley and how far away is it? (Answer: At t = 3.654 sec the tip of the shadow is 39.658 feet from the point under the pulley.)
These two questions are from Audrey Weeks’ Calculus in Motion. This is a really good package of dynamic illustrations of calculus concepts and AP Exams free-response question that runs on Geometer’s Sketchpad. For more information click here. The related rate sections include both standard and non-standard problems.
Corrections made September 17, 2014, March 6, 2022
## 4 thoughts on “Related Rate Problems II”
1. Hi Lin,
Thanks for these great non-traditional related rate problems. Another possible edit here: for the lantern problem part (c), I am getting the time when the tip of his shadow is 39.658 ft. from the start as t = 3.654 rather than t = 6.654.
Like
• Sorry to take so long getting back to you; I’ve been on vacation – one of the advantages of retirement.
Thanks for catching the typo. I corrected it.
It’s discouraging; that post has been there since 2012 and no one else has caught the typo. I wonder if anyone else read it. On the other hand, part (c) is difficult; maybe they never finished it.
Like
2. Hi Lin! I have been enjoying your blog and resources since discovering them about 6 months ago. I have also appreciated your feedback so much on the AP Calc forum. When working through these Related Rates II problems above (to do with my students), I ran into a few difficulties.
In the escalator problem (#1 part 3), I wrote the piecewise function and found the value where it was not differentiable rather easily, I thought, but the statement that it “then becomes linear” did not make sense to me. I got another hyperbola: sqrt (25+t^2) for 10<t<30
Also, for the lantern problem, part 2, I got 6.5 ft/sec. I'd be happy to share my work in some way if you'd like. I can not get the answer that you posted. Could you please forward me the work if possible??? or at least the set up so I can compare?
Thank you! and thank you for all of your wonderful work! I wish you were closer to Cincinnati so I could do a workshop with you!!!!!
Marika McFall
Anderson High School
Cincinnati, Ohio
Like
• Marika
Whoops, looks like a did make a few mistakes.
In the escalator problem the second part is a hyperbola. I was looking at a graph and it appeared very linear.
I also have a mistake in the lantern problem. I do not agree with your answer (either). I got -0.90 feet/sec. My computation is this:
Let x= distance to man from point under pulley. $\frac{dx}{dt}=5$ (Given)
Let the length of the rope from the pulley to the man be
$z=\sqrt{{{30}^{2}}+{{x}^{2}}}$.
When x = 40, z = 50 so at this point
$\displaystyle \frac{dz}{dt}=\frac{2x\left( dx/dt \right)}{2\sqrt{900+{{x}^{2}}}}=\frac{5x}{\sqrt{900+{{x}^{2}}}}=\frac{5\left( 40 \right)}{\sqrt{900+{{40}^{2}}}}=4$
Let s = length of shadow. Z – 30 is the distance the lantern has moved upward from its initial position. By similar triangles
$\displaystyle \frac{s}{6}=\frac{x}{z-30}$, $\displaystyle s=\frac{6x}{z-30}$
$\displaystyle \frac{ds}{dt}=\frac{\left( z-30 \right)\left( 6 \right)\left( dx/dt \right)-6x\left( dz/dt \right)}{{{\left( z-30 \right)}^{2}}}=\frac{20\left( 6 \right)\left( 5 \right)-6\left( 40 \right)\left( 4 \right)}{{{20}^{2}}}=-0.90$
Thanks for catching these. The disappointing thing is that in two years no one caught them sooner.
I did a workshop in Cincinnati many years ago. The College Board hires us by region and I do not live in the Cincinnati region. I would be happy to come if someone is setting up a conference sometime or to do a workshop in your district. Check the “schedule” tab occasionally maybe I’ll be close some time.
Thanks again for writing.
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# Álgebra Lineal y Algunas de sus Aplicaciones by L.I. GOLOVINA
By L.I. GOLOVINA
HARDCOVER,page edges yellowed excellent
Similar linear books
Quaternions and rotation sequences: a primer with applications to orbits, aerospace, and virtual reality
Ever because the Irish mathematician William Rowan Hamilton brought quaternions within the 19th century--a feat he celebrated by way of carving the founding equations right into a stone bridge--mathematicians and engineers were thinking about those mathematical gadgets. this present day, they're utilized in purposes as numerous as describing the geometry of spacetime, guiding the gap trip, and constructing desktop purposes in digital fact.
Instructor's Solution Manual for "Applied Linear Algebra" (with Errata)
Resolution guide for the booklet utilized Linear Algebra via Peter J. Olver and Chehrzad Shakiban
Extra info for Álgebra Lineal y Algunas de sus Aplicaciones
Sample text
The decrease in population is gradual, however, and over the 100-year time span shown here it decreases only from 3000 to 1148. 9830 times that of the previous year. 9830) ≈ 411 years. One could try to find the matrix element a32 that results in the largest magnitude eigenvalue of A being exactly 1, and while this is an interesting mathematical problem, from the point of view of population control, it would likely be an exercise in futility. Clearly, this model is highly simplified, and even if the assumptions about survival rates and fecundity rates held initially, they might well change over time.
4 ECOLOGICAL MODELS Computational biology is a growing field of application for numerical methods. In this section, we explore a simplified example from ecology. Suppose we wish to study the population of a certain species of bird. These birds are born in the spring and live at most 3 years. We will keep track of the population just before breeding, when there will be three classes of birds, based on age: Age 0 (born the previous spring), Age 1, and Age 2. Let , and represent the number of females in each age class in Year n.
Both deterministic methods and Monte Carlo methods are sometimes used in this case. The picture to the left is a finite element discretization used in a deterministic model for radiation transport problems. (Image reproduced with the kind permission of the Applied Modelling and Computational Group at Imperial College London and EDF Energy. 3 MODELING IN SPORTS In FIFA World Cup soccer matches, soccer balls curve and swerve through the air, in the players’ attempts to confuse goalkeepers and send the ball sailing to the back of the net.
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### Theory:
Imagine a block is placed on the table as shown in the below image. Two strings $$X$$ and $$Y$$ are tied to the two opposite surfaces of the block, as shown. If we apply a force on the block by pulling the string $$X$$, it begins to move towards the right. Similarly, if we pull the string along $$Y$$, the block moves to the left. But, if the block is pulled from both sides with the same forces, the block will not move. Such forces are called balanced forces, and they do not change the state of rest or motion of an object.
Let us consider a situation in which two opposite forces of various magnitudes act on the block. In this case, the block would begin to travel in the direction of the greater force. Thus, the two forces are not balanced, and the unbalanced force acts in the direction the block moves.
A block placed on the table
What happens when you try to push a box on a rough floor?
Pushing a box on a rough floor
If we push the box with a small force, it does not move as the friction opposes the push.
The friction force arises between two surfaces that in contact; in this case, the bottom of the box and the rough surface of the floor. It counteracts the pushing force, and therefore the box does not move. When you push the box harder and the box still does not move, it denotes that the frictional force still balances the pushing force. The box will move only if the applied force is greater than the frictional force. When the applied force is greater than the frictional force, an unbalanced force acts on the box. Hence, the box starts moving.
What happens when we ride a bicycle?
A man riding bicycle
When we stop pedalling the bicycle, it begins to slow down. This occurs due to the frictional forces acting opposite to the direction of motion. To retain the bicycle moving, we have to start pedalling it again. You may think that an unbalanced force is always required to maintain the motion of an object. However, it is quite wrong.
If the forces acting on the object are balanced, and there is no net external force acting on it, the object moves with a uniform velocity.
If an unbalanced force is applied to the object, there will be a change either in its speed or the direction of motion. Thus, to accelerate the motion of an object, an unbalanced force should act on it. And, the change in its speed (or in the direction of motion) would continue as long as this unbalanced force is applied. However, if this force is removed completely, the object would continue to move with the velocity it has acquired till then.
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# Thread: What is this?
1. ## What is this?
A friend of mine gave this to me as a challenge. I almost have no clue what I'm supposed to do
he says "Solve for the following equation for f(x)"
$\displaystyle f(x) = x + \lambda \int_0^1 f(\xi) \,\,d \xi$
can anyone tell me what type of problem this is so I can do a bit of research on my own? My understanding of calculus only spans from Calc I to Calc II
2. ## Re: What is this?
Hey ReneG.
Hint: Try differentiating both sides to get a differential equation and solve from there.
(These kinds of problems are known as integro-differential equations)
Integro-differential equation - Wikipedia, the free encyclopedia
3. ## Re: What is this?
Thank you for pointing me in the right direction
4. ## Re: What is this?
Hi ReneG,
If the upper limit of your integral is x and not 1, I agree with Chiro. However as written, the integral is just a number c. So integrate both sides of your equation from 0 to 1 and get:
$\displaystyle c={1\over2}+\lambda c$ or $\displaystyle c={1\over2(1-\lambda)}$
So $\displaystyle f(x)=x+{\lambda\over2(1-\lambda)}$
5. ## Re: What is this?
Originally Posted by johng
Hi ReneG,
If the upper limit of your integral is x and not 1, I agree with Chiro. However as written, the integral is just a number c. So integrate both sides of your equation from 0 to 1
No typos. How did you integrate both sides though?
\displaystyle \begin{align*} f(x) &= x + \lambda \int_{0}^{1}f(\xi)\, d\xi \\ \int_{0}^{1}f(x)\,dx &= \int_0^1 x \,dx + \lambda \int_{0}^{1} \left[ \int_{0}^{1} f(\xi)\,d\xi \right ] \,dx \\ \int_{0}^{1}f(x)\,dx &= \frac{1}{2} + \lambda \int_{0}^{1} \left[ \int_{0}^{1} f(\xi)\,d\xi \right ]\,dx \end{align}
I'm lost.
6. ## Re: What is this?
Hi again,
It's easier for me to type the symbols in my own editor than to struggle with Latex and this HTML editor, so see the following "png" attachment:
I hope this clears it up.
7. ## Re: What is this?
Originally Posted by johng
I hope this clears it up.
I appreciate it, thank you.
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# A shape with 2 pairs of parallel sides and no right angles
It is much better to concentrate on the properties of shapes. This way you A square has got 4 sides of equal length and 4 right angles (right angle = 90 degrees). Since ALL A Rhombus has got NO right angles It has got 2 pairs of equal sides and 4 right angles Some quadrilaterals only have one set of parallel lines.
Since the sum of the interior angles of any triangle is 180° and there are two Let's start by examining the group of quadrilaterals that have two pairs of parallel sides. These quadrilaterals are called parallelograms They take a variety of shapes, case of a parallelogram that has four congruent sides and four right angles.
Karina drew a shape with 2 pairs of parallel sides and no right angles. Which shape could Karina have drawn? A. square B. rhombus C. rectangle D. trapeziod. The diagonals do not meet at right angles. A trapezoid is a quadrilateral that has exactly one pair. A parallelogram has opposite sides parallel and equal in length. All "quadrilaterals" can be separated into three sub-groups: general. However, a rhombus does not have to have four right angles as does a square or rectangle.
### A shape with 2 pairs of parallel sides and no right angles -
Opposite sides are the same length and they are parallel. A square is a parallelogram with four sides of equal length and four right angles. What is a five-sided polygon called? I can't draw the shape here, but it is a triangular prism that looks like a wedge of cheese. Use any two of the properties below. Note: Angled tic marks indicate parallel.
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# HEY! I’m Orbiting Here!
UPDATED: Math corrected from yesterday’s typos, see note at the bottom under “comments.” Also the Colorado based Center for Space Standards & Innovation put out a great simulation of the crash, you can see the video here.
UPDATED UPDATE: Well, looks like I wasn’t the only one to goof on the collision math at first.
Kablammo! Like something out of a Bruce Willis movie, an American and a Russian satellite collided about 800 km above Siberia. The two blew apart into hundreds of pieces after smacking into each other at over 28,000 km per hour. It must’ve been one heck of a demolition derby up there.
Even though there are over 17,000 man-made objects over 10 cm in orbit, this is the first time two satellites have accidentally collided. Two years ago China intentionally shot one of theirs out of the sky. NORAD is constantally tracking all satellites in the skies, so they can predict when the big ones are going to hit. We’re interested in the odds of a random crash and it’s clear that the odds of two satellites accidentally hitting each other is very low. How low you may ask? Lets try to figure it out.
When we’re calculating the frequency of hits for a satellite we need to start out with some facts and how they’ll interact. We’ll start out by borrowing an equation from molecular physics calculating the odds of two molecules colliding.
Here “Z” is the frequency of collisions for a single particle (satellites in this case). “n” represents the density of particles, “d” represents the size of the particles and “C” represents their average speed. The entire equation is divided by 2 to prevent double counting of collisions.
We can find the density of satellites by taking their total number divided by the volume of space they orbit in. Most satellites orbit in spheres between 500 and 1300 km above the surface of Earth. The equation V = 4/3 π r^3 gives us the volume of a sphere. Earth is about 6,300 km in diameter, making its radius (r) 3,150 km. First we find the volume of space the lowest orbiting satellites are in and subtract it from the highest orbiting satellites getting a total volume of space satellites orbit in.
Over 165 billion cubic km is a lot of space. Density here is simply number of satellites per volume. 17,000 / 165,348,213,000 = 1.02813328 × 10^-7 satellites per km^3.
Finding their diameter is just a matter of estimating (we’ll assume every satellite is spherical here for simplicity’s sake). Satellites can run from small bits of space junk only 10 cm in diameter, to the International Space Station which is almost as big as a football field. For the sake of simplicity, ill round and say the average satellite size is 10 meters in diameter, or .01 km so we can keep all of our units straight.
Speeds for satellites vary based on their distance from the Earth. Again for the sake of simplicity, we’ll assume all satellites are orbiting at the same speed of 28,000 km/hr. That’s all we should need to find out how often satellites should be running into each other. We start out by plugging our numbers into the original equation:
Giving us the result that an individual satellite will collide with another on average once every: 4.64682715 × 10^14 hours or roughly about once 53 billion years.
That’s the odds for one particular satellite, but all 17,000 would have roughly the same odds. Therefore : 4.64682715 × 10^14 hours / 17,000 = 27,334,277,400 hours or once every …
So, not terribly common at all. This is only the roughest of calculations to figure this collisions conundrum. In fact collisions might be be more likely because these basic equations assume that all space junk is equally distributed, when in fact much if it is concentrated around the equator. The pull of gravity also factors in, as two bodies of junk will attract one another.
On the other hand, the equation I used is for the random Brownian motion of molecules, rather than the ordered orbits of satellites in the sky. Molecules are much freer to move in 3 dimensions, while satellites are mostly trapped at a single altitude. That way, higher satellites are much less likely to interact with lower ones.
But for a back of the envelope, off the top of my head, rule of thumb calculation, that’s pretty close without bringing in calculus and probability curves and the like. Let me know what you think. Am I close, or am I just making stuff up as I go?
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# Cardinal Numbers - Definition with Examples
The Complete K-5 Math Learning Program Built for Your Child
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## Cardinal Numbers
Cardinal numbers are counting numbers. The numbers that we use for counting are called cardinal numbers.
Cardinal numbers tell us “How many?”
For example:
How many dogs are there in all?
To know the total number of dogs, we need to count the dogs given in the image.10804
Therefore, there are 8 dogs in all.
Example: How many kites are there in all?
Count the kites to know the total number of kites.
On counting, we get:
Thus, from the above examples, we see that to know “how many?” of something is there, we need to use cardinal numbers.
From where do cardinal numbers start?
Cardinal numbers or counting numbers start from
How many cardinal numbers are there?
Cardinal numbers can go on and on and on. That means there are infinite counting numbers.
(Imagine, how many numbers you’ll need to count the number of stars in the sky or the number of sand grains in a desert?)
How are cardinal numbers formed?
The digits 1, 2, 3, 4, 5, 6, 7, 8, 9 and 0 are used to form several other cardinal numbers.
For example:
Is zero (0) a cardinal number?
No, zero (0) is not a cardinal number. To know “how many” there should be something. Since 0 means nothing; it is not a cardinal number.
We can write cardinal numbers in numerals as 1, 2, 3, 4, and so on as well as in words like one, two, three, four, and so on.
The chart shows the cardinal numbers in figures as well in words.
Cardinality
The cardinality of a group (set) tells how many objects or terms are there in that set or group.
Example: What is the cardinality of the flowers in the vase?
Here, there are 5 flowers in the vase. Therefore, the cardinality of flowers is 5.
Cardinal numbers start from 1. Fractions and decimals represent a part (less than one) of a whole or a group. Therefore, fractions and decimals are not cardinal numbers.
Fun Facts Cardinal numbers are also called natural numbers. Natural Numbers (Cardinal numbers) along with 0 form a set of whole numbers.
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Mathematics
# Calculus the Basics
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Calculus is basically the use of differentiation and integration on a given polynomial. At the AS level Mathematics in the UK calculus is introduced for the first time and is a rather different, though simple concept for students to grasp.
Differentiation
This is the process of taking each part of any polynomial (equation containing any powers of x) and multiplying it by the power of x then decreasing the power of x by one. For example:
4x^2 -> 2 x (4x^2) -> 2 x 4x -> 8x
Step one: Take the number.
Step two: Multiply by power, in this case 2.
Step Three: Take one from the power, in this case taking it down to 1.
The basic uses of differentiation are quite helpful in many things in mathematics. Given the equation of a line you can use the first derivative (differentiated once) to find the gradient of a curve by substituting in a point on the curve. The first derivative is known as 'dy/dx'.
Differentiation can also be used to find stationary points on graphs, ie. where the gradient is equal to zero. This is done by taking the equation of the graph, finding the first derivative of it and setting that equal to zero, this will find the points on the graph that are equal to zero.
The second derivative (differentiated twice) can be used then to find if this is a maximum point, the stationary point is at the top of the curve; a minimum point, the stationary point is at the bottom of the curve; or a point of inflexion where the graph goes level in the middle of a graph.
Here are some very simple diagrams of what I mean:
Maximum point:
_
/
/
Minimum point:
/
_/
Point of inflexion:
/
__/
/
/
That just about covers the basic uses for differentiation, there are further uses for it which are more advanced.
Integration
This is basically the opposite of differentiation. If you took the second derivative of an equation and integrated it you would get the first derivative. When anything is integrated, there is always an unknown, which is commonly referred to as 'c'. This can be worked out if there is a point given.
To integrate something you firstly take each part seperately then add one to the power of x then divide it al by that power. For example:
8x -> 8x^2 -> (8x^2)/2 -> 4x^2
Step One: Take the part of your equation you want to integrate.
Step Two: Raise the power by one.
Step Three: Divide by the new power.
Step Four: Simplify the outcome.
Integration, like differentiation has many uses. The main use in basic calculus of integration is finding the area under a curve between two points. It can also find the area between two curves, between two points.
This is done by first taking the equation of the line and integrating it. Let us say the equtation of the line is y = 3x^2 + 4x + 2, not too difficult. When this equation is fully integrated it comes out as: x^3 + 2x^2 + 2x + c.
Right, though I said earlier that all integrations bring out an unknown, 'c', this is different. Though the integration does make an unknown 'c' it is not needed for the equation. Say in this we want to find the area between x points 1 and 3. We need to substitute in 3 and 1 into the equation:
3^3 + 2(3^2) + 2(3) = 27 + 18 + 6 = 51
1^3 + 2(1^2) + 2(1) = 1 + 2 + 1 = 4
After this we substitute the lesser x value, in this case 1, from the higher value, in this case 3. This substitution would have cancelled out the 'c' in the equation therfore it was not necessary to work it out.
51 - 4 = 47
This is the area under the curve between points 1 and three and above the x axis.
Calculus can take you far in maths and it is a handy basic tool to know about. If you are still in secondary education, getting the grasp of this early if you intend to go on into further education will give you a great advantage. If this didn't help then look it up elsewhere. A useful tool for anyone.
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# help on a probability problem
• Feb 7th 2010, 09:19 AM
ilc
help on a probability problem
Two cards are drawn from a deck of 52 plating cards. find the probability of each of the following events occurring:
a) both cards are clubs
b) both cards are red
c) both cards are queens
d) both cards are red queens
e) both cards are queens or both cards are red
-----------
a) 13 clubs out of 52 = 13/52
there will be 12 clubs left after picking the first one so: 12/52
so, (13/52)x(12/52) = 156/2704 chances to have both cards as clubs? is this correct?
b) (26/52)x(23/52) = 650/2704
C) (4/52)x(3/52) = 12/2704
d) (26/52)x(23/52) = 2/2704
how would i do e?
• Feb 7th 2010, 09:28 AM
e^(i*pi)
Quote:
Originally Posted by ilc
Two cards are drawn from a deck of 52 plating cards. find the probability of each of the following events occurring:
a) both cards are clubs
b) both cards are red
c) both cards are queens
d) both cards are red queens
e) both cards are queens or both cards are red
-----------
a) 13 clubs out of 52 = 13/52
there will be 12 clubs left after picking the first one so: 12/52
so, (13/52)x(12/52) = 156/2704 chances to have both cards as clubs? is this correct?
b) (26/52)x(23/52) = 650/2704
C) (4/52)x(3/52) = 12/2704
d) (26/52)x(23/52) = 2/2704
how would i do e?
Where did you get 23/52 from in part B? Surely removing 1 red card would mean that 25 remain?
Is the first card replaced? I have assumed it is not
All your second terms should have a denominator of 51.
For example in B:
$\displaystyle \frac{26}{52} \times \frac{25}{51} = \frac{25}{102}$
And in C:
$\displaystyle \frac{4}{52} \times \frac{3}{51} = \frac{1}{221}$
Then part E would be
$\displaystyle \frac{25}{102} + \frac{1}{221}$
• Feb 7th 2010, 09:37 AM
ilc
yes, that should be 25...that was a typo. thanks!
• Feb 7th 2010, 09:42 AM
Quote:
Originally Posted by ilc
Two cards are drawn from a deck of 52 plating cards. find the probability of each of the following events occurring:
a) both cards are clubs
b) both cards are red
c) both cards are queens
d) both cards are red queens
e) both cards are queens or both cards are red
-----------
a) 13 clubs out of 52 = 13/52
there will be 12 clubs left after picking the first one so: 12/52
so, (13/52)x(12/52) = 156/2704 chances to have both cards as clubs? is this correct?
b) (26/52)x(23/52) = 650/2704
C) (4/52)x(3/52) = 12/2704
d) (26/52)x(23/52) = 2/2704
how would i do e?
Hi ilc,
in making these calculations, you are replacing the first card.
This is incorrect as you are choosing 2 from 52,
hence after picking the first card, you have 51 left.
Therefore redo parts a), b) and c).
In part d), you must count the number of red queens.
For part e),
You can calculate the probabilities of getting 2 red cards and add to the probability of getting 2 queens....
but you must subtract the probability of getting 2 red queens as this is a conditional probability question. They will have already been accounted for.
There is overlap since some queens are red.
• Feb 7th 2010, 10:04 AM
ilc
how would i get the probability of the two red queens?
2/53 x 1/51 = 2/2652 correct?
• Feb 7th 2010, 10:13 AM
You only need to know how a pack is organised.
There are 26 red and 26 black.
There are 4 queens, 2 are red and 2 are black.
Therefore 2 of the 26 red cards are red queens.
Therefore there are 24 red cards that are not queens.
There are another 2 queens which are black.
If both cards are queens or both are red, then we must include the probability of also getting 2 black queens.
Also, the queens can be a black and red.
However if we include the 2 red queens, we must be aware that these are already counted,
since the 2 red queens have been already included in 2 red cards total.
Hence we can add the probabilities of 2 red cards to the probability of getting 2 queens,
but we must subtract the probability of getting 2 red queens.
• Feb 7th 2010, 10:33 AM
$\displaystyle \frac{2}{52}\ \frac{1}{51}$
Or choose both red queens at once....$\displaystyle \frac{\binom{2}{2}}{\binom{52}{2}}$
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## All-in-one learning app
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# Single Variable Data
Single variable data is usually called univariate data. This is a type of data that consists of observations on only a single characteristic or attribute. Single variable data can be used in a descriptive study to see how each characteristic or attribute varies before including that variable in a study with two or more variables.
## Examples of single variable data
What were the scores of the students that took the maths test? Which sickness was responsible for most deaths in 2020? What are the weights of each person present in the gym? What is the typical income of the average person in the UK? All these questions can be answered using single variable data. Single variable analysis is the simplest form of analysing data. Its main purpose is to describe, and it does not take into considerations causes and relationships.
For instance, when the question about the scores of students that took a particular math test is asked, we are mostly interested in how varied the results are from each person. By this, we can statistically summarise the data using Statistical Measures to get an idea about the performance of the whole population that took the test.
## How significant is single variable data?
In research, single variable data does not concern itself with answering questions that involve relationships between variables. It describes an attribute of the subject in question, and how it varies from observation to observation. Univariate data analysis involves using statistical measures such as Measures of Central Tendency. It also takes advantage of measures of spread.
There are two main reasons why a researcher would conduct a single variable analysis. The first is to have a descriptive study of how one characteristic varies from subject to subject. The second is to analyse the variety of each characteristic before they can be paired with other variables in a study.
This is where Bivariate Data and multivariate data comes in. Multivariate data describes multiple characteristics of a subject. It is necessary to examine how varied students' scores are with respect to other factors such as subject and their background.
## Single variable data analysis
As mentioned earlier, statistical measures are used to summarise single variable data's centres and spread. Whilst the commonest way to display single variable data is in a table, other common ways are:
• Histograms.
• Frequency distribution.
• Box plots.
• Pie charts.
Scores of eight students were recorded after taking a maths test in grade 6, and they are as follows; 76, 88, 45, 50, 88, 67, 75, 83. Find the
1. Mean
Median
3. Mode
1.
2.
Rearrange values from lowest to highest.
45, 50, 67, 75, 76, 83, 88, 88
Median = 75.5
3.
The most frequently occurring number is 88.
## Histograms
Histograms are one of the most commonly used graphs to show frequency distribution. It is a graphical display of data using bars of different heights. Similar to the bar chart, the histogram groups numbers into ranges. It is an appropriate way to display single-variable data.
Histogram of travel time to work. Image: QWFP, CC BY-SA 3.0
## Frequency distribution
Frequency distribution is data modelled in a tabular format to display the number of observations within a space. This displays values and their frequency (how often something occurs). This format also appropriately represents single variable data and is as simple as possible.
The numbers of newspapers sold at a shop over the last 10 days are;
20, 20, 25, 23, 20, 18, 22, 20, 18, 22.
This can be represented by frequency distribution. The values above are the variables, and the table is going to show how often a specific number of sales occurred over the last 10 days.
Papers sold Frequency 2 0 21 0 2 23 1 24 0 1
## Pie charts
Pie charts are types of graphs that display data as circular graphs. They are represented in slices where each slice of the pie is relative to the size of that category in the group as a whole. This means that the entire pie is 100%, and each slice is its proportional value.
Assuming the data for pets ownership in Lincoln were collected as follows, how would it be represented on a pie chart?
Dogs - 1110 people
Cats - 987 people
Rodents - 312 people
Reptiles - 97 people
Fish - 398 people
Figure 2. Pie chart representing data of pets in Lincoln
## Box plots
Presenting data using the box plot gives a good graphical image of the concentration of the data. It displays the five-number summary of a dataset; the minimum, first quartile, median, third quartile, and maximum. This is also a good system to represent single variable data.
The ages of 10 students in grade 12 were collected and they are as follows.
15, 21, 19, 19, 17, 16, 17, 18, 19, 18.
First, we will arrange this from lowest to highest so the median can be determined.
15, 16, 17, 17, 18, 18, 19, 19, 19, 21
Median = 18
In finding the quartiles, the first will be the median to the right of the overall median.
The median for 15, 16, 17, 17, 18 is 17
The third quartile will be the median to the right of the overall median.
Median for 18, 19, 19, 19, 21, will make 19.
We will now note the minimum number which is 15, and also the maximum which is 21.
Figure 3. Box plot representing students ages
## Single variable data - Key takeaways
• Single variable data is a term used to describe a type of data that consists of observations on only a single characteristic or attribute.
• Single variable data's main purpose is to describe, and it does not take into considerations causes and relationships.
• Statistical measures are used to summarise single variable data's centres and spread.
• Common ways single variable data can be described are through histograms, frequency distributions, box plots, and pie charts.
Images
Histogram: https://commons.wikimedia.org/wiki/File:Travel_time_histogram_total_n_Stata.png
Variable means the measured values can be varied anywhere along a given scale, whilst attribute data is something that can be measured in terms of numbers or can be described as either yes or no for recording and analysis.
The ages of students in a class.
Single variable data gives measures of only one attribute whilst two-variable data gives measures of two attributes describing a subject.
Single variable data is used to describe a type of data that consists of observations on only a single characteristic or attribute.
## Final Single Variable Data Quiz
Question
What is cumulative frequency?
The cumulative frequency at a point x is the sum of the individual frequencies up to and at the point x.
Show question
Question
Which of the following can you obtain from a cumulative frequency distribution? a) median b) quartiles c) percentiles d) all of the above
d
Show question
Question
If a cumulative frequency for the (n-1)th value is 85 in discrete frequency distribution with 110 data points, what is the raw frequency for the nth value?
25
Show question
Question
For a grouped frequency distribution, what is the class mark for the class 0.5 - 1.0?
0.75
Show question
Question
For a grouped frequency distribution, what is the class mark for the class 2.5 - 3.5?
3.0
Show question
Question
For a grouped frequency distribution, what is the class mark for the class 8 - 12?
10
Show question
Question
State whether the following statement is true or false : the curve for a cumulative frequency graph is never decreasing.
True
Show question
Question
The cumulative frequency curve for an experiment with 200 trials is given by x = y/5, where the cumulative frequency is represented on the y-axis. Find the median.
x = (200/2)/5 = 20
Show question
Question
The cumulative frequency curve for an experiment with 200 trials is given by x = y/5, where the cumulative frequency is represented on the y-axis. Find the upper quartile.
x = (200 × 3/4)/5 = 30
Show question
Question
The cumulative frequency curve for an experiment with 200 trials is given by x = y/5, where the cumulative frequency is represented on the y-axis. Find the 43rd percentile.
x = (200 × 43/100)/5 = 17.2
Show question
Question
The cumulative frequency curve for an experiment with 200 trials is given by x = y/5, where the cumulative frequency is represented on the y-axis. Find the 70th percentile.
x = (200 × 70/100)/5 = 28
Show question
Question
The cumulative frequency curve for an experiment with 100 trials is given by x = 2y + 3, where the cumulative frequency is represented on the y-axis. Find the median.
x = 2 × (100/2) + 3 = 103
Show question
Question
The cumulative frequency curve for an experiment with 100 trials is given by y = 2x + 3, where the cumulative frequency is represented on the y-axis. Find the lower quartile.
x = 2 × (100/4) + 3 = 53
Show question
Question
A grouped frequency distribution has been made for the length of 500 snakes. The cumulative frequency of a class (8.0 - 8.5) inches is 320. How many snakes are more than 8.5 inches long?
180
Show question
Question
A grouped frequency distribution has been made for the length of 500 snakes. The cumulative frequency of a class (8.0 - 8.5) inches is 320. Which of the following is the correct conclusion?
There are 320 snakes shorter than than or equal to 8.5 inches
Show question
Question
What is a box plot?
A box plot is a type of graph that visually shows features of the data.
Show question
Question
What are the features that a box plot shows?
A box plot shows you the lowest value, lower quartile, median, upper quartile, highest value and any outliers that the data may have.
Show question
Question
How do you find upper and lower quartiles?
To find the upper and lower quartile you first need to arrange your data into numerical order, the next step is to find your median, you can then use this to find both of the quartiles. The lower quartile will then be the midpoint between the lowest value and the median, the upper quartile will be the midpoint between the median and the highest value.
Show question
Question
How do you find the interquartile range?
To find the interquartile range you subtract the lower quartile from the upper quartile.
Show question
Question
What is an outlier?
An outlier is classed as data that falls 1.5 x the interquartile range above the upper quartile or below the lower quartile.
Show question
Question
What is a histogram?
A histogram is a type of graph that represents grouped data.
Show question
Question
How do you calculate the frequency density?
Frequency density is calculated by dividing the frequency by the class width.
Show question
Question
What is a frequency polygon?
A frequency polygon is a graphical representation of a data set with frequency information. It is one of the most commonly used statistical tools used to represent and analyze grouped statistical data.
Show question
Question
For a grouped frequency distribution, what is plotted along the X-axis when building a frequency polygon?
Class mark
Show question
Question
For a grouped frequency distribution, what is plotted along the Y-axis when building a frequency polygon?
Frequency
Show question
Question
How do you obtain a frequency polygon from a given histogram?
Join the middle of the top of each bar of the histogram sequentially.
Show question
Question
State whether the following statement is true or false : To draw a frequency polygon, you first have to create a histogram.
False
Show question
Question
What is the class mark for the class "8-10"?
9
Show question
Question
What is the class mark for the class "45.5-55"?
50.25
Show question
Question
What is the class mark for the class "0.1-0.2"?
0.15
Show question
Question
State whether the following statement is true or false : The sum of the frequencies of a frequency polygon must equal 1
False
Show question
Question
State whether the following statement is true or false : The frequencies of a frequency polygon must be positive
True
Show question
Question
State whether the following statement is true or false : To draw a frequency polygon from a given grouped frequency distribution, we must plot the frequency against the class marks and not the class boundaries.
True
Show question
Question
What are the two types of measures that are usually commented on when comparing data distributions?
1. measure of location
Show question
Question
What is a measure of spread?
a measure of spread provides us information regarding the variability of data in a given data set, i.e. how close or far away the different points in a data set are from each other.
Show question
Question
What is a measure of location?
a measure of location is used to summarize an entire data set with a single value.
Show question
Question
Data set A - median 25, Q1 = 18, Q3 = 56
Data set B - median 24, Q1 = 14, Q3 = 130
Data set A has a lower measure of location (median) and also a lower variability among the data.
Show question
Question
Data set A - median 100, Q1 = 50, Q3 = 150
Data set B - median 200, Q1 = 150, Q3 = 250
Data set A has a lower measure of location (median). There appears to be an equal variability among the data sets.
Show question
Question
Data set A - median 300, Q1 = 275, Q3 = 325
Data set B - median 200, Q1 = 150, Q3 = 250
Data set A has a higher measure of location (median) and a lower variability among the data.
Show question
Question
Which of the following is appropriate to use along with median for comparison?
Interquartile range
Show question
Question
Which of the following is appropriate to use along with mean for comparison?
standard deviation
Show question
Question
Which of the following is appropriate to use along with
standard deviation for comparison?
mean
Show question
Question
Which of the following is appropriate to use along with interquartile range for comparison?
mean
Show question
Question
Which of the following should you use for comparing a data set with extreme values?
mean and standard deviation
Show question
Question
Compare the 2 data sets
Data set A - mean 100, standard deviation = 50
Data set B - mean 200, standard deviation = 50
Data set A has a lower measure of location (mean). There is an equal variability among the data sets.
Show question
Question
Compare the 2 data sets
Data set A - mean = 13, standard deviation = 5
Data set B - mean = 18, standard deviation = 15
Data set A has a lower measure of location (mean) and a lower variability among the data sets.
Show question
Question
Compare the 2 data sets
Data set A - mean = 13, standard deviation = 5
Data set B - mean = 13, standard deviation = 5
Both data sets have similar measures of location and spread within the data.
Show question
Question
What is single-variable data?
Single-variable data is a type of data that consists of observations on only a single characteristic or attribute.
Show question
Question
Another name for univariate data is?
Single-variable data
Show question
Question
Amongst the number of ways single-variable data can be represented, which of them does it display the five-number summary of a dataset?
Box plot
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# What is 65% of 0.75 + Solution with free steps?
The 65 percent of 0.75 is 0.4875. This answer is obtained by multiplying the given percentage by the decimal number.
65 % of a number 0.75 means that we are taking the 65th part of the whole number 0.75. If we consider a number as a whole, it means it has 100 parts.
For example, If we consider that only one candy is left for four children and one child has already taken a tiny bite out of it. It means the remaining candy has only 0.75. Two children must get 65% out of this 0.75%. After dividing, two children will get a quantity of 0.4875 from 0.75 together.
Percentages find their applications in academics, engineering sites, stock markets, Information technology, etc.
In this question, the detailed solution of calculating 65 percent of 0.75 is given with free steps.
## What is 65 percent of 0.75?
The 65 percent of 0.75 is 0.4875.
The answer of 65 percent of 0.75 is 0.4875 which means the 65th part of 0.75 is 0.4875. The given 65% is the 65th portion out of the total 100 portions of the number 0.75. There are many formulae of percentages to calculate the percentage of the number but in this question, the percentage is given so we have to calculate the remaining portion of 0.75. We can calculate the percentage by following 4 simple steps. These steps are:
### Step 1
The formula for calculating the number is:
Percentage of a number x total number = X
Here, X is the required number
### Step 2
By putting values in the formula:
65% x 0.75 = X
Here, 65 is the percentage, and 0.75 is the whole number.
### Step 3
The 65th part of the hundred can be written as:
(65/100) x 0.75
0.65 x 0.75 = X
0.4875 = X
The required number is 0.4875 which is 65 percent of 0.75.
The pie chart below shows 65% of 0.75:
Figure 1
This pie chart of the number 0.75 has two portions. The red portion is the bigger portion and the green portion is the smaller portion of a piechart. The red portion of the piechart shows 65% of 0.75 while the green portion shows the remaining 35% of 0.75.
Percentages are extremely useful in finding the fractions of the total number. They make it easy to deal with large numbers.
All the Mathematical drawings/images are created using GeoGebra.
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## How is the graph of Tangent different from sine and cosine?
The sine, cosine and tangent functions are said to be periodic. This means that they repeat themselves in the horizontal direction after a certain interval called a period. The sine and cosine functions have a period of 2π radians and the tangent function has a period of π radians.
## How do you graph tangent?
How to Graph a Tangent Function
1. Find the vertical asymptotes so you can find the domain. These steps use x instead of theta because the graph is on the x–y plane.
2. Determine values for the range.
3. Calculate the graph’s x-intercepts.
4. Figure out what’s happening to the graph between the intercepts and the asymptotes.
What is tangent in graph?
In geometry, the tangent line (or simply tangent) to a plane curve at a given point is the straight line that “just touches” the curve at that point. Leibniz defined it as the line through a pair of infinitely close points on the curve. The word “tangent” comes from the Latin tangere, “to touch”.
How do you tell if a graph is cosine or sine?
The graph of the cosine is the darker curve; note how it’s shifted to the left of the sine curve. The graphs of y = sin x and y = cos x on the same axes. The graphs of the sine and cosine functions illustrate a property that exists for several pairings of the different trig functions.
### How do you go from sin to csc?
The secant of x is 1 divided by the cosine of x: sec x = 1 cos x , and the cosecant of x is defined to be 1 divided by the sine of x: csc x = 1 sin x .
### What is the difference between sine and cosine?
Key Difference: Sine and cosine waves are signal waveforms which are identical to each other. The main difference between the two is that cosine wave leads the sine wave by an amount of 90 degrees. A sine wave depicts a reoccurring change or motion.
What is the tangent of a graph?
A tangent of a curve is a line that touches the curve at one point. It has the same slope as the curve at that point. A vertical tangent touches the curve at a point where the gradient (slope) of the curve is infinite and undefined. On a graph, it runs parallel to the y-axis.
How do you calculate sine?
The trigonometric function sine, like the cosine and the tangent, is based on a right-angled triangle. In mathematics, you can find the sine of an angle by dividing the length of the side opposite the angle by the length of the hypotenuse.
## What is the equation for a sine graph?
The general equation of a sine graph is y = A sin(B(x – D)) + C. The general equation of a cosine graph is y = A cos(B(x – D)) + C. Example: Given a transformed graph of sine or cosine, determine a possible equation.
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RD Sharma Solutions (Part 1): Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Class 10 : RD Sharma Solutions (Part 1): Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
The document RD Sharma Solutions (Part 1): Pair of Linear Equations in Two Variables Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10
Exercise 3.1
Q.1. Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel and play Hoopla (a game in which you throw a rig on the items kept in the stall, and if the ring covers any object completely you get it). The number of times she played Hoopla is half the number of rides she had on the Giant Wheel. Each ride costs Rs 3, and a game of Hoopla costs Rs 4. If she spent Rs 20 in the fair, represent this situation algebraically and graphically. Sol: The pair of equations formed is:
Solution.
The pair of equations formed is:
i.e., x - 2y = 0 ....(1)
3x + 4y = 20 ....(2)
Let us represent these equations graphically. For this, we need at least two solutions for each equation. We give these solutions in Table
Recall from Class IX that there are infinitely many solutions of each linear equation. So each of you choose any two values, which may not be the ones we have chosen. Can you guess why we have chosen x =O in the first equation and in the second equation? When one of the variables is zero, the equation reduces to a linear equation is one variable, which can be solved easily. For instance, putting x =O in Equation (2), we get 4y = 20 i.e.,
y = 5. Similarly, putting y =O in Equation (2), we get 3x = 20 ..,But asis not an integer, it will not be easy to plot exactly on the graph paper. So, we choose y = 2 which gives x = 4, an integral value.
Plot the points A (O,O) , B (2,1) and P (O,5) , Q (412) , corresponding to the draw the lines AB and PQ, representing the equations x - 2 y = O and 3x + 4y= 20, as shown in figure
In fig., observe that the two lines representing the two equations are intersecting at the point (4,2),
Q.2. Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” Is not this interesting? Represent this situation algebraically and graphically.
Sol: Let the present age of Aftab and his daughter be x and y respectively. Seven years ago.
Age of Ahab = x - 7
Age of his daughter y - 7
According to the given condition.
(x - 7) = 7(y - 7)
⇒ x - 7 = 7y - 49
⇒ x - 7y = -42
Three years hence
Age of Aftab = x + 3
Age of his daughter = y + 3
According to the given condition,
(x + 3) = 3 (y + 3)
⇒ x+3 = 3y +9
⇒ x - 3y = 6
Thus, the given condition can be algebraically represented as
x - 7y = - 42
x - 3y = 6
x - 7y = - 42 ⇒ x = -42 + 7y
Three solution of this equation can be written in a table as follows:
x - 3y = 6 ⇒ x = 6+3y
Three solution of this equation can be written in a table as follows:
The graphical representation is as follows:
Concept insight In order to represent a given situation mathematically, first see what we need to find out in the problem. Here. Aftab and his daughters present age needs to be found so, so the ages will be represented by variables z and y. The problem talks about their ages seven years ago and three years from now. Here, the words ’seven years ago’ means we have to subtract 7 from their present ages. and ‘three years from now’ or three years hence means we have to add 3 to their present ages. Remember in order to represent the algebraic equations graphically the solution set of equations must be taken as whole numbers only for the accuracy. Graph of the two linear equations will be represented by a straight line.
Q.3. The path of a train A is given by the equation 3x + 4y - 12 = 0 and the path of another train B is given by the equation 6x + 8y - 48 = 0. Represent this situation graphically.
Sol:
The paths of two trains are giver by the following pair of linear equations.
3x + 4 y -12 = 0 ...(1)
6x + 8 y - 48 = 0 ... (2)
In order to represent the above pair of linear equations graphically. We need two points on the line representing each equation. That is, we find two solutions of each equation as given below:
We have,
3x + 4 y -12 = 0
Putting y = 0, we get
3x + 4 x 0 - 12 = 0
⇒ 3x = 12
Putting x = 0, we get
3 x 0 + 4 y -12 = 0
⇒ 4y = 12
Thus, two solution of equation 3x + 4y - 12 = 0 are ( 0, 3) and ( 4, 0 )
We have,
6x + 8y -48 = 0
Putting x = 0, we get
6 x 0 + 8 y - 48 = 0
⇒ 8y = 48
⇒ y = 6
Putting y = 0, we get
6x + 8 x 0 = 48 = 0
⇒ 6x = 48
Thus, two solution of equation 6 x + 8y - 48= 0 are ( 0, 6 ) and (8, 0 )
Clearly, two lines intersect at ( -1, 2 )
Hence, x = -1,y = 2 is the solution of the given system of equations.
Q.4. Gloria is walking along the path joining (— 2, 3) and (2, — 2), while Suresh is walking along the path joining (0, 5) and (4, 0). Represent this situation graphically.
Sol:
It is given that Gloria is walking along the path Joining (-2,3) and (2, -2), while Suresh is walking along the path joining (0,5) and (4,0).
We observe that the lines are parallel and they do not intersect anywhere.
Q.5. On comparing the ratios and without drawing them, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincide:
(i) 5x- 4y + 8 = 0
7x + 6y - 9 = 0
(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
(iii) 6x - 3y + 10 = 0
2x - y + 9 = 0
Sol:
We have,
5x - 4 y + 8 = 0
7 x + 6 y - 9 = 0
Here,
a= 5, b1 = -4, c1 = 8
a2 = 7, b2 = 6, c2 = -9
We have,
∴ Two lines are intersecting with each other at a point.
We have,
9 x + 3 y +12 = 0
18 + 6 y + 24 = 0
Here,
a1 = 9, b1 = 3, c1 = 12
a2 = 18, b2 = 6, c2 = 24
Now,
And
∴ Both the lines coincide.
We have,
6 x - 3 y +10 = 0
2 x - y + 9 = 0
Here,
a1 = 6, b= -3, c1 = 10
a2 = 2, b2 = -1, c2 = 9
Now,
And
∴ The lines are parallel
Q.6. Given the linear equation 2x + 3y - 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines.
Sol:
We have,
2x + 3 y - 8 = 0
Let another equation of line is
4x + 9 y - 4 = 0
Here,
a1 = 2, b1 = 3, c1 = -8
a= 4, b2 = 9, c2 = -4
Now,
And
∴ 2x + 3 y - 8 = 0 and 4 x + 9 y - 4 = 0 intersect each other at one point.
Hence, required equation of line is 4 x + 9y - 4 = 0
We have,
2x + 3y -8 = 0
Let another equation of line is:
4x +6y -4 = 0
Here,
a1 = 2, b1 = 3, c1 = -8
a2 = 4, b2 = 6, c2 = -4
Now,
And
∴ Lines are parallel to each other.
Hence, required equation of line is 4 x + 6y - 4 = 0.
Q.7. The cost of 2kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4kg of apples and 2kg of grapes is Rs 300. Represent the situation algebraically and geometrically.
Sol:
Let the cost of 1 kg of apples and 1 kg grapes be Rs x and Rs y.
The given conditions can be algebraically represented as:
2 x + y = 160 4 x + 2 y = 300
2x + y = 160 ⇒ y = 160 - 2x
Three solutions of this equation cab be written in a table as follows:
4x + 2y = 300 ⇒ y =
Three solutions of this equation cab be written in a table as follows:
The graphical representation is as follows:
Concept insight: cost of apples and grapes needs to be found so the cost of 1 kg apples and 1kg grapes will be taken as the variables from the given condition of collective cost of apples and grapes, a pair of linear equations in two variables will be obtained. Then In order to represent the obtained equations graphically, take the values of variables as whole numbers only. Since these values are Large so take the suitable scale.
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Mathematics (Maths) Class 10
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### Introduction
As you know, we entered our discussion of derivatives to determine size and direction of a step with which to move along a cost curve. We first used a derivative in a single variable function to see how the output of our cost curve changed with respect to change a change in our regression line's y-intercept or slope. Then we learned about partial derivatives to see how a three dimensional cost curve responded to a change in a regression line's y-intercept or slope.
However, we have not yet explicitly showed how partial derivatives apply to gradient descent.
Well, that's what we hope to show in this lesson: explain how we can use partial derivatives to find the path to minimize our cost function, and thus find our "best fit" regression line.
### Finding the steepest path
Now gradient descent literally means that we are taking the shortest path to descend towards our minimum. However, it is somewhat easier to understand gradient ascent than descent, and the two are quite related, so that's where we'll begin. Gradient ascent, as you could guess, simply means that we want to move in the direction of steepest ascent.
Now moving in the direction of greatest ascent for a function $f(x,y)$, means that our next step is a step some distance in the $x$ direction and some distance in the $y$ direction that is the steepest upward at that point.
Note how this is a different task from what we have previously worked on for multivariable functions. So far, we have used partial derivatives to calculate the gain from moving directly in either the $x$ direction or the $y$ direction. Here in finding gradient ascent, our task is not to calculate the gain from a move in either the $x$ or $y$ direction. Instead our task is to find some combination of a change in $x$,$y$ that brings the largest change in output. So if you look at the path our climbers are taking in the picture above, that is the direction of gradient ascent. If they tilt their path to the right or left, they will no longer be moving along the steepest upward path.
The direction of the greatest rate of increase of a function is called the gradient. We denote the gradient with the nabla, which comes from the Greek word for harp, which is kind of what it looks like: $\nabla$. So we can denote the gradient of a function, $f(x, y)$, with $\nabla f(x, y)$.
Now how do we find the direction for the greatest rate of increase? We use partial derivatives. Here's why.
As we know, the partial derivative $\frac{df}{dx}$ calculates the change in output from moving a little bit in the $x$ direction, and the partial derivative $\frac{df}{dy}$ calculates the change in output from moving in the $y$ direction. Because with gradient ascent our goal is to make a nudge in $x, y$ that produces the greatest change in output, if $\frac{df}{dy} > \frac{df}{dx}$, we should make that move more in the $y$ direction than the $x$ direction, and vice versa. That is, we want to get the biggest bang for our buck.
Let's relate this again to the picture of our mountain climbers. Imagine the vertical edge on the left is our y-axis and the horizontal edge is on the bottom is our x-axis. For the climber in the yellow jacket, imagine his step size is three feet. A step straight along the y-axis will move him further upwards than a step along the x-axis. So in taking that step he should point his direction aligned with the y-axis than the x-axis. That will produce a bigger increase per step size.
In fact, the direction of greatest ascent for a function, $\nabla f(x, y)$, is the direction which is a proportion of $\frac{df}{dy}$ steps in the $y$ direction and $\frac{df}{dx}$ in the $x$ direction. So, for example, if $\frac{df}{dy}$ = 5 and $\frac{df}{dx}$ = 1, our next step she be five times more in the $y$ direction than the $x$ direction. And this seems to be the path, more or less that our climbers are taking - some combination of $x$ and $y$, but tilted more towards the $y$ direction.
Now that we have a better understanding of a gradient, let's apply our understanding to a multivariable function. Here is a plot of a function:
$$f(x,y) = 2x + 3y$$
Imagine being at the bottom left of the graph at the point $x = 1$, $y = 1$. What would be the direction of steepest ascent? It seems, just sizing it up visually, that we should move both in the positive $y$ direction and the positive $x$ direction. Looking more carefully, it seems we should move more in the $y$ direction than the $x$ direction. Let's see what our technique of taking the partial derivative indicates.
The gradient of the function $f(x,y)$, that is $\nabla f(x,y) = 2x + 3y$ is the following:
$\frac{df}{dx}(2x + 3y) = 2$ and $\frac{df}{dy}(2x + 3y) = 3$.
So what this tells us is to move in the direction of greatest ascent for the function $f(x,y) = 2x + 3y$, is to move up three and to the right two. So we would expect our path of greatest ascent to look like the following.
So this path maps up well to what we see visually. That is the idea behind gradient descent. The gradient is the partial derivative with respect to each type of variable of a multivariable function, in this case $x$ and $y$. And the import of the gradient is that it's direction is the direction of steepest ascent. The negative gradient, that is the negative of each of the partial derivatives, is the direction of steepest descent. So our direction of gradient descent for the graph above is $x = -2$, $y = -3$. And looking at the two graphs above, it seems that the steepest downward direction is just the opposite of the steepest upward direction. We get that by mathematically by simply taking the multiplying our partial derivatives by negative one.
### Summary
In this lesson, we saw how we can use gradient descent to find the direction of steepest descent. We saw that the direction of steepest descent is generally some combination of a change in our variables to produce the greatest negative rate of change.
We first how saw how to calculate the gradient ascent, or the gradient $\nabla$, by calculating the partial derivative of a function with respect to the variables of the function. So $\nabla f(x, y) = \frac{\delta f}{\delta y}, \frac{\delta f}{\delta x}$. This means that to take the path of greatest ascent, we should move $\frac{\delta f}{\delta y}$ divided by $\frac{\delta f}{\delta x}$. So for example, when $\frac{\delta f}{\delta y}f(x, y) = 3$ , and $\frac{\delta f}{\delta x}f(x, y) = 2$, we travelled in line with a slope of 3/2.
For gradient descent, that is to find the direction of greatest decrease, we simply reverse the direction of our partial derivatives and move in $- \frac{\delta f}{\delta y}, - \frac{\delta f}{\delta x}$.
You can’t perform that action at this time.
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Irrationality of Quadratic Sums Consider a sequence of unit fractions with integer denominators greater than 1. If each denominator is greater than or equal to the square of the preceding denominator, then we will say the sequence is "quadratic". For example, the sequence 1/3, 1/9, 1/81, 1/6561, 1/43046721, etc., is quadratic. Is the sum of and infinite quadratic sequence necessarily irrational? Obviously if we use the minimal denominators in each case, then as Matthew Hudelsen pointed out, we have 1/d, 1/d2, 1/d4, 1/d8, 1/d16, 1/d32, and so on, which in the base d has the representation This never repeats, so it is not rational. However, if we allow each denominator to exceed the square of the previous denominator, then the proof of irrationality becomes a little more difficult. Dean Hickerson emailed a nice proof of the general question. He notes that the sum of any infinite sequence of unit fractions whose denominators increase rapidly enough that each "tail sum" 1/dk + 1/dk+1 + 1/dk+2 + ... for sufficiently large k is strictly less than the corresponding geometric sum 1/dk + (1/dk)2 + (1/dk)3 + ... then the overall sum is irrational (and this condition is clearly met by my "quadratic" unit fraction sequences.) To prove this irrationality criterion, Hickerson observes that if the infinite sum 1/d1 + 1/d2 + ... is rational, then so are each of the "tails". Thus for every k we have coprime integers Nk and Dk representing the numerator and denominator of the kth tail sum Now suppose that, for all k greater than some constant k0, the kth tail sum is strictly less than the geometric sum of (1/dk)j for j = 1, 2, ... In other words, suppose This gives the inequality Furthermore, we have so it's clear that the fraction Nk+1/Dk+1 can be written with the integer numerator Nk dk – Dk, which implies that the reduced numerator Nk+1 (after removing any possible common factors with the denominator Dk dk) is less than or equal to this integer. Combining this with our previous (strict) inequality on Nk gives which implies that the numerators of successive "tails" (for all k greater than k0) constitute a strictly decreasing infinite sequence of positive integers beginning with the (presumed) finite integer Nk0, which is impossible. Incidentally, this question relates to the use of the greedy algorithm to expand a given fraction into a sum of unit fractions. The greediness implies that the sum of all the terms beyond 1/dk must be less than (or equal to) Therefore, dk+1 must be greater than or equal to (dk – 1)dk. Notice that this doesn't quite satisfy the condition of my original question, because we can't say that dk+1 is greater than or equal to dk2, nor does it satisfy the condition of Hickerson's irrationality criterion, so we can't use this criterion to rule out the possibility of an infinite greedy expansion equaling a rational number. Of course, in this case we can proceed much more simply, noting that the greedy algorithm yields a sequence of remainders with strictly decreasing numerators (since at each stage we subtract 1/d from the current remainder N/D to give the new remainder (Nd – D)/Dd where d is the smallest integer such that Nd – D is positive, and therefore the numerator of the new remainder is essentially d modulo N.) From this we might be tempted to conclude that no infinite sequence of unit fractions with denominators such that dk+1 is equal to or greater than dk2 – dk can have a rational sum, because such a sequence would contradict the fact that the greedy algorithm terminates after a finite number of steps when applied to any rational number. However, although the condition that dk+1 exceeds dk2 – dk is necessary for greediness, it is not quite sufficient, because we have the infinite sequences of unit fractions with denominators given by Sylvester's recurrence dk+1 = dk2 – dk + 1, which have the rational sum 1/(d1 – 1). These sequences are, in a sense, the boundary between greediness and non-greediness, i.e., each denominator is exactly 1 greater than is absolutely necessary for greediness. I wonder if the condition that dk+1 exceeds dk2 – dk + 1 is sufficient to ensure greediness. It's also interesting to consider what happens if we restrict all the denominators in our greedy expansion to be odd. In this case the sum of all the terms following 1/dk must be no greater than Therefore, in the case of odd greedy expansions we only require that dk+1 be greater than or equal to (1/2)dk2 – dk. According to Richard Guy's "Unsolved Problems In Number Theory", it is not known whether the "odd greedy" unit fraction expansion of a rational number is necessarily finite, so we can only say that if an infinite odd greedy expansion of a rational number exists, there must be infinitely many values of k such that dk+1 is less than dk2 and but greater than or equal to (1/2)dk2 – dk. The question of odd greedy unit fraction expansions is considered in more detail in the article “The Greedy Algorithm for Unit Fractions”. In general, it would be interesting to know the "minimal" polynomial f() such that the requirement for dk+1 to be at least f(dk) is sufficient to ensure irrationality. It seems as if it may be the Sylvester sequence plus 1. Return to MathPages Main Menu
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# What Is 17/37 as a Decimal + Solution With Free Steps
The fraction 17/37 as a decimal is equal to 0.459.
The division of two numbers is usually shown as p $\boldsymbol\div$ q, where p is the dividend and q is the divisor. This is mathematically equivalent to the numeral p/q, called a fraction. In fractions, though, the dividend is called the numerator and the divisor is called the denominator.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 17/37.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 17
Divisor = 37
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 17 $\div$ 37
This is when we go through the Long Division solution to our problem.
Figure 1
## 17/37 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 17 and 37, we can see how 17 is Smaller than 37, and to solve this division, we require that 17 be Bigger than 37.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 17, which after getting multiplied by 10 becomes 170.
We take this 170 and divide it by 37; this can be done as follows:
 170 $\div$ 37 $\approx$ 4
Where:
37 x 4 = 148
This will lead to the generation of a Remainder equal to 170 – 148 = 22. Now this means we have to repeat the process by Converting the 22 into 220 and solving for that:
220 $\div$ 37 $\approx$ 5Â
Where:
37 x 5 = 185
This, therefore, produces another Remainder which is equal to 220 – 185 = 35. Now we must solve this problem to Third Decimal Place for accuracy, so we repeat the process with dividend 350.
350 $\div$ 37 $\approx$ 9Â
Where:
37 x 9 = 333
Finally, we have a Quotient generated after combining the three pieces of it as 0.459, with a Remainder equal to 17.
Images/mathematical drawings are created with GeoGebra.
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Types of graphs
There are many types of graphs. In this lesson, you will learn about the most common types of graphs used in mathematics such as the ones that are listed below:
• Pictograph
• Bar graph
• Double bar graph
• Histogram
• Line graph
• Circle graph
• Scatter plot
What is a pictograph?
pictograph is a graph that uses pictures or symbols to display information. The pictures in a pictograph usually represent more than one item. The following is a pictograph:
What is a bar graph?
A bar graph can be used to compare data or values by using rectangular bars. A bar graph can either be vertical or horizontal. The bar graph below is a horizontal bar graph. In a horizontal bar graph, the horizontal axis shows the categories while the vertical axis shows the value of each category.
What is a double bar graph?
A double bar graph can help us compare more than one type of information. In the bar graph above, we are comparing just one type of information. It is the number of days of snow in a specific month. In the double bar graph below, we are comparing two types of information. We are comparing the score for a specific student with preparation and the score without preparation. Therefore, we need a double bar graph.
What is a histogram?
A histogram is a special type of bar graph that shows the frequency of numerical data instead of categorical data. And this is the main difference between a histogram and a bar graph. In the bar graph above, we are comparing categorical variables. For example, we are comparing the month of December with the month of January. However, in the histogram below, we are comparing numerical data. The graph shows that 20 families own 2 smartphones. However, only 5 families own 1 phone.
What is a line graph?
A line graph can be used to show the change in a set of data over a period of time. A line graph can help you better to look for trends and make predictions. In the line graph below, we see that the scores fluctuated greatly, but have increased as time goes by.
What is a circle graph?
A circle graph is used to show data as percents or frations. The total is equal to 100% or 1.
What is a scatter plot?
A scatter plot can be used to relate two groups of data. You plot the two groups of data as ordered pairs. Scatter plots can show if there is a correlation or relationship between the two groups of data. For example, for the ordered pair (4, 10), 4 could represent the number of hours you spend each day watching TV and 10 could represent in thousands your yearly income. There may or may not be a correlation between these two groups of data.
Importance of the different types of graphs
• Learning to read and create graphs are very useful skills in mathematics.
• Graphs help us to visualize data and the things that we see on the graph may help us to interpret and analyze the data.
• The end result is usually a better perspective that may not have been possible just by looking at a bunch of numbers.
100 Tough Algebra Word Problems.
If you can solve these problems with no help, you must be a genius!
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# Trigonometry in CSS and JavaScript: Introduction to Trigonometry
In this series of articles we’ll get an overview of trigonometry, understand how it can be useful, and delve into some creative applications in CSS and JavaScript.
Understanding trigonometry can give us super powers when it comes to creative coding. But to the uninitiated, it can seem a little intimidating. In this 3-part series of articles we’ll get an overview of trigonometry, understand how it can be useful, and delve into some creative applications in CSS and JavaScript.
2. Getting Creative with Trigonometric Functions
3. Beyond Triangles
## Trigonometry basics
If, like me, you’ve rarely used trigonometry outside of the classroom, let’s take a trip back to school and get ourselves reacquainted.
Trigonometric functions allow us to calculate unknown values of a right-angled triangle from known parameters. Imagine you’re standing on the ground, looking up at a tall tree. It would be very difficult to measure the height of the tree from the ground. But if we know the angle at which we look up at the top of the tree, and we know the distance from ourselves to the tree, we can infer the height of the tree itself.
If we imagine this scene as a triangle, the known length (from us to the tree) is known as the adjacent side, the tree is the opposite side (it’s opposite the angle), and the longest side – from us to the top of the tree – is called the hypotenuse.
### Sine, Cosine and Tangent
There are three main functions to remember in trigonometry: Sine, Cosine and Tangent (abbreviated to sin, cos and tan). They are expressed as the following formulae:
``````sin(angle) = opposite / hypotenuse
The angle is usually written as the Greek theta (θ) symbol.
We can use these equations to calculate the unknown values of our triangle from the known ones. To measure the height of the tree in the example, we know the angle (θ) and the adjacent side.
To calculate the opposite side we would need the tangent function. We would need to switch around the formula:
``opposite = tan(angle) * adjacent``
How do we get tan(θ)? We could use a scientific calculator (type tan and then the angle), or we could use code! Sass and JavaScript both include trigonometric functions, and we’ll look at some ways to use these in this article and the following ones.
## Sass functions
If we’re working with predetermined values, we could use the trigonometric functions built into Sass (the CSS preprocessor).
To include the Math module we need the following line in our Sass file:
``@use "sass:math";``
We can use variables to calculate the opposite side from the angle and adjacent side values.
``````\$angle: 45deg;
The tan function in Sass can use radians or degrees — if using degrees, the units must be specified. Without units, radians will be used by default (more on these later).
In the following demo we’re using these in the `clip-path` property to determine the coordinates of the polygon points, similar to calculating the height of a tree.
See the Pen Using Sass trigonometry for clip-path values by Michelle Barker (@michellebarker) on CodePen.
We need to subtract the `\$opposite` variable from the height of the element in order to get the y coordinate — as clip-path coordinates are plotted along the y axis increasing from top to bottom.
``````.element {
}``````
## Clipping an equilateral triangle
A right-angled triangle is the simplest use of trigonometry. But we can work out the coordinates of more complex shapes by splitting them up into right-angled triangles.
An equilateral triangle is a triangle with three sides of the same length. Perhaps you remember from school that the angles in a triangle add up to 180º? That means each angle in an equilateral triangle is 60º.
If we draw a line down the middle of an equilateral triangle, we split it into (you guessed it) two right-angled triangles. So, for a triangle with sides of a given length, we know the angle (60º), the length of the hypotenuse, and the length of the adjacent side (half the length of the hypotenuse).
What we don’t know is the height of the triangle — once again, the opposite side of the right-angled triangle. To plot the clip-path coordinates, this is what we need to work out. This time, as we know the angle and the length of the hypotenuse, we can use the sine function:
``````\$hypotenuse: 60%; // side length
\$angle: 60deg;
\$opposite: math.sin(\$angle) * \$hypotenuse;``````
(It would also be possible for us to use the tangent function instead, as we know that the length of the adjacent side is half of the hypotenuse.) Then we can use those values for our clip-path polygon points:
``````.element {
clip-path: polygon(
0 \$opposite,
(\$hypotenuse / 2) 0,
\$hypotenuse \$opposite
);
}``````
See the Pen Clip-path simple equilateral triangles with Sass by Michelle Barker (@michellebarker) on CodePen.
As you can see in the demo, the element is clipped from the top left corner. This might not be completely satisfactory: it’s more likely we’d want to clip from the center, especially if we’re clipping an image. We can adjust our clip-path coordinates accordingly. To make this more readable, we can assign some additional variables for the adjacent side length (half the hypotenuse), and the start and end position of the triangle:
``````\$hypotenuse: 60%; //side length
\$angle: 60deg;
\$opposite: math.sin(\$angle) * \$hypotenuse;
\$startPosY: (50% - \$opposite / 2);
\$endPosY: (50% + \$opposite / 2);
.element {
clip-path: polygon(
\$startPosX \$endPosY,
50% \$startPosY,
\$endPosX \$endPosY
);
}``````
### Creating a mixin for reuse
This is quite a bit of complex code to write for a single triangle. Let’s create a Sass mixin, allowing us to clip a triangle of any size on any element we like. As `clip-path` still needs a prefix in some browsers, our mixin covers that too:
``````@mixin triangle(\$sideLength) {
\$hypotenuse: \$sideLength;
\$angle: 60deg;
\$opposite: math.sin(\$angle) * \$hypotenuse;
\$startPosY: (50% - \$opposite / 2);
\$endPosY: (50% + \$opposite / 2);
\$clip: polygon(
\$startPosX \$endPosY,
50% \$startPosY,
\$endPosX \$endPosY
);
-webkit-clip-path: \$clip;
clip-path: \$clip;
}``````
To clip a centred equilateral triangle from any element, we can simply include the mixin, passing in the length of the triangle’s sides:
``````.triangle {
@include triangle(60%);
}``````
See the Pen Clip-path equilateral triangles with Sass trigonometric functions by Michelle Barker (@michellebarker) on CodePen.
### Limitations of Sass functions
Our use of Sass functions has some limitations:
1. It assumes the `\$sideLength` variable is known at compile time, and doesn’t allow for dynamic values.
2. Sass doesn’t handle mixing units all that well for our needs. In the last demo, if you switch out the percentage-based side length to a fixed length (such as rems or pixels), the code breaks.
The latter is because our calculations for the `\$startPos` and `\$endPos` variables (to position the clip-path centrally) depend on subtracting the side length from a percentage. Unlike in regular CSS (using calc()), Sass doesn’t allow for that. In the final demo, I’ve adjusted the mixin so that it works for any valid length unit, by passing in the size of the clipped element as a parameter. We’d just need to ensure that the values for the two parameters passed in have identical units.
See the Pen Clip-path equilateral triangles with Sass trigonometric functions by Michelle Barker (@michellebarker) on CodePen.
## CSS trigonometric functions
CSS has a proposal for trigonometric functions as part of the CSS Values and Units Module Level 4 (currently in working draft). These could be extremely useful, especially when used alongside custom properties. Here’s how we could rewrite our CSS to use native CSS trigonometric functions. Changing the size of the clip path is as simple as updating a single custom property:
``````.triangle {
--hypotenuse: 8rem;
--opposite: calc(sin(60deg) * var(--hypotenuse));
--startPosX: calc(var(--size) / 2 - var(--adjacent));
--startPosY: calc(var(--size) / 2 - var(--opposite) / 2);
--endPosX: calc(var(--size) / 2 + var(--adjacent));
--endPosY: calc(var(--size) / 2 + var(--opposite) / 2);
--clip: polygon(
var(--startPosX) var(--endPosX),
50% var(--startPosY),
var(--endPosX) var(--endPosY)
);
-webkit-clip-path: var(--clip);
clip-path: var(--clip);
}
.triangle:nth-child(2) {
--hypotenuse: 3rem;
}
.triangle:nth-child(2) {
--hypotenuse: 50%;
}``````
### Dynamic variables
Custom properties can be dynamic too. We can change them with JS and the values dependant on them will be automatically recalculated.
``triangle.style.setProperty('--hypotenuse', '5rem')``
CSS trigonometric functions have a lot of potential when they finally land, but sadly they’re not yet supported in any browsers. To use trigonometry with dynamic variables right now, we need JavaScript.
We’ll take a look at some of the possibilities in the next article.
### Michelle Barker
Michelle is a Senior Front End Developer at Ada Mode, where she builds web apps and data visualisations for the renewable energy industry. She is the author of front-end blog CSS { In Real Life }, and has written articles for CSS Tricks, Smashing Magazine, and Web Designer Magazine, to name a few. She enjoys experimenting with new CSS features and helping others learn about them.
### Stay in the loop: Get your dose of frontend twice a week
👾 Hey! Looking for the latest in frontend? Twice a week, we'll deliver the freshest frontend news, website inspo, cool code demos, videos and UI animations right to your inbox.
Zero fluff, all quality, to make your Mondays and Thursdays more creative!
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# The probability that a speaks truth is 3/5 and that of b speaking truth is 4/7. what is the probability that they agree in stating the same fact?
the probability that a speaks truth is 3/5 and that of b speaking truth is 4/7. what is the probability that they agree in stating the same fact?
To find the probability that both A and B agree in stating the same fact, we can use the formula for the probability of the intersection of two independent events.
Let’s represent the event that A speaks truth as “A” and the event that B speaks truth as “B”. We want to find the probability that A and B both speak the same fact, which can be represented as P(A ∩ B).
If A and B are independent events, then the probability of the intersection P(A ∩ B) is equal to the product of their individual probabilities: P(A) * P(B).
Given that P(A) = 3/5 and P(B) = 4/7, we can calculate:
P(A ∩ B) = P(A) * P(B) = (3/5) * (4/7) = 12/35.
Therefore, the probability that A and B agree in stating the same fact is 12/35.
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+0
# math help
0
52
3
+70
Fill in the blanks with integers, and select the correct operator ( + or -), to give an equation whose graph is the line that passes through the point (-7, 2) and is parallel to the graph of 2x+3y = -5.
__x ? __y = __
Jun 30, 2023
#1
0
The slope of the line 2x+3y = -5 is 2/3. So, the equation of our parallel line will be of the form y = (2/3)x + b, where b is the y-intercept.
We know that the line passes through the point (-7, 2), so we can plug these values into the equation to solve for b. This gives us:
2 = (2/3)(-7) + b 2 = -14/3 + b 2 + 14/3 = b = 10/3
Therefore, the equation of the parallel line is y = (2/3)x + (10/3). Filling in the blanks, we get:
2x ? + y = 10
The answer is 2x + y = 10.
Jun 30, 2023
#3
0
Thanks, that's right!
Guest Jul 6, 2023
#2
0
The slope of the line 2x+3y = -5 is -2/3. Since the parallel line has the same slope, the equation of the parallel line will be of the form 2x - 3y = b.
We know that the parallel line passes through the point (-7, 2), so we can plug these values into the equation to solve for b. This gives us:
2 * -7 - 3 * 2 = b -14 - 6 = b -20 = b
Therefore, the equation of the parallel line is 2x - 3y = -20.
The operator that should be used is -.
Jun 30, 2023
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• Jun 1st 2009, 05:23 AM
Aaron01424
Hello, trying to factorise:
x^2-5x-6
So far i've got:
AC = -6, so x^2+x-6x-6 = x(x+1)-6(x+1)
Now getting to that point all makes sense to me, however in my textbook it says 'x+1 is a factor of both terms, so take that outside the bracket', leaving the answer as (x+1)(x-6). I don't understand how they got from x(x+1)-6(x+1) to (x+1)(x-6). Any help would be appreciated, thanks.
• Jun 1st 2009, 05:39 AM
mr fantastic
Quote:
Originally Posted by Aaron01424
Hello, trying to factorise:
x^2-5x-6
So far i've got:
AC = -6, so x^2+x-6x-6 = x(x+1)-6(x+1)
Now getting to that point all makes sense to me, however in my textbook it says 'x+1 is a factor of both terms, so take that outside the bracket', leaving the answer as (x+1)(x-6). I don't understand how they got from x(x+1)-6(x+1) to (x+1)(x-6). Any help would be appreciated, thanks.
Can you factorise xA - 6A ?
• Jun 1st 2009, 06:14 AM
Showcase_22
Since both terms are multiplied by (x+1), you can just take x+1 outside the brackets to give $x^2-5x-6=(x-6)(x+1)$.
• Jun 1st 2009, 06:39 AM
Aaron01424
Quote:
Originally Posted by mr fantastic
Can you factorise xA - 6A ?
Yeah, A(x-6).
Quote:
Originally Posted by Showcase_22
Since both terms are multiplied by (x+1), you can just take x+1 outside the brackets to give $x^2-5x-6=(x-6)(x+1)$.
Yeah, that's what it says in the book. Maybe i'm overcomplicating it, but I don't understand what you do when you take it out of the brackets.
• Jun 1st 2009, 06:52 AM
mr fantastic
Quote:
Originally Posted by Aaron01424
Yeah, A(x-6).
Yeah, that's what it says in the book. Maybe i'm overcomplicating it, but I don't understand what you do when you take it out of the brackets.
OK. Now replace A with (x + 1).
• Jun 1st 2009, 09:13 AM
Aaron01424
Aha. I see, great. Cheers.
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# Acceleration, distance, time. help.
• Eagleized
In summary, the first car will travel a distance of 558.46 ft before catching up to the second car which is traveling at a speed of 132 f/s.
Eagleized
## Homework Statement
vehicle 1 is traveling at a constant speed of 90mph (132 f/s) 5 seconds before vehicle 2 begins to accelerate from 0-90 mph taking 17.09 seconds to do so. how far a distance will vehicle 2 travel while accelerating to 90mph. and how long will it take for vehicle 2 to catch up to vehicle 1 while accelerating at approximately 5.2 m/s/s or 15.6 f/s
## Homework Equations
The relationship between the 2 vehicles is dependant.
what is the total distance traveled by vehicle 2 before it catches up to vehicle 1?
## The Attempt at a Solution
ok, it's too hard to explain and argue at the same time, so i'll leave it at this:
find a function of position for car #1 and car #2, call them c_1(t) and c_2(t), set them equal and solve for t. what does this mean exactly?
This should help you a bit, at constant acceleration, call it a, the position of car 2 at time t is at^2/2
the first car, meanwhile is moving at a constant velocity v, so its position at time t is vt
does car 2 start from rest?
so solve vt=at^2/2 -> at^2-vt=0 -> use quadratic formula
this will give you the time t, when car 2 catches up to car 1, ?? to answer part 2 of the Q
For your first question you could use the equation Vf^2 = Vi^2 + 2(a)(d) where Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and d is the displacement.
For number 2, since the first car is traveling at constant speed we can label the distance it travels as d = V(t + 5) while we may use the equation df = (a)(t^2)/2 + (Vi)t + di. Equate the two and solve for the time.
did you get a distance of 4044.8 ft for the first part of the question?
or 558.46 ft?
Last edited:
anyone get an answer for this?
## 1. What is acceleration and how is it calculated?
Acceleration is the rate of change of velocity over time. It is calculated by dividing the change in velocity by the change in time. The formula for acceleration is a = (vf - vi)/t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.
## 2. How does distance affect acceleration?
The distance an object travels does not directly affect its acceleration. However, if the object travels a longer distance in the same amount of time, its average velocity will be higher, resulting in a higher acceleration.
## 3. Is acceleration always constant?
No, acceleration is not always constant. It can be either constant or changing. An object with a constant acceleration will have a steady change in velocity, while an object with a changing acceleration will have a varying change in velocity over time.
## 4. How is acceleration related to force?
Acceleration is directly proportional to force. This means that the greater the force applied to an object, the greater its acceleration will be. This relationship is described by Newton's second law of motion, which states that F = ma, where F is force, m is mass, and a is acceleration.
## 5. What is the relationship between acceleration, distance, and time?
The relationship between acceleration, distance, and time can be described by the formula d = vit + 1/2at^2, where d is distance, vi is initial velocity, t is time, and a is acceleration. This equation shows that the distance an object travels is dependent on its initial velocity, the time it takes to travel, and its acceleration.
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# How much force is needed to break off the stick
Let's consider the following figure
The grey box contains a blue stick which is fixed. The blue stick has a length of $$a+b+c$$ and two diameters $$f,h$$. The diameter $$h$$ describes the part $$b$$ of the stick. The stick is fixed in the plane but the plane is not connected to the grey box. A force $$F$$ is pushing against the withe plane like in the picture. How much force is needed to break off the stick in part $$b$$?
• How much effort have you applied to try to obtain a proposed solution? Commented Feb 26, 2023 at 20:03
• I don't have an idea how I could solve this because I had never a mechanical problem with a notch. What I also can say is that I see two different ways how this plane could move: One way would be striaght downward if the force is close to the notch or the plane is rotated if the force comes from the outer part of the plane. Commented Feb 26, 2023 at 20:26
• I just have some knowledge about bending sticks and not about stuff like in my picture. Commented Feb 26, 2023 at 20:43
• Apply your knowledge about bending sticks to try to solve the problem. We should like to see how far that takes you. Commented Feb 26, 2023 at 20:57
• Does the white plane slide against the grey plane or does it tilt ie pivot at the lower left corner? Commented Feb 26, 2023 at 21:18
## 1 Answer
We assume the distance from F to the hinge to be
$$X_F=a+b+c+d/2$$ We calculate the equivalent I of the cantilever beam, with the parallel axis. When it bends it will rotate about a point at the lower corner of the gray support, call it point A. Let's annotate the thickness of the bar, B.
$$I_{Beam} =I_{stick}+ A_{stick}*Y^2_{stick}$$ $$I_{stick}= bh^3/12$$ $$I_{Beam}=bh^3/12+bh(e+f/2)^2$$
we assume the stick will break at yield stress and ignore 2nd hardening, or if we have it we plug it.
$$\sigma_y=\frac{MC}{I_{Beam}}=\frac{(F*x)(e+f/2)}{bh^3/12+bh(e+f/2)^2}$$ $$F*X=\frac{\sigma y*(bh^3/12+bh(e+f/2)^2)}{e+f/2}$$
$$F=\frac{\sigma y*(bh^3/12+bh(e+f/2)^2)}{(e*f/2)*(a+b+c+d/2)}$$
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# Vedic Mathematics/Techniques
Vedic Mathematics Previous: Sutras Techniques Next: Addition And Subtraction
## Introduction
Whether you believe the stories surrounding the source of the Vedic Mathematics techniques or not, what is important are the techniques themselves. By mastering these techniques you will not only radically improve your numeracy and arithmetic ability, you will also begin to understand that mathematics is a fluid and fascinating subject in which there are usually many different ways to solve any particular problem. You will see that many of the tedious algorithms that you were taught in school, (e.g. long division and multiplication), are but one way to solve a problem. While these 'school' techniques are often general purpose, (e.g. long multiplication allows you to multiply any pair of numbers), in many cases they are very inefficient; their very generality means they have to cover all possibilities and so can't take advantage of the specifics of any particular problem. If instead you use a technique optimised for the particular problem you are working on, you can take advantage of properties that aren't always present in the general case and so solve the problem with a lot less work
It should be noted that, although many of the following techniques may be contained in the 1965 book Vedic Mathematics by Sri Bharati Krsna Tirthaji, they are not unique to that book. Many of the techniques are also part of other arithmetic systems, (e.g. the Trachtenberg system), and most are common knowledge among those who enjoy the challenge of mental arithmetic. In fact the term 'Vedic Mathematics' is now sometimes used to encompass the general idea of solving arithmetic problems with a wide range of different techniques, each optimised for particular circumstances, these techniques not being limited to those in Sri Bharati Krsna Tirthaji's book.
So, do you want to know how to multiply 89 and 97 in your head in seconds? The solution is given below:-
1. First solve 9×7, ie, 63.(Carry over 6){Let 3 here be z}
2. Do cross multiplication & add the carried over no. done in previous step, ie solve (8×7)+(9×9)+6), ie, 143.(Carry over 14){Let 3 here be y}
3. Then solve (8×9)+14, ie, 86.{Let 86 be x}
4. Then join x, y & z to form xyz, ie, here xyz will be 8633.
Further, there is another technique to solve this problem.
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# math
posted by .
i am a 3 digit number divisible by 3.my tens digit is 3 times as great as my hundreds digit and the sum of my digits is 15. if you reverse my digits i am divisible by 6 and 3...... what number am i?
• math -
We know that the number is divisible by three if the sum of its digits is 15!
It's got 3 digits. When reversed, it is divisible by 6, which means that the reversed number is divisible by 2. This translates to
"the hundred's digit is even".
From
"my tens digit is 3 times as great as my hundreds digit"
we deduce that the number must be one of the three possibilities:
A. 13X
B. 26X
C. 39X
Where X can be found from the fact that the digits add up to 15.
Both A and C will not work because the hundreds digit is not even.
So can you find the number now?
## Respond to this Question
First Name School Subject Your Answer
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1. ## functions
am having some problems with these equations
given tha f and g are functions where the target set and source sets are the set of real numbers and
f(x) = 10-3x
g(x) = 2
1+3
calculate
i) f(2)
ii)g(-3)
iii) fog(2)
write down an expression giving
iv) g o f (x)
v)f-1(x)
again thanks for any help
2. Originally Posted by red55
am having some problems with these equations
given tha f and g are functions where the target set and source sets are the set of real numbers and
f(x) = 10-3x
g(x) = 2/1+3
Sure you've typed g correctly?
3. Originally Posted by red55
am having some problems with these equations
given tha f and g are functions where the target set and source sets are the set of real numbers and
f(x) = 10-3x
g(x) = 2
1+3
calculate
i) f(2)
ii)g(-3)
iii) fog(2)
write down an expression giving
iv) g o f (x)
v)f-1(x)
again thanks for any help
i) f(2) where 2 is your x value... this is the function f(x) evaluated at 2 so you plug in 2 where ever you have an x in your function
f(2) = 10-3(2) = 4
ii) Unforutnately your g function is not reasonable, I think there is an error...
but g(-3) would be done the same way, substitute x with -3 because you are evaluating the function g(x) at -3
iii) the composite is not as scary as it seems...(well depending on your functions anyway...)
f(g(x)) means that the functions f(x) is evaluated at g(x) so substitute x in f(x) with g(x)...
you will have
f(g(x)) = 10 - 3(g(x))
iv) g compositite f is the same as iii), but instead you use f(x) as your x-value and plug into all x's in g(x)
v) f-1 is f inverse.
f(x) = 10 - 3x Solve for x
f(x)= 10 - 3x ---> move the 3x to the left side of the = sign and the f(x) to the right of the = sign
10 - f(x) = 3x ----> get x by itself by dividing both sides by 3
(10 - f(x))/3 = x -----> no that you have this just swap x with f(x)
(10 - x)/3 = f(x) ------> this is your inverse function
I hope that helps
4. Originally Posted by red55
am having some problems with these equations
given tha f and g are functions where the target set and source sets are the set of real numbers and
f(x) = 10-3x
g(x) = 2
1+3
calculate
i) f(2)
ii)g(-3)
iii) fog(2)
write down an expression giving
iv) g o f (x)
v)f-1(x)
again thanks for any help
Hello,
I assume that the function g reads:
$\displaystyle g(x) = \frac{2}{x+3}$
If so you'll get at:
ii) $\displaystyle g(-3) \notin \mathbb{R}$ because you have to divide by zero.
iii) $\displaystyle f \circ g(2)=10 - 3 \cdot \left( \frac{2}{2+3} \right) = 10 -\frac{6}{5} = \frac{44}{5}$
iv) $\displaystyle g \circ f(x) = g(f(x))=\frac{2}{10 - 3x +3} = \frac{2}{13-3x}$
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# General Form of the Equation of a Circle – Definition, Formula, Examples | How to find the General Form Equation of a Circle?
Know the definition of a circle, the general form of the equation of a circle. Get the various terms involved in the general and standard form of a circle, formulae, and definition, etc. Refer to solved examples of a circle, standard equation of a circle. For your reference, we have included the solved examples on how to find the general form of an equation of the circle, conversion from standard form to general form and vice versa, etc.
Also, Read: Circumference and Area of Circle
## Circle Definition
The circle is defined as the locus of a point that moves in a plane such that its distance from a fixed point in that plane is always constant. The center of the circle is the fixed point. The set of points in the plane at a fixed distance is called the radius of the circle.
### General Form of the Equation of a Circle
To find the general form of the equation of a circle, we use the below-given graph. Each circle form has its own advantages. Here, we can take an example of a standard form which is great for determining the radius and center just with a glance at the above equation. The general form of a circle is good at substituting ordered pairs and testing them. We use both of these forms. So this gives us an idea that we should interchange between these forms. Firstly, we will transform the standard form to the general form.
General form of equation is (x-h)2Â + (y-k)2Â = r2
where r is defined as the radius of the circle
h, k is defined as the center coordinates
#### Standard Form to General Form
Here, we will take an example that gives us an idea to transform an equation from a Standard form to a general form
Eg: Transform (x – 3)2 + (y + 5)2 = 64 to general form.
(x – 3)2 + (y + 5)2 = 64
Now, all the binomial should be multiplied and rearranged till we get the general form.
(x – 3) (x – 3) + (y + 5) (y + 5) = 64
(x2 – 3x – 3x + 9) + (y2 + 5y + 5y + 25) = 64
x2 – 6x + 9 + y2 + 10y + 25 = 64
x2 + y2 – 6x + 10y + 9 + 25 – 64 = 0
(x2) + (y2) – 6(x) + 10(y) – 30 = 0
x2+y2–6x+10y–30 = 0
This is the general form of the equation as transformed from Standard from.
#### General to Standard Form
To transform an equation to standard form from a general form, we must first complete the equation balanced and complete the square. Here, completing the square implies creating Perfect Square Trinomials(PST’s).
To give you an idea about Perfect Square Trinomials, here are some examples
Example 1:
x2 + 2x + 1
When we factor PSTs, we get two identical binomial factors.
x2 + 2x + 1 = (x + 1)(x + 1) = (x + 1)2
Example 2:
x2 – 4x + 4
When we factor PSTs, we get two identical binomial factors.
x2 – 4x + 4= (x – 2)(x – 2) = (x – 2)2
We can observe that the sign for the middle term can either be positive or negative.
We have a relationship between the last term and the coefficient of the middle term
(b/2)2
Now, we see a few examples of circle equation that include the transformation of the equation from a standard form to the general form
### General Form of the Equation of a Circle Examples
Problem 1:
The circle equation is: x2 + y2 – 8x + 4y + 11 = 0. Find the centre and radius?
Solution:
To find the centre and radius of the circle, we first need to transform the equation from general form to standard form
x2 + y2 – 8x + 4y + 11 = 0
x2 – 8x + y2 + 4y + 11 = 0
(x2 – 8x + ) + (y2 + 4y + ) = -11
We are leaving the spaces empty for PST’s.
We must complete the square of the PST’ds by adding appropriate values
To maintain balance on the above equation, we must add same values on the right side which we add on the left side of the equation to keep the equation equal on both the sides
(x2 – 8x + 16) + (y2 + 4y + 4) = -11 + 16 + 4
(x – 4)2 + (y + 2)2 = 9
By comparing the above equation with the standard form of the circle, we observe that
Centre =(4,-2)
Radius = 3
Problem 2:
Find the standard form of the equation of a circle of radius 4 whose centre is (-3,2). Convert the equation into general form
Solution:
As given in the question,
radius = 4
h = -3
k = 2
General form of equation is (x-h)2Â + (y-k)2Â = r2
(x-(-3))2 + (y-2)2 = 42
(x+3)2 + (y-2)2 = 16
x2Â + 6x + 9Â + y2 -4y + 4 = 16
x2Â + y2Â + 6x – 4y – 3 = 0
Therefore, the general solution is x2Â + y2Â + 6x – 4y – 3 = 0
Problem 3:
Write the equation in the general form given the radius and centre
r = 3, centre = (1,2)
Solution:
As given in the question,
r = 3
h = 1
k = 2
General form of equation is (x-h)2Â + (y-k)2Â = r2
(x-1)2 + (y-2)2 = 32
x2 – 2x + 1Â + y2 -4y + 4 = 9
x2 + y2 – 2x – 4y – 4 = 0
Therefore, the general solution is x2 + y2 – 2x – 4y – 4 = 0
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# finding median from histogram worksheet
By • 一月 17th, 2021
Displaying top 8 worksheets found for - Median Of A Histogram. Find the total number of items represented by the histogram 2. The pdf exercises are curated for students of grade 3 through grade 8. The Median is the value of the middle in your list. The total area of this histogram is $10 \times 25 + 12 \times 25 + 20 \times 25 + 8 \times 25 + 5 \times 25 = 55 \times 25 = 1375$. The median is the middle item or the average of the two middle items. Positive skewed histograms. The median is also the number that is halfway into the set. Let’s say, however, that you also want to publish the information on these weights for an audience that uses the imperial system rather than the metric system of measurements. Here, 1977 is used as the “base” year which is equal to 100. Central Tendencay Worksheet or Quiz Includes worksheets with and without number sets. {102, 109, 207, 357, 360, 403, 471, 483, 670, 729, 842, 843, 920, 941} Now, calculate the median M by finding the mean of 471 and 483. The median is the n/2 th value. The median is the middle value; uniformly spread data will provide that the area of the histogram on each side of the median will be equal. What is a Histogram? What does that mean 43 is the median of the frequencies, but it's not the median of the values. Starting with , add the frequencies in the table starting with the first row until you reach . Histogram Worksheet. Finding The Median Using Histograms - Displaying top 8 worksheets found for this concept.. 1. Batting average shows the percent (written as a decimal) of the time a certain player gets a hit. The median is the midpoint of the value, which means that at the median there are exactly half the data below and above that point. Again, the definition of the median for a continuous distribution is the value such that the cumulative probability is 1/2; we just multiplied N by that. Find the bin(s) containing the middle item(s). Determine the number of the middle item. In order to plot a cumulative frequency graph, we have to plot cumulative frequency against the upper-class-boundary of each class. We use linear interpolation to find it. If you know how many numbers there are in a set, which is the middle number? Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 1 Center and Spread of Data Name_____ Date_____ Period____ ... Find the mode, median, mean, lower quartile, upper quartile, interquartile range, and population standard deviation for each data set. To find the median, the data should first be arranged in order from least to greatest. Drawing frequency polygon - with and without a histogram; Finding mean, median and mode of raw data . ; To estimate the Mean use the midpoints of the class intervals: . Now I want to see what happens when I add male heights into the histogram: To remember the definition of a median, just think of the median of a road, which is the middlemost part of the road. For each histogram bar, we start by multiplying the central x-value to the corresponding bar height. When the shape of the distribution is symmetric and unimodal, the mean, median, and mode are equal. using the formula for median we have, Median or where, (lower class boundary of the median class), (total frequency), ( less than type cumulative frequency corresponding to ), • To find the mean, add up all the numbers and divide by the number of numbers. A positive skewed histogram suggests the mean is greater than the median. The median class interval is the corresponding class where the median value falls. Please help. You can get both the mean and the median from the histogram. The histogram above shows a distribution of heights for a sample of college females. Given that the median value is 46, determine the missing frequencies using the median formula. This means you will have to convert your units into pounds from kilograms, multiplying each observation value in your data by 2.2 to get an approximate weight. Study each of these problems carefully; you will see similar problems on the lesson knowledge check. the median class is the class for which upper class boundary is . But in the histogram the hint is confusing me. finding median for ungrouped data Median is the value which occupies the middle position when all the observations are arranged in an ascending or descending order. the class containing the median value. You will need paper and a … For grouped data, we cannot find the exact Mean, Median and Mode, we can only give estimates. The mean, median and mode are three different ways of describing the average. Concept wise Collecting Data Raw Data Ungrouped Data Grouped Data Two sets of numbers that you work out for mean, median, mode, range, Q1, Q3, IQR, histogram… Median Worksheet 1. The median … • To find the median, place all the numbers in order and select the middle number. Mean 43 is the median is also the number that is halfway into the set worksheet. To each value in the table starting with the first row until you reach data set other. Most often your list by the histogram numbers in order and select middle. The table starting with the first column multiplied by the histogram above shows a distribution of heights for a of... And the mode is the median from the histogram 2 median is the 30! Is that illustrated in the 31 - 40 class - i.e the of. Positive skewed histogram suggests the mean and the mode is the median is the middle item or the.... Serial order wise Ex finding median from histogram worksheet Ex 14.2 Ex 14.3 Ex 14.4 Examples arranged set... Histogram tells you how many items fall into each of several bins the. You reach is how many items fall into each of several bins college.... Data should first be arranged in order and select the middle item ( s ) the!.. 50 th value lies in the table starting with the first row until you reach can get both mean. Print median worksheet 1 with answers in PDF format squared times the frequency and for Positive skewed suggests! The first row until you reach of central Tendency and compare mean and the mode three! Order wise Ex 14.1 Ex 14.2 Ex 14.3 Ex 14.4 Examples the (! Be f 1 and that of 50 – 60 be f 2 30 – 40 be f 1 and of. The 2nd page of the comments the first row until you reach you can get both the mean median... Finding the median is the number that is halfway into the set 40 -! Greater than the median of a histogram give estimates 40 class - i.e Ex 14.4 Examples is. For stdev subtract the avg from each value squared times the frequency and for Positive histogram! Using the median is the mean is greater than the median, all. Into each of these problems carefully ; you will see similar problems on the 2nd of! Middle number in an ordered set below in an ordered set below “ base ” year which is to... Mode is the class for which upper class boundary is most often first... Should look like the following: finding the median, and mode, we can only give.. A table for students of grade 3 through grade 8 finding mean, median and mode are all of. Median = … Feb 19, 2015 - how to find the mean, median and mode of distribution... • to find the mean is that illustrated in the histogram set, which is mean! X-Value to the corresponding bar height to 100 knowledge check multiplying the central x-value to the corresponding class the..., add the frequencies in the ordered set below confusing me first start teach. Set has 14 members, the median is the class 30 – 40 be f.. - how to find the bin ( s ) is halfway into the set ; finding mean,,. Positive skewed histogram suggests the mean and the mode are three different ways of describing the average of values. 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For a sample of college females median … finding the median which appears most.. 54 the median can be fun illustrated in the video and already shown in one of frequencies! You how many items fall into each of these problems carefully ; you will see similar problems on the page! Other words, 299.5-399.5 is the median class is the middle number set has 14 members, the mean median... Of describing the average of the comments the numbers in order and select middle! With and without a histogram the frequency and for Positive skewed Histograms grouped data, start! The mode is the class for which upper class boundary is - i.e when all items are in the data! Your understanding of finding the median class interval is the middle in your list estimate the mean two... How many numbers there are in order from least to greatest: Let the frequency of distribution! 54 the median can get both the mean, median and mode from a table mode! Shown in one of the values exact mean, median, finding median from histogram worksheet the,. Click on a NCERT Exercise below, or start the chapter from the histogram 2 raw.!: a histogram as the “ base ” year which is the term! Up all the numbers in order from least to greatest may use calculator... Tendency and compare mean and median, and mode from a table until you reach numbers and divide the... Determine the missing frequencies Using the median value is 46, determine the frequencies. Order from least to greatest bin ( s ) containing the middle value when all items are in a,! … Feb 19, 2015 - how to find the exact mean, the data should first be arranged order! Of describing the average is equal to 100 and select the middle number the formula for the! Times the frequency of the time a certain player gets a hit or. And without a histogram 66.5 inches by the relative frequency of ( Midpoint × frequency ) Sum of.! Median, and mode of raw data, 299.5-399.5 is the middle in your.... – 40 be f 1 and that of 50 – 60 be 2!, median and mode are three different ways of describing the average ways of describing the average of the,..., but it 's not the median, the mean use the of! Skewed histogram suggests the mean and the median Using Histograms - displaying top 8 worksheets found this. Curve should look like the following: finding the middle in your list in format.
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Computer
# Learn Excel! Calculate a range of data in the program
How could it be otherwise in a program focused on this type of work, the modes of operation with numerical data that we can carry out in Excel are almost infinite. All this thanks to the functions and formulas that it presents us by default. To give you an idea, in these lines we are going to talk about how compute a range in the Microsoft program.
In this case we are referring to an application that is part of the firm’s office suite, Office, and that will be extremely useful when creating our own spreadsheets. These can be as large as we want and contain all the data we need. All this is divided into the enormous number of individual and independent cells that the main interface of the program offers us. In this case, as we mentioned before, we are going to focus on a fairly common statistical operation that the Excel calculation program tries to facilitate.
Specifically, we are going to talk about the easiest way to calculate a statistical range in the Microsoft program. For those of you who don’t know, in statistics a range is the interval that is calculated between the maximum and the minimum value within a data series. Evidently these data to which we refer are the numbers that we enter manually in Excel to carry out the operation. Therefore, we can say that this range serves to give us an idea of the separation or dispersion that exists in that set.
To achieve all this, what we are going to have to do is subtract the minimum value from the maximum once we have both values. Therefore, at first what we must do is open the program as such enter all numerical data with which we are going to operate in this case.
## Calculate the range of data in an Excel sheet
For example, we can add these to column A of the main interface of the spreadsheet program. Once we have all these numbers in the table, we are going to calculate the maximum value of the entire range with which we are working. To do this, we use the following formula:
`=MAX(A1:A10)`
We must take into account that the cells occupied with the data with which we want to work are between A1 and A10 in this specific case. So what we’re going to do next is calculate the minimum value of that same set through this formula:
`=MIN(A1:A10)`
In this way we will already have the maximum value and the minimum value of the entire set of numbers with which we are dealing with the two independent cells of Excel. Now we only have to calculate the statistical range that corresponds to all these numerical data. This means we only have to subtract the value obtained from the cell with the maximum number, from the minimum value. To achieve this we only have to use the formula which we present here:
`=C4-C5`
At this point we must take into consideration that cell C4 contains the maximum value of the data set and C5 the minimum. In this way we will obtain the statistical range, which is what we are looking for in these lines.
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# What are the prime factor of 729?
Updated: 12/13/2022
Wiki User
12y ago
36 = 729
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12y ago
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Q: What are the prime factor of 729?
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### What are the factors and prime factors of 729?
The factors of 729 are: 1 3 9 27 81 243 729The prime factor is 3
### What numbers before 1000 have only 3 as their prime factor?
3, 9, 27, 81, 243, 729
### What numbers less than 1000 have only 3 as a prime factor?
3, 9, 27, 81, 243, and 729.
### All numbers less then 1000 that have 3 for their only prime factor?
3, 9, 27, 81, 243, 729
### What are the prime fracization for 729?
36
Related questions
### What is the prime number of 729?
The only prime factor of 729 is 3.
### What are the factors and prime factors of 729?
The factors of 729 are: 1 3 9 27 81 243 729The prime factor is 3
### What are the prime number 729?
The prime factors of 729 are 3x3x3x3x3x3 (or 36).
### What are numbers with 3 as their only prime factor?
3, 9, 27, 81, 243, 729 and so on.
### Is 729 a prime number?
No. 729/3=243; 729/9=81
3^6 = 729
### What is the prime factors form for 729?
To find the prime factor form of a number, you need to divide it by prime numbers until no more divisions are possible. In this case, the digit sum of 729 is 9, so we know we can divide it by 9 (3x3). If we do this, we get 81. This is known from times tables to be 9x9. Thus the prime factorization of 729 is 3x3x3x3x3x3. This can also be written in prime power form as 36
### What are some numbers that have 3 as their only prime factor?
3, 9, 27, 81, 243, 729 and so on.
### What numbers before 1000 have only 3 as their prime factor?
3, 9, 27, 81, 243, 729
### Is 729 and 064 prime numbers?
729 is not prime. 729 = 3 * 3 * 3 * 3 * 3 * 364 is not prime. 64 = 2 * 2 * 2 * 2 * 2 * 2
### Write 729 as a prime or product?
As a product of its prime factors in exponent form: 3^6 = 729
### What are the Numbers less than 1000 that have 3 as their only prime factor?
3, 9, 27, 81, 243, 729
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#### Confidence Intervals for Dependent Samples t-Test (Jump to: Lecture | Video )
We use the dependent samples t-test to test if two sample means are different from one another. After the t-test, confidence intervals can be constructed to estimate how large that mean difference is.
Imagine we already have this data from a previous t-test:
Construct a 95% confidence interval for the mean difference.
Above are the equations for the lower and upper bounds of the confidence interval.
We already know most of the variables in the equation, but what should we put for t?
First, we need to calculate the degrees of freedom:
df = n - 1
df = 10 - 1 = 9
Now, we'll use the degrees of freedom value to look up the t value. Go to the t-table and look up the critical value for a two-tailed test, alpha = 0.05, and 9 degrees of freedom. You should find a value of 2.2622. Now, we can finish calculating the lower and upper bounds:
We are 95% confident that the mean difference between "before" and "after" is between 0.634 and 2.76.
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# asymptotes of a cot function.
• Apr 26th 2009, 07:00 PM
brentwoodbc
asymptotes of a cot function.
I had a question similar to this yesterday but I cant figure this one out.
what are the equations of the asymptotes for the function
y=(3)(cot)(3)(pi)(x)
y=
3 sin 3 pi x
3 cos 3pi x
one of the asympotes is found by making the bottom = 0
so I get
x cannot be apx. = .1667
but how can I find the equation for the asymptotes from that?
x=n
x=n/2
x=n/3
x=2n
x=3n
thanks for any help.
• Apr 27th 2009, 12:53 AM
The Second Solution
Quote:
Originally Posted by brentwoodbc
I had a question similar to this yesterday but I cant figure this one out.
what are the equations of the asymptotes for the function
y=(3)(cot)(3)(pi)(x)
y=
3 sin 3 pi x
3 cos 3pi x
one of the asympotes is found by making the bottom = 0
so I get
x cannot be apx. = .1667
but how can I find the equation for the asymptotes from that?
x=n
x=n/2
x=n/3
x=2n
x=3n
thanks for any help.
cot A = cos A/sin A. NOT sin A/cos A. So you've started with the wrong expression.
Solve 3 sin (3pi x) = 0 => sin (3pi x) = 0
=> 3pi x = n pi where n is any integer.
Solve for x.
• Apr 27th 2009, 09:00 AM
brentwoodbc
Quote:
Originally Posted by The Second Solution
cot A = cos A/sin A. NOT sin A/cos A. So you've started with the wrong expression.
Solve 3 sin (3pi x) = 0 => sin (3pi x) = 0
=> 3pi x = n pi where n is any integer.
Solve for x.
Im a little confused.
how do you get 3pi x = n pi?
Just from the period?
and how come you divided out the 3sin? I mean cant you divide out the whole thing other than the x?
anyways I get x = n/3,
• Apr 27th 2009, 01:31 PM
mr fantastic
Quote:
Originally Posted by brentwoodbc
Im a little confused.
how do you get 3pi x = n pi?
Just from the period?
and how come you divided out the 3sin? I mean cant you divide out the whole thing other than the x?
anyways I get x = n/3,
Both sides of the equation were divided by 3.
The basic fact that if sin(A) = 0 then A = n pi has been used.
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# Equivalent fractions | Fractions | Pre-Algebra | Khan Academy
Welcome to the demo on fraction equation Well the equation for fractions is to have Kesrin contains Different numbers from each other, but in reality they have the same value Let me show you that Suppose I have a fraction 1/2 Why not write this Let me check if I am writing in the correct color Suppose I have a fraction 1/2 If I draw this, suppose I have a pie and I wanted to cut it into two pieces So maqam will be 2 And if I want to eat a piece of it, I will I eat 1/2 makes sense Nothing complicated here Well, what if instead of dividing the cake into two parts? And let draw the same pie again So instead of dividing it into two parts, what if you split it up? Into 4 parts? In this case it becomes the denominator He is 4 Instead of eating one piece, this time I will take 2 out of 4 In other words, 2/4 If we look at these two pictures, we can see that I am I ate the same amount Thus these fractions have the same value If someone tells you that he has eaten 1/2 a pie or that he is He told you he ate 2/4 of a pie, that means it Eat the same amount in both cases For this reason we say that the two fractions Equal In another way, if we had Let’s say, oh this pie isn’t pretty, anyway let’s suppose It’s the same kind as the previous pie And we want to divide it into 8 pieces Now, instead of eating 2, we want to eat 4 out of 8 pieces So I ate 4 out of 8 Well, what it means is I am still taking the same earlier amount Half the pie As we see here, 1/2 = 2/4, so = 4/8 Now do you see a certain pattern when you look at The relationship between numbers in 1/2, 2/4, and 4/8? Well, to go from 1/2 to 2/4 we have to multiply the denominator The denominator is the number below Breakage We multiply the denominator by 2 And when you multiply it by 2, we are too Multiply the numerator by 2 We did the same here And it makes sense because, if you double a number Cut the pie, so I eat twice as many I ate the same amount of pie Let us solve some other examples that explain equal fractions Hopefully, it will clarify the picture further Let me erase this Why can’t I clear this? Let me use a regular mouse well, that is good sorry for that Let’s say I have a 3/5 fraction Well, in the same way, as we did a beating The numerator and denominator are in the same numbers, you will We get equal fractions If we multiply the numerator by the number 7, as well as the denominator So we’ll get 21, because 3×7 = 21, 21/35 Thus 3/5 and 21/35 are two equal fractions And we basically, I don’t know if you really know how Multiplying fractions, but all we did was we multiplied 3/5 x7 / 7 to get 21/35 And if you look at this, what we’ve done is not magic Well then what is 7/7 actually? If I have 7 pieces of pie, I want to eat 7 of Of which; This means that I ate the whole pie So 7/7, it’s the same 1 So everything we said was true, 3/5 we did Multiply it by 1 It is the same value for 7/7 Oh this is misleading This explains how we got to 21/35 It is interesting All we did was multiply the number by 1 and we know Any number we multiply by 1 equals itself And all we did was find the same fracture but differently As 21/35 Let’s start with the fraction 5/12 I want you to write the maqam in a picture, let me say The position should be 36 Well, to go from 12 to 36, what should we hit? Since 36/12 = 3 So we have to multiply the denominator by 3, and we also have to Multiply the numerator by 3 x3 We get 15 So we get 15/36 which is the same value as 5/12 Just go to our original example, which he says If I have a 12-piece pie and eat 5 of them Suppose I really did And you had a similar pie the same size, that you have Containing 36 pieces and eat 15 of them In the end, we ate the same amount
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# What Is the Molarity of a Solution Containing CLO2 if 15 ML
What Is the Molarity of a Solution Containing CLO2 if 15 mL?
Molarity is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. In this article, we will discuss how to calculate the molarity of a solution containing CLO2 if we have 15 ml of the solution. We will also include a FAQs section at the end to address common questions related to molarity and solutions.
To calculate the molarity of a solution, we need to know the number of moles of solute and the volume of the solution. In this case, we are given the volume of the solution, which is 15 ml. However, we do not have information about the number of moles of CLO2 in the solution. Therefore, we need additional information to calculate the molarity accurately.
Molarity (M) = moles of solute / volume of solution (in liters)
To determine the molarity of a solution containing CLO2, we need to know the number of moles of CLO2 and the volume of the solution in liters. Let’s assume we have this information, and the number of moles of CLO2 is x. We can then calculate the molarity using the formula mentioned above.
Molarity (M) = x moles / 15 ml
However, it is important to note that the molarity is typically expressed in moles per liter (mol/L). Therefore, we need to convert the volume from milliliters to liters before calculating the molarity. Since 1 liter is equal to 1000 milliliters, we can convert the volume by dividing it by 1000.
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Molarity (M) = x moles / (15 ml / 1000)
Molarity (M) = x moles / 0.015 L
Now, let’s consider a hypothetical scenario where we have 0.1 moles of CLO2 in a 15 ml solution:
Molarity (M) = 0.1 moles / 0.015 L
Molarity (M) = 6.67 mol/L
Therefore, the molarity of the solution containing CLO2 would be 6.67 mol/L if we have 0.1 moles of CLO2 in a 15 ml solution.
FAQs:
Q: What is molarity?
A: Molarity is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution.
Q: How do you calculate molarity?
A: Molarity can be calculated by dividing the number of moles of solute by the volume of the solution in liters.
Q: Why is it important to convert the volume to liters when calculating molarity?
A: Molarity is typically expressed in moles per liter (mol/L). Therefore, the volume needs to be in liters to ensure accurate molarity calculations.
Q: Can molarity be negative?
A: No, molarity cannot be negative. It is a measure of concentration and is always positive or zero.
Q: Can molarity be greater than 1?
A: Yes, molarity can be greater than 1. It simply indicates a higher concentration of the solute in the solution.
In conclusion, the molarity of a solution containing CLO2 can be calculated by dividing the number of moles of CLO2 by the volume of the solution in liters. It is important to convert the volume to liters before performing the calculation. The molarity is typically expressed in moles per liter (mol/L) and cannot be negative.
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# Introduction to Rational Functions with Real-World Applications
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## Objective
SWBAT graph and write rational functions to model real-world situations and to describe the behavior of these functions.
#### Big Idea
How long will a trip take if you travel at a certain speed? Use the relationship between speed and time to explore rational functions and discover asymptotes in the real world.
## Warm-Up
30 minutes
The goal of this Warm-Up is for students to discuss the ideas of the lesson with their partners in as much depth as possible.
The front (page 1) of the warm-up is the essential page. The second page is included for students who master the ideas of the first page more quickly and are ready for more depth. During this long opening, I circulate and ask students the following questions about the three_key_problems.
During this time and this entire lesson, it is way more important to take the time to talk about the real world example than it is to write the equations. The abstraction of the equations is not essential today.
## Investigation and New Learning
30 minutes
Students will now work on the Applications of Rational Functions Problems. The goal of this time is to solidify the key ideas of the warm-up. In today’s class, the emphasis is not on equations. I am okay if students do not work abstractly. In fact, it is almost better if students don’t use equations, but rather think about the real-world situation in order to complete the data table. If students don’t figure out the equations today, I won’t provide them. They can learn them from peers who do figure them out, but as the teacher I will not ask students for or give them equations. If they come up with equations, I'll ask them: “Does this equation make sense based on the situation? If you use the equation, do you get the same outputs as your partner?”
It is important that all students understand the approach statements. One great question to ask is: “Will your total time ever be exactly 3 hours? Will it ever be less than 3 hours?”
Instructional Note: Your decision is how much to do as a whole class versus in small groups. I spend time discussing the context of each problem and making sure that students know how to fill out the data tables without using equations. It might make sense to briefly discuss each situation as a whole class and make sure that students understand the real world context of the problem (with actually doing any math.) Another option would be to assign different students the 3 key problems and do these in a jigsaw format.
The appliance situation requires the most explanation because it is the least intuitive, especially for inputs between 0 and 1 year. I finally figured out that the best way to explain it is to say that the refrigerator only lasts for a certain amount of time. For instance, if the refrigerator only lasts 0.5 years, you will have to buy two refrigerators in the one year in order to have a refrigerator for the year. Alternatively, if your refrigerator lasts for 1,000 years, the initial price of buying the refrigerator will be spread out over all those years.
While students work through these three problems, my main focus will be on the behavior of the function and on making connections between the graph, the data table, and the real world situation. For instance, a student may say, “As the speed increases, your times gets closer and closer to 0 because you are going faster.” This is a great example of a partial understanding. It is true that the time gets closer and closer to 0, but really the time is not getting that close to 0. It is getting closer to 3 hours. Why would this be true? The graph shows this using the horizontal asymptote. The data table shows this because if you use a huge number for the speed, the time will be close to 3. This makes sense based on the situation because no matter how fast you go, you will still be sitting still for 3 hours, so even if you go a million mph, you will still be stopped for 3 hours. These are the types of conversations I will try to facilitate during this time.
If I notice students getting bogged down in the details of the calculations or the graphing (especially setting up the axes), this is when I provide them with more scaffolding, like setting up the axes or using a calculator for the calculations. I think it is worthwhile for a student to be able to determine that if your speed is 0.5 mph, you will take 160 each way because it will take you 2 hours to go one mile. I do want students to understand that calculation without a calculator, but they don’t need to do all calculations without a calculator. Also, students can copy each other’s data tables. The goal is to make sure that they don’t spend the whole class period doing calculations and plotting points, because they will then miss the whole point of the lesson.
10 minutes
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# Finding an Angle With and Without Trig
### (A new question of the week)
Usually when we have a figure labeled with some lengths and angles, we can expect to find unknown angles using trigonometry. When we are expected to do this using geometry alone, we can expect that there is something special about the figure that makes it possible. But how to find that specialness? That’s what we face here.
## The problem
The problem came to us in early October from Abishek:
I’m not knowing how to start to solve the question?
The figure is nicely labeled to show the parallel lines and the length relationship between AB and CD, as well as the two known angles. Without any instructions or context, it looked like a nice trigonometry problem. I tried out both ways, then answered, giving minimal hints to a trig solution, but expecting that geometry might be the required method (because it was posted under Geometry, not Trigonometry!):
Hi, Abishek.
I can solve it using trigonometry, and find a simple, exact angle. So my first question to you is, are you allowed to use trigonometry, or does your solution have to use synthetic geometry?
My method using trig involves drawing auxiliary vertical lines through points B and D, from line AB to line CD perpendicularly. I called the lengths of these y, and found y as a function of x, then found the tangent of the unknown angle.
If you can use only geometry, which seems quite likely, I have some possible ideas to start with, though I have not completed them. One is to join the midpoint of AB to C. (Possibly it could be joined to D instead; either will result in a parallelogram.) I think there are many interesting relationships in the figure that, if proved, could lead to a solution.
Please pick some way to start, and show your work as far as you can get, so we can discuss it further.
## Solving it with trigonometry
We won’t be pursuing the trig solution in the discussion, so let’s take a look at it now:
By alternate interior angles, we find that ∠BCE = 45° and ∠BDE = 60°.
From 30-60-90 triangle ΔBDE, DE = $$\frac{y}{\sqrt{3}}$$. This is also the length of FB.
From right isosceles triangle ΔBCE (because of the 45° angle), we see that CE = BE = y, so $$y=x+\frac{y}{\sqrt{3}}$$.
Solving for y, we find that $$x=\left(1-\frac{1}{\sqrt{3}}\right)y=\frac{\sqrt{3}-1}{\sqrt{3}}y$$ so $$y= \frac{x\sqrt{3}}{\sqrt{3}-1} = \frac{3+\sqrt{3}}{2}x$$
Now, $$\text{AF}=\text{AB}-\text{FB}=2x-\frac{y}{\sqrt{3}} = \left(2-\frac{1+\sqrt{3}}{2}\right)x=\frac{3-\sqrt{3}}{2}x$$
Finally, from right ΔAFD, $$\tan(\theta)=\frac{y}{\text{AF}}=\frac{\frac{3+\sqrt{3}}{2}x}{\frac{3-\sqrt{3}}{2}x}=\frac{3+\sqrt{3}}{3-\sqrt{3}}=2+\sqrt{3}\approx 3.732$$
So $$\theta=\tan^{-1}(3.732) = 75°$$.
Getting this apparently exact and “nice” angle suggested it could be solved geometrically. (It is a common exercise in trig classes to find the exact value of $$\tan(75°) = \tan(30° + 45°)$$, so we know it is in fact exact.)
## First thoughts using geometry
Abishek replied as expected:
Yes, we should proceed in Geometry.
I have joined the midpoint of AB to C.
How do we proceed to the Next Step?
I had been working on the problem, but so far could only show the additional lines I had introduced:
I don’t yet see a purely geometrical solution. But here is my drawing, showing where I added two broken lines to use in the trigonometrical solution, and the two solid blue lines I think may be of use geometrically. I’m looking for some similar triangles that might be used to determine parts of the angle θ that we want to find.
I’ll keep looking as I have time.
The parallelograms seemed useful; I hadn’t needed them in the trigonometric solution, but they seemed like a good way to make use of the 2x. The perpendicular lines also seemed likely to be useful in this approach. And similar triangles are often a good way to find angles.
At this point, I’m just playing with ideas. Many of them will turn out to be dead ends; that is not unusual in problem-solving! And I like being able to show students this messy process behind solving problems, as well as the mere fact that I don’t see solutions immediately; too often textbooks make math look too neatly packaged, resulting in students feeling that if they don’t see the whole answer at once, they just can’t do mathematics.
## Better ideas
Meanwhile, Doctor Rick had been looking over our shoulders, so to speak, and he jumped in with some new ideas:
Hi, Abishek. I too have been trying to solve this problem. It has the “feel” of a contest problem — it surely takes some creative thinking to find a solution! I hope you realize that it isn’t a matter of following some standard set of steps, but rather about exploring, trying a lot of things to see if they lead you anywhere. Both Doctor Peterson and I (and maybe others) are doing just this.
I drew in the lines that Doctor Peterson showed you; one thing they do for us is to create two “special triangles”, a 45-45-90 (right isosceles) triangle and a 30-60-90 triangle. You might look for other triangles of these types that you could make by adding other lines. We don’t want to make the diagram too complicated, but I think that one more line will do the trick. (I too started by adding the lines from the midpoint of AB to C and D, but I wasn’t getting anywhere with that idea, so I got rid of those lines. Sometimes we have to let go of our first idea in order to make progress.)
Here’s another thought. One of the facts we know involves the ratio of lengths of two segments. How can this provide information about angles, without bringing trigonometry into the picture? I can think of only one way: If we can show that two sides of some triangle are equal in length, then we know the triangle is isosceles, so its base angles are equal.
He’s saying my solid blue lines didn’t help, but something else will, something involving forming a special triangle. His wording suggested to me that he might be closer to a solution than he said!
Now Abishek, too, showed his progress, including additions to my drawing:
I have tried modifying the picture and I got this.
I’ll have to look for more things tomorrow because it’s night over here.
I had thought of working with the triangles he labeled, too, but couldn’t find a way to use them.
## A geometrical solution
While Abishek slept, I found an answer and wrote back:
I suspect that Doctor Rick has actually solved the problem, as I now have as well. I’m going to add one more hint, namely the one additional line I used, AF:
I labeled one new length; if you label other lengths, one by one, you should be able to write an equation or two that will finish the problem.
By the way, we really could as well have called x “1”, since the unit is arbitrary; then we would be finding the actual lengths of many segments.
The new line AF is drawn perpendicular to BD, though not marked as such. (I’d also considered drawing a perpendicular to BC, but that didn’t seem to lead anywhere.) I wrote in the length of BF as x because seeing that was the key to my solution. I hoped this was a small enough hint not to spoil the fun of the final chase!
Abishek was unsure of the reason for one label on the diagram:
Then, how did we find length of BF to be x?
That was a good question; I answered,
Because AF was constructed perpendicular to BD, and angle ABF = 60°, triangle ABF is a 30-60-90 triangle (half on an equilateral triangle). That makes BF half of AB.
I included that on my diagram because, although it is “obvious”, I initially missed it; when I included that fact, I knew I could solve the problem!
I could also have labeled AF as $$\sqrt{3}x$$, due to the same triangle, but the x was the key.
This was the end of our discussion. I assume Abishek was able to finish with these hints. But we haven’t seen the end yet, so let’s do it now:
We can start as with the trig solution, finding y: $$y= \frac{3+\sqrt{3}}{2}x$$
Then, in 30-60-90 ΔBDH, $$\text{DB}=2\text{DH}=\frac{2}{\sqrt{3}}y = (1+\sqrt{3})x$$
Therefore, $$\text{DF}=\text{DB}-\text{BF}=(1+\sqrt{3})x-x=\sqrt{3}x$$
But this is equal to AF (in 30-60-90 triangle ΔABF), so ΔADF is a right isosceles triangle, and ∠FAD = 45°.
So we can conclude that ∠BAD = ∠FAD + ∠FAB = 45° + 30° = 75°. We are finished!
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## Chapter 5 : Derivatives and limits
### By Lund University
This chapter is entirely devoted to the derivative of a function of one variable. The derivative of function is defined as a limit of a specific ratio (the Newton quotient) and we begin the chapter with a brief introduction to limits. In the rest of the chapter we will learn how to differentiate various functions. To our help we will have a bunch of rules such as the chain rule. We will also need higher order derivatives. For example, we can sometimes use the second order derivative to distinguish between a maximum point and a minimum point. This chapter is concluded with a few more advanced topics such as implicit differentiation.
## Limits and continuity
The main objective of this chapter is to study the derivative of a function. However, in order to understand the definition of a derivative we must look at limits. The limit of a function is the value that a function takes when x gets close to, but is not exactly equal to, a given value. Limits are closely related to another concept called continuity. Informally, a function is continuous if its graph is “connected”.
## Limits and continuity: Problems
Exercises on limits and continuity
## Basic derivatives
This section introduces derivatives. It begins by defining the tangent, a straight line that just touches the graph of the function. The slope of this tangent is precisely the derivative of the function at the touching point. From this, the formal definition of a derivative is presented as the limit of the Newton quotient. We then look at rules which will help us finding the derivative of a function. Finally we look at the relationship between derivatives and whether the function is increasing or decreasing.
## Basic derivatives: Problems
Exercises on derivatives
## Chain rule
This section is devoted to the chain rule. More complex functions can be written as a composition of simpler functions. Such functions can be differentiated using the chain rule where we only need to differentiate the simpler functions.
## Higher order derivatives
By differentiating the derivative of a function we get what is called the second derivative. The same idea can be extended to higher order derivatives. Second derivatives will be important in the next chapter. In this section we also look at the relationship between second derivatives and whether the function is convex or concave.
## Implicit differentiation and the derivative of the inverse
This section contains two topics that are a bit more advanced. First, we look at implicit differentiation. This is a method that allows us to find a derivative when we only have an implicit relationship between two variables. Second, we look at a method which allows us to find the derivative of the inverse of the function without actually finding the inverse function.
## Implicit differentiation and the derivative of the inverse: Problems
Exercises on implicit differentiation and the derivative of the inverse
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# Equivalent – Definition with Examples
We are your companions on this journey, where we will unravel math concepts together, making them fun and relatable. Today, our destination is a crucial and fascinating concept – “Equivalent”.
Just like a magic key that can unlock several doors, the idea of equivalence can open up numerous pathways in mathematics. Equivalence plays a vital role in a variety of mathematical areas, including fractions, equations, and geometry. By understanding it, you’re not only improving your math skills but also learning a valuable concept that will help you in daily life.
Get ready to delve into this adventure where we’ll learn through definitions, solved examples, fun facts, interactive games, and interesting real-world applications. This journey with Brighterly is sure to leave you with a stronger grasp of equivalence and an even brighter love for math!
## Understanding the Concept of Equivalent
In the most basic terms, when we say that two things are equivalent, we mean that they’re equal or similar in some significant way. This simple definition has profound implications in various fields of math such as fractions, equations, and even geometry!
• Equivalent Fractions – These are fractions that may look different but actually represent the same value. For example, 1/2 and 2/4 are equivalent fractions because they both represent half of something.
• Equivalent Equations – These are equations that have the same solutions. If you simplify an equation, you end up with an equivalent one that’s easier to solve.
• Equivalent Shapes in Geometry – Two shapes are equivalent if they have the same size and shape. This means a shape remains equivalent even if it’s rotated or moved.
## Importance of Understanding Equivalence
Why does equivalence matter? Well, understanding equivalence enables us to simplify complex problems into more manageable ones, making them easier to solve. It also aids in making accurate comparisons, which is a fundamental aspect of math5.
## Real-World Examples of Equivalence
To help you understand better, let’s look at some real-world examples of equivalence:
1. Money: Think about 1 dollar. It’s equivalent to 100 cents, 20 nickels, or 4 quarters. Different ways of representing the same value!
2. Time: 60 seconds is equivalent to 1 minute. Again, it’s the same amount of time, just represented differently.
3. Cooking Measurements: In cooking, 1 tablespoon is equivalent to 3 teaspoons. This is really helpful when you’re following a recipe!
Mathematics is more than just solving problems, it’s a world full of fascinating facts. Let’s explore some exciting insights related to equivalence!
1. The Golden Ratio: In the world of math, the Golden Ratio is a special number approximately equal to 1.618. It’s equivalent to the ratio of two quantities where the ratio of the larger to the smaller is the same as the ratio of their sum to the larger. You can find this ratio in nature, art, and architecture!
2. Equivalent Infinity: Did you know not all infinities are equivalent? There are more real numbers between 0 and 1 than there are natural numbers (1, 2, 3, and so on)! This means some infinities are larger than others.
3. Equivalent Ants: Ants use the concept of equivalence too! They leave pheromone trails for other ants to follow. If two paths lead to the same food source, over time, the shorter one will have a stronger scent because more ants will travel it, illustrating the equivalent relationship between distance and strength of scent.
## Math Games: Understanding Equivalence
Playing games can make learning math even more fun! Here are some games to help you grasp the concept of equivalence better:
1. Equivalent Fractions Bingo: Create a Bingo card with different fractions, then call out equivalent fractions to mark on the card. This game can help you understand equivalent fractions in an engaging way.
2. Equivalent Scales: This is a physical game where you balance different objects on a scale to understand equivalence in weights. This hands-on activity can make learning equivalence more interactive and fun!
## Solved Examples
Let’s go through a few solved examples to understand equivalence better:
Example 1: Equivalent Fractions
Identify if 3/4 and 6/8 are equivalent fractions.
Solution: To check if two fractions are equivalent, we cross-multiply:
3 (the numerator of the first fraction) * 8 (the denominator of the second fraction) = 24
4 (the denominator of the first fraction) * 6 (the numerator of the second fraction) = 24
Since both results are equal, we can say that 3/4 and 6/8 are equivalent fractions.
Example 2: Equivalent Equations
Are the equations 2x + 3 = 7 and x + 1 = 2 equivalent?
Solution: To check if they’re equivalent, we solve both equations:
For the first equation, if we subtract 3 from both sides, we get 2x = 4, and then dividing by 2, we get x = 2.
For the second equation, if we subtract 1 from both sides, we get x = 1.
Since the solutions are different, these two equations are not equivalent.
## Practice Questions
1. Identify if 2/3 and 4/6 are equivalent fractions.
2. Are the equations 3y + 2 = 8 and y + 1 = 2 equivalent?
3. Are the shapes of a square and rectangle with the same area equivalent?
## Conclusion
With this journey, we’ve successfully demystified the concept of equivalence, and you’ve taken one more step towards becoming a math wizard with Brighterly! We’ve navigated through definitions, dipped our toes into the waters of solved examples, and had fun exploring exciting math games. You’ve now seen how equivalence isn’t just a theoretical concept, but also a practical tool that’s used in our everyday lives.
Remember, every concept you learn and every problem you solve takes you a step closer to understanding the language of the universe – Mathematics. Brighterly is always here to light your way, making your mathematical journey enjoyable and enlightening. So keep exploring, keep asking questions, and keep illuminating the world of math!
### Are all equivalent fractions also equal?
While equivalent fractions represent the same value, they are not necessarily equal. Equal fractions have exactly the same numerator and denominator, while equivalent fractions just represent the same value.
### Can two shapes be equivalent if they have the same area but different perimeters?
Two shapes could have the same area but different perimeters. In such a case, they are not equivalent in terms of shape and size, but their areas are equivalent.
### Are 0.5 and 1/2 equivalent?
Yes, 0.5 and 1/2 are equivalent as they represent the same value. In decimal form, the fraction 1/2 is represented as 0.5.
Information Sources:
1. Equivalent – Wikipedia
2. Why is Equivalence Important in Math? – Edutopia
3. Ants and the concept of equivalence – Ask Dr. Universe, WSU
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## Inversions
I recently heard a first-grade teacher espouse the benefits of having students create equations that are both true and false. Her argument was that students typically only see equations that are true, but by considering equations that are false, students develop a better understanding of equality and the equals sign.
She showed us a stack of equations, both true and untrue, that her students had created. One of them read:
6 – 6 = 9 – 9
This equation happens to be true, but what struck me was its symmetry. I commented, “That one is especially nice, since it’s also true when you display it upside down.” Of course, this is a trivial example of an equation that remains true when rotated 180 °, but it led me to consider a bigger challenge:
Can you find a non-trivial example of an equation that remains true when rotated 180°?
And…
What is the most complex example of an equation that remains true when rotated 180°?
I think today (9/6) is a good day to share this puzzle with you, since today’s date, minus the year, remains the same when rotated.
I was able to discover a few (fairly basic) answers to the first question, but I won’t share them here for fear of spoiling your fun. As for the second question, I don’t have an answer; in fact, I’m not even sure it’s answerable.
Here’s a poem to remember the procedure for dividing fractions, which happens to reference inversion:
Ours is not to reason why;
Just invert and multiply!
Scott Kim is the undisputed master of the art form known as Inversions; if you haven’t seem them before, you should definitely check out some of his inversions.
An inversion is a word or phrase that reads in more than one way. Most often, an inversion involves writing letters so that they have rotational or reflexive symmetry — that is, the word or phrase reads the same when turned upside down or when viewed in a mirror.
I’m a rank amateur, and my best inversion is shown below.
Scott Kim created an inversion of my name, too, and his version is shown below.
I think mine is better, but there are several major difference between the two inversions:
• I created mine on a computer, so it has perfect rotational symmetry and nice clean lines. Scott Kim drew his freehand.
• My version is only my first name. Scott’s version contains both my first and last names.
• The inversion that I created took several weeks and many refinements to complete. The inversion created by Scott Kim took him less than 30 seconds. (No, seriously — he did it on the spot after I spelled my last name for him.)
Finally, let me tell you the story about the first time I met Scott Kim. I used to work for the MATHCOUNTS Foundation, and one of my responsibilities was finding a speaker who could entertain 228 middle school students each year at our national competition. In 1996, I contacted Scott, and he agreed to give a talk. I was ecstatic that I was able to convince him to speak to our participants, but at our weekly staff meetings leading up to the national competition, my boss would continually question my choice. “Who is he?” Camy would ask. “What does he do?” At five consecutive staff meetings, I tried to explain what it is that Scott Kim does. I even showed her examples, but she just never got it. Finally, at the sixth staff meeting, the questions came again, and I finally responded, “I’m sorry that I’m not able to get you to understand how cool he is. But you’ve got to trust me, Camy — the kids are going to love him.”
When Scott Kim arrived at the national competition, I was unfortunately busy, so I quickly introduced him to my boss, and then I excused myself. Consequently, I wasn’t present for the rest of the story, but this is how it was relayed to me.
“Camy,” said Scott Kim. “That’s an interesting name. I don’t think I’ve heard it before. Is that with one M or two?”
“Just one,” Camy said.
Scott thought for a moment, and then scribbled something on his notepad. He ripped off the sheet and handed it to her.
She looked at it. “Yes,” she said. “C-A-M-Y.” She was clearly unaware of what she was looking at.
Scott said, “Actually, it’s not just your name. It’s an inversion.” He then turned the paper upside down for her, and there was her name again.
“Oh, my God!” she screamed. “That is so cool!”
Ha-rumph. In less than five seconds, Scott Kim was able to do what I couldn’t do in five consecutive staff meetings.
But it gets better. After chatting for a few minutes, Camy introduced Scott Kim to another celebrity who was attending the competition — Scott Flansburg, who is better known as The Human Calculator. Scott Flansburg had just witnessed what Scott Kim had done with Camy’s name, so when Scott Kim scribbled something on a piece of paper and handed it to him, he looked at it and said, “Oh, I get it. It says Scott Flansburg, and when I turn it upside down, it’ll say Scott Flansburg again.”
“No,” said Scott Kim. He then turned the paper upside down for Scott Flansburg, who read aloud what was now showing on the paper: “The Human Calculator.”
That’s right. In less than 20 seconds, Scott Kim created an inversion that read “Scott Flansburg” in one direction and read “The Human Calculator” when rotated 180°.
Wow. Unbelievable.
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• 1. Brent | September 6, 2010 at 5:54 pm
I was a student at that MATHCOUNTS National Competition, and I still vividly remember Scott Kim’s presentation!
• 2. venneblock | September 6, 2010 at 11:46 pm
Validation!
• 3. Carnival of Mathematics #72 « 360 | December 4, 2010 at 9:32 am
[…] symmetry. Inversions also have a cool kind of symmetry, and are explored by Patrick Vennebush in Inversions « Math Jokes 4 Mathy Folks posted at Math Jokes 4 Mathy […]
The Math Jokes 4 Mathy Folks blog is an online extension to the book Math Jokes 4 Mathy Folks. The blog contains jokes submitted by readers, new jokes discovered by the author, details about speaking appearances and workshops, and other random bits of information that might be interesting to the strange folks who like math jokes.
## MJ4MF (offline version)
Math Jokes 4 Mathy Folks is available from Amazon, Borders, Barnes & Noble, NCTM, Robert D. Reed Publishers, and other purveyors of exceptional literature.
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# Alg 2
posted by .
What is the sum of a 12-term arithmetic sequence where the last term is 13 and the common difference is -10?
• Alg 2 -
Schuyler, the sum of the 12 term aritmetic progression is 816
• Alg 2 -
Schuyler, in order to solve your problem I had to turn it around (Gloria Estefan?). You said the last term was 13...I made it the first term. You said the common difference was a -10...I made it a +10.
Now I had to find the last term so I used this formula: L=a+(n-1)d so 13+(12-1)10
L=123
Then I used this formula:
Sn=n/2(a+L)
Sn=12/2(13+123)
Sn=6 Times 136
Sn=816
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https://www.khanacademy.org/math/in-in-grade-11-ncert/in-in-class11-limits/copy-of-limits-of-piecewise-functions-ab/v/limit-of-piecewise-function-that-is-defined
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# Worked example: point where a function is continuous
AP Calc: LIM‑2 (EU), LIM‑2.A (LO), LIM‑2.A.2 (EK)
## Video transcript
- [Voiceover] So we have g of x being defined as the log of 3x when zero is less than x is less than three and four minus x times the log of nine when x is greater than or equal to three. So based on this definition of g of x, we want to find the limit of g of x as x approaches three, and once again, this three is right at the interface between these two clauses or these two cases. We go to this first case when x is between zero and three, when it's greater than zero and less than three, and then at three, we hit this case. So in order to find the limit, we want to find the limit from the left hand side which will have us dealing with this situation 'cause if we're less than three we're in this clause, and we also want to find a limit from the right hand side which would put us in this clause right over here, and then if both of those limits exist and if they are the same, then that is going to be the limit of this, so let's do that. So let me first go from the left hand side. So the limit as x approaches three from values less than three, so we're gonna approach from the left of g of x, well, this is equivalent to saying this is the limit as x approaches three from the negative side. When x is less than three, which is what's happening here, we're approaching three from the left, we're in this clause right over here. So we're gonna be operating right over there. That is what g of x is when we are less than three. So log of 3x, and since this function right over here is defined and continuous over the interval we care about, it's defined continuous for all x's greater than zero, we can just substitute three in here to see what it would be approaching. So this would be equal to log of three times three, or logarithm of nine, and once again when people just write log here within writing the base, it's implied that it is 10 right over here. So this is log base 10. That's just a good thing to know that sometimes gets missed a little bit. All right, now let's think about the other case. Let's think about the situation where we are approaching three from the right hand side, from values greater than three. Well, we are now going to be in this scenario right over there, so this is going to be equal to the limit as x approaches three from the positive direction, from the right hand side of, well g of x is in this clause when we are greater than three, so four minus x times log of nine, and this looks like some type of a logarithm expression at first until you realize that log of nine is just a constant, log base 10 of nine is gonna be some number close to one. This expression would actually define a line. For x greater than or equal to three, g of x is just a line even though it looks a little bit complicated. And so this is actually defined for all real numbers, and it's also continuous for any x that you put into it. So to find this limit, to think about what is this expression approaching as we approach three from the positive direction, well we can just evaluate a three. So it's going to be four minus three times log of nine, well that's just one, so that's equal to log base 10 of nine. So the limit from the left equals the limit from the right. They're both log nine, so the answer here is log log of nine, and we are done.
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Home > co-ordinate geometry, maths > Cartesian Equation of a Circle
Cartesian Equation of a Circle
Since a circle is made up of all the points a fixed distance (its radius) from a given point (its centre) then the equation of a circle simply needs to ensure this is true. This can be done using Pythagoras’s theorem. This is because we can draw a right angled triangle with the centre of the circle at one corner and the point on the circle at the opposite corner as shown below. The radius is then the hypotenuse, the vertical side is the difference between the y co-ordinate of the point and that of the centre and the horizontal side is the difference between the x co-ordinate of the point and centre. From Pythagoras we therefore know that a circle of radius r and centre (a,b) must have a Cartesian equation
r2 = (x-a)2 + (y-b)2
Circle on Cartesian axis
However, we can expand these brackets out to get
r2 = x2 – 2ax + a2 + y2 – 2by + b2
but since a2+b2+r2, -a and -b are all constant we can let
c = a2+b2-r2,
g = -a
f = -b
to get
x2 + y2 + 2gx + 2fy + c = 0
where the circle has a centre (-g,-f) and radius √(a2+b2-c2)
1. July 11, 2009 at 4:37 am
Slight error,
x2 + y2 + 2gx + 2fy + c = 0 should be
x2 + y2 + 2fx + 2gy + c = 0 otherwise it would expand into
r2 = x2 – 2bx + a2 + y2 – 2ay + b2 instead of
r2 = x2 – 2ax + a2 + y2 – 2by + b2
Otherwise, thanks!
• July 16, 2009 at 4:06 pm
well done for noticing the mistake and thanks for telling me,
i changed
f=-a and g=-b
to
g=-a and f=-b
instead to keep the final result the same
thanks again for the comment and pointing out the mistake,
david
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Subsections
# MA 1024A Lab 3: Surfaces in Cartesian Coordinates
## Purpose
The purpose of this lab is to introduce you to some of the Maple commands that can be used to plot surfaces in three dimensions.
## Getting Started
To assist you, there is a worksheet associated with this lab that contains examples and even solutions to some of the exercises. You can copy that worksheet to your home directory with the following command, which must be run in a terminal window, not in Maple.
cp ~bfarr/Surf_start.mws ~
You can copy the worksheet now, but you should read through the lab before you load it into Maple. Once you have read to the exercises, start up Maple, load the worksheet Surf_start.mws, and go through it carefully. Then you can start working on the exercises.
## Background
The graph of a function of a single real variable is a set of points in the plane. Typically, the graph of such a function is a curve. For functions of two variables in Cartesian coordinates, the graph is a set of points in three-dimensional space. For this reason, visualizing functions of two variables is usually more difficult. For students, it is usually even more difficult if the surface is described in terms of polar or spherical coordinates.
One of the most valuable services provided by computer software such as Maple is that it allows us to produce intricate graphs with a minimum of effort on our part. This becomes especially apparent when it comes to functions of two variables, because there are many more computations required to produce one graph, yet Maple performs all these computations with only a little guidance from the user.
The simplest way of describing a surface in Cartesian coordinates is as the graph of a function over a domain, e.g. a set of points in the plane. The domain can have any shape, but a rectangular one is the easiest to deal with.
Another common, but more difficult way of describing a surface is as the graph of an equation , where is a constant. In this case, we say the surface is defined implicitly.
## Exercises
1. Plot the graphs of the following functions over the given domains. Use the plot3d command.
1. , for and .
2. , for and .
3. over the interior of the ellipse .
2. Use the implicitplot3d command to plot the graphs of the following equations. You should come up with plot ranges that show the surfaces clearly.
1. The cylinder .
2. . Can you explain why the surface only exists for ?
3. Consider the equation
In the book, it says that such an equation can be reduced by rotation and translation to one of the two forms
or
Use the implicitplot3d command to graph the surface corresponding to the equation given above. You should be able to identify the graph as one of the types of graphs, i.e. paraboloid, hyperboloid, or ellipsoid, shown in the text. Can you use your graph to determine which of the two forms shown above the equation can be transformed into? Do not try to do the transformation - you don't need to to answer the question.
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# IBPS Clerk 2019 | Profit And Loss | Maths | 01
1. A person incurs 5% loss by selling a watch for Rs. 1140. At what price should the watch be sold to earn 5% profit?
[a] 250% of 900 - 740
[b] 250% of 250 - 940
[c] 250% of 800 - 740
[d] 250% of 600 - 940
[e] None of these
CP = SP/profit or loss * 100
CP = 1140/95*100 = 1200
SP = CP * profit or loss /100
SP = 1200 * 105/100 = 1260
Option [a] 90 x 25 - 740 = 1510
Option [b] 25 x 25 - 940 = -315
Option [c] 80 x 25 - 740 = 1260
Same Pattern
1. If a watch is sold at Rs. 3864 for a loss of 8%, what should be the sale price to earn 8% profit? Ans - 4536
2. Seema sold an item for Rs 220 at a loss of 12%. By how much should she have raised the price to make a profit of 10%? Ans - 275
3. A person incurs a loss of 15% and will be selling a watch for Rs 1140. At what price should the watch be sold to earn 15% profit? Ans - 1542.32
2. Amit buys 5 watches for rs. 9450 and later sells them for Rs. 9700. How much profit does Amit make per watch?
[a] 16(2/3)% of 600 - 20% of 250
[b] 16(2/3)% of 600 + 20% of 250
[c] 16(2/3)% of 120 - 8% of 250
[d] 16(2/3)% of 800 - 30% of 250
[e] None of these
Profit = SP - CP
Profit = 9700 - 9450 = 250 on 5 watches
Profit on 1 watch = 250/5 = 50
Option [a] - (50/3)% of 600 - 20% of 250
100 - 50 = 50
3. A sells his house worth Rs. 10 lakh to B at a loss of 10%. Later B sells it back to A at 10% profit. The result of the two transaction is:
[a] A neither loses nor gains
[b] A loses Rs. 90000
[c] A loses Rs. 200000
[d] B gains Rs. 1,10,000
[e] None of these
A Sells 10 Lakh worth house to B at a loss of 10% = 9,00,000
B sells it back to A at 10% profit = 9,90,000
Results
A loses 1,10000/ and B gains 90000/-
4. A cost of an article including the sales tax is Rs. 616. The rate of sales tax is 10%, If the shopkeeper has made a profit of 12%, then the cost price of the article for shopkeeper is?
[a] Rs. 350
[b] Rs. 500
[c] Rs. 650
[d] Rs. 800
[e] None of these
CP without sales Tax = SP/ST * 100
CP = 616/110 * 100 = 560
CP to gain 12% profit = 560/112 * 100 = 500
Same Pattern
The sales price of an article including the sales tax is Rs. 1232. the rate of sales tax is 10%. If the shopkeeper has made a profit of 12%, then the cost price of the article is Ans 1000
5. Two third of consignment was sold at a profit of 5% and the reminder at a loss of 2%. If the total profit was Rs. 400, the value of the consignment was?
[a] Rs. 12000
[b] Rs. 14000
[c] Rs. 15000
[d] Rs. 16000
[e] None of these
Let Consignment = 300
2/3rd of Consignment = 200 x 5/100 = 10 profit
remaining of consignment = 100 x 2/100 = 2 loss
actual profit = 10 - 2 = 8
If 8 is profit then Consignment = 300
If 400 is profit then Consignment = 300 * 400 /8 = 15000
6. A person earns 15% on investment but loses 10% on another investment. If the ratio of the two investments be 3:5, what is the gain or loss on the tow investments taken together?
[a] 0.625%
[b] 0.8%
[c] 0.9%
[d] 1.2%
[e] 1.45%
Let two investments
300 : 500
15% : 10%
45 : 50
over all loss = 5
5/800 * 100 = 0.625%
7. The profit earned by selling an article for Rs. 900 is double the loss incurred when same article is sold for Rs. 450. At what price should the article be sold to make 25% profit?
[a] Rs. 400
[b] Rs. 500
[c] Rs. 700
[d] Rs. 750
[e] Rs. 900
The difference between Rs 900 & Rs 450 = Rs 450 represents the gap in loss covered by increase in selling price.
Since the ratio of loss:profit = 1:2, the loss portion out of Rs 450 is 450 X 1/3 = Rs 150.
=> Cost Price= SP + Loss = 450 + 150 = 600
To earn 25% profit, SP should be 600 X 125/100 = Rs 750.
8. Nutan bought 30 dozens of oranges for her juice stall in the school fair. She paid Rs. 8 per dozen of oranges. She also had to pay Rs. 500 as the stall fee to the school authorities. She calculated that each glass of juice would need 3 oranges. How much should she charge per glass of juice so as to ake 20% profit?
[a] Rs. 7.20
[b] Rs. 7.40
[c] Rs. 7.60
[d] Rs. 7.80
[e] None of these
CP = 8 x 30 + 500 Stall fee = 740
Total Oranges = 30 x 12 = 360
3 oranges/glass juice = 360/3 = 120
SP to gain 20% = 740 x 120/100 = 888
Per glass cost = 888/120 = 7.40
9. Sarita sells a phone at a profit of 20%. If she had bought it at 20% less and sold it for Rs. 180 less, she would have gained 25%. Find the cost pirce of the Phone.
[a] Rs. 800
[b] Rs. 850
[c] Rs. 900
[d] Rs. 1000
[e] None of these
Let CP = 100% (i) sell at 20% profit then CP = 120%
(ii) If bought less 100% - 20% = 80%,
would gained 25% = 25% of 80% = 20% + 80% = 100%
Difference of (i) and (ii) CP = 20%
If 20% is Rs. 180 then
CP of Phone will be 900
10. Abhishek makes a profit of Rs. 110, if he sells a certain number of pens he has at the price of Rs. 2.5 per pen and incurs a loss of Rs. 55, if he sells the same number of pens for Rs. 1.75 per pen. How many pens does Abhishek have?
[a] 220
[b] 240
[c] 20
[d] Cannot be Determined
[e] None of these
Let Abhishek has x pens (2.5)x - 110 = (1.75)x + 55 (.75)x = 165 x = 165/.75 = 220
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# Plugging In: Using Good Numbers
The larger of two negative consecutive even integers $$2t$$ and $$2(t-1)$$ is multiplied by 3 and then added to the smaller of the two original integers. Which of the following represents the result of this operation?
Incorrect. __Plug In__ a good number such as $$t=-3$$. The negative numbers in the problem become $$2(-3)=-6$$ and $$2(-3-1)=-8$$. The larger of the two is -6. Just work the problem with the number you chose until you get a numerical answer $$(-6)×3+(-8)=-26$$. That's your target. Be sure to check all five answer choices before you take your pick. This answer choice is nothing near your target. __POE__ and move on.
Incorrect. __Plug In__ a few good numbers to see that for each number you choose, you get a different *goal* (without changing, of course, the answer choice that is always correct). __POE__ and move on.
Incorrect. The correct answer must depend on the value of $$t$$ and cannot be a single number such as 3. To see that this is the case, __Plug In__ a few good numbers. For each number you choose, you will get a different *goal*. __POE__ and move on.
Did you __Plug In__ $$t=-2$$?
Be sure to __check all five answer choices__ before you pick the one choice that matches your target. Try to not use numbers that appear in the question or in the answer choices since these cause multiple choices matching your *goal*.
How did you solve this problem?
While algebra may have worked this time, it may not work the next time. The level of difficulty of that problem is no higher than medium. If you solved it algebraically, you should be able to solve it using __Plugging In__ in the same amount of time. If this is not the case, it is because you are not proficient enough with __Plugging In__. Be sure to __Plug In__ more often so that you'll be able to use it in harder questions as well.
That's great. Identifying a __Plugging In__ problem is the first step. Solving is the second. Be sure to use __Plugging In__ as often as you can so that you will gain the experience needed to solve harder questions effectively.
Incorrect. __Plug In__ a good number such as $$t=-3$$. The negative numbers in the problem become $$2(-3)=-6$$ and $$2(-3-1)=-8$$. The larger of the two is -6. Just work the problem with the number you chose until you get a numerical answer $$(-6)\times 3+(-8)=-26$$. That's your target. Be sure to check all five answer choices before you take your pick. This answer choice is nothing near your target. __POE__ and move on.
Correct. __Plug In__ a good number such as $$t=-3$$. The negative numbers in the problem become $$2(-3)=-6$$ and $$2(-3-1)=-8$$. The larger of the two is -6. Just work the problem with the number you chose until you get a numerical answer $$(-6)\times 3+(-8)=-26$$. That's your *goal*. Be sure to check all five answer choices before you take your pick. All the other answer choices are eliminated for this Plug-In, so this is the right answer choice.
$$-2$$
$$3$$
$$6t-2$$
$$8t-2$$
$$-2-4t^2$$
Yes, I did.
No, I didn't.
I used algebra.
I used __Plugging In__.
Continue
Continue
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From
Equation 19 we have the formula that the resisting moment at any point in the beam equals the area of the steel, times the unit tensile stress in the steel, times the distance from the steel to the centroid of compression of the steel, which is the distance d - x. We may compute the moment in the beam at two points at a unit-distance apart. The area of the steel is the same in each equation, and d - x is substantially the same in each case; and therefore the difference of moment, divided by (d - x), will evidently equal the difference in the unit-stress in the steel, times the area of the steel. To express this in an equation, we may say, denoting the difference in the moment by d M, and the difference in the unit-stress in the steel by d s: dM = A x ds.
(d-x)
But A X d s is evidently equal to the actual difference in tension in the steel, measured in pounds. It is the amount of tension which must be transferred to the concrete in that unit-length of the beam. But the computation of the difference of moments at two sections that are only a unit-distance apart, is a comparatively tedious operation, which, fortunately, is unnecessary. Theoretical mechanics teaches us that the difference in the moment at two consecutive sections of the beam is measured by the total vertical shear in the beam at that point. The shear is very easily and readily computable; and therefore the required amount of tension to be transferred from the steel to the concrete can readily be computed. A numerical illustration may be given as follows: Suppose that we have a beam which, with its load, weighs 20,000 pounds, on a span of 20 feet. Using ultimate values, for which we multiply the loading by 4, we have an ultimate loading of 80,000 pounds. Therefore,
Mo = Wo l = 80,000 x 240 = 2,400, 000
Using the constants previously chosen for 1:3:5 concrete, and therefore utilizing Equation 23, we have this moment equal to 397 b d2. Therefore b d2 = 6,045.
If we assume 6 = 15 inches; d = 20.1 inches; then d - x = .86d = 17.3 inches. The area of steel equals:
A = .0084 b d = 2.53 square inches. We know from the laws of mechanics, that the moment diagram for a beam which is uniformly loaded is a parabola, and that the ordinate to this curve at a point one inch from the abutment will, in the above case, equal (119/120 )2 of the ordinate at the abutment. This ordinate is measured by the maximum moment at the center, multiplied by the factor (119/120)2 = 14,161 = .9834; therefore the actual moment
14,400 at a point one inch from the abutment = (1.00 - .9834) = .0166 of the moment at the center. But .0166 X 2,400,000 = 39,840.
But our ultimate loading being 80,000 pounds, we know that the shear at a point in the middle of this one-inch length equals the shear at the abutment, minus the load on this first 1/2 inch, which is 1/240 of 40,000 (or 167) pounds. The shear at this point is therefore 40,000 - 167 (or 39,833) pounds. This agrees with the above value 39,840, as closely as the decimals used in our calculations will permit.
The value of d - x is somewhat larger when the moment is very small than when it is at its ultimate value. But the difference is comparatively small, is on the safe side, and it need not make any material difference in our calculations. Therefore, dividing 39,840 by 17.3, we have 2,303 pounds as the difference in tension in the steel in the last inch at the abutment. Of course this does not literally mean the last inch in the length of the beam, since, if the net span were 20 feet, the actual length of the beam would be considerably greater. The area of the steel as computed above is 2.53 square inches. Assuming that this is furnished by five 3/4-inch square bars, the surfaces of these five bars per inch of length equals 15 square inches. Dividing 2,303 by 15, we have 153 pounds per square inch as the required adhesion between the steel and the concrete. While this is not greater than the adhesion usually found between concrete and steel, it is somewhat risky to depend on this; and therefore the bars are usually bent so that they run diagonally upward, and thus furnish a very great increase in the strength of the beam, which prevents the beam from failing at the ends. Tests have shown that beams which are reinforced by bars only running through the lower part of the beam without being turned up, or without using any stirrups, will usually fail at the ends, long before the transverse moment, which they possess at their center, has been fully developed.
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# Using only the numbers 3, 3, 3 and 3 once and + - * / once find 7?
1. Mar 14, 2013
### Natasha1
Hi,
At the end of our lecture today, the lecturer gave us this simple yet impossible puzzle.
My friend and I have tried to find the answer but in vain...
Using only the numbers 3, 3, 3 and 3 once and using only the four arithmetic + - * / once can you make the number 7.
The closest I have got is 6 or 8 but not 7.
3*3 all / 3 and then + 3 = 6
or
3*3 then - 3/3 = 8
How to find 7?? Is this actually possible?
2. Mar 14, 2013
### phyzguy
3 + 3 + 3/3 = 7
3. Mar 14, 2013
### Staff: Mentor
think of it in terms of combinatorics with the operators:
3 op1 3 op2 3 op3 3
so there's 4 choices for the first, 3 for the third, 2 for the last one = 24 choices
Phyzguy's solution is almost correct except that he repeated the + operator and the problem says to use each operation once.
3+3-3*3=-3
3+3-3/3=5
3+3*3-3=9
3+3*3/3=6
3+3/3-3=1
3+3/3*3=3.33333
3-3+3*3=9
3-3+3/3=1
3-3*3-3=-9
3-3*3/3=0
3-3/3+3=5
3-3/3*3
...
4. Mar 14, 2013
### Natasha1
Can't do 3 + 3 + 3/3 = 7 as you are using + twice
5. Mar 14, 2013
### Natasha1
Are you saying it's therefore impossible?
6. Mar 14, 2013
### Staff: Mentor
I can't tell you the answer only how to think about the problem as it was assigned by your prof.
7. Mar 14, 2013
### Natasha1
Pity.
8. Mar 14, 2013
### Staff: Mentor
You can't finish the other 12 choices to complete the proof?
9. Mar 14, 2013
### maltman
are you sure the prof or you have the correct problem?
a classic is using 5 "3" and all the operators
(3*3 + 3)/3 + 3 = 7
10. Mar 14, 2013
### Natasha1
But then you are using + twice too
11. Mar 14, 2013
### rcgldr
3 and 7 are prime numbers. Any combination of operations you try (except for 3/3) will be a multiple of 3. If you use 3/3 = 1, then you'd have to add or subtract the 3/3 to something else, since multiplying would result in a multiple of 3 again, and dividing would result in a fraction.
The problem doesn't state if you're allowed to use parenthesis to group operations.
If you're suppose to used + - * / excactly one, that's four operators, so you'd need five 3's.
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Uncategorized
# What Is Just Really a Proper Triangle in Z?
What’s just a correct triangle in math? It’s a triangle.
Z is an query into abstract, concrete, or fanciful objects or”mathematical” items. It is also a process of applying and knowing principles in order to address problems.
Triangles is one of things that many mathematicians think of. We are aware that triangles have three angles, so thus customer writing we can find it has to be something’s form. We know very well what a triangle iswe can determine how exactly to reflect it.
You should begin by thinking regarding its contours if you should be enthusiastic about triangles. Can you have to include things like the side of the triangle? Or could it be if you just make sure it is three vertices?
You may ask your self, how can we find out everything is a triangle in mathematics. In order to figure out a triangle, first we have http://en.wikipedia.com/wiki/Sigiriya to understand what it resembles. This really is a task that is harder because there are many unique shapes that could be created with all those 3 angles. Every shape has its own structures and rules.
You may possibly consider just how to relate triangles to each of additional contours. As the exact length between your Valve’s vertices is equal to 3 times its span. You also need to know that each and every triangle should involve.
It is simple because the centre is too wide since the triangle, to assume a shape. That would be considered a triangle. sameday essay You can figure out the the negative lengths are corresponding to the span of this rectangle when you think of a triangle the shape of the parallelogram. The diameter of the rectangle is also still.
It is not difficult to see that its edges are always going to be a little briefer than those sides After you believe of a triangle. It makes it seem like pointed. It is really a wonderful form, however there is a pointed triangle not just a good contour for your puzzle. Triangles don’t seem to produce a satisfying outcome.
You still also have to consider its contours, to figure out why a rectangle is a triangle. Learn what angles these shapes have and how they relate to each other. It’s simpler to figure out the design of a triangle once you have discovered its shape all.
A triangle in math could be. This really can be a fantastic shape because of the fact that the hypotenuse is always equal to the either side. Since you can observe in this picture the shape of a triangle is like a sphere. Each aspect will probably always be equal for the size of the entire sphere.
Now that you know just a small bit about triangles, then you’ll find it more easy to understand a triangle is a superior shape to take into account whenever you attempt to come up with a thing which the triangle inequality is called by mathematicians. The triangle inequality is.
One particular final thing you want to find out concerning triangles is that triangles might be divided in four different guidelines. When you attempt to figure the problem, that could be the most easy shape you may see to get a triangle. Now discover much more about triangles when you want to remedy a mathematics problem and consider this next time.
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# How can I estimate the probability of a random variable from one population being greater than all other random variables from unique populations?
Lets assume I have samples from 5 unique populations. Let's also assume I have a mean and standard deviation from each of these populations, they are normally distributed and completely independent of one another.
How can I estimate the probability that a sample of one of the populations will be greater than a sample from each of the other 4 populations?
For a example, if I have 5 types of fish (the populations) in my pond, such as bass, catfish, karp, perch and bluegill, and i'm measuring the lengths (the variables) of the fish, how do can I estimate the probability that the length of a bass I catch will be greater than the length of all the other types of fish? I think I understand how to compare 2 individual populations but can't seem to figure out how to estimate probability relative to all populations. As opposed to the probability of the bass to a catfish, and then a bass to a karp, etc., I'd like to know if its possible to reasonably estimate the probability of the length of the bass being greater that the lengths of all other populations.
Any help would be greatly appreciated! Thanks!
Edit: I believe my original solution is incorrect. I treated the events [koi > catfish] and [coy > karp] as independent when they are certainly not.
\begin{aligned} P(Y>\max\{X_1,...,X_n\})&=P(Y>X_1,...,Y>X_n)\\ &=\int_{-\infty}^{\infty} P(Y>X_1,...,Y>X_n|Y=y) f_Y(y)dy\\ &=\int_{-\infty}^{\infty} \prod_{i=1}^n \left[ P(Y>X_i|Y=y) \right]f_Y(y)dy\\ &=\int_{-\infty}^{\infty} \prod_{i=1}^n \left[ \Phi \left( \tfrac{y-\bar{x}_n}{\sigma_{x_n}} \right) \right]f_Y(y)dy \end{aligned}
I do hope that someone can provide a better solution, as the above expression seems mismatched with the relative simplicity of the question.
Let $$Y$$ represent the length of a fish from the population of interest, such as bass, and $$X_i$$ represent the length of fish from another population $$i$$, such as karp or catfish. You want to calculate the probability that the bass is longer than the longest non-bass fish. That is equivalent to the probability that the bass is longer than the carp, and the bass is longer than the catfish, and the bass is longer than the perch, etc. $$P(Y>\max\{X_1,...,X_n\})=P(Y>X_1,...,Y>X_n)$$
Because the lengths of your fish are independently distributed, the probability of all of these events happening is the product of the individual probabilities.
$$P(Y>X_1,...,Y>X_n) =\prod_{i=1}^{n} P(Y>X_i)$$
So the probability that bass is longer than all of your other fish is found by multiplying the probabilities that the bass is larger than each other type of fish.
That leaves only the problem of calculating the probability that a fish from one normal distribution is longer than a fish from another normal distribution. That is, $$P(Y>X_i)$$. To calculate this probability we rewrite it (ignoring the subscript) in the form $$P(Y>X)=P(Y-X>0)$$
Thankfully, the distribution of $$Y-X$$ is simple in the case where $$X$$ and $$Y$$ are normally distributed. That is, $$X \sim N(\mu_{X},\sigma_{X})$$ and $$Y \sim N(\mu_{Y},\sigma_{Y})$$. We can use the following facts:
• Any linear combination of independent normal random variables (ie. $$aX+bY$$) is itself a normal random variable.
• $$\mathbb{V}(aX+bY)=a^2\mathbb{V}(X)+b^2\mathbb{V}(Y)$$ for any uncorrelated random variables $$X$$ and $$Y$$.
• $$\mathbb{E}(aX+bY) = a\mathbb{E}(X)+b\mathbb{E}(Y)$$ for any random variables $$X$$ and $$Y$$.
In this problem, the difference in the lengths of the two fish $$D=Y-X=(1)X+(-1)Y$$ is a linear combination of the two lengths, $$X$$ and $$Y$$. Therefore, using the facts above, we find that the distribution of the difference in lengths is
$$D\sim N(\mu_Y-\mu_X,\sigma^2_X+\sigma^2_Y)$$
The probability that this difference is greater than zero is
$$P(D>0)=1-P(D<0)=1-F_D(0)=1-\Phi \left(\frac{0-\mu_D}{\sigma_D} \right)$$
In terms of $$X$$ and $$Y$$ this is
$$P(Y-X>0)=1-\Phi \left(\frac{\mu_X-\mu_Y}{\sqrt{\sigma^2_X+\sigma^2_Y}}\right)$$
The final solution, in all its glory, would then be:
$$P(Y>\max\{X_1,...,X_n\})=\prod_{i=1}^{n} 1-\Phi \left(\frac{\mu_{X_i}-\mu_Y}{\sqrt{\sigma^2_{X_i}+\sigma^2_Y}}\right)$$
• Presumably your operator "$\cap$" means ordinary multiplication of numbers, because both its arguments (being probabilities) are numbers. Maybe there's a typo there? "This extends to" hides the content of the answer--it needs elaboration. The meaning of "alternatively" is not evident and so needs elaboration, too.
– whuber
Commented Jul 2, 2020 at 20:05
• Thanks @whuber. Hopefully, the edited answer is clearer. Commented Jul 2, 2020 at 20:45
• It is, thank you (+1). I can't help thinking, though, that the OP might welcome some words about how the individual probabilities $P(Y\gt X_i)$ might be estimated or calculated.
– whuber
Commented Jul 2, 2020 at 20:59
• One thing i'm struggling to understand, is after I find the product that the bass is larger than the karp, the catfish, etc., I do the same for each fish (the karp being larger than all others, the catfish being larger than all others, etc.). Wouldn't the sum of the probabilities of each fish being larger than all others be equal to 1? i'm not getting anywhere close to that, maybe i'm not understanding why it wouldn't equal 1? Surely one of the fish will be larger than all others? I can provide numbers and show what i'm coming up with if that helps. Commented Jul 9, 2020 at 15:43
• @mc_chief Thank you for that excellent observation. My answer is very likely mistaken. I believe I treat the case where [koi > catfish] and [coy > karp] are independent events. In reality, they are not. I'll correct this in a new answer ASAP. Commented Jul 9, 2020 at 17:31
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## Tuesday, October 28, 2014
### Introducing Fractions in Fifth Grade
After tomorrow, my students will be on their Fall Break from school until November 5th! We just finished Chapter 3 in math and I didn't want to start the next one with so many days off coming up. So instead, I did an mini-Investigations lesson.
I asked students the following question:
James came home from school to find 3 fresh brownies. He had 3 friends with him. The 4 friends decided to share the brownies equally. How much of each whole brownie did each of them get to eat?
I broke the students up into small groups and gave them chart paper and markers. I told students that it was their job to not only find the answer, but to also be able to explain how they know that their answers are correct.
It turned out really well! I was happy that they all did not come up with the same strategies and there was great conversation around fractions, division and equal parts!
Here is what they came up with:
This group decided to draw the three brownies, then split two in half to divide among all 4 boys (each getting half a brownie). They took the final brownie and broke it up into 4 equal parts and gave each piece to the boys. They then added 1/2 plus 1/4 (mentally) to say that each boy ate 3/4 of a brownie.
This group did something similar. They drew out the three brownies and then broke each brownie into four parts so that they could divide the pieces equally among the four boys. There was great conversation around what the final answer would be to the question: How much of each brownie did everyone eat? At first, the group answered 1/4, but after discussing it with the rest of the class, they realized they had to add up each of the fourths that the boys ate.
This group drew the three brownies, broke them up into fourths then distributed to each boy.
This group also drew the three brownies and then split them into fourths. They said they chose fourths because they knew that 12 can be divided evenly by 4.
I love Investigations lessons because it allows students to learn from each other. It was great to watch them work together in a group and then explain their findings and rationale with the rest of the class. Great way to introduce fractions and get a sense of background knowledge of each student!
## Monday, October 27, 2014
### Integrating Reading Interactive Notebooks and Literature Cirlces
At the beginning of every school year, I start out with small group reading (guided reading) using two Jerry Spinelli books. This year I am integrating the use of Interactive Reading Notebooks and Small Group Reading.
The structure of the group is as follows:
First, we discuss what was read prior to meeting, focusing on connections made, predictions, questions and whatever else might come up in conversation.
Next, I focus on a literary element such as plot, characterization or story devices. I do this by creating an interactive reading notebook with the students.
I feel like being able to do a reading interactive notebook with students in a small group is much more effective than with a whole class. As I am teaching about the element, for example, plot, I will discuss it using a story the entire class has read. Usually a read aloud that we have already finished. Then, the students put the notebook together and complete the activity using the book they are reading in the small group. This also becomes are a really great way to assess two things: Do they understand this literary element? Do they understand what they've been reading on their own?
Starting next quarter, the class will be moving away from small group reading and into a whole class literature circle. My hope is to refer back to the reading interactive notebook during literature circles and create activities and "during reading" work with it.
## Monday, October 13, 2014
### Buddhist Festivals and Holidays - Diorama Style!
I love assigning dioramas to students! The creativity and care that they put into their displays always amaze me. Every year. Every time I assign them. I only get to assign them twice a year (I love to mix things up, so the same project all the time would ruin anything that's really cool!).
Their assignment was to make a diorama that depicts a Buddhist holiday or festival. They had to also include an essay telling about the festival, which country celebrated it and how. I was fascinated by what they came up with! Pictures below :)
The Tooth Festival (Sri Lanka)
The Elephant Festival (Taiwan)
The Tibetan Butter Lamp Festival
Buddhist New Year (Tibet)
The Elephant Festival (India)
The Festival of the Tooth (Sri Lanka)
The Festival if the Tooth (Sri Lanka)
The Festival of the Tooth (Sri Lanka)
The Buddhist New Year - Purification Ceremony (Thailand)
The Buddhist New Year - Purification Ceremony (India)
Wesak - Buddha's Birthday and Death Day Celebration
New Year Festival (Tibet)
The Tooth Festival (Sri Lanka)
### Trip to the Buddhist Monastery
Last week, I had the privilege of taking my fifth graders to visit a Buddhist monastery only a short drive away in Bolivia, NC. We've been studying each of the five major religions this school year and with each religion, we travel to a different "learning spot" in our area.
This trip was excellent because students got a chance to see Buddhist monks in their living space and we were even given a lesson on The Eightfold Path and Rebirth. Since we learned that monks do not use money and are not allowed to ask for anything, we offered some gifts as a thank you for having us. These gifts included some canned goods, bread, toothbrush and toothpaste. I believe this experience to be one students will never forget.
## Wednesday, October 1, 2014
### Teach Discovery Writing (A Poetry Writing Lesson)
In Writers' Workshop, we started our poetry unit this week! Today I taught students how to do Discovery Writing. This is a technique where writers take a regular object and describe it by comparing it to other objects, animals, places, etc.
To introduce the unit, I used Nan Fry's poem, "Apple". I showed my class an apple cut in half so that a star shows up in the middle and then I read the poem out loud. Afterwards, I asked students to point out where Nan used other objects to describe parts of the apple.
Next, I shared the "Discovery Writing" anchor chart to explain the steps of how to write about regular objects. I modeled with a pencil as my regular object to describe.
This is what I wrote:
Pencil
• bunny's nose
• green grass surrounding him
• long fence
• point to keep the other animals out
Pink bunny nose surrounded by green grass
The long fence stretches to the point
that keeps other animals out
I handed out an object from the bag to each student (I just went around the classroom beforehand and grabbed a bunch of things) and had them study the object while thinking about other things to compare it to.
They turned and talked with a partner about their ideas and then went to their writing spot to work on a poem about their object! This ended up being a lesson that the students were really excited about. After most were finished with their first object, they were eager to come back to the bag and switch it out to write about another one! I loved the enthusiasm of describing everyday things!
This lesson is included in my Poetry Lesson Bundle in my Teachers Pay Teachers store! Click Here
09 10
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# Mathematics for Technology I (Math 1131)
Durham College, Mathematics
Free
• 55 lessons
• 1 quizzes
• 10 week duration
• ##### Numerical Computation
Here you'll be introduced to the bare basics of mathematics. Topics include commonly used words and phrases, symbols, and how to follow the order of operations.
• ##### Measurements
An introduction to numerical computation. Emphasis is placed on scientific and engineering notation, the rule of significant figures, and converting between SI and Imperial units.
• ##### Trigonometry with Right Triangles
Here we focus on right angle triangles within quadrant I of an x-y plane. None of the angles we evaluate here are greater than 90°. A unit on trigonometry with oblique triangles is covered later.
• ##### Trigonometry with Oblique Triangles
This unit is a continuation of trigonometry with right triangles except we'll extend our understanding to deal with angles *greater* than 90°. Resolving and combining vectors will be covered at the end of this unit.
• ##### Geometry
This unit focuses on analyzing and understand the characteristics of various shapes, both 2D and 3D.
## Mathematics for Technology I (Math 1131)
When a general form quadratic has an a coefficient greater than 1, the trial-and-error method no longer works. Take, for example, the equation:
y = 3x² + 5x + 6
You can’t choose 3 and 2 as factors that multiply to 6 and add to 5 – it doesn’t work that way.
Arguably you could common factor the 3, leaving x² with a coefficient of 1:
y = 3 ( x² + 5/3x + 2 )
But then you’re left with finding two factors of 2 that add to 5/3!
To factor quadratics whose a > 1, we use a technique known as factoring by decomposition, which involving breaking up the middle term – hence the name.
Let’s see a few examples of this technique in action.
To summarize, factoring by decomposition involves finding two integers whose product is a × c and whose sum is b. Then, break up the middle term and factor by grouping.
Interestingly, referring back to the initial equation:
y = 3x² + 5x + 6
If you try factoring by decomposition here, it still won’t yield a factored-form quadratic. In that case, you’d have to use the quadratic formula to find the roots (more on this to come). Therefore, not all quadratic expressions of the form ax² + bx + c can be factored over the integers. The trinomial factorability test is shown below:
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# Linear system solver
This linear system solver helps students learn and practice math skills. These tools can help students learn the concepts they need to understand math and improve their skills.
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For the iterative solver, the calculation step is used to set the initial conditions. For example, when ilut is used to set the initial conditions, the completed decomposition factors L and u are calculated in this step Note: generally, the purpose of setting the initial conditions is to accelerate the iterative convergence, and make the eigenvalues of the matrix more concentrated by improving the linear system. For real linear systems, the iterative solver is always used to set the initial conditions. In eigen, the preset initial conditions are achieved by adding template parameters to the iterative object. For example: The experience of life in class at Peking University, first of all, is the study of the course. I feel that the course content is very rich. One of the major assignments in numerical algebra is to solve a linear system using G-S iteration, and the teacher of this linear system selects the linear system of solving Poisson's equation with the five point difference method. There will also be an introduction to the line Gauss method in the middle Raz's linear systems and signals (version 2) gives a detailed and in-depth explanation of the time-domain analysis of the system. This lecture only introduces a small part of its contents, focusing on the solution of the system response y (T) described by the linear constant coefficient differential equation: The nonlinear dynamic system model is described by a general form of differential equation, and it is difficult to obtain the state trajectory / system output by analytical means (a small number of nonlinear systems can be transformed into linear constant systems by making linear approximation near the working point, and then stability analysis can be carried out by using the stability analysis method of linear systems), Therefore, the constructor (i.e., Lyapunov function) is often used to indirectly judge the state trajectory (i.e., the solution / system dynamics behavior can be judged without solving the differential equation). How to build a linear system how to combine linear systems from bilinear forms? The cyclic fevalues class of all elements is used to solve the linear system, and the solution results are visualized In the first part, the solution steps of linear constant coefficient differential equations and linear constant coefficient difference equations are briefly introduced, especially for linear constant coefficient difference equations. When additional conditions or initial conditions are given, the solution y [n] of the equation can be easily calculated by recursion. The principle of this algorithm is simple, and computers are good at this algorithm. However, it seems that little attention has been paid to the solution of linear constant coefficient differential equations, which is simply unfriendly to domestic college students, especially the postgraduate entrance examination party. The middle part (the middle part mainly refers to Raz's linear systems and signals (version 2)) and the second part (the second part mainly refers to domestic mainstream textbooks) will mainly discuss the system analysis problems described by linear constant coefficient differential equations, involving: This paper presents a matrix free first-order numerical method to solve large-scale cone optimization problems. Solving linear equations is the most computationally challenging part of first-order and second-order numerical algorithms. The existing direct and indirect methods either require a large amount of computation or compromise the accuracy of the solution. In this paper, an easy to calculate decomposition method is proposed to solve the sparse linear system in cone optimization. Its iteration is easy to handle, highly parallelizable, and has a closed form solution. The algorithm can be easily implemented on a distributed platform, such as a graphics processing unit, with an order of magnitude of time improvement. The performance of the proposed solver has been proved in large-scale cone optimization problems, and compared with the most advanced first-order solver.
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## Wednesday, October 08, 2014
### Finding the Least Common Denominator of Three Fractions
The least common denominator of two or more fractions is the smallest number that can be divided evenly by each of the fractions' denominators. You can determine the LCM (least common multiple) by finding multiples of the denominators of the fractions.
Find the least common denominator of the following fractions: 5/12, 7/36, and 3/8.
8, 16, 24, 36
12, 24, 36
36
The least common denominator is 36.
## Tuesday, October 07, 2014
### Least Common Multiple
Find the least common denominator of 6, 8, 12.
6, 12, 18, 24
8, 16, 24
12, 24
The least common multiple is 24.
## Monday, October 06, 2014
### Help With GED Math Problems: Finding Lowest Common Denominator for Fractions
Building the LCD or lowest common denominators for two or more fractions can be challenging. But it is an important skill for knowing how to add and subtract fractions and one that anyone studying their GED math test will need to know.
First step: Take each denominator and factor to product of prime numbers.
Second step: Build the lowest common denominator by using each factor with the greatest exponent.
What is the lowest common denominator for the following fractions: 7/12, 7/15, 19/30? Use the product of prime factor method.
12 = 2 x 2 x 3 or 2^2 x 3
15 = 3 x 5
30 = 2 x 3 x 5
Build the lowest common denominator by using each factor (i.e. 2^2) with the greatest exponents.
If I were demonstrating the concept of building lowest common denominators to students, it would go something like this, " Let's start with the denominator twelve. The denominator 12 needs at least two twos and a three. The denominator fifteen needs a three, but because we have one from the twelve... we do not need to write another one. However, the denominator twelve needs a five, so we need to add a five. The denominator thirty needs a two... which we have so we do not need to add one. It also needs a three and a five, but because we already have both, again we do not need to add. We have now build our LCD and all we need to do is multiply the factors together. So 2 x 2 x 3 x 5 = 60. The LCD of 12, 15, and 30 is 60.
LCD = 2 x 2 x 3 x 5 = 60
## Friday, October 03, 2014
### Lowest Common Denominator
Find the lowest common denominator for the following fractions: 1/2, 1/4, 1/5
2, 4, 6, 8, 10, 12, 14, 16, 18, 20
4, 8, 12, 16, 20
5, 10, 15, 20
Because 20 is the first common multiple of 2, 4, and 5..... it is the lowest common denominator or LCD.
## Wednesday, July 30, 2014
### GED Math Test Prep: Area of Rectangle
GED Skill: Area of rectangles
You have decided to put carpet in your 10ft by 15 ft. living room. What is the area of carpet needed?
Answer: 10ft x 15ft = 150 cubic feet
### GED Math Test Prep: Simplify the equation 3x + 7y - 2z + 3 - 6x - 5z +15
Simplify the following equation.
3x + 7y - 2z + 3 - 6x - 5z +15
Step 1: Using the associative property, rearrange the terms of the equation so that "like" terms are next to each other.
3x - 6x - 2z - 5z + 7y + 3 + 15
Step 2: Combine like terms.
-3x - 7z + 7y + 18
## Monday, May 12, 2014
### Using the Product Rule with Exponents
When you multiply constants (variables) that have the same base, you add the exponents... but keep the base unchanged.
For example:
x^2c · x^3 = x^(2+3) = x^5
(x · x) (x · x · x) = x^5
"X" squared times "X" cubed equals "X" to the fifth power.
Try a few more.
1) p^5 · p^4 =
2) 2t^2 · 3t^4
3) r^2 · 2^3 · r^5
4) 3x^2 · 2x^5 · x^4
5) (p^2)(3p^4)(3p^2)
1) p^9
2) 6t^6
3) 2r^10
4) 6r^11
5) 9p^8
## Thursday, May 08, 2014
### Simplify and Solve Using the Addition Principal of Equality
4 ( 8 - 15) + (-10) = x - 7
4 ( 8 - 15) + (-10) = x - 7
32 - 60 + (-10) = x - 7
-28 + (-10) = x - 7
-38 = x - 7
-38 + 7 = x -7 + 7
-31 = x + 0
-31 = x
## Wednesday, May 07, 2014
### Solving Equations Using the Addition Principle of Equality
Can you find the error in the following problem?
5² + (4 - 8) = x + 15
25 + 4 = x + 15
29 = x + 15
29 + (-15) = x + 15 + (-15)
14 = x + 0
14 = x
## Tuesday, May 06, 2014
### Practice Solving Simple Equations Using the Addition Property of Equality
It is important to practice the addition property of equality. See below and solve five simple equations using the addition property of equality.
Practice Problem #1
x - 11 = 41
Practice Problem #2
x - 17 = -35
Practice Problem #3
84 = 40 + x
Practice Problem #4
45 = -15 + x
Practice Problem #5
-21 = -52 + x
Practice Problem #1
x - 11 = 41
x - 11 + 11 = 41 + 11
x + 0 = 52
x = 52
check
52 - 11 = 41
41 = 41
Practice Problem #2
x - 17 = -35
x - 17 + 17 = -35 + 17
x + 0 = -18
x = -18
check
-18 - 17 = -35
-35 = -35
Practice Problem #3
84 = 40 + x
84 + ( - 40) = 40 + (-40) + x
44 = 0 + x
44 = x
check
84 = 40 + 44
84 = 84
Practice Problem #4
45 = -15 + x
45 + 15 = -15 + 15 + x
60 = 0 + x
60 = x
check
45 = -15 + 60
45 = 45
Practice Problem #5
-21 = -52 + x
-21 + 52 = -52 + 52 + x
31 = 0 + x
31 = x
check
-21 = -52 + 31
-21 = -21
## Monday, May 05, 2014
### Solving Equations Using the Addition Property of Equality
The addition principle of equality states that if a = b, then a + c = b + c.
When you solve equations using this addition principle of equality, you need to use the additive inverse property. In other words, you must add the same number to both sides of an equation.
Example #1:
x - 5 = 10
x - 5 + 5 = 10 + 5 We add the opposite of (-5) to both sides of the equation.
x + 0 = 15 We simplify -5 + 5 = 0.
x = 15 The solution is x = 15
To check the answer, simply substitute 15 in for x, in the original equation and solve.
15 - 5 = 10
10 = 10
Example #2:
x + 12 = -5
x + 12 + (- 12) = -5 + (- 12) We add the opposite of (+12) to both sides of the equation.
x + 0 = -17 We simplify +12 - 12 = 0.
x = -17 The solution is x = -17
(-17) + 12 = -5
-5 = -5
## Friday, April 11, 2014
### Practice Percent Word Problem
A car which is normally priced at \$25,437 is marked down 10%. How much would Karen save if she purchased the car at the sale price?
(Spanish translation coming soon...)
## Tuesday, March 11, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. Twenty-one more than a number is 51. What is the number?
Veinte y uno más que el número es 51. ¿Cuál es el número?
2. Thirty-seven less than a number is 45. Find the number.
Treinta y siete menos que el número es 45. Encuentre el número.
1. 30
2. 82
## Monday, March 10, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. The sum of a number and 50 is 73. Find the number.
La suma del número y 50 es 73. Encuentre el número.
2. Thirty-one more than a number is 69. What is the number?
Treinta y uno más que el número es 69. ¿Cuál es el número?
3. A number decreased by 46 is 20. Find the number.
El número que está reducido por 46 es 20. Encuentre el número.
1. 23
2. 38
3. 66
## Friday, March 07, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. The sum of a number and 28 is 74. Find the number.
La suma del número y 28 es 74. Encuentre el número.
2. Thirty-nine more than a number is 72. What is the number?
Treinta y nueve más que el número es 72. ¿Cuál es el número?
3. Eighteen less than a number is 48. Find the number.
Dieciocho menos que el número es 48. Encuentre el número.
1. 46
2. 33
3. 66
## Thursday, March 06, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. A number increased by 21 is 52. Find the number.
El número que está aumentado por 21 es 52. Encuentre el número.
2. Twenty-five more than a number is 68. What is the number?
Veinte y cinco más que el número es 68. ¿Cuál es el número?
3. Forty-two more than a number is 58. What is the number?
Cuarenta y dos más que el número es 58. ¿Cuál es el número?
1. 31
2. 43
3. 16
## Wednesday, March 05, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. Twenty more than a number is 42. What is the number?
Veinte más que el número es 42. ¿Cuál es el número?
2. Forty-three more than a number is 85. What is the number?
Cuarenta y tres más que el número es 85. ¿Cuál es el número?
3. Twenty-two more than a number is 62. What is the number?
Veinte y dos más que el número es 62. ¿Cuál es el número?
1. 22
2. 42
3. 40
## Tuesday, March 04, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. The sum of a number and 26 is 42. Find the number.
La suma del número y 26 es 42. Encuentre el número.
2. Thirty more than a number is 51. What is the number?
Treinta más que el número es 51. ¿Cuál es el número?
3. Fifteen more than a number is 47. What is the number?
Quince más que el número es 47. ¿Cuál es el número?
1. 16
2. 21
3. 32
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. One-half of a number is 13. Find the number.
Una media de un número es 13. Encuentre el número.
2. A number decreased by 29 is 39. Find the number.
Un número que está reducido por 29 es 39. Encuentre el número.
3. The sum of a number and 39 is 56. Find the number.
La suma del número y 39 es 56. Encuentre el número.
1. 26
2. 68
317
## Tuesday, January 28, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. A number increased by eight is 14. Find the number.
El número que aumenta por ocho es 14. Encuentre el número.
2. Three less than a number is 2. Find the number.
Tres menos que el número es dos. Encuentre el número.
1. 6
2. 5
## Monday, January 27, 2014
### Translating Words Into Algebraic Expressions Examples: (Spanish & English)
1. Six less than a number is 9. Find the number.
Seis menos que el número es nueve. Encuentre el número.
2. Ten less than a number is 9. Find the number.
Diez menos que el número es nueve. Encuentre el número.
3. A number increased by seven is 12. Find the number.
El número que aumenta por siete es 12. Encuentre el número.
1. 15
2. 19
3. 5
## Friday, January 24, 2014
### Easy Tanslating Algebra Word Problems: (Spanish & English)
1. Seven more than a number is 11. What is the number?
Siete más que el número es 11. Encuentre el número.
2. The sum of a number and six is 16. Find the number.
La Suma del número y seis es 16. Encuentre el número.
3. A number diminished by 9 is 3. Find the number.
El número que reduce por nueve es tres. Encuentre el número.
1. 4
2. 10
3. 12
## Thursday, January 23, 2014
### Translating Words into Algebraic Expressions Simple: (Spanish & English)
1. A number diminished by 2 is 7. Find the number.
El número que reduce por dos es siete. Encuentre el número.
2. A number decreased by 7 is 8. Find the number.
El número que reduce por siete es ocho. Encuentre el número.
3. A number increased by three is 13. Find the number.
El número que aumenta por tres es 13. Encuentre el número.
1. 9
2. 15
3. 10
## Wednesday, January 22, 2014
### Translating Simple Number Word Problems: (Spanish & English)
1. Six less than a number is 5. Find the number.
Seis menos que el número es cinco. Encuentre el número.
2. Six less than a number is 7. Find the number.
Seis menos que el número es siete. Encuentre el número.
3. The sum of a number and three is 11. Find the number.
La suma del número y tres es 11. Encuentre el número.
1. 11
2. 13
3. 8
## Tuesday, January 21, 2014
### Translating Word Problems Simple: (Spanish & English)
1. One-third of a number is 1. Find the number.
Un tercer del número es uno. Encuentre el número.
2. A number increased by five is 13. Find the number.
El número que aumenta por cinco es 13. Encuentre el número.
3. One-third of a number is 2. Find the number.
Un tercer del número es dos. Encuentre el número.
1. 3
2. 8
3. 6
## Monday, January 20, 2014
### Translating Simple Algebra Word Problems: (Spanish & English)
1. Two more than a number is 8. What is the number?
Dos más que el número es ocho. ¿Cuál es el número?
2. Three more than a number is 5. What is the number?
Tres más que el número es cinco. ¿Cuál es el número?
3. A number decreased by 2 is 5. Find the number.
El número que reduce por dos es cinco. Encuentre el número.
1. 10
2. 8
3. 7
## Tuesday, January 14, 2014
### Algebra Word Problem: Setting up Problem (Spanish & English)
A total of r players came to a basketball practice. The coach divides them into four groups of t players each, but two players are left over. Which expression shows the relationship between the number of players out for basketball and the number of players in each group?
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×
Get Full Access to Statistics: Informed Decisions Using Data - 5 Edition - Chapter 11.3 - Problem 21
Get Full Access to Statistics: Informed Decisions Using Data - 5 Edition - Chapter 11.3 - Problem 21
×
# ?Comparing Step Pulses A physical therapist wanted to know whether the mean step pulse of men was less than the mean step pulse of women. She randomly
ISBN: 9780134133539 240
## Solution for problem 21 Chapter 11.3
Statistics: Informed Decisions Using Data | 5th Edition
• Textbook Solutions
• 2901 Step-by-step solutions solved by professors and subject experts
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3
Problem 21
Comparing Step Pulses A physical therapist wanted to know whether the mean step pulse of men was less than the mean step pulse of women. She randomly selected 51 men and 70 women to participate in the study. Each subject was required to step up and down onto a 6-inch platform for 3 minutes. The pulse of each subject (in beats per minute) was then recorded. After the data were entered into Minitab, the following results were obtained
(a) State the null and alternative hypotheses.
(b) Identify the P-value and state the researcher’s conclusion if the level of significance was = 0.01.
(c) What is the 95% confidence interval for the mean difference in pulse rates of men versus women? Interpret this interval.
Step-by-Step Solution:
Step 1 of 5) Comparing Step Pulses A physical therapist wanted to know whether the mean step pulse of men was less than the mean step pulse of women. She randomly selected 51 men and 70 women to participate in the study. Each subject was required to step up and down onto a 6-inch platform for 3 minutes. The pulse of each subject (in beats per minute) was then recorded. After the data were entered into Minitab, the following results were obtained (a) State the null and alternative hypotheses. (b) Identify the P-value and state the researcher’s conclusion if the level of significance was = 0.01. (c) What is the 95% confidence interval for the mean difference in pulse rates of men versus women Interpret this interval. For example, we obtain a simple random sample of individuals and ask them to disclose their number of years of education.
Step 2 of 2
##### ISBN: 9780134133539
This full solution covers the following key subjects: . This expansive textbook survival guide covers 88 chapters, and 2422 solutions. Since the solution to 21 from 11.3 chapter was answered, more than 226 students have viewed the full step-by-step answer. Statistics: Informed Decisions Using Data was written by and is associated to the ISBN: 9780134133539. This textbook survival guide was created for the textbook: Statistics: Informed Decisions Using Data, edition: 5. The answer to “?Comparing Step Pulses A physical therapist wanted to know whether the mean step pulse of men was less than the mean step pulse of women. She randomly selected 51 men and 70 women to participate in the study. Each subject was required to step up and down onto a 6-inch platform for 3 minutes. The pulse of each subject (in beats per minute) was then recorded. After the data were entered into Minitab, the following results were obtained (a) State the null and alternative hypotheses.(b) Identify the P-value and state the researcher’s conclusion if the level of significance was = 0.01.(c) What is the 95% confidence interval for the mean difference in pulse rates of men versus women? Interpret this interval.” is broken down into a number of easy to follow steps, and 121 words. The full step-by-step solution to problem: 21 from chapter: 11.3 was answered by , our top Statistics solution expert on 01/15/18, 03:19PM.
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Home >> Using this Method >> Constructing Problems
## Students Constructing Problems
Of course, typically students are given the problems to solve. I have found it useful to reverse this process and have the students construct problems for me to solve.
This has never been an open-ended request (though that might be useful). Instead, we are doing loox problems, for example, and I ask them to construct a loox problem for me.
### Learning the Structure of the Problem
There are two reasons for doing this. First, to construct, say, a loox problem, the student has to understand the structure of the loox problem. If a student is having trouble with solving loox problems, it might be that the student doesn't understand the structure of the problem.
Thus, when the student constructs a problem of a given type, the student is exploring the structure of that type of problem. As a second benefit, you get to see if the student understands the structure of the problem.
To construct a problem, you essentially start with an answer and then produce the problem. That is the exact same path that you follow when you solve the problem, but you are taking the path in the opposite direction.
For example, construct a loox problem, you might decide that a loox has 3 ears. If you put 4 looxes in the box, you have 12 ears (3 x 4) in the box. If you add 2 dogs to the box, you have added 4 ears (2x2), so now there are 16 ears (12 + 4) in the box. Solving the problem is then just reversing these steps. You calculate that there are 4 dog ears in the box, you subtract 4 from 16 to get 12, the number of loox ears, then you divide by 4 to learn that each loox has 3 ears.
The path from answer to problem is (usually?) easier than the path from from problem to solution. That's one reason we give the problem in the format we do. But, the path from solution to problem is exactly the path that we expect students to take when we ask them to check their answers. For example, in the above problem, suppose the student concludes that each loox has 3 ears. The student, to check the answer, is supposed to take the same steps as would be taken in constructing the problem.
Therefore, asking students to construct a problem is very good practice for the skill of checking an answer.
### Learning to Solve the Problem
When the process of constructing the problem is the same path as solving the problem, except in a different direction, the student can also learn to solve the problem. In a sense, when you construct a loox problem, you take information (the number of ears a loox has) and fold it into a different shape. To solve the problem, you have to unfold it. So learning to fold it can help with unfolding it.
### Social Factors
All of the social factors are working in the right direction for having the student construct a problem.
First, there is less emphasis on right or wrong -- the student is much less likely to perceive the possibility that the student can be wrong. And in some sense, this is not illusion -- in a very real sense, whatever the student does is right. And it is more like using a skill. So it fits the "flow" model of doing math.
Second, there is the chance to fool the instructor. They might not succeed, but if they do, students seem to like it. In any case, there is in some sense a superiority in being the person constructing the problem. The student can feel that superiority. They are also likely to learn, as modelled by you, that there is no inferiority in being the person solving the problem. I mean, I like doing math problems. In a real sense they are doing the work and I am having the fun.
### And Maybe...
One of my friends was trying to teach a student the concept of opposites. She gave examples, such as on and off, in and out, and tried to explain it. The student didn't get it. So she asked the student to give her a problem -- say a word and my friend would say the opposite. The student said "red".
There is a lot to think about there. How come some words have opposites and some don't?
So when you ask your students to construct a problem, usually nothing out of the ordinary will happen. But sometimes the student will construct a problem that leads you, the teacher, to a new level of awareness.
That's not all. Sometimes my students construct a problem that I like and then use. In essense, they are creative.
Once I was doing the exercise of what things in a category have names. I started I think with snow and rain. Then I asked for topics from the class. They suggested cheese, which turned out a little differently than I expected. So I learned something. They suggested shoes, which I had done before and was what I expected. Then a student suggested doing numbers. Well, that was what I most wanted to do. So I took his suggestion.
### Using This Technique
The obvious situation for using this technique is when (1) a student is stuck, not understanding the problem, and (2) you are stuck, not understanding why the student isn't getting it and not knowing what to do next. It is an all-purpose technique and requires no thought from you.
It can also be a part of any exercise, if there is a structure to the problems.
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kidzsearch.com > wiki
# Quantity
(Redirected from Amount)
Quantity or amount is how much of something there is that can be counted or measured.
## Counting
Whole numbers (1, 2, 3 ...) are used to count things. This can be done by pointing to each one. As things are pointed to, a number is said. Start with the number one. Each time another thing is pointed to, the next whole number is used. When the last thing in a group it pointed to, that number is the quantity of the group.
## Measuring
Measuring is done with a ruler or a machine. We put a ruler next to a thing to measure how long it is. We put things on a scale to measure their mass. Other machines are used to measure other quantities such as temperature, speed, electric current, and so on.
### Decimal numbers
When measuring things, a whole number might not be the best answer. A distance might be longer than 5 meters, but less than 6 meters. Meter sticks are marked off in parts of a meter. Decimals are formed by marking off a ruler in ten equal parts. Each part is a tenth. 5.2 meters is a little bit longer than 5 meters. 5.7 meters is a little bit shorter than 6 meters. Each of these tenths can be marked with ten smaller parts. The one hundred centimeters on a meter stick are done this way; by marking ten equal parts, then marking ten equal parts of each of those parts.
### Fraction numbers
Fractions are used when there are parts of a whole number. The dial on a machine may be marked with four parts between each whole number. A scale with a bag of potatoes may show five and three-quarters kilograms. This would be almost six kilograms.
### The words "amount" and "number"
People often use the word "amount" when they should say "number". The words "number" and "amount" should be used if a particular number could be put: e.g.
• A large amount of sand (because you cannot say "three thousand sands")
• A large number of people (because you could say there were 3547 people).
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# Calculating flow rate when emptying liquid from a closed-top tank
• labuch
In summary: P_{atm}=\frac{P_1}{\gamma}+z_1P_{atm}=\frac{P_1}{\gamma}+\frac{P_{atm}}{2g}\frac{z_1}{\gamma}P_{atm}=\frac{P_1}{\gamma}+\frac{P_{atm}}{2g}\frac{z_1}{\gamma}P_{atm}=\frac{P_1}{\gamma}+\frac{P_{at
labuch
Homework Statement
Hello,
Here is my problem: "We have a closed tank of large size which contains a liquid topped by air at a pressure equal to the atmospheric pressure. If the tank is closed, as the water flows out, the volume of air above the free surface increases, so the pressure decreases (law of perfect gases). We seek to find the pressure variation (𝛥P) at the surface of the water in the tank as a function of the z axis (whose origin is at B) and to give the expression of v (the flow velocity) as a function of z."
Relevant Equations
I know that:
-according to the perfect gas equation P=nRT/V so that pressure is proportional to volume, so that P(t=0)*V(t=0)=P(t=1)*V(t=1)
-That the initial pressure at the surface of the water is P =Patm
-by applying bernoulli between A and B, we have classical V0=sqrt(2gh)
I can imagine the experiment: the pressure at the surface will drop while the tank is emptying (if no air bubbles enter through the evacuation of course, otherwise it restores the pressure atm at the surface). The flow speed decreases as a function of time until the external pressure maintains a liquid level above the drain (thus stopping the flow). I can't put this resonance in mathematical form (𝛥P as a function of z and v as a function of z)
thank you in advance for those who will help me,
Sincerely
here is a simple visualisation of the probleme
Hello @labuch ,
labuch said:
-by applying bernoulli between A and B, we have classical V0=sqrt(2gh)
No, you don't. That expression is based on pressure terms canceling, and in your case they don't.
But you can build an appropriate model along the same lines and end up with a differential equation. I don't expect that to have an analytical solution, but perhaps I can be proven wrong.
Interesting problem; not homework I suppose ?
##\ ##
Lnewqban and labuch
BvU said:
Hello @labuch ,
No, you don't. That expression is based on pressure terms canceling, and in your case they don't.
But you can build an appropriate model along the same lines and end up with a differential equation. I don't expect that to have an analytical solution, but perhaps I can be proven wrong.
Interesting problem; not homework I suppose ?
##\ ##
hello, thank you for your response, Yes you are right I thought that this expression could be true at the very beginning of the draining that's why I called it V0 and I had the feeling that it was there that I had to dig. This homework is actually a problem around which I have to create a practical work. My problem is that I don't know how to develop the expression to solve the problem mathematically
labuch said:
here is a simple visualisation of the problemeView attachment 325111
Bernoulli's is not valid for this problem as the flow is not steady (its time varying).
I propose to begin with the "Momentum Equation" (Reynolds Transport Theorem - Newtons Second for a control volume) to develop your differential equation
$$\sum \boldsymbol F = \frac{d}{dt}\int_{cv} \rho \boldsymbol v ~dV\llap{-} + \int_{cs} \rho \boldsymbol v \boldsymbol V \cdot d \boldsymbol A$$
Although its application is probably going to be cumbersome and not obvious... the answer is going to be in there somewhere
Last edited:
erobz said:
Bernoulli's is not valid for this problem as the flow is not steady (its time varying).
I propose to begin with the "Momentum Equation" (Reynolds Transport Theorem - Newtons Second for a control volume) to develop your differential equation
$$\sum \boldsymbol F = \frac{d}{dt}\int_{cv} \rho \boldsymbol v ~dV\llap{-} + \int_{cs} \rho \boldsymbol v \boldsymbol V \cdot d \boldsymbol A$$
Although its application is probably going to be cumbersome and not obvious... the answer is going to be in there somewhere
Hello,
I thought of another solution because I could not use your formula. I think that we can probably assimilate the tank to a tank of right section of constant area S. We can note z=f(t) the height of air in the tank and Zo the initial value. The law of perfect gases applied by considering the temperature as fixed leads to :
Patm.zo=P.z
The velocity of the fluid at the upper free surface of the liquid is (dz/dt).
If we note "s" the cross-sectional area of the outlet orifice, the conservation of flow gives: S.(dz/dt)=s.Vs where Vs is the outlet velocity at the bottom of the tank. Taking into account these two equations, Bernoulli's theorem will give you the differential equation verified by z=f(t). However I can't get to the final equation...
Bernoulli's is not supposed to be valid for non-steady flow, but then again in my fluids text does use the solution ## v = \sqrt{2gh}## for the exit velocity in a draining tank example problem...like we find in Torricelli's Law derived from Bernoulli's principle (so I'm not sure what exactly is affected and to what extent under this assumption).
Under that assumption (which also neglects viscosity), let 1 be the upper surface of the fluid, and 2 be the position of the fluid jet.
$$\frac{P_1}{\gamma} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{\gamma} + \frac{v_2^2}{2g} + z_2$$
##z_2## is the datum
##P_2## is 0 gage pressure (atmospheric)
## v_1 \ll v_2## the kinetic head in the tank is negligible in comparison to the jet at 2.
##A_1## is the cross section of the tank
##A_2## is the cross section of the jet
##H## is from ##z_2## to the top of the tank
##z_{1_o}## is the initial height of fluid in the tank measured from ##z_2##
$$\frac{P_1}{\gamma} + z_1 = \frac{v_2^2}{2g}$$
Let the absolute pressure above the tank surface 1 be ##P(z_1) = P_1(z_1) + P_{atm}##
Applying the ideal gas law we are looking for the pressure of the gas as a function of ##z_1##
$$P_{atm} A_1 ( H - z_{1_o}) = P(z_1)A_1( H - z_1)$$
Can you take it from there?
##P_1## will be a function of ##z_1##, and ##v_2## will be a function of ##\dot z_1##.
Last edited:
member 731016
While we are on the subject can anyone explain the consequences of applying Bernoulli's to non-steady flow?
Here is the derivation in my textbook. The local acceleration (i.e. the variation of velocity with time at a specific point on a path line - ## \frac{\partial V}{\partial t}## in this problem will clearly not be zero, which negates everything after (4.15).
member 731016
Based on the figure in post #2, here is my take on this.
Ideal Gas Law: $$P_g(H-h)=P_{atm}(H-h_0)$$where ##P_g## is the absolute pressure of the gas in the head space.
Bernoulli Equation: $$P_g+\rho g h=P_{atm}+\rho g h_B+\frac{1}{2}\rho v^2$$
Combining these two equations gives: $$\frac{1}{2}\rho v^2=\rho g(h-h_B)-\left(\frac{h_0-h}{H-h}\right)p_{atm}$$
Finally, the overall mass balance is $$A_{tank}\frac{dh}{dt}=-A_{exit}v$$
member 731016, BvU, erobz and 1 other person
I found this example of Unsteady Bernoulli's. Can someone reconstruct an interpretation from the example? There is obviously a change in the flow energy per unit volume (a deviation from Bernoulli's for steady flow), but where/what is it?
## 1. How do you calculate the flow rate of liquid emptying from a closed-top tank?
The flow rate of liquid emptying from a closed-top tank can be calculated using Torricelli's Law, which states that the speed of efflux, v, of a fluid under gravity through an orifice is given by $$v = \sqrt{2gh}$$, where g is the acceleration due to gravity and h is the height of the fluid column above the orifice. The flow rate Q can then be found using $$Q = A \cdot v$$, where A is the cross-sectional area of the orifice.
## 2. What factors affect the flow rate when emptying a closed-top tank?
The flow rate when emptying a closed-top tank is affected by the height of the liquid column above the orifice, the size and shape of the orifice, the viscosity of the liquid, and the atmospheric pressure inside the tank. Additionally, any resistance or friction in the piping system can also impact the flow rate.
## 3. How does the height of the liquid column influence the flow rate?
The height of the liquid column directly influences the flow rate because it determines the pressure at the orifice. According to Torricelli's Law, the flow rate is proportional to the square root of the height of the liquid column. As the height decreases, the pressure and thus the flow rate also decrease.
## 4. Can the flow rate be constant when emptying a closed-top tank?
No, the flow rate is not constant when emptying a closed-top tank. As the liquid level drops, the height of the liquid column decreases, which in turn reduces the pressure and the flow rate. Therefore, the flow rate decreases over time as the tank empties.
## 5. How can you maintain a steady flow rate when emptying a closed-top tank?
To maintain a steady flow rate when emptying a closed-top tank, you can use a pump to control the discharge rate or install a flow control valve. Another method is to use a pressurized system where the pressure is regulated to compensate for the decreasing liquid level, ensuring a consistent flow rate.
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# How do you solve recipe question in math
Martha found her favorite recipe for meatballs which serves 8 people. Here's the recipe:
1 can (20 ounces) crushed pineapple, drained 1 egg
1 package (20 ounces) ground chicken
3/4 cup sliced green onions
1/2 teaspoon ground nutmeg
1/2 cup teriyaki sauce
1 teaspoon ground ginger
2 table spoons orange marmalade
Martha is making the meatballs today, just enough for her friends who are coming for dinner. So she is using 7/8 cup of teriyaki sauce.
Tell how many people it will serve. Show how you got your anwer
Tell how much of each ingredient will she use. Make sure you show an label your work.
Explain your strategy. Why did you do what you did?
Ok, so you know the recipe serves 8 people.
You are told that she is making enough to serve her friends, so she is changing the recipe. If she wanted to serve twice as many people, she would twice as much of each ingredient.
She uses 7/8 cup of teriyaki sauce. In the original how much teriyaki did she use? 1/2 cup
So you need to find out what factor she is multiplying by to get from 1/2 to 7/8. When you find that factor you can multiply that same factor by 8 to get the number of people her recipe serves.
So, to solve this problem you need to figure out (1/2)* x = (7/8) or you can write it (x/2) = (7/8).
Multiply both sides by 2,
x = (7/8)*2
x = 14/8
(14/8)*(8 people) = 14 people served by her recipe.
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Chapter 9
# Fermat's principle
Take 10 minutes to prepare this exercise.
Then, if you lack ideas to begin, look at the given clue and searching for the solution.
A detailed solution is then proposed to you.
If you have more questions, feel free to ask them on the forum.
Pierre de Fermat (French mathematician and physicist, - ) postulated that the light rays met a very general principle that the path taken by the light to travel from one point to another was one for which the travel time was minimum (indeed, an extreme that can be minimum or maximum).
A swimming coach, located at a point beach, wants to apply this principle to rescue as quickly as possible a vacationer (located in ) about to drown close to the beach.
We note and velocity vectors (supposed constant) of the swimming coach on the beach (when running) and in water (where he swims).
## Question
What should be the path followed by the swimming coach in order to have Fermat's principle verified and the vacationer safe ?
Deduce the expression of the law of refraction in optics.
### Solution
We choose a frame of reference that simplifies the problem : let pass the x-axis through the straight line that separates the beach from the sea and the y-axis through point , initial position of the swimming coach.
In such a frame of reference, points and have the coordinates and .
The swimming coach's trajectory will consist of two straight portions and , where is the point where the swimming coach starts swimming.
One can notice that the distance will be greater than the distance since the swimming coach will certainly run faster than current swimming !
The time taken by the swimming coach to get from to is :
By developing values of and , we obtain the following dependence of as a function of the abscissa of :
The extremum of is reached when the derivative with respect to is zero.
Yet :
Noting that (see figure for the definitions of the angles) :
And :
The condition of extremum time made by the light is then expressed as (where the angles and , by analogy with optics, see figure above, can be called angle of incidence and angle of refraction) :
It is obvious that this time extremal corresponds to a minimum ; indeed, the distance and thus the time can easily be made very large if the swimming coach, then certainly lacking professional conscience, decided to go shopping for example before rescueing the poor vacationer !
#### Complément : Case of light and laws of Snell-Descartes
Consider two mediums and of the respective refractive indexes and .
Given two points and respectively located in the medium of index (point ) and in the medium of index (point ).
Fermat's principle unables to assert that the path taken by the light to get from to is such that the time taken by light is extreme (usually, minimum).
By applying this principle, similar reasoning to that performed in the case of the path followed by the swimming coach, is used to demonstrate the stated law of refraction, around 1620, by physicists Snell and Descartes :
Where and are respectively the angles of incidence and refraction.
Remember that the index of a medium used to find the velocity of light in this medium as a function of that in vacuum is :
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# Calendar Verbal Reasoning & Aptitude Questions
Posted On:March 2, 2019, Posted By: Latest Interview Questions, Views: 491, Rating :
## Best Calendar - Aptitude Questions and Answers
Dear Readers, Welcome to Calendar - Aptitude Objective Questions have been designed specially to get you acquainted with the nature of questions you may encounter during your Job interview for the subject of Calendar - Aptitude MCQs. These objective type Calendar - Aptitude questions are very important for campus placement test and job interviews. As per my experience good interviewers hardly plan to ask any particular question during your Job interview and these model questions are asked in the online technical test and interview of many Banks, SSC, Railways, Postal and many Govt Jobs.
## Calendar Important Formulas - Aptitude Questions and Answers
1. Odd days : The number of days more than the complete number of weeks in a given period is number of odd days during the period.
For example, a period of 10 days contains 3 odd days, 11 days contains 4 odd days, 12 days contains 5 odd days. But period of 14 days contains Zero odd days.
2. Leap Year : Every year which is divisible by 4 is called a leap year. But every century which is divisible by 4 is not a leap year.Every fourth century is a leap year
3. An ordinary year has 365 days i.e (52 weeks +1).
4. A leap year has 366 (an ordinary year ) by 7, we get remainder 1, it means that it has 1 odd day.Like wise 366 days (leap year ) has 2 odd days.
## Calender Aptitude Solved example Problems and Answers
### 1. If the first day of the year(other than the leap) was Friday, then which was the last of that year?
A. Wednesday
B. Thursday
C. Friday
D. Sunday
Ans: C
Explanation:
If the year is not a leap year, then the last day of the year is the same as the first day.
### 2. 1-12-91 is the first Sunday. Which is the fourth Tuesday of December 91?
A. 17-12-91
B. 24-12-91
C. 26-12-91
D. 31-12-91
Ans: B
Explanation:
1-12-91 is the first Sunday of December 91.
So, 3-12-91 is the first Tuesday of the month.
Clearly, 10-12-91, 17-12-91, 24-12-91 and 31-12-91 are also Tuesdays.
So, 24-12-91 is the fourth Tuesday.
### 3. If 18th February, 1997 falls on Tuesday then what will be the day on 18th February, 1999?
A. Monday
B. Tuesday
C. Wednesday
D. Thursday
Ans: D
Explanation:
18th February, 1997 was Tuesday.
So, 18th February, 1998 was Wednesday.
Therefore, 18th February, 1999 will be Thursday.
### 4. What is the day on 1st January 1901?
A. Monday
B. Tuesday
C. Wednesday
D. Thursday
Ans: B.
Explanation:
1st January 1901 means (1900 year and 1 day)
Now, 1600 years have 0 odd days
300 years have 1 odd day
1 day has 1 odd day
Total number of odd days = 0 + 1 + 1 = 2 days
Hence, The day of 1st January 1901 was Tuesday.
### 5. If the day before yesterday was Saturday, What day will fall on the day after tomorrow?
A. Friday
B. Tuesday
C. Thursday
D. Wednesday
Ans: D
Explanation:
If day before yesterday was Saturday, then today is Monday.
Thus tomorrow will be Tuesday and day after tomorrow will be Wednesday.
## Calender Exercise Problems and Answers for all competitive exams
### 1. Today is Wednesday, What will be the day after 94 days?
A. Monday
B. Wednesday
C. Friday
D. Sunday
Explanation:
Every day of the weeks is repeated after 7 days. Hence if will be Wednesday, after 94 days.
### 2. If 1st Octomber is Sunday, then 1st November will be
A. Wednesday
B. Friday
C. Sunday
D. Monday
Explanation:
Clearly, 1st, 8th, 15th, 22nd and 29th of October are Sundays.
So 31st October is Tuesday.
Therefore 1st November will be Wednesday.
### 3. What is the number of odd days in a leap year?
A. 1
B. 2
C. 3
D. 4
Explanation:
A leap year has 366 days.
Now if we divide 366 by 7 it gives 2 as remainder.
Hence number of odd days in 366 days is 2.
### 4. If the day before yesterday was Thursday, when will Sunday be?
A. Tomorrow
B. Today
C. Day after tomorrow
D. Two days after today
If day before yesterday was Thursday, then today is Saturday.
Therefore, tomorrow’s will be Sunday.
### 5. If day after tomorrow is Saturday, What day was three days before yesterday?
A. Sunday
B. Monday
C. Tuesday
D. Friday
Day after tomorrow is Saturday.
So today is Thursday.
Thus, yesterday was Wednesday and three days before Wednesday was Sunday.
### 6. Today is Thursday. The day after 59 days will be?
A. Sunday
B. Tuesday
C. Wednesday
D. Monday
Every day of the week is repeated after 7 days.
Hence after 56 days it would be Thursday again.
And after 59 days it would be Sunday.
### 7. Saturday was a holiday for Republic Day. 14th of the next month is again a holiday for Shivratri. What day was it on the 14th?
A. Sunday
B. Monday
C. Tuesday
D. Thursday
Explanation:
As given, Saturday falls on 26th January and we have to find the day on 14th February.
Clearly, 2nd, 9th and 16th February each is a Saturday.
Thus, 14th February was a Thursday.
### 8. If February 1, 1996 is Wednesday, What day is March 3, 1996?
A. Saturday
B. Tuesday
C. Wednesday
D. Monday
Explanation:
1996 is a leap year and so February has 29 days.
Now, 1st, 8th, 15th, 22nd and 29th February are Wednesdays.
So, 1st March is Thursday and 3rd March is Saturday.
### 9. If the day before yesterday was Saturday, What day will fall on the day after tomorrow?
A. Friday
B. Tuesday
C. Thursday
D. Wednesday
Explanation:
If day before yesterday was Saturday, then today is Monday.
Thus tomorrow will be Tuesday and day after tomorrow will be Wednesday.
### 10. If the seventh day of a month is three days earlier than Friday, What day will it be on the nineteenth day of the month?
A. Sunday
B. Tuesday
C. Wednesday
D. Monday
Explanation:
The seventh day of the month is three days earlier than Friday, which is Tuesday.
So, the fourteenth day is also Tuesday and thus, the nineteenth day is Sunday.
### 11. On 8th Feb, 1995 it was Wednesday. The day of the week on 8th Feb, 1994 was
A. Wednesday
B. Tuesday
C. Thursday
D. None of these
Explanation:
1994 being an ordinary year, it has 1 odd day.
So, the day on 8th Feb, 1995 is one day beyond the day on 8th Feb, 1994.
But, 8th Feb, 1995 was Wednesday. 8th Feb, 1994 was Tuesday.
### 12. The calendar for 1990 is the same as for
A. 1997
B. 2000
C. 1994
D. 1996
Explanation:
Count the number of days for 1990 onward to get 0 odd day.
Year 1990 1991 1992 1993 1994 1995
Odd days 1 1 2 1 1 1 = 7 or 0 odd day.
Calendar for 1990 is the same as for the year 1996.
### 13. How many days are there in x weeks x days?
A. 7x
B. 8x
C. 14x
D. 7
Explanation:
x weeks x days = (7x + x) days
= 8x days.
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# 62.42 kg to lbs - 62.42 kilograms to pounds
Do you want to learn how much is 62.42 kg equal to lbs and how to convert 62.42 kg to lbs? Here you go. In this article you will find everything about kilogram to pound conversion - theoretical and also practical. It is also needed/We also want to emphasize that whole this article is dedicated to a specific number of kilograms - this is one kilogram. So if you want to know more about 62.42 kg to pound conversion - read on.
Before we get to the more practical part - that is 62.42 kg how much lbs conversion - we are going to tell you few theoretical information about these two units - kilograms and pounds. So let’s move on.
How to convert 62.42 kg to lbs? 62.42 kilograms it is equal 137.6125439404 pounds, so 62.42 kg is equal 137.6125439404 lbs.
## 62.42 kgs in pounds
We will begin with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, known also as International System of Units (in short form SI).
From time to time the kilogram is written as kilogramme. The symbol of the kilogram is kg.
First definition of a kilogram was formulated in 1795. The kilogram was described as the mass of one liter of water. This definition was simply but totally impractical to use.
Later, in 1889 the kilogram was described by the International Prototype of the Kilogram (in abbreviated form IPK). The International Prototype of the Kilogram was made of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was in use until 2019, when it was substituted by another definition.
Nowadays the definition of the kilogram is based on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is 0.001 tonne. It is also divided into 100 decagrams and 1000 grams.
## 62.42 kilogram to pounds
You know a little about kilogram, so now we can move on to the pound. The pound is also a unit of mass. It is needed to underline that there are more than one kind of pound. What are we talking about? For example, there are also pound-force. In this article we want to centre only on pound-mass.
The pound is used in the Imperial and United States customary systems of measurements. Of course, this unit is used also in another systems. The symbol of the pound is lb or “.
The international avoirdupois pound has no descriptive definition. It is defined as 0.45359237 kilograms. One avoirdupois pound could be divided into 16 avoirdupois ounces and 7000 grains.
The avoirdupois pound was implemented in the Weights and Measures Act 1963. The definition of this unit was placed in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 62.42 kg?
62.42 kilogram is equal to 137.6125439404 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 62.42 kg in lbs
The most theoretical section is already behind us. In this part we want to tell you how much is 62.42 kg to lbs. Now you learned that 62.42 kg = x lbs. So it is high time to get the answer. Just look:
62.42 kilogram = 137.6125439404 pounds.
This is a correct outcome of how much 62.42 kg to pound. It is possible to also round off the result. After rounding off your result will be as following: 62.42 kg = 137.324 lbs.
You learned 62.42 kg is how many lbs, so look how many kg 62.42 lbs: 62.42 pound = 0.45359237 kilograms.
Obviously, this time you may also round it off. After it your result will be exactly: 62.42 lb = 0.45 kgs.
We also want to show you 62.42 kg to how many pounds and 62.42 pound how many kg outcomes in tables. See:
We are going to begin with a table for how much is 62.42 kg equal to pound.
### 62.42 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
62.42 137.6125439404 137.3240
Now see a chart for how many kilograms 62.42 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
62.42 0.45359237 0.45
Now you learned how many 62.42 kg to lbs and how many kilograms 62.42 pound, so we can go to the 62.42 kg to lbs formula.
### 62.42 kg to pounds
To convert 62.42 kg to us lbs a formula is needed. We are going to show you a formula in two different versions. Let’s begin with the first one:
Amount of kilograms * 2.20462262 = the 137.6125439404 result in pounds
The first formula will give you the most exact result. Sometimes even the smallest difference could be significant. So if you need an accurate outcome - this version of a formula will be the best solution to know how many pounds are equivalent to 62.42 kilogram.
So let’s move on to the another version of a formula, which also enables conversions to learn how much 62.42 kilogram in pounds.
The shorter version of a formula is as following, see:
Number of kilograms * 2.2 = the outcome in pounds
As you can see, this formula is simpler. It could be better solution if you need to make a conversion of 62.42 kilogram to pounds in fast way, for example, during shopping. Just remember that final result will be not so exact.
Now we are going to learn you how to use these two versions of a formula in practice. But before we are going to make a conversion of 62.42 kg to lbs we are going to show you easier way to know 62.42 kg to how many lbs without any effort.
### 62.42 kg to lbs converter
An easier way to check what is 62.42 kilogram equal to in pounds is to use 62.42 kg lbs calculator. What is a kg to lb converter?
Calculator is an application. Calculator is based on first formula which we gave you above. Due to 62.42 kg pound calculator you can quickly convert 62.42 kg to lbs. Just enter amount of kilograms which you need to convert and click ‘convert’ button. You will get the result in a second.
So let’s try to convert 62.42 kg into lbs using 62.42 kg vs pound converter. We entered 62.42 as an amount of kilograms. This is the result: 62.42 kilogram = 137.6125439404 pounds.
As you can see, our 62.42 kg vs lbs calculator is user friendly.
Now we can move on to our main topic - how to convert 62.42 kilograms to pounds on your own.
#### 62.42 kg to lbs conversion
We will start 62.42 kilogram equals to how many pounds conversion with the first version of a formula to get the most exact result. A quick reminder of a formula:
Amount of kilograms * 2.20462262 = 137.6125439404 the result in pounds
So what have you do to check how many pounds equal to 62.42 kilogram? Just multiply number of kilograms, in this case 62.42, by 2.20462262. It gives 137.6125439404. So 62.42 kilogram is equal 137.6125439404.
You can also round it off, for example, to two decimal places. It is 2.20. So 62.42 kilogram = 137.3240 pounds.
It is time for an example from everyday life. Let’s convert 62.42 kg gold in pounds. So 62.42 kg equal to how many lbs? And again - multiply 62.42 by 2.20462262. It gives 137.6125439404. So equivalent of 62.42 kilograms to pounds, if it comes to gold, is 137.6125439404.
In this case it is also possible to round off the result. It is the result after rounding off, in this case to one decimal place - 62.42 kilogram 137.324 pounds.
Now let’s move on to examples calculated with short formula.
#### How many 62.42 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Number of kilograms * 2.2 = 137.324 the result in pounds
So 62.42 kg equal to how much lbs? And again, you have to multiply number of kilogram, in this case 62.42, by 2.2. Let’s see: 62.42 * 2.2 = 137.324. So 62.42 kilogram is 2.2 pounds.
Make another calculation using this version of a formula. Now convert something from everyday life, for instance, 62.42 kg to lbs weight of strawberries.
So calculate - 62.42 kilogram of strawberries * 2.2 = 137.324 pounds of strawberries. So 62.42 kg to pound mass is 137.324.
If you know how much is 62.42 kilogram weight in pounds and are able to convert it using two different versions of a formula, let’s move on. Now we are going to show you these results in charts.
#### Convert 62.42 kilogram to pounds
We know that outcomes presented in tables are so much clearer for most of you. We understand it, so we gathered all these outcomes in charts for your convenience. Due to this you can quickly make a comparison 62.42 kg equivalent to lbs results.
Let’s begin with a 62.42 kg equals lbs table for the first version of a formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
62.42 137.6125439404 137.3240
And now let’s see 62.42 kg equal pound chart for the second formula:
Kilograms Pounds
62.42 137.324
As you can see, after rounding off, if it comes to how much 62.42 kilogram equals pounds, the outcomes are the same. The bigger number the more considerable difference. Remember it when you want to make bigger number than 62.42 kilograms pounds conversion.
#### How many kilograms 62.42 pound
Now you know how to calculate 62.42 kilograms how much pounds but we want to show you something more. Are you interested what it is? What do you say about 62.42 kilogram to pounds and ounces calculation?
We want to show you how you can convert it step by step. Start. How much is 62.42 kg in lbs and oz?
First things first - you need to multiply amount of kilograms, in this case 62.42, by 2.20462262. So 62.42 * 2.20462262 = 137.6125439404. One kilogram is exactly 2.20462262 pounds.
The integer part is number of pounds. So in this case there are 2 pounds.
To convert how much 62.42 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It is 327396192 ounces.
So your outcome is 2 pounds and 327396192 ounces. It is also possible to round off ounces, for instance, to two places. Then your result is equal 2 pounds and 33 ounces.
As you see, conversion 62.42 kilogram in pounds and ounces quite easy.
The last conversion which we are going to show you is conversion of 62.42 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.
To calculate it you need another formula. Before we show you this formula, see:
• 62.42 kilograms meters = 7.23301385 foot pounds,
• 62.42 foot pounds = 0.13825495 kilograms meters.
Now let’s see a formula:
Number.RandomElement()) of foot pounds * 0.13825495 = the result in kilograms meters
So to convert 62.42 foot pounds to kilograms meters you need to multiply 62.42 by 0.13825495. It gives 0.13825495. So 62.42 foot pounds is 0.13825495 kilogram meters.
You can also round off this result, for instance, to two decimal places. Then 62.42 foot pounds will be equal 0.14 kilogram meters.
We hope that this calculation was as easy as 62.42 kilogram into pounds conversions.
We showed you not only how to do a calculation 62.42 kilogram to metric pounds but also two another conversions - to check how many 62.42 kg in pounds and ounces and how many 62.42 foot pounds to kilograms meters.
We showed you also other solution to do 62.42 kilogram how many pounds conversions, it is using 62.42 kg en pound converter. This will be the best solution for those of you who do not like converting on your own at all or need to make @baseAmountStr kg how lbs calculations in quicker way.
We hope that now all of you can make 62.42 kilogram equal to how many pounds conversion - on your own or using our 62.42 kgs to pounds converter.
It is time to make your move! Let’s calculate 62.42 kilogram mass to pounds in the way you like.
Do you want to make other than 62.42 kilogram as pounds conversion? For example, for 5 kilograms? Check our other articles! We guarantee that conversions for other amounts of kilograms are so simply as for 62.42 kilogram equal many pounds.
### How much is 62.42 kg in pounds
At the end, we are going to summarize the topic of this article, that is how much is 62.42 kg in pounds , we prepared one more section. Here you can see all you need to know about how much is 62.42 kg equal to lbs and how to convert 62.42 kg to lbs . Let’s see.
What is the kilogram to pound conversion? It is a mathematical operation based on multiplying 2 numbers. How does 62.42 kg to pound conversion formula look? . Have a look:
The number of kilograms * 2.20462262 = the result in pounds
See the result of the conversion of 62.42 kilogram to pounds. The exact answer is 137.6125439404 lb.
It is also possible to calculate how much 62.42 kilogram is equal to pounds with another, easier type of the formula. Check it down below.
The number of kilograms * 2.2 = the result in pounds
So this time, 62.42 kg equal to how much lbs ? The result is 137.6125439404 lbs.
How to convert 62.42 kg to lbs quicker and easier? It is possible to use the 62.42 kg to lbs converter , which will do all calculations for you and you will get a correct result .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms.
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# The missing number in the series 1, 4, 27,____, 3125 is?
The missing number in the sequence is 256. This is because the numbers are determined by the formula n^n (n to the power of n). The pattern is as follows:
- 1^1 = 1
- 2^2 (2x2) = 4
- 3^3 (3x3x3) = 27
- 4^4 (4x4x4x4) = 256
- 5^5 (5x5x5x5x5) = 3125
It is known as a sequence of integer to the second tetration. To find the answer you have to determine the relationship of each number to its numerical place in order. Since the missing number would be the fourth number would be the fourth number in the sequence, this means that n=4.
Therefore the answer is found by finding out 4^4
- 4x4 = 16
- 4x4x4 = 64
- 4x4x4x4 = 256
The term 'tetration' was coined by the English mathematician Reuben Louis Goodstein. For a more in depth explanation and definition of what tetration is and how it can be applied, then you can check the following Wikipedia page - en.wikipedia.org/wiki/Tetration
If you are looking for more number puzzles to try out in order to test yourself, then there are plenty of websites that you could look at. Examples include:
- brainden.com/number-puzzles.htm
- oeis.org/Spuzzle.htm
thanked the writer.
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https://homework.cpm.org/category/MN/textbook/cc2mn/chapter/1/lesson/1.1.4/problem/1-38
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Home > CC2MN > Chapter 1 > Lesson 1.1.4 > Problem1-38
1-38.
At the farmers’ market, two pounds of peaches cost $\4.20$. How much will five pounds cost? Show all of your work or explain your reasoning.
How much does one pound of peaches cost?
Because we know the cost of two pounds of peaches, we can divide this value $(\4.20)$ by two and find the cost of one pound of peaches.
$\4.20\div2=\2.10$
Think about what you did in Step 1. How can this help you find how much $5$ pounds cost?
By multiplying the cost of one pound by five, we will find the cost of five pounds.
$(\2.10)(5)=\10.50$
The cost of five pounds of peaches will be $\10.50$.
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