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# Properties of inequalities pdf
Home Forums Chat Properties of inequalities pdf
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inequalities worksheet pdf
subtraction property of inequality
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properties of inequalities worksheet
Solving One-Step Inequalities. Division Property for Inequalities. The image cannot be displayed. Your computer may not have enough memory to open the
“Some problems in algebra lead to inequalities instead of equations. Note: Another method would be to apply properties of equations to all parts of the
By definition a ? b if either a = b or a<b, and a ? b if either a = b or a>b. Properties of Inequalities. Let a, b and c be real numbers. 1) If a<b, then a + c<b + c and
Example 1. Find the solution set to the following inequality. Do your work in a step-by-step manner so as to demonstrate the order axioms and properties used.
the first and third columns. Write an inequality symbol that makes the statement true. Repeat this four times to complete the four rows in the table. ? Perform the
Properties of Inequality Handout. Inequality Symbols : Greater Than. Greater Than or Equal To. (The line underneath the Greater Than sign indicates also Equal
Properties of Inequality. For real numbers a, b and c: 1. If a < b, then a + c < b + c,. 2. If a < b and if c > 0, then ac < bc,. 3. If a < b and if c < 0, then ac > bc.
We solve quadratic and higher degree inequalities very much like we solve quadratic and higher degree using the Zero Product Property. To accomplish that
You will study. • the properties of inequality. • how to solve inequalities by using inverse operations. • how to solve inequalities with variables on both sides.
The algebraic properties listed apply given a, b, and c are real numbers. This is Property. Equality. Inequality. Multiplicative. Property of Zero a · 0 = 0 = 0 · a.
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Mind Stretchers
“Mind Stretchers,” Family Home Evening Resource Book (1997), 281
The following activities are all quiet puzzle activities. They may use pictures, geometric shapes, mathematics, or words. Family members will find that these puzzles require careful observation. The activities are designed for elementary children, teenagers, and adults.
Materials Needed
Pencils
Copies for everyone of the worksheets you will need for the games you choose.
Preparation
Assign a family member to become thoroughly acquainted with all of the puzzles and decide which ones would be appropriate for family members to do as a family activity. He should understand the instructions given for the puzzles and the possible solutions.
Activity
As a family, try at least two of the following puzzles. Follow the instructions on the worksheets and then check your answers with those on the answer sheets.
“How Many Squares Do You See?”
1. “How Many Squares Do You See?” Ask family members how many squares they see. Have them number the squares on their paper if they would like. If they find sixteen or seventeen, they have found the number that most people find. Let them look again to see how many squares they can see, and then show the answer sheet.
2. “What Is This Thing Called Love?” In the heart, there are thirty-four hidden words that describe what love is. See how many you can find. Words run in all directions, left to right, right to left, top to bottom, bottom to top, and diagonally. Circle them as you find them. Then look at the answer sheet to see how many you missed.
Answers: accepting, bond, communicating, loyalty, happiness, exciting, joyful, caring, trusting, fulfilling, receiving, giving, forgiving, understanding, tender, lovely, belonging, respect, natural, sharing, ageless, open, warm, nice, patient, faith, alive, always, hope, true, real, forever, right, wed.
3. “The Tricky T.” Make patterns for these shapes by tracing them onto lightweight paper. Then use the patterns to make cardboard cutouts. Number them, and keep all the numbered sides facing you.
See if you can make the four pieces fit together to make a capital T.
This T is exactly the same size as the puzzle pieces when they’re put together properly.
4. “Lots of Triangles.” There are thirty-five triangles in this pentagon. Can you find them all?
5. “Division.” There are seven tennis balls inside this square. Can you divide up the square so that each tennis ball is left in its own compartment without any others—by using only three straight lines?
6. “Snatch a Match.” Arrange twelve used matches to make four equal squares as shown. By moving only three matches, try to make three equal squares.
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# Find the half range cosine fourier series expansion for $f(x)=(x-1)^2,\quad 0<x<1$.
Find the half range cosine fourier series expansion for $$f(x)=(x-1)^2,\quad 0
and hence deduce that $$\pi^2=8\left(\frac 1 {1^2}+\frac 1 {3^2}+\frac 1 {5^2}+\ldots\right)\tag{1}$$
My work
I have derived that the expansion is $$f(x)=\frac 1 3+\sum\limits_{n=1}^\infty \frac 4 {n^2\pi^2} \cos n\pi x$$ But I could not deduce that $$(1)$$.
• Set $x=0$ then $x=1$ then subtract. – Alexey Burdin Jun 18 '15 at 5:18
• but in x=0 and 1 the function is not defined know...then how can i do this??@AlexeyBurdin – David Jun 18 '15 at 5:57
Define $f$ as a periodic function of $1$ on $\mathbb{R}$ as $$f(x)=(x−1)^2, \hspace{5 mm} 0\le x<1$$ Set $x=0$, then $$f(0)=1=\frac 1 3+\sum\limits_{n=1}^\infty \frac 4 {n^2\pi^2}$$
So we have $$\dfrac{2}{3}=\sum\limits_{n=1}^\infty \frac 4 {n^2\pi^2} \hspace{4 mm} \text{and} \hspace{4 mm} \dfrac{\pi^2}{6}=\sum\limits_{n=1}^\infty \dfrac1{n^2}$$
And set $x\to1^{-}$, then $$\lim\limits_{x\to1^{-}}f(x)=0=\frac 1 3+\sum\limits_{n=1}^\infty \frac 4 {n^2\pi^2} \cos n\pi$$
Thus \begin{align} \dfrac{\pi^2}{12}&=\sum\limits_{n=1}^\infty \dfrac1 {(2n-1)^2}-\sum\limits_{n=1}^\infty \dfrac1 {(2n)^2} \\ &=\sum\limits_{n=1}^\infty \dfrac1 {(2n-1)^2}-\frac1{4}\sum\limits_{n=1}^\infty \dfrac1 {n^2} \\ &=\sum\limits_{n=1}^\infty\dfrac1 {(2n-1)^2}-\dfrac{\pi^2}{24} \end{align}
So $$\sum\limits_{n=1}^\infty\dfrac1 {(2n-1)^2}=\dfrac{\pi^2}{8}$$
• how it hold even the fuction is not defined on 1 and 2. – David Jun 18 '15 at 6:39
• $f$ becomes a periodic function of $1$ so it repeats on $[1,2]$ from $[0,1]$, and that is why you get $f(x)=\frac 1 3+\sum\limits_{n=1}^\infty \frac 4 {n^2\pi^2} \cos n\pi x$. – hermes Jun 18 '15 at 6:44
• ya,,, thanks for the comment – David Jun 18 '15 at 7:02
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How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?
# Merry-go-round Problem | ISI-B.Stat Entrance | TOMATO 104
Try this Merry-go-round Problem based on the combinatorics from TOMATO useful for ISI B.Stat Entrance.
## Merry-go-round Problem | ISI B.Stat Entrance | Problem - 104
Four married couples are to be seated in a merry-go-round with 8 identical seats. In how many ways can they be seated so that:
i) males and females seat alternately, and
ii) no husband seats adjacent to his wife?
• 8
• 12
• 16
• 20
### Key Concepts
combinatorics
probability
Number theory
Answer: $12$
TOMATO, Problem 104
Challenges and Thrills in Pre College Mathematics
## Try with Hints
There are $8$ persons......$W_1,W_2,W_3,W_4,M_1,M_2,M_3,M_4$.Given that males and females seat alternately & no husband seats adjacent to his wife.Let us assume that .$W_1$ is the wife of $M_1$,$W_2$ is the wife of $M_3$ and the similar for others.....
Therefore $M_1$ can not be seat beside or after $W_1$.similar for others.can you draw a circular form ...?
Can you now finish the problem ..........
$W_1$ can sit in two seats either in the seat in left side figure or in the seat in
right side figure. In left side figure when $W_1$ is given seat then $W_4$ can sit in
one seat only as shown and accordingly $W_2$ and $W_3$ can also take only one
seat. Similarly, right side figure also reveals one possible way to seat. So
there are two ways to seat for every combination of Men
Can you finish the problem........
Now, Men can arrange themselves in (4 – 1)! = 6 ways. So number of ways = $2 \times 6$= $12$.
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# Homework Help: Algerbraic logarithm method question
1. Sep 20, 2009
### Cosmicon
First of all, greetings to the scientific community here at Physics Forums.
The following set of equations is given:
y^(x-3y) = x^2
x^(x-3y) = y^8
With the next assumption given: x-3y is unequal to (-4).
My attempt was to isolate the y variable in both of the equations so that i will be able to devide the the equations and recieve a simple x^a = b equations. This did not help, because the powers of the x's included the y, with no option to get rid of it.
I Would like your assistance with finding the shortest, most efficient, non-logarithmic way to solve this question, asap.
Last edited: Sep 20, 2009
2. Sep 20, 2009
### Hurkyl
Staff Emeritus
Why haven't you taken the logarithm of the equations? Isn't that the method you said you were asking about in the title?
3. Sep 20, 2009
### Cosmicon
You are right it is a possibility. I did not fully mention my request here in the topic.
I am searching for an additional method to logarithm, by modifying the powers in the two equations.
Just tackled with another question with great similarity:
x^(2x+y) = y^4
y^(2x+y) = x^16
I noticed the identity of the powers in the left sides of the equations, but even after immediatly deviding the two equations, I get stuck.
What would be the most efficient way to solve this with no log?
Last edited: Sep 20, 2009
4. Sep 20, 2009
### g_edgar
It is the method of logarithm, you just don't write log ...
$x^{(2x+y)^2} = y^{4(2x+y)} = x^{16\cdot 4} = x^{64}$. Therefore $(2x+y)^2 = 64$ (that's where we really did logarithms without mentioning the word), so assuming $x>0, y>0$ we have $2x+y=8$. Then $x^8 = y^4$ so $x^2=y$. Substitute in the equation: $2x+x^2=8$ then solve the quadratic equation. The only positive solution is $x=2$ so then $y=4$.
I assumed x and y are positive so that irrational exponents would be defined. But to solve for integer solutions, you could allow negative x,y since integer powers still make sense.
5. Sep 20, 2009
### Cosmicon
edgar, using which formula did you make this calculation:
$x^{(2x+y)^2} = y^{4(2x+y)} = x^{16\cdot 4} = x^{64}$
6. Sep 22, 2009
### g_edgar
Start with $x^{(2x+y)} = y^{4}$, raise both sides to power $2x+y$ , get $x^{(2x+y)^2} = y^{4(2x+y)}$.
Start with $y^{(2x+y)} = x^{16}$, raise both sides to the power $4$ , get $y^{4(2x+y)} = x^{16\cdot 4}$ .
OK?
7. Sep 23, 2009
### Cosmicon
yes, I now understand this technique. very useful and widespread in problems involving equations.
*According to this, when solving the quadratic equation, two answers apply: x=2, and x=-4.
why is x=-4 (y=16) not valid in this case?
** It is also important to mention that another solution in this form of questions:
when the two bases are equal to 1, the powers can be placed with any suggested number.
Therefore: when x=1, y=1.
Last edited: Sep 23, 2009
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Student Manual Math Ready Unit 8 Lesson 4 29 Task 12 Independent Practice for
# Student manual math ready unit 8 lesson 4 29 task 12
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• 71% (7) 5 out of 7 people found this document helpful
This preview shows page 242 - 250 out of 283 pages.
Student Manual Math Ready . Unit 8 . Lesson 4
29 Task 12: Independent Practice for Lesson 3 Two golfers kept track of their scores over the course of a year. Their scores are given below. (All courses played were 72 par) Player 1: 85, 80, 78, 83, 83, 88, 72, 75, 79, 79 Player 2: 72, 71, 90, 95, 70, 88, 78, 78, 70, 90 Who is the better golfer and why? Explain your reasoning. (Note: In golf, the lower score wins!) Student Manual Math Ready . Unit 8 . Lesson 4
30 Task #13: Explore Give an example of a set of five positive numbers whose median is 10 and whose mean is larger than 10. Student Manual Math Ready . Unit 8 . Lesson 5
31 Task #14: Mean and Median Investigation Part 1: Consider the set of numbers {10, 15, 25, 30, 30, 50, 55, 55, 60, 80} a. Find the mean. b. Find the median. Part 2: Now, replace the 80 with 800 to create a new set {10, 15, 25, 30, 30, 50, 55, 55, 60, 800} c. Find the mean. d. Find the median. e. What effect did changing 80 to 800 have on the mean and the median? Student Manual Math Ready . Unit 8 . Lesson 5
32 Task #15: Guided Notes Definition An is any data point that is more than 1.5 times the Inner Quartile Range (IQR) away from either the lower or upper quartile. {39, 51, 36, 39, 90, 55, 32, 61, 45, 53, 14} 1. Complete the Five Number Summary for the data set above. minimum = lower quartile (Q 1 ) = median = upper quartile (Q 3 ) = maximum = 2. Calculate the Inner Quartile Range (IQR). IQR = Q 3 – Q 1 = 3. Check for outliers. Q 1 – (1.5 x IQR ) = Q 3 + (1.5 x IQR ) = 4. Does this data set contain any outliers? If so, list them. Student Manual Math Ready . Unit 8 . Lesson 5
33 Task #16: Celestial Bodies and Home Prices 1. Calculate the mean and median for these distances. Would the typical distance of these celestial bodies best be communicated using the mean or the median? Why? 2. What impact do the very large values in the data set have on the mean? 3. Suppose that a sample of 100 homes in the metropolitan Phoenix area had a median sales price of \$300,000. The mean value of these homes was \$1,000,000. Explain how this could happen. Why might the median price be more informative than the mean price in describing a typical house price? Student Manual Math Ready . Unit 8 . Lesson 5 Object Distance in light years Moon 0.000000038 Venus 0.0000048 Jupiter 0.000067 Mars 0.0000076 Mercury 0.0000095 Syrius 8.6 Canopus 310 Saturn 0.00014
34 Task #17: Lesson 5 Exit Ticket Suppose the mean annual income for a sample of one hundred Minneapolis residents was \$50,000. Do you think the median income for this sample would have been greater than, equal to, or less than \$50,000? Explain? Student Manual Math Ready . Unit 8 . Lesson 5
35 Student Manual Math Ready . Unit 8 . Lesson 6 Task A statistically-minded state trooper wondered if the speed distributions are similar for cars traveling northbound and for cars traveling southbound on an isolated stretch of interstate highway. He uses a radar gun to measure the speed of all northbound cars and all southbound cars passing a particular location during a fifteen minute period.
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# math
posted by .
solve the system by elimination.
6x-6y-4z=-10
-5z+4y-z=-12
2x+3y-2z=9
• math -
Multiply the last equation (both sides) by 2. That gives you
4x + 6y - 4z = 18
6x - 6y - 4z = -10
That gives you
10x -8z = 8
So far, we have eliminated the variable y.
Use similar tricks to get rid of x or z.
I believe your second equation is typed incorrectly. Did you mean
-5x +4y -z = -12?
## Similar Questions
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2. ### College Algebra
Please help! x-2y+z=7 2x+y-z=0 3x+2y-2z=-2 a. Solve the above system of equations using Gaussian Elimination or Gauss-Jordan Elimination. You must show row operations. b. Solve the above system of equations using Cramer's Rule.
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Solve the system using elimination. 2x – 2y = –8 x + 2y = –1 And this one: Solve the system using elimination. 3x – y = 28 3x + y = 14 Pleease help me :(
8. ### Math Homework
Hello! I need help with these math questions, thanks! :) 1.) Solve the system by substitution. 5x + 2y = -17 x = 3y 2.) Solve the system by substitution. x/3 + y = 4/3 -x + 2y = 11 3.) Solve the system by elimination. 2x - 3y = 0 2x …
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# David number
Jana and David train the addition of the decimal numbers so that each of them will write a single number and these two numbers then add up. The last example was 11.11. David's number had the same number of digits before the decimal point, the Jane's number also. David's number was written in different numbers, Jane's number had exactly two digits the same. Find the largest possible number David could write.
Result
D = 0.9
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
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## Next similar examples:
1. Motion
If you go at speed 3.7 km/h, you come to the station 42 minutes after leaving train. If you go by bike to the station at speed 27 km/h, you come to the station 56 minutes before its departure. How far is the train station?
2. Excavation
Mr. Billy calculated that excavation for a water connection dig for 12 days. His friend would take 10 days. Billy worked 3 days alone. Then his friend came to help and started on the other end. On what day since the beginning of excavation they met?
3. Diophantus
We know little about this Greek mathematician from Alexandria, except that he lived around 3rd century A.D. Thanks to an admirer of his, who described his life by means of an algebraic riddle, we know at least something about his life. Diophantus's youth l
4. Brick weight
The brick weighs 2 kg and a half bricks. How much does one brick weigh?
5. Garden
Area of a square garden is 6/4 of triangle garden with sides 56 m, 35 m, and 35 m. How many meters of fencing need to fence a square garden?
6. Pool
If water flows into the pool by two inlets, fill the whole for 8 hours. The first inlet filled pool 6 hour longer than second. How long pool take to fill with two inlets separately?
7. Forestry workers
In the forest is employed 56 laborers planting trees in nurseries. For 8 hour work day would end job in 37 days. After 16 days, 9 laborers go forth? How many days are needed to complete planting trees in nurseries by others, if they will work 10 hours a d
8. Bonus
Gross wage was 527 EUR including 16% bonus. How many EUR were bonuses?
9. Troops
Route is long 147 km and the first day first regiment went at an average speed 12 km/h and journey back 21 km/h. The second day went second regiment same route at an average speed 22 km/h there and back. Which regiment will take route longer?
10. Motion problem
From Levíc to Košíc go car at speed 81 km/h. From Košíc to Levíc go another car at speed 69 km/h. How many minutes before the meeting will be cars 27 km away?
11. Logic
A man can drink a barrel of water for 26 days, woman for 48 days. How many days will a barrel last between them?
12. Monkey
Monkey fell in 23 meters deep well. Every day it climbs 3 meters, at night it dropped back by 2 m. On what day it gets out from the well?
13. Store
One meter of the textile were discounted by 2 USD. Now 9 m of textile cost as before 8 m. Calculate the old and new price of 1 m of the textile.
14. One three
We throw two dice. What is the probability that max one three falls?
15. Trapezoid MO
The rectangular trapezoid ABCD with right angle at point B, |AC| = 12, |CD| = 8, diagonals are perpendicular to each other. Calculate the perimeter and area of the trapezoid.
16. Three cats
If three cats eat three mice in three minutes, after which time 260 cats eat 260 mice?
17. Root
The root of the equation ? is: ?
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# 6.6: The Product & Quotient Theorems
Difficulty Level: At Grade Created by: CK-12
Learning Objectives
• Determine the quotient theorem of complex numbers in polar form.
• Determine the product theorem of complex numbers in polar form.
• Solve everyday problems that require you to use the product and/or quotient theorem of complex numbers in polar form to obtain the correct solution.
## The Product Theorem
Multiplication of complex numbers in polar form is similar to the multiplication of complex numbers in standard form. However, to determine a general rule for multiplication, the trigonometric functions will be simplified by applying the sum/difference identities for cosine and sine. To obtain a general rule for the multiplication of complex numbers in polar from, let the first number be r1(cosθ1+isinθ1)\begin{align*}r_1(\cos \theta_1 + i \sin \theta_1)\end{align*} and the second number be r2(cosθ2+isinθ2)\begin{align*}r_2(\cos \theta_2 + i \sin \theta_2)\end{align*}. The product can then be simplified by use of three facts: the definition i2=1\begin{align*}i^2 = -1\end{align*}, the sum identity cosαcosβsinαsinβ=cos(α+β)\begin{align*}\cos \alpha \cos \beta - \sin \alpha \sin \beta = \cos(\alpha + \beta)\end{align*}, and the sum identity sinαcosβ+cosαsinβ=sin(α+β)\begin{align*}\sin \alpha \cos \beta + \cos \alpha \sin \beta = \sin (\alpha + \beta)\end{align*}.
Now that the numbers have been designated, proceed with the multiplication of these binomials.
r1(cosθ1+isinθ1)r2(cosθ2+isinθ2)r1r2(cosθ1cosθ2+icosθ1sinθ2+isinθ1cosθ2+i2sinθ1sinθ2)r1r2[(cosθ1cosθ2sinθ1sinθ2)+i(sinθ1cosθ2+cosθ1sinθ2)]r1r2[cos(θ1+θ2)+isin(θ1+θ2)]\begin{align*}& r_1(\cos \theta_1 + i \sin \theta_1) \cdot r_2(\cos \theta_2 + i \sin \theta_2)\\ & r_1r_2(\cos \theta_1 \cos \theta_2 + i \cos \theta_1 \sin \theta_2 + i \sin \theta_1 \cos \theta_2 + i^2 \sin \theta_1 \sin \theta_2)\\ & r_1r_2[(\cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2) + i (\sin \theta_1 \cos \theta_2 + \cos \theta_1 \sin \theta_2)]\\ & r_1r_2[\cos(\theta_1 + \theta_2) + i \sin(\theta_1 + \theta_2)]\end{align*}
Therefore:
r1(cosθ1+isinθ1)r2(cosθ2+isinθ2)=r1r2[cos(θ1+θ2)+isin(θ1+θ2)]\begin{align*}r_1(\cos \theta_1 + i \sin \theta_1) \cdot r_2(\cos \theta_2 + i \sin \theta_2)=r_1r_2[\cos (\theta_1+\theta_2)+i \sin(\theta_1+\theta_2)]\end{align*}
## Quotient Theorem
Division of complex numbers in polar form is similar to the division of complex numbers in standard form. However, to determine a general rule for division, the denominator must be rationalized by multiplying the fraction by the complex conjugate of the denominator. In addition, the trigonometric functions must be simplified by applying the sum/difference identities for cosine and sine as well as one of the Pythagorean identities. To obtain a general rule for the division of complex numbers in polar from, let the first number be r1(cosθ1+isinθ1)\begin{align*}r_1(\cos \theta_1 + i \sin \theta_1)\end{align*} and the second number be r2(cosθ2+isinθ2)\begin{align*}r_2(\cos \theta_2 + i \sin \theta_2)\end{align*}. The product can then be simplified by use of five facts: the definition i2=1\begin{align*}i^2 = -1\end{align*}, the difference identity cosαcosβ+sinαsinβ=cos(αβ)\begin{align*}\cos \alpha \cos \beta + \sin \alpha \sin \beta = \cos(\alpha - \beta)\end{align*}, the difference identity sinαcosβcosαsinβ=sin(αβ)\begin{align*}\sin \alpha \cos \beta - \cos \alpha \sin \beta = \sin (\alpha - \beta)\end{align*}, the Pythagorean identity, and the fact that the conjugate of cosθ2+isinθ2\begin{align*}\cos \theta_2 + i \sin \theta_2\end{align*} is cosθ2isinθ2\begin{align*}\cos \theta_2 - i \sin \theta_2\end{align*}.
r1(cosθ1+isinθ1)r2(cosθ2+isinθ2)r1(cosθ1+isinθ1)r2(cosθ2+isinθ2)(cosθ2isinθ2)(cosθ2isinθ2)r1r2cosθ1cosθ2icosθ1sinθ2+isinθ1cosθ2i2sinθ1sinθ2cos2θ2i2sin2θ2r1r2(cosθ1cosθ2+sinθ1sinθ2)+i(sinθ1cosθ2cosθ1sinθ2)cos2θ2+sin2θ2r1r2[cos(θ1θ2)+isin(θ1θ2)]\begin{align*}& \frac{r_1(\cos \theta_1+i \sin \theta_1)}{r_2(\cos \theta_2+i \sin \theta_2)}\\ & \frac{r_1(\cos \theta_1+i \sin \theta_1)}{r_2(\cos \theta_2+i \sin \theta_2)} \cdot \frac{(\cos \theta_2-i \sin \theta_2)}{(\cos \theta_2-i \sin \theta_2)}\\ & \frac{r_1}{r_2} \cdot \frac{\cos \theta_1 \cos \theta_2-i \cos \theta_1 \sin \theta_2 + i \sin \theta_1 \cos \theta_2-i^2 \sin \theta_1 \sin \theta_2}{\cos^2 \theta_2-i^2 \sin^2 \theta_2}\\ & \frac{r_1}{r_2} \cdot \frac{(\cos \theta_1 \cos \theta_2+ \sin \theta_1 \sin \theta_2) + i (\sin \theta_1 \cos \theta_2-\cos \theta_1 \sin \theta_2)}{\cos^2 \theta_2 + \sin^2 \theta_2}\\ & \frac{r_1}{r_2}[\cos (\theta_1-\theta_2)+i \sin(\theta_1-\theta_2)]\end{align*}
In general:
r1(cosθ1+isinθ1)r2(cosθ2+isinθ2)=r1r2[cos(θ1θ2)+isin(θ1θ2)]\begin{align*}\frac{r_1(\cos \theta_1+i \sin \theta_1)}{r_2(\cos \theta_2+i \sin \theta_2)}=\frac{r_1}{r_2}[\cos(\theta_1-\theta_2)+i \sin(\theta_1-\theta_2)]\end{align*}
## Using the Product and Quotient Theorems
The following examples illustrate the use of the product and quotient theorems.
Example 1: Find the product of the complex numbers 3.61(cos56.3+isin56.3)\begin{align*}3.61(\cos 56.3^\circ + i \sin 56.3^\circ)\end{align*} and 1.41(cos315+isin315)\begin{align*}1.41(\cos 315^\circ + i \sin 315^\circ)\end{align*}
Solution: Use the Product Theorem, r1(cosθ1+isinθ1)r2(cosθ2+isinθ2)=r1r2[cos(θ1+θ2)+isin(θ1+θ2)]\begin{align*}r_1(\cos \theta_1 + i \sin \theta_1) \cdot r_2(\cos \theta_2 + i \sin \theta_2) = r_1r_2[\cos (\theta_1 + \theta_2) + i \sin (\theta_1 + \theta_2)]\end{align*}.
\begin{align*}& \quad 3.61(\cos 56.3^\circ + i \sin 56.3^\circ) \cdot 1.41(\cos 315^\circ + i \sin 315^\circ)\\ & = (3.61)(1.41)[\cos(56.3^\circ + 315^\circ) + i \sin(56.3^\circ + 315^\circ)\\ & = 5.09(\cos 371.3^\circ + i \sin 371.3^\circ)\\ & = 5.09(\cos 11.3^\circ + i \sin 11.3^\circ)\end{align*}
\begin{align*}^*\end{align*}Note: Angles are expressed \begin{align*}0^\circ \le \theta \le 360^\circ\end{align*} unless otherwise stated.
Example 2: Find the product of \begin{align*}5 \left(\cos \frac{3\pi}{4}+i \sin \frac{3\pi}{4} \right ) \cdot \sqrt{3} \left (\cos \frac{\pi}{2}+i \sin \frac{\pi}{2} \right )\end{align*}
Solution: First, calculate \begin{align*}r_1r_2=5 \cdot \sqrt{3}=5\sqrt{3}\end{align*} and \begin{align*}\theta =\theta_1+\theta_2=\frac{3\pi}{4}+\frac{\pi}{2}=\frac{5\pi}{4}\end{align*}
\begin{align*}5\sqrt{3} \left (\cos \frac{5\pi}{4}+i \sin \frac{5\pi}{4} \right )\end{align*}
Example 3: Find the quotient of \begin{align*}(\sqrt{3}-i) \div (2- i2\sqrt{3})\end{align*}
Solution: Express each number in polar form.
\begin{align*}& \sqrt{3}-i && 2-i2\sqrt{3}\\ & r_1=\sqrt{x^2+y^2} && r_2=\sqrt{x^2+y^2}\\ & r_1=\sqrt{(\sqrt{3})^2+(-1)^2} && r_2 = \sqrt{(2)^2+(-2\sqrt{3})^2}\\ & r_1=\sqrt{4}=2 && r_2=\sqrt{16}=4\end{align*}
\begin{align*}& \frac{r_1}{r_2}=.5\\ & \theta_1=\tan^{-1}\left(\frac{-1}{\sqrt{3}}\right) && \theta_2=\tan^{-1}\left(\frac{-2\sqrt{3}}{2}\right) && \theta=\theta_1-\theta_2\\ & \theta_1=5.75959 \ rad. && \theta_2=5.23599 \ rad. && \theta=5.75959-5.23599\\ &&&&& \theta=0.5236\end{align*}
Now, plug in what we found to the Quotient Theorem.
\begin{align*}\frac{r_1}{r_2}[\cos(\theta_1-\theta_2)+i \sin(\theta_1-\theta_2)]=.5(\cos 0.5236+i \sin 0.5236)\end{align*}
Example 4: Find the quotient of the two complex numbers \begin{align*}28 \angle 35^\circ\end{align*} and \begin{align*}14 \angle 24^\circ\end{align*}
Solution:
\begin{align*}& \text{For} \ 28 \ \angle 35^\circ && \text{For} \ 14 \ \angle 24^\circ && \frac{r_1}{r_2}=\frac{28}{14}=2\\ & r_1=28 && r_2=14 && \theta=\theta_1-\theta_2\\ & \theta_1=35^\circ && \theta_2=24^\circ && \theta=35^\circ-24^\circ=11^\circ\end{align*}
\begin{align*} \frac{r_1 \angle \theta_1}{r_2 \angle \theta_2} &= \frac{r_1}{r_2} \angle (\theta_1-\theta_2)\\ &=2 \angle 11^\circ\end{align*}
## Points to Consider
• We have performed the basic operations of arithmetic on complex numbers, but we have not dealt with any exponents or any roots of complex numbers. How might you calculate \begin{align*}(x + yi)^2\end{align*} or \begin{align*}\sqrt{r\angle \theta}\end{align*}?
• How might you calculate the \begin{align*}n^{th}\end{align*} power or root of a complex number?
## Review Questions
1. Multiply together the following complex numbers. If they are not in polar form, change them before multiplying.
1. \begin{align*}2 \angle 56^\circ, 7 \angle 113^\circ\end{align*}
2. \begin{align*}3(\cos \pi + i \sin \pi), 10 \left(\cos \frac{5\pi}{3}+i \sin \frac{5\pi}{3}\right)\end{align*}
3. \begin{align*}2+3i, -5+11i\end{align*}
4. \begin{align*}6-i, -20i\end{align*}
2. Part c from #1 was not in polar form. Mulitply the two complex numbers together without changing them into polar form. Which method do you think is easier?
3. Use the Product Theorem to find \begin{align*}4 \left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)^2\end{align*}.
4. The electric power (in watts) supplied to an element in a circuit is the product of the voltage \begin{align*}e\end{align*} and the current \begin{align*}i\end{align*} (in amps). Find the expression for the power supplied if \begin{align*}e=6.80 \angle 56.3^\circ\end{align*} volts and \begin{align*}i=7.05 \angle -15.8^\circ\end{align*} amps. Note: Use the formula \begin{align*}P = ei\end{align*}.
5. Divide the following complex numbers. If they are not in polar form, change them before dividing. In
1. \begin{align*}\frac{2 \angle 56^\circ}{7 \angle 113^\circ}\end{align*}
2. \begin{align*}\frac{10 \left(\cos \frac{5\pi}{3}+i\sin \frac{5\pi}{3}\right)}{5(\cos \pi+i \sin \pi)}\end{align*}
3. \begin{align*}\frac{2+3i}{-5+11i}\end{align*}
4. \begin{align*}\frac{6-i}{1-20i}\end{align*}
6. Part c from #5 was not in polar form. Divide the two complex numbers without changing them into polar form. Which method do you think is easier?
7. Use the Product Theorem to find \begin{align*}4\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)^3\end{align*}. Hint: use #3 to help you.
8. Using the Quotient Theorem determine \begin{align*}\frac{1}{4cis \frac{\pi}{6}}\end{align*}.
1. \begin{align*}2 \angle 56^\circ, 7 \angle 113^\circ=(2)(7) \angle (56^\circ+113^\circ)=14 \angle 169^\circ\end{align*}
2. \begin{align*}3(\cos \pi+i \sin \pi), 10 \left(\cos \frac{5\pi}{3}+i \sin \frac{5\pi}{3}\right)=(3)(10)cis \left(\pi+\frac{5\pi}{3}\right)=30cis \frac{8\pi}{3}=30cis \frac{2\pi}{3}\end{align*}
3. \begin{align*}2+3i, -5+11i \rightarrow \ \text{change to polar}\end{align*}\begin{align*}& x=2, y=3 && x=-5, y=11\\ & r=\sqrt{2^2+3^2}=\sqrt{13} \approx 3.61 && r=\sqrt{(-5)^2+11^2}=\sqrt{146} \approx 12.08\\ & \tan \theta =\frac{3}{2} \rightarrow \theta=56.31^\circ && \tan \theta=-\frac{11}{5} \rightarrow \theta=114.44^\circ\end{align*}\begin{align*}(3.61)(12.08) \angle (56.31^\circ+114.44^\circ)=43.61 \angle 170.75^\circ\end{align*}
4. \begin{align*}6-i, -20i \rightarrow \ \text{change to polar}\end{align*}
\begin{align*}& x=6, y=-1 && x=0, y=-20\\ & r=\sqrt{6^2+(-1)^2}=\sqrt{37} \approx 6.08 && r=\sqrt{0^2+(-20)^2}=\sqrt{40}=20\\ & \tan \theta=-\frac{1}{6} \rightarrow \theta=350.54^\circ && \tan \theta=\frac{-20}{0}=und \rightarrow \theta=270^\circ\end{align*} \begin{align*}(6.08)(20) \angle (350.54^\circ+270^\circ)=121.6 \angle 620.54^\circ=121.6 \angle 260.54^\circ\end{align*}
1. Without changing complex numbers to polar form, you mulitply by FOIL-ing. \begin{align*}(2+3i)(-5+11i)=-10+22i-15i+33i^2=-10-33+7i=-43+7i\end{align*} The answer is student opinion, but they seem about equal in the degree of difficulty.
2. \begin{align*}4\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)^2 &= 4\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right) \cdot 4\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)\\ &= 16\left(\cos \left(\frac{\pi}{4}+\frac{\pi}{4}\right)+i \sin \left(\frac{\pi}{4}+\frac{\pi}{4}\right)\right)\\ &= 16\left(\cos \frac{\pi}{2}i \sin \frac{\pi}{2}\right)\end{align*}
3. \begin{align*}P=(6.80)(7.05) \angle (56.3^\circ-15.8^\circ), P=47.9 \angle 40.5^\circ watts\end{align*}
1. \begin{align*}\frac{2 \angle 56^\circ}{7 \angle 113^\circ}=\frac{2}{7} \angle (56^\circ-113^\circ)=\frac{2}{7} \angle -57^\circ\end{align*}
2. \begin{align*}\frac{10\left(\cos \frac{5\pi}{3}+i \sin \frac{5\pi}{3}\right)}{5(\cos \pi+i \sin \pi)}=2 \left(\cos \left(\frac{5\pi}{3}-\pi\right)+i \sin \left(\frac{5\pi}{3}-\pi\right)\right)=2\left(\cos \frac{2\pi}{3}+i \sin \frac{2\pi}{3}\right)\end{align*}
3. \begin{align*}\frac{2+3i}{-5+11i} \rightarrow \ \text{change each to polar}.\end{align*}\begin{align*}& x=2, y=3 && x=-5, y=11\\ & r=\sqrt{2^2+3^2}=\sqrt{13} \approx 3.61 && r=\sqrt{(-5)^2+11^2}=\sqrt{146} \approx 12.08\\ & \tan \theta = \frac{3}{2} \rightarrow \theta = 56.31^\circ && \tan \theta=-\frac{11}{5} \rightarrow \theta=114.44^\circ\end{align*}\begin{align*}\frac{3.61}{12.08} \angle (56.31^\circ-114.44^\circ)=0.30 \angle - 58.13^\circ\end{align*}
4. \begin{align*}\frac{6-i}{1-20i} \rightarrow \ \text{change both to polar}\end{align*}\begin{align*}& x=6, y=-1 && x=1, y=-20\\ & r=\sqrt{6^2+(-1)^2}=\sqrt{37} \approx 6.08 && r=\sqrt{1^2+(-20)^2}=\sqrt{401}=20.02\\ & \tan \theta=-\frac{1}{6} \rightarrow \theta=350.54^\circ && \tan \theta =\frac{-20}{1} \rightarrow \theta=272.68^\circ\end{align*}\begin{align*}\frac{6.08}{20.02} \angle (350.54^\circ-272.86^\circ)=0.304 \angle 77.68^\circ\end{align*}
4. \begin{align*}\frac{2+3i}{-5+11i}=\frac{2+3i}{-5+11i} \cdot \frac{-5-11i}{-5-11i}=\frac{-10-22i-15i+33}{25+121}=\frac{23-37i}{146}\end{align*} Again, this is opinion, but in general, using the polar form is “easier.”
5. \begin{align*}[4\left(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\right)]^3=[4\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)]^2 \cdot 4\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right).\end{align*} From #3, \begin{align*}[4\left(\cos \frac{\pi}{4}i \sin \frac{\pi}{4}\right)]^2=16\left(\cos \frac{\pi}{2}i \sin \frac{\pi}{2}\right)\end{align*}. So, \begin{align*}[4\left(\cos \frac{\pi}{4}i \sin \frac{\pi}{4}\right)]^3 &= 16\left(\cos \frac{\pi}{2}i \sin \frac{\pi}{2}\right) \cdot 4 \left(\cos \frac{\pi}{4}i \sin \frac{\pi}{4}\right)\\ &= 64\left(\cos \frac{3\pi}{4}i \sin \frac{3\pi}{4}\right)\end{align*}
6. Even though 1 is not a complex number, we can still change it to polar form. \begin{align*}1 \rightarrow x=1, y=0\end{align*} \begin{align*}r &= \sqrt{1^2+0^2}=1 \qquad \\ \tan \theta &= \frac{0}{1}=0 \rightarrow \theta = 0^\circ\end{align*} \begin{align*}\text{So}, \frac{1}{4cis\frac{\pi}{6}}=\frac{1cis0}{4cis\frac{\pi}{6}}=\frac{1}{4}cis \left(0-\frac{\pi}{6}\right)=\frac{1}{4}cis\left(-\frac{\pi}{6}\right).\end{align*}
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## About JK and T Flip-Flop Diagrams
written by: shankar • edited by: KennethSleight • updated: 8/23/2013
Learn the basic construction of JK & T flip-flops, their logic diagrams, characteristic tables and characteristic equations.
• slide 1 of 3
### JK Flip-Flop
In the previous article we discussed RS and D flip-flops. Now we'll lrean about the other two types of flip-flops, starting with JK flip flop and its diagram.
A JK flip-flop has two inputs similar to that of RS flip-flop. We can say JK flip-flop is a refinement of RS flip-flop. JK means Jack Kilby, a Texas instrument engineer who invented IC. The two inputs of JK Flip-flop is J (set) and K (reset). A JK flip-flop is nothing but a RS flip-flop along with two AND gates which are augmented to it.
The flip-flop is constructed in such a way that the output Q is ANDed with K and CP. This arrangement is made so that the flip-flop is cleared during a clock pulse only if Q was previously 1. Similarly Q’ is ANDed with J and CP, so that the flip-flop is cleared during a clock pulse only if Q’ was previously 1.
When J=K=0
When both J and K are 0, the clock pulse has no effect on the output and the output of the flip-flop is the same as its previous value. This is because when both the J and K are 0, the output of their respective AND gate becomes 0.
When J=0, K=1
When J=0, the output of the AND gate corresponding to J becomes 0 (i.e.) S=0 and R=1. Therefore Q’ becomes 0. This condition will reset the flip-flop. This represents the RESET state of Flip-flop.
When J=1, K=0
In this case, the AND gate corresponding to K becomes 0(i.e.) S=1 and R=0. Therefore Q becomes 0. This condition will set the Flip-flop. This represents the SET state of Flip-flop.
When J=K=1
Consider the condition of CP=1 and J=K=1. This will cause the output to complement again and again. This complement operation continues until the Clock pulse goes back to 0. Since this condition is undesirable, we have to find a way to eliminate this condition. This undesirable behavior can be eliminated by Edge triggering of JK flip-flop or by using master slave JK Flip-flops.
The characteristic table explains the various inputs and the states of JK flip-flop.
• slide 2 of 3
### T Flip-Flop
T flip-flops are similar to JK flip-flops. T flip-flops are single input version of JK flip-flops. This modified form of JK flip-flop is obtained by connecting both inputs J and K together. This flip-flop has only one input along with Clock pulse. These flip-flops are called T flip-flops because of their ability to complement its state (i.e.) Toggle. So they are called as Toggle flip-flop.
When T=1 and CP=1, the flip-flop complements its output, regardless of the present state of the Flip-flop. In this case the next state is the complement of the present state.
When T=0, there is no change in the state of the flip-flop (i.e.) the next state is same as the present state of the flip-flop. From the characteristic table and characteristic equation it is quite evident that when T=0, the next sate is same as the present state.
• slide 3 of 3
### Applications Of Flip-Flops
• Counters
• Frequency Dividers
• Shift Registers
• Storage Registers
These are the various types of Flip-flops which are being used in Digital electronic circuits and the applications of Flip-flops are as specified above.
Image and Content Courtesy:
“DIGITAL LOGIC DESIGN" by Morris Mano
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>> << Ndx Usr Pri JfC LJ Phr Dic Rel Voc !: wd Help Dictionary
11. Classification (Sets and Propositions)
The list -.+./t appended to any classification table t will yield a complete classification table, and the function defined below therefore completes a classification table. The function tab ensures that a scalar or vector argument is treated as a one-rowed table.
``` c=: complete=: (] , (+./ {. ,:)@:-.@:(+./))@:tab
tab=: ,:^:(0:>.2:-#@\$)
c 0 0 1,:0 1 0
0 0 1
0 1 0
1 0 0
c 1 0 1,:0 1 0
1 0 1
0 1 0
(c 1 0 1);(c c 1 0 1);(c 0);(c 1)
+-----+-----+-+-+
|1 0 1|1 0 1|0|1|
|0 1 0|0 1 0|1| |
+-----+-----+-+-+
```
A function that yields a single boolean list is called a proposition; its result is a one-way classification called a set. The classification can, of course, be completed by the complementary set. For example:
``` p1=: 2&<: *. <&5 Set defined by interval
p1a=: (2:<:]) *. (]<5:) Alternative definition
p2=: = <. Set of integers
a=: 2 %~ i. 11
(],p1,p1a,p2,(p1+.p2),:(p1*.p2)) a
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
0 0 0 0 1 1 1 1 1 1 0
0 0 0 0 1 1 1 1 1 1 0
1 0 1 0 1 0 1 0 1 0 1
1 0 1 0 1 1 1 1 1 1 1
0 0 0 0 1 0 1 0 1 0 0
list=: 1 : 'x # ]' Adverb to list elements of set
((p1 list);(p2 list);((p1*.p2)list)) a
+-----------------+-----------+-----+
|2 2.5 3 3.5 4 4.5|0 1 2 3 4 5|2 3 4|
+-----------------+-----------+-----+
```
>> << Ndx Usr Pri JfC LJ Phr Dic Rel Voc !: wd Help Dictionary
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Pvillage.org
# How is flow rate affected by pipe size?
## How is flow rate affected by pipe size?
At any given flow rate, flow velocity is inversely proportional to the t cross sectional area of the pipe. Smaller pipes will lead to higher flow speeds; larger pipes, will lead to slower flow speeds.
## How do you calculate maximum flow in a pipe?
Flow rate is the volume of fluid per unit time flowing past a point through the area A. Here the shaded cylinder of fluid flows past point P in a uniform pipe in time t. The volume of the cylinder is Ad and the average velocity is ¯¯¯v=d/t v ¯ = d / t so that the flow rate is Q=Ad/t=A¯¯¯v Q = Ad / t = A v ¯ .
How many gallons per minute will flow through a 1.5 inch pipe?
Assume Average Pressure. (20-100PSI) About 12f/s flow velocity
1″ 1.00-1.03″ 37 gpm
1.25″ 1.25-1.36″ 62 gpm
1.5″ 1.50-1.60″ 81 gpm
2″ 1.95-2.05″ 127 gpm
How many GPM can a 3/4 pipe flow?
Water Flow (GPM/GPH) based on Pipe Size and Inside/Outside Diameters
Assume Average Pressure (20-100PSI). About 12 f/s flow velocity
Pipe Size (Sch. 40) I.D. (range) GPM (w/ min. PSI loss & noise)
1/2″ 0.5 – 0.6″ 14
3/4″ 0.75 – 0.85″ 23
1″ 1 – 1.03″ 37
### How many GPM is a 2 inch pipe?
Metric PVC Pipe
Assume Average Pressure. (20-100PSI) About 12f/s flow velocity
1.25″ 1.25-1.36″ 62 gpm
1.5″ 1.50-1.60″ 81 gpm
2″ 1.95-2.05″ 127 gpm
2.5″ 2.35-2.45″ 190 gpm
### What is maximum flow through a pipe?
For normal liquid service applications, the acceptable velocity in pipes is 2.1 ± 0.9 m/s (7 ± 3 ft/s) with a maximum velocity limited to 2.1 m/s (7 ft/s) at piping discharge points including pump suction lines and drains.
How do you calculate flow rate through pipe?
The flow rate depends on the area of the pipe or channel that the liquid is moving through, and the velocity of the liquid. If the liquid is flowing through a pipe, the area is A = πr2, where r is the radius of the pipe. For a rectangle, the area is A = wh where w is the width, and h is the height.
What is the formula for water flow in a pipe?
Water Flow Rates for Pipe Sizes over a Range of Diameters with the Hazen Williams Formula. For flow of water under pressure in a circular pipe, the Hazen Williams formula shown above can be rewritten into the following convenient form: in U.S. units: Q = 193.7 C D2.63 S0.54 , where: Q = water flow rate in gal/min (gpm) D = pipe diameter in ft
## What is maximum velocity of water in pipe?
Steel Pipes & Maximum Water Flow Capacity Pipe Size (in) Maximum Flow (gal.min) Velocity (ft/s) Head Loss (ft H20/100ft, m/100m) 45 3.9 75 4.1 130 3.9 250 4.0 800 4.0 1600 3.8 3000 4.0 4700 4.0 6000 4.0 8000 3.5 10000 3.0 12000 2.4 18000 2.1 Water ‐ Delivery Flow Velocity Normal Pipe Size Water Delivery Velocity ft/s 3.5 3.6 3.8 4 4.7 5.5 6.5 8.5 Inch 1
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0
# Which times tables have 45 in?
Updated: 9/22/2023
Wiki User
11y ago
1, 3, 5, 9, 15 & 45.
Wiki User
11y ago
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Q: Which times tables have 45 in?
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Related questions
### What times tables does 45 go into?
How about 9 times 5 = 45
### What numbers are in the 3 and 5 times tables?
15, 30, 45, 60 and so on.
### Is 52 in the 9 times tables?
No, it is not 9 x 5 = 45 9 x 6 = 54
### Why are times tables called times tables?
Because they are tables of the numbers that are the result of "times"-ing a number.
### 45℅ of a number is 45. What is the number?
You can find out at Proofhouse.com in the Colt tables.
### What are the 5 and 9 times tables?
5 times tables:5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100.9 times tables:9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, 108.(:Get how it goes? You can also use the "Calculator" application that is on every computer. I hope I helped.(:
### Is 777 in the 7 times tables?
Yes - 777/7 = 111
Times tables
### Rohan have some books and some tables if he puts 9 books into each table he will have two table left vacant if he puts 6 books on each table he will have 3 books left over find the numbers of books?
There are 45 books and 7 tables because 9 times 5 = 45 leaving two tables vacant or instead 6 times 7 = 42 leaving 3 books left over.
### What numbers are in the 5 times tables and 11 times tables?
55 and its multiples. 1, 5, and 55 are all in both the 5 times and 11 times tables.
### 4 times tables?
4,8,12,16,20,24,28,32,36,40,44,48,52,56,60,64,68,72,76,80,84,88,92,96,100
### If a number ends in 5 does it mean it is in the 5 times table?
It will be in the 5 times table but it may be in other times tables like 45 is in the 5 times table and the 9 times table. Also if it ends in a 0 it will be in the 5 times table
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0 sold
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Item description
Finding the Surface Area and Volume of Candy Project
Do your students know the surface area and volume formulas for many 3d shapes? Do they understand how these can be used and where there might be some differences between their mathematical solutions and actual data? For this project I have the students get into four groups. I hand out one candy item to each group and have them fill out the first page of the assignment. For volume of candy they will determine the diameter/radius, height and more in order to estimate the volume of all candies of that type. They will do the same for the wrapper size (ignoring the extra wrapper piece that may have been used to tie it over. I do all measurements with the item still wrapped if it is wrapped but this can easily be altered.
Once groups have all of the candies completed for both surface area and volume, I hand out the containers. One to each group. They then find the surface area (paint) required for the item and the space that the object can contain (volume). Students will need to use the same units when measuring all objects (ex. centimeters).
The third page has the student pick a candy and a container. They do the math to determine how many should fit (mathematically) and then ask for a bunch of that candy to see how many actually fit. I then put the formula for percent error up on the board and ask them to calculate their error.
The last page asks them to wrap up their findings and make some conjectures about what they learned and saw. A sample answer key is provided which mostly contains the formulas needed for a given shape as any candy and any container can be used for this. I typically pick a cylinder, sphere, rectangular prism and a cone for both candy and container.
Skills Required:
• Apply surface area and volume to different shapes
• Analyze how the real world affects their results
Resource is Great for:
• Middle School
• High School
• Geometry
• Summative Assessment
• Projects
• Percent error
• Real life application
• Surface Area and Volume applications
Includes:
• Printable PDF
• Student charts and write up page for hypothesis and conclusion
• 4 student pages
• Sample Rubric (1-5 scale)
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# Filter Arguments in CALCULATE - SQLBI.
Excel: Calculate Based on Multiple Conditions. This page is an advertiser-supported excerpt of the book, Power Excel 2010-2013 from MrExcel - 567 Excel Mysteries Solved. If you like this topic, please consider buying the entire e-book. Problem: COUNTIF and SUMIF have been around since Excel 97. Whenever someone learns how to use these functions, they inevitably come up with a situation where.
Learn how to calculate the amount of time between two dates with precision in Excel. Using the YEARFRAC function in Excel. Using the YEARFRAC function, you can calculate the precise difference between two dates because unlike other methods that return an integer result, this function returns a decimal result to indicate fractions of a year. The YEARFRAC function, however, requires a bit more.
## How To Calculate Future Value in Excel - Excel Functions.
So today, in this post, I'd like show you how to calculate ratio using 4 different ways. Let's get started and make sure to download this sample file from here to follow along. 4 Formulas to Calculate Ratio in Excel. Simple Divide Method; GCD Function; SUBSTITUTE and TEXT; Using Round Function; 1. Calculate Ratio by using Simple Divide Method. We can use this method when the larger value is.The Excel Combin and Combina functions both calculate a number of combinations of a set of objects. However, the two functions differ in that the Combin function does not count repetitions whereas the Combina function does count repetitions. For example, in a set of 3 objects, a, b, c, how many combinations of 2 objects are there? The Combin function returns the result 3 (combinations: ab.List of values that Excel substitutes in the input cell, B3. Use a two-variable data table to see how different values of two variables in one formula will change the results of that formula. For example, you can use a two-variable data table to see how different combinations of interest rates and loan terms will affect a monthly mortgage payment.
How to Calculate Probability of Combinations? How to do you calculate probability of combinations? Probability Calculation. Probability is a branch of mathematics used to measure the likeliness of an event to occur within the total number of possibilities. Consider for example, in tossing a coin once, the probability of getting the head is one.How to Count Unique and Distinct Values in Excel. The unique values are the ones that appear only once in the list, without any duplications. The distinct values are all the different values in the list. In this example, you have a list of numbers ranging from 1-6. The unique values are the ones that appear only once in the list, without any duplications. The distinct values are all the.
The COMBIN Function is an Excel Math and Trigonometry functio Functions List of the most important Excel functions for financial analysts. This cheat sheet covers 100s of functions that are critical to know as an Excel analyst n. The function will calculate the number of combinations without repetitions for a given number of items. It was.
This tutorial demonstrates how to use the Excel COMBIN Function in Excel to calculate the number of ways a given number of items can be ordered. COMBIN Function Overview. The COMBIN function calculates the number of combinations for a given number of items. To use the COMBIN Excel Worksheet Function, select a cell and type: (Notice how the formula inputs appear) COMBIN Function Syntax and.
A data table is a range of cells in which you can change values in some in some of the cells and come up with different answers to a problem. A good example of a data table employs the PMT function with different loan amounts and interest rates to calculate the affordable amount on a home mortgage loan. Experimenting with different values to observe the corresponding variation in results is a.
I need a way to generate all possible UNIQUE combinations in sets of four from a list I have (preferably using MS Access or else Excel). For example, I know that finding unique combinations of two letters from a choice of ABC will give: AB, AC and BC. In my case, I want to find all unique combinations, for example, of four letters from a set of 26 letters.
Excel; Theorems; Possible Outcomes Calculator. The chances of an event to occur is called as the possible outcome. Consider, you toss a coin once, the chance of occurring a head is 1 and chance of occurring a tail is 1. Hence, the number of possible outcomes is 2. Selecting items from a set without considering the order is called as combination. If the order of selection is considered, it is.
If you don't want to do the summation by hand you can calculate it easily in Excel. It's just x from the previous comparison plus n-1. It's just x from the previous comparison plus n-1. Cite.
To calculate permutations in Excel, we can use the PERMUTE formula. Same thing, we're going to have 4 that are available to choose from. If we're going to permute those into 2 spaces. So, we're making pairs. Then I can do a 2, and we have 12 permutations. If we write out all of these combinations Alex and Barry, Alex Cathy, Alex Denise, Barry Cathy, Barry Denise, Cathy Denise. Those are our.
I want to find how many different combinations of the items there are. Notice: The combinations will always have 1 item from every list; The 1st item in every combination will be chosen from the 1st list, the 2nd from the 2nd etc. So, basically the first 5 combinations would be: apples - a - black - 1; apples - a - black - 2; apples - a - white - 1.
I have 4 columns with 4 different parameters, if that's the right word. Each parameter has about 3-5 variables. What I want to do is I want to create ALL possible combinations of the 4 different parameters while maintaining the different columns. So let's say I have the following as an example.
I have tried to calculate ic50 from mtt results from excel curve fitting and also different formula provided in some papers, both of their results are not matching!!! can anybody guide me what may.
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# Why is ${x^{\frac{1}{2}}}$ the same as $\sqrt x$?
Why is ${x^{\frac{1}{2}}}$ the same as $\sqrt x$?
I'm currently studying indices/exponents, and this is a law that I was told to accept without much proof or explanation, could someone explain the reasoning behind this.
Thank you.
• How do you define $x^{\frac12}$? As $\exp(\frac12\ln x)$? Commented Oct 23, 2013 at 20:28
• You would have to tell us how you define $x^{\frac12}$, but you will probably have $\left(x^{\frac12}\right)^2=x^1=x$. Commented Oct 23, 2013 at 20:29
• Here is a wonderful explanation of what you're looking for.. Commented Oct 23, 2013 at 20:29
• Hmm I think I tagged this wrong, it's more to do with the law of exponents, but as I was studying exponential functions this popped into my head. Commented Oct 23, 2013 at 20:29
• The "exponential function" referred to by the tag is the function I mentioned in my answer that takes $x$ to $e^x$. It's not really appropriate for your question.
– MJD
Commented Oct 23, 2013 at 21:03
When $m$ and $n$ are integers, we have the important law that $$x^m\cdot x^n =x^{m+n}$$
We'd like this law to continue to hold when we define $x^\alpha$ for fractional $\alpha$, unless there's a good reason it shouldn't. If we do want it to continue to hold for fractional exponents, then whatever we decide that $x^{1/2}$ should mean, it should obey the same law: $$x^{1/2}\cdot x^{1/2} = x^{1/2+1/2} = x^1 = x$$
and so $x^{1/2} = \sqrt x$ is the only choice.
Similarly, what should $x^0$ mean? If we want the law to continue to hold, we need $$x^0\cdot x^n = x^{0+n} = x^n$$ and thus the only consistent choice is $x^0 = 1$. And again, why does $x^{-1} = \frac1x$? Because that's again the only choice that preserves the multiplication law, since we need $x^{-1}\cdot x^{1} = x^{-1+1} = x^0 = 1$.
But there is more to it than that. Further mathematical developments, which you may not have seen yet, confirm these choices. For example, one shows in analysis that as one adds more and more terms of the infinite sum $$1 + x + \frac{x^2}2 + \frac{x^3}6 + \frac{x^4}{24} + \cdots$$ the sum more and more closely approaches the value $e^x$, where $e$ is a certain important constant, approximately $2.71828$. One can easily check numerically that this holds for various integer values of $x$. For example, when $x=1$, and taking only the first five terms, we get $$1 + 1 + \frac12 + \frac16 + \frac1{24}$$
which is already $2.708$, quite close to $e^1$, and the remaining terms make up the difference. One can calculate $e^2$ by this method and also by straightforward multiplication of $2.71828\cdot2.71828$ and get the same answer.
But we can see just by inspection that taking $x=0$ in this formula gives $e^0 = 1$ because all the terms vanish except the first. And similarly, if we put in $x=\frac12$ we get approximately $1.648$, which is in fact the value of $\sqrt e$.
If it didn't work out this way, we would suspect that something was wrong somewhere. And in fact it has often happened that mathematicians have tried defining something one way, and then later developments revealed that the definition was not the right one, and it had to be revised. Here, though, that did not happen.
• Excellent explanation, thank you! Commented Oct 23, 2013 at 20:38
• Can I give you some more points somehow? Commented Oct 23, 2013 at 20:52
• Your thanks are my reward.
– MJD
Commented Oct 23, 2013 at 21:00
• @Assad: One of the choices when starting a bounty is "reward a good answer." You must wait 2 days before you can start a bounty, however. Commented Oct 23, 2013 at 22:33
• Why can't it be $-\sqrt{x}$. Is it defined non- negative? Commented Sep 26, 2015 at 18:08
For $x\geq 0$ and by definition $\sqrt x$ is the positive real $y$ such that $y^2=x$ and since $$\left(x^p\right)^q=x^{pq}$$ then for $p=\frac1 2$ and $q=2$ we see that $x^{1/2}=\sqrt x$
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# If x3
Firstly, equating x2 – x to 0 to find the zeros we get:
x (x – 1) = 0
x = 0 or x – 1 = 0
x = 0 or x = 1
As x3 + x2 – ax + b is divisible by x2 – x
The zeros o x2 – x will satisfy x3 + x2 – ax + b
Hence, (0)3 + 02 – a (0) + b = 0
b = 0
Also,
(1)3 + 12 – a (1) + 0 = 0
a = 2
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# Dynamics HW problem giving me a real headache
Hi everyone at PF. Thanks for taking time to read my HW problem. It's really neat that you are willing to take time from your busy schedules to help out students such as myself. I really apprecciate it!
## Homework Statement
A particle moving along a straight line decelerates according to a= -kv, where k is a constant and v is velocity. If it’s initial velocity at time t=0 is v0=4 m/s and its velocity at time t=2 s is v = 1 m/s, determine the time T and corresponding distance D for the particle speed to be reduced to one-tenth of its initial value
Answer : T = 3.32 s, D = 5.19 M
## Homework Equations
I think I figured out the value of the constant (k). However, I have no idea where to go from there. Below is my work on how I got to that point. I have tried integrating the new acceleration equation but have not had success further to find answers that are correct. Its very possible I'm not even on the right track so if anyone could please point me in the right direction I would be grateful.
## The Attempt at a Solution
equation: a=-kv ;; a=-k(dv/dt) ;; 1/v(dv) = -kdt ;; ln|v| = -kt +c
With that, solving for C we can use: ln|4| = -kt +c @ t=0 ;; ln|4| = 1.3863 = C
Then solve for k using: ln|1| = -kt + 1.3863 ;; k=0.69315
So from here the acceleration at t=0 and t=2 can be solved by using the velocities
a (@ t=0; v=4) = -.69315(4) = -2.7726m/s^2
a (@ t=2; v=1) = -.69315(1) = -.69315m/s^2
No idea where to go from here. :(
Related Introductory Physics Homework Help News on Phys.org
kuruman
Homework Helper
Gold Member
## The Attempt at a Solution
equation: a=-kv ;; a=-k(dv/dt) ;; 1/v(dv) = -kdt ;; ln|v| = -kt +c
Does this last equation make sense? You are doing physics not math, so when you integrate, there need to be limits of integration on both sides. What are those limits of integration? Furthermore, in physics, arguments of logarithms, sines, cosines, etc. have to be dimensionless. Yours isn't.
collinsmark
Homework Helper
Gold Member
Does this last equation make sense? You are doing physics not math, so when you integrate, there need to be limits of integration on both sides. What are those limits of integration? Furthermore, in physics, arguments of logarithms, sines, cosines, etc. have to be dimensionless. Yours isn't.
Actually, in my opinion it's a matter of preference. You could evaluate the definite integral using initial conditions as limits, or evaluate the indefinite integral and later solve for the arbitrary constant using initial conditions (the latter is what the OP did). Both are fine approaches.
[Another edit: Oh, and regarding the units (the OP's arguments weren't dimensionless). It can be shown that the units sort of pop out with the arbitrary constant. Using different units for v merely cause the arbitrary constant to reflect the difference. It's a property of log(ab) = loga + logb.]
equation: a=-kv ;; a=-k(dv/dt) ;; 1/v(dv) = -kdt ;; ln|v| = -kt +c
With that, solving for C we can use: ln|4| = -kt +c @ t=0 ;; ln|4| = 1.3863 = C
Then solve for k using: ln|1| = -kt + 1.3863 ;; k=0.69315
Okay, so far so good.
Eventually, you'll probably want to do some algebra and solve for v as a function of t. But you can solve for that later if you wish.
So from here the acceleration at t=0 and t=2 can be solved by using the velocities
a (@ t=0; v=4) = -.69315(4) = -2.7726m/s^2
a (@ t=2; v=1) = -.69315(1) = -.69315m/s^2
No idea where to go from here. :(
Wait, what? Acceleration? :uhh:
You don't need to find acceleration in order to solve this problem.
You already have a full relationship between velocity and time. And you already know what the initial v0 is (4 m/s). So just plug in 0.1v0 in for v, and solve for t!
For the next step you'll need to integrate to find distance, x(t) (complete with some algebra too). And you'll end up with another arbitrary constant if you use the indefinite integral. It's assumed in the problem statement that it's looking for the change in distance starting from t = 0. So that implies that x = 0 at t = 0.
[Edit: customwelds, I just noticed that in one of your listed equations, you had "a = -k(dv/dt)." I think you meant, dv/dt = -kv. I assume that was just a typo/simple mistake, because later your relationship turns out to be correct. And also, welcome to Physics Forums!]
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# Ring question
1. May 27, 2009
### samkolb
1. The problem statement, all variables and given/known data
Let R be a ring and suppose there exists a positive even integer n such that x^n = x for
every x in R. Show that -x = x for every x in R.
2. Relevant equations
3. The attempt at a solution
I solved the case where n = 2.
Let x be in R.
(x+x)^2= x+x = 2x,
(x+x)^2 = 4x^2 = 4x.
So 4x = 2x and 2x = 0. Done.
I tried using this same method when n = 4 and got nowhere.
2. May 27, 2009
### VKint
Let $$n = 2k$$. What's $$(-x)^{2k}$$?
(By the way, the proof you have for $$n = 2$$ doesn't work for noncommutative rings. The above hint suggests a method that does. Can you see why?)
3. May 28, 2009
### samkolb
Thanks for the hint. That works. But why does my proof for n=2 not work for noncommutative rings? Since the only terms in the expansion of (x+x)^2 are powers of x, I don't think I ever used commutativity.
4. May 28, 2009
### VKint
Yeah...you're right. I read your work as $$(x + x)^2 = (2x)^2 = 2^2 x^2 = 4 x^2$$; my only point was that in a noncommutative ring, $$(ab)^k \neq a^k b^k$$ in general. However, I suppose it's true that $$[(n \cdot 1) b]^k = (n^k \cdot 1) b^k$$ for natural numbers $$n$$.
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# How long is 21.4 centimeters in inches?
## Conversion formula
The conversion factor from centimeters to inches is 0.39370078740157, which means that 1 centimeter is equal to 0.39370078740157 inches:
1 cm = 0.39370078740157 in
To convert 21.4 centimeters into inches we have to multiply 21.4 by the conversion factor in order to get the length amount from centimeters to inches. We can also form a simple proportion to calculate the result:
1 cm → 0.39370078740157 in
21.4 cm → L(in)
Solve the above proportion to obtain the length L in inches:
L(in) = 21.4 cm × 0.39370078740157 in
L(in) = 8.4251968503937 in
The final result is:
21.4 cm → 8.4251968503937 in
We conclude that 21.4 centimeters is equivalent to 8.4251968503937 inches:
21.4 centimeters = 8.4251968503937 inches
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 inch is equal to 0.11869158878505 × 21.4 centimeters.
Another way is saying that 21.4 centimeters is equal to 1 ÷ 0.11869158878505 inches.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that twenty-one point four centimeters is approximately eight point four two five inches:
21.4 cm ≅ 8.425 in
An alternative is also that one inch is approximately zero point one one nine times twenty-one point four centimeters.
## Conversion table
### centimeters to inches chart
For quick reference purposes, below is the conversion table you can use to convert from centimeters to inches
centimeters (cm) inches (in)
22.4 centimeters 8.819 inches
23.4 centimeters 9.213 inches
24.4 centimeters 9.606 inches
25.4 centimeters 10 inches
26.4 centimeters 10.394 inches
27.4 centimeters 10.787 inches
28.4 centimeters 11.181 inches
29.4 centimeters 11.575 inches
30.4 centimeters 11.969 inches
31.4 centimeters 12.362 inches
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My teacher wants us to solve this using excel. I'll leave out the extra stuff. "Your radar provides the following information about the incoming warhead at time of detection (t=0 seconds): vertical hight= 490,000 feet (about 93 miles) with an incoming vertical velocity= -4,107 feet/second. You know the following information about your Patriot missiles located at sea level: initial upwards velocity if 6,115 ft/second. You also know that each Chinese missile is typically designed to explode at 3 miles (about 16,000 feet) above the ground... armed with this information and using quadratic mathematics and MS Excel, answer the following questions: 1. If your missiles require a minimum of 10 seconds to launch from t=0, calculate the time and altitude of earliest interception. 2. What is the latest time (as measured from t=0) that you can launch the Patriots and still expect to intercept the missile...?" He wants us to put everything, including graphs, in a word document with the answers to the questions. I'd like to know how to get the functions to enter into excel. The early and late Patriot launches are confusing...
asked Nov 27, 2016
Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: To avoid this verification in future, please log in or register.
The mathematical equation for the warhead's height after time t seconds is h=490000-4107t. At t=0, h=490000.
The intercepting missiles are delayed by 10 seconds so the equation for their height is h=6115(t-10) so that when t=10, h=0. The warheads and interceptors are assumed to moving at uniform velocity, so graphically the height equations for both missiles are straight lines. Where they cross represents interception, which must take place before the warheads explode at h=16000. t=(490000-16000)/4107=115.41secs.
To find out the earliest time for interception we have 6115(t-10)=490000-4107t, when the missile and interceptor are at the same height. 6115t-61150=490000-4107t; 10222t=551150, t=53.92 secs. (h=268,550.56 ft approx.)
The latest time for interception is t=115.41 secs when the missiles explode. To reach a height of 16000 feet it takes the interceptor 16000/6115=2.62 seconds. Add on the 10 seconds to launch=12.62 secs. It takes the missile 115.41 secs to reach 16000 feet so the interceptor must be launched no later than t=115.41-12.62=102.79 seconds.
To illustrate this using Excel we could draw the two graphs initially, h being the vertical axis and t the horizontal: h=490000-4107t and h=6115(t-10). This represents the earliest time of interception. Then we need to "slide" h=6115(t-10) to the right to cross the other graph where h=16000. Where this shifted graph intercepts the t axis should correspond to 102.79 seconds. Some scaling down will be necessary for h perhaps measuring height in 1000's or 10000's of feet. The t axis should be scaled to accommodate about 120 seconds and of course the origin is on the extreme left.
If a graph is not used, Excel formula should be used to create a table of h and t values suitably spaced.
=IF(A1<10,0,6115*(A1-10)) can be used as a formula for the interceptor.
=IF(490000-4107*A1<16000,16000,490000-4107*A1) can be used as a formula for the missile. The graph can be inserted using the tabulated data.
Cell A1 contains t, so column A contains all the values of t you want to use. B1 and C1 correspondingly contain data for h for the two lines.
answered Nov 27, 2016 by Top Rated User (425,140 points)
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# NCERT Exemplar Solutions Class 8 Mathematics Solutions for Rational Numbers - Exercise in Chapter 1 - Rational Numbers
Question 93 Rational Numbers - Exercise
A 117\frac{1}{3} m long rope is cut into equal pieces measuring 7\frac{1}{3}m each. How many such small pieces are these?
From the question it is given that,
The length of the rope = 117\frac{1}{3} m
=\frac{(117\times3+1)}{3}=\frac{352}{3}\ m
Then length of each piece measures = 7\frac{1}{3} m=\frac{22}{3}m
So, the number of pieces of the rope
=\frac{total\ length\ of\ the\ rope}{length\ of\ each\ piece}=\frac{352}{3}\div\frac{22}{3}
=\frac{352}{3}\times\frac{3}{22}=\frac{16}{1}\times\frac{1}{1}
= 16
Hence, number of small pieces cut from the 117\frac{1}{3} m long rope is 16.
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# A swimming pool is 40m long and 15m wide. Its shallow and deep ends are 1.5 m and 3m deep respectively. If the bottom of pool slopes uniformly, find the amount of water in litres required, find the amount of water in litres required to fill the pool.
1
by mathbubble8
2015-01-19T18:26:02+05:30
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
So we can consider the pool is in a shape of a cuboid + a half cuboid with same dimension
so total volume = 3/2V = 3/2 x 40 x 1.5 x 15 = 1350 m³ = 1350000 L (1 m³ = 1000 L)
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The coordinates of two points are A(-2,6) & B(9,3). Find the coordinates of the point C on the X-axis such that AC=BC.
A(-2,6) B (9,3)
We need to find C on x-axis ==> C (x,0)
such that :
AC = BC
AC= sqrt(0-6)^2 + (x+2)^2]= sqrt[(36 + (x+2)^2]
BC= sqrt[(0-3)^2+ (x-9)^2]= sqrt[(9 + (x-9)^2]
==> AC = BC
==> sqrt(36+(x+2)^2]= sqrt[(9+ (x-9)^2]
square both sides:
==> 36 + (x+2)^2 = 9 + (x-9)^2
==> 36 + x^2 +4x + 4 = 9 + x^2 -18x + 81
Now group similars:
==> 22x -50 = 0
==> x= 50/22= 25/11
Then the point C is (25/11, 0)
Approved by eNotes Editorial Team
Let the point C be at (x,y).
AC--> D^2 = (-2 - x)^2 + (6 - y)^2 = x^2 + 4x + 4 + y^2 - 12y + 36
BC--> D^2 = (x - 9)^2 + (y - 3)^2 = x^2 - 18x + 81 + y^2 - 6y + 9
AC = BC
x^2 + 4x + 4 + y^2 - 12y + 36 = x^2 - 18x + 81 + y^2 - 6y + 9
11x - 25 - 3y = 0
The point C lies on this line.
On the x-axis, y= 0. Therefore:
11x - 25 = 0
x = 25/11
Approved by eNotes Editorial Team
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# Thread: Difference between an IVP and BVP
1. ## Difference between an IVP and BVP
FOlks I have googled with this entry 'difference between ivp and bvp' but did not find any comprehensive information.
The question is a part of past exam paper.
Explain the difference between an IVP and BVP in the context of PDE's giving explicit examples of each.
Thanks
2. Difference between an Initial Value Problem(IVP) and a Boundary Value Problem(BVP):
- IVP: all the values needed to solve the particular problem are specified at a single point.
- BVP: all the values needed to solve the particular problem are specified at different points.
3. Originally Posted by bugatti79
FOlks I have googled with this entry 'difference between ivp and bvp' but did not find any comprehensive information.
The question is a part of past exam paper.
Explain the difference between an IVP and BVP in the context of PDE's giving explicit examples of each.
Thanks
Dear bugatti79,
Well I think this will clarify your doubts, Boundary value problem - Wikipedia, the free encyclopedia.
4. The terms "initial value" and "boundary value" come from physics application where, for example, determining the motion of an object under a given force where we are typically given the position and speed at t= 0 (of course, it not necessary that the point at which we are given all additional conditions be 0) while problems where we are given values at two or or more points are problems where we are looking for the values inside a boundary.
There is an important mathematical difference between these kinds of problems: if we have a second order differential equation with given initial values whether or not a unique solution exists depends only on the equation, not the initial conditions, while whether or not there exists a solution to the same equation, with given boundary conditions, depends upon the boundary conditions as well.
example: the intial value problem y''+ y= 0, y(0)= A, y'(0)= B has the unique solution y(t)= A cos(t)+ B sin(t). But the boundary value problem y''+ y= 0, y(0)= 0, y(pi)= 1 has NO solution while the bvp y''+ y= 0, y(0)= 0, y(pi)= 0 has an infinite number of solutions.
Think of it this way: you are firing a rifle down a range. An initial value problem would be where you fix the rifle at a given point and raise the barrel at a given angle. When you pull the trigger the bullet will follow a specific path. A boundary value problem would be where you fix the rifle at a given point and want to hit a target at a given distance down the range, by adjusting the barrel's angle. You might NOT be able to do that at all- the target might not be within the rifles range. Or there might not be a unique solution-if the target is within range, there can be two different angles that will give the same distance.
5. Originally Posted by HallsofIvy
The terms "initial value" and "boundary value" come from physics application where, for example, determining the motion of an object under a given force where we are typically given the position and speed at t= 0 (of course, it not necessary that the point at which we are given all additional conditions be 0) while problems where we are given values at two or or more points are problems where we are looking for the values inside a boundary.
There is an important mathematical difference between these kinds of problems: if we have a second order differential equation with given initial values whether or not a unique solution exists depends only on the equation, not the initial conditions, while whether or not there exists a solution to the same equation, with given boundary conditions, depends upon the boundary conditions as well.
example: the intial value problem y''+ y= 0, y(0)= A, y'(0)= B has the unique solution y(t)= A cos(t)+ B sin(t). But the boundary value problem y''+ y= 0, y(0)= 0, y(pi)= 1 has NO solution while the bvp y''+ y= 0, y(0)= 0, y(pi)= 0 has an infinite number of solutions.
Think of it this way: you are firing a rifle down a range. An initial value problem would be where you fix the rifle at a given point and raise the barrel at a given angle. When you pull the trigger the bullet will follow a specific path. A boundary value problem would be where you fix the rifle at a given point and want to hit a target at a given distance down the range, by adjusting the barrel's angle. You might NOT be able to do that at all- the target might not be within the rifles range. Or there might not be a unique solution-if the target is within range, there can be two different angles that will give the same distance.
Excellent! Thats pretty clear! Thanks to all.
6. Originally Posted by HallsofIvy
The terms "initial value" and "boundary value" come from physics application where, for example, determining the motion of an object under a given force where we are typically given the position and speed at t= 0 (of course, it not necessary that the point at which we are given all additional conditions be 0) while problems where we are given values at two or or more points are problems where we are looking for the values inside a boundary.
There is an important mathematical difference between these kinds of problems: if we have a second order differential equation with given initial values whether or not a unique solution exists depends only on the equation, not the initial conditions, while whether or not there exists a solution to the same equation, with given boundary conditions, depends upon the boundary conditions as well.
example: the intial value problem y''+ y= 0, y(0)= A, y'(0)= B has the unique solution y(t)= A cos(t)+ B sin(t). But the boundary value problem y''+ y= 0, y(0)= 0, y(pi)= 1 has NO solution while the bvp y''+ y= 0, y(0)= 0, y(pi)= 0 has an infinite number of solutions.
Think of it this way: you are firing a rifle down a range. An initial value problem would be where you fix the rifle at a given point and raise the barrel at a given angle. When you pull the trigger the bullet will follow a specific path. A boundary value problem would be where you fix the rifle at a given point and want to hit a target at a given distance down the range, by adjusting the barrel's angle. You might NOT be able to do that at all- the target might not be within the rifles range. Or there might not be a unique solution-if the target is within range, there can be two different angles that will give the same distance.
Second bugatti79's comments. Very clear. Nice analogy.
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# differences between Initial value problem and Boundary value problem
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Related Articles
Honaker Prime Number
• Last Updated : 17 Nov, 2020
Honaker Prime Number is a prime number P such that the sum of digits of P and sum of digits of index of P is a Prime Number.
Few Honaker Prime Numbers are:
131, 263, 457, 1039, 1049, 1091, 1301, 1361, 1433, 1571, 1913, 1933, 2141, 2221,…
### Check if N is a Honaker Prime Number
Given an integer N, the task is to check if N is a Honaker Prime Number or not. If N is an Honaker Prime Number then print “Yes” else print “No”.
Examples:
Input: N = 131
Output: Yes
Explanation:
Sum of digits of 131 = 1 + 3 + 1 = 5
Sum of digits of 32 = 3 + 2 = 5
Input: N = 161
Output: No
Approach: The idea is to find the index of the given number and check if sum of digits of index and N is the same or not. If it is same then, N is an Honaker Prime Number and print “Yes” else print “No”.
## C++
`// C++ program for the above approach` `#include ` `#define limit 10000000` `using` `namespace` `std;` `int` `position[limit + 1];` `// Function to precompute the position` `// of every prime number using Sieve` `void` `sieve()` `{` ` ``// 0 and 1 are not prime numbers` ` ``position[0] = -1, position[1] = -1;` ` ``// Variable to store the position` ` ``int` `pos = 0;` ` ``for` `(``int` `i = 2; i <= limit; i++) {` ` ``if` `(position[i] == 0) {` ` ``// Incrementing the position for` ` ``// every prime number` ` ``position[i] = ++pos;` ` ``for` `(``int` `j = i * 2; j <= limit; j += i)` ` ``position[j] = -1;` ` ``}` ` ``}` `}` `// Function to get sum of digits` `int` `getSum(``int` `n)` `{` ` ``int` `sum = 0;` ` ``while` `(n != 0) {` ` ``sum = sum + n % 10;` ` ``n = n / 10;` ` ``}` ` ``return` `sum;` `}` `// Function to check whether the given number` `// is Honaker Prime number or not` `bool` `isHonakerPrime(``int` `n)` `{` ` ``int` `pos = position[n];` ` ``if` `(pos == -1)` ` ``return` `false``;` ` ``return` `getSum(n) == getSum(pos);` `}` `// Driver Code` `int` `main()` `{` ` ``// Precompute the prime numbers till 10^6` ` ``sieve();` ` ``// Given Number` ` ``int` `N = 121;` ` ``// Function Call` ` ``if` `(isHonakerPrime(N))` ` ``cout << ``"Yes"``;` ` ``else` ` ``cout << ``"No"``;` `}`
## Java
`// Java program for above approach ` `class` `GFG{ ` `static` `final` `int` `limit = ``10000000``; ` `static` `int` `[]position = ``new` `int``[limit + ``1``]; ` ` ` `// Function to precompute the position ` `// of every prime number using Sieve ` `static` `void` `sieve() ` `{ ` ` ``// 0 and 1 are not prime numbers ` ` ``position[``0``] = -``1``; ` ` ``position[``1``] = -``1``; ` ` ` ` ``// Variable to store the position ` ` ``int` `pos = ``0``; ` ` ``for` `(``int` `i = ``2``; i <= limit; i++) ` ` ``{ ` ` ``if` `(position[i] == ``0``) ` ` ``{ ` ` ` ` ``// Incrementing the position for ` ` ``// every prime number ` ` ``position[i] = ++pos; ` ` ``for` `(``int` `j = i * ``2``; j <= limit; j += i) ` ` ``position[j] = -``1``; ` ` ``} ` ` ``} ` `} ` `// Function to get sum of digits` `static` `int` `getSum(``int` `n)` `{` ` ``int` `sum = ``0``;` ` ``while` `(n != ``0``) ` ` ``{` ` ``sum = sum + n % ``10``;` ` ``n = n / ``10``;` ` ``}` ` ``return` `sum;` `}` `// Function to check whether the given number` `// is Honaker Prime number or not` `static` `boolean` `isHonakerPrime(``int` `n)` `{` ` ``int` `pos = position[n];` ` ``if` `(pos == -``1``)` ` ``return` `false``;` ` ``return` `getSum(n) == getSum(pos);` `}` `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ``// Precompute the prime numbers till 10^6` ` ``sieve();` ` ``// Given Number` ` ``int` `N = ``121``;` ` ``// Function Call` ` ``if` `(isHonakerPrime(N))` ` ``System.out.print(``"Yes\n"``);` ` ``else` ` ``System.out.print(``"No\n"``); ` `} ` `} ` `// This code is contributed by shubham `
## Python3
`# Python3 program for the above approach ` `limit ``=` `10000000` `position ``=` `[``0``] ``*` `(limit ``+` `1``)` `# Function to precompute the position ` `# of every prime number using Sieve ` `def` `sieve(): ` ` ` ` ``# 0 and 1 are not prime numbers ` ` ``position[``0``] ``=` `-``1` ` ``position[``1``] ``=` `-``1` ` ``# Variable to store the position ` ` ``pos ``=` `0` ` ``for` `i ``in` `range``(``2``, limit ``+` `1``):` ` ``if` `(position[i] ``=``=` `0``): ` ` ` ` ``# Incrementing the position for ` ` ``# every prime number` ` ``pos ``+``=` `1` ` ``position[i] ``=` `pos ` ` ` ` ``for` `j ``in` `range``(i ``*` `2``, limit ``+` `1``, i):` ` ``position[j] ``=` `-``1` `# Function to get sum of digits ` `def` `getSum(n): ` ` ``Sum` `=` `0` ` ` ` ``while` `(n !``=` `0``): ` ` ``Sum` `=` `Sum` `+` `n ``%` `10` ` ``n ``=` `n ``/``/` `10` ` ` ` ``return` `Sum` `# Function to check whether the given` `# number is Honaker Prime number or not ` `def` `isHonakerPrime(n): ` ` ``pos ``=` `position[n] ` ` ` ` ``if` `(pos ``=``=` `-``1``): ` ` ``return` `False` ` ` ` ``return` `bool``(getSum(n) ``=``=` `getSum(pos))` `# Driver code` `# Precompute the prime numbers till 10^6 ` `sieve() ` `# Given Number ` `N ``=` `121` `# Function Call ` `if` `(isHonakerPrime(N)): ` ` ``print``(``"Yes"``)` `else``:` ` ``print``(``"No"``)` `# This code is contributed by divyeshrabadiya07`
## C#
`// C# program for above approach ` `using` `System;` `class` `GFG{ ` `static` `readonly` `int` `limit = 10000000; ` `static` `int` `[]position = ``new` `int``[limit + 1]; ` ` ` `// Function to precompute the position ` `// of every prime number using Sieve ` `static` `void` `sieve() ` `{ ` ` ``// 0 and 1 are not prime numbers ` ` ``position[0] = -1; ` ` ``position[1] = -1; ` ` ` ` ``// Variable to store the position ` ` ``int` `pos = 0; ` ` ``for` `(``int` `i = 2; i <= limit; i++) ` ` ``{ ` ` ``if` `(position[i] == 0) ` ` ``{ ` ` ` ` ``// Incrementing the position for ` ` ``// every prime number ` ` ``position[i] = ++pos; ` ` ``for` `(``int` `j = i * 2; j <= limit; j += i) ` ` ``position[j] = -1; ` ` ``} ` ` ``} ` `} ` `// Function to get sum of digits` `static` `int` `getSum(``int` `n)` `{` ` ``int` `sum = 0;` ` ``while` `(n != 0) ` ` ``{` ` ``sum = sum + n % 10;` ` ``n = n / 10;` ` ``}` ` ``return` `sum;` `}` `// Function to check whether the given number` `// is Honaker Prime number or not` `static` `bool` `isHonakerPrime(``int` `n)` `{` ` ``int` `pos = position[n];` ` ``if` `(pos == -1)` ` ``return` `false``;` ` ``return` `getSum(n) == getSum(pos);` `}` `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` ` ``// Precompute the prime numbers till 10^6` ` ``sieve();` ` ``// Given Number` ` ``int` `N = 121;` ` ``// Function Call` ` ``if` `(isHonakerPrime(N))` ` ``Console.Write(``"Yes\n"``);` ` ``else` ` ``Console.Write(``"No\n"``); ` `} ` `} ` `// This code is contributed by 29AjayKumar`
Output:
```No
```
Reference: https://oeis.org/A033548
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
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• question_answer If a number is multiplied by three-fourth of itself, the value thus obtained is 10800. What is that number? A) 210 B) 180 C) 120 D) 160
Let the number be $x.$ $\therefore$ $x\frac{3x}{4}=10800$ $\Rightarrow$ ${{x}^{2}}=\,\frac{10800\,\,\,\,4}{3}=14400$ $\therefore$ $x=\sqrt{14400}\,\,=120$
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ME3122E - Tutorial Solution 4
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Problem Set 4-Solutions
1. Hot exhaust gases used in a finned-tube cross-flow heat exchanger heat 2.5kg/s of water from 35 to 85oC. The gases [cp = 1.09 kJ/kg.K] enter at 200oC and leave at 93oC. The overall heat-transfer coefficient is 180W/m2K. Calculate the area of the heat exchanger using (a) the LMTD approach and (b) the effectiveness – NTU method. (c) If the water flow rate is reduced by half, while the gas flow rate is maintained constant along with the fluid inlet temperatures. Calculate the percentage reduction in heat transfer as a result of this reduced flow rate. Assume that the overall heat transfer coefficient remains the same. [Ans: (a) 37.8 m2 (b) 37.8 m2 (c) 15%.]
Solution: a) 1 2 1 2 2 1 1 1 1 2 2 1
LMTD LMTD
2
37.8m
U=180W/m2K Water t1=350C m =2.5 Hot gases T1=2000C t2 =850C T2 =930C A T Th1 200 Th2 93 Tc1 85 Tc2 35 1 2
(2)
b) min max min max min 2
g g g g w w g
c) min max min min max 2 2
( )
g g w w g g g g g g
(3)
2. A shell-and-tube heat exchanger operates with two shell passes and four tube passes. The shell fluid is ethylene glycol, which enters at 140oC and leaves at 80oC with a flow rate of 4500 kg/h. Water flows in the tubes, entering at 35oC and leaving at 850C. The overall heat-transfer coefficient for this arrangement is 850 W/m2K. Calculate the flow rate of water required and the area of the heat exchanger. [Ans: 0.984 kg/s, 5.24 m2].
The flow rate of glycol to the exchanger is reduced in half with the entrance temperatures of both fluids remaining the same. What is the water exit
temperature under these new conditions, and by how much is the heat-transfer rate reduced? [Ans: 70.9oC, 28.2%]
Solution: 3 max min min max
g g w w w g g
min 2o 2
A T 140 80 85 35 1 2 Ethylene glycol water
(4)
min min min max 1 2 2 2
( )
g g g g g g o g g g
1 1
water w w
(5)
3. A small steam condenser is designed to condense 0.76 kg/min of steam at 85 kN/m2 with cooling water at 10oC. The exit water temperature is not to exceed 57oC. The overall heat-transfer coefficient is 3400 W/m2K. Calculate the area required for a double-pipe heat exchanger. Use both LMTD and effectiveness-NTU methods. [Ans: 0.145m2]. Solution: g 2 fg f 1 2 LMTD 1 2 2 i) Steam is condensed at 85KN/m h =2.27 MJ/kg 0.76 Q = m h 2270 = 28.75kW 60 T ln( / ) (95-10)-(85-57) = 58.3 ln(85 / 38) Q=UA LMTD 28.75 A= 0.145 3.4 58.3 T T T T C m 2 1 1 1 max min c c c min min min ii) C , T T 57 10 = 0.553 T T 95 10
For heat exchanger with steam condensing, 1
=0.553=1 , giving NTU =0.805 UA NTU= 0.805, Q= (57 10) 28753 611.77 Su h NTU NTU steam C water e e C C C
2
bstituting qives A=0.145m
10 57 Steam Tsat =950C A 1 2 water A A T
(6)
4. A shell-and-tube heat exchanger consists of 135 thin-walled tubes in a double-pass arrangement, each of 12.5 mm diameter with a total surface area 47.5 m2. Water (the tube-side fluid) enters the heat exchanger at 15oC and 6.5 kg/s and is heated by exhaust gas entering at 200oC and 5 kg/s. The gas may be assumed to have the properties of atmospheric air, and the overall heat transfer coefficient is approximately 200 W/m2. What are the gas and water outlet temperatures? Assuming fully developed flow, what is the tube-side convection coefficient? [Ans: To(water) = 41.6oC, h = 2320 W/m2K.]
Solution: i) 2 2 min max min max
g pg w pw g w
2 2 2 1 1 g min
g g g w w
A T 2000C Tg2 Tw1 15 1 2 Gas water
(7)
ii) bw 2 3 6 0.8 0 w
d d
.4 w 2
(8)
5. A single-pass cross-flow heat exchanger uses hot exhaust gases (mixed) to heat water (unmixed) from 30oC to 80oC at a rate of 3 kg/s. The exhaust gases, having thermophysical properties similar to air, enter and exit the exchanger at 225oC and 100oC, respectively. If the overall heat transfer coefficient is 200W/m2K, estimate the required surface area. [Ans: 35.2m2].
Solution:
2 1 LMTD
3 / ; 4.2 / ,
1.005 /
Use LMTD Method, counter flow equivalent all the temperatures are known
q=m ( ) 3 4.20 (80 30) 630kJ/s (225-80)-(100-30) T = 103 ln(145/70) R(or Z)= w w g w w w w m kg s c KJ kgK c KJ kgK c T T C LMTD 2 min 225-100 2.5 80-30 80 30 0.26 225 30
for single pass, crossflow hx, 1mix, 1unmixed, correction factor F=0.92
Hence
q=630=UAF T =0.200 A 0.92 103
giving A=33.2m
method just as straight forward, C P NTU
1 2 1 2 max min max , C C Calculate C 225 110 0.64 225 30 , , gas g g water w w g g g w C m c C m c T T T T
from chart get NTU hence A
A T 2250C 100 80 30 1 2 Gas water
(9)
6. The oil in an engine is cooled by air in a cross-flow heat exchanger where both fluids are unmixed. Atmospheric air enters at 30oC and 0.53 kg/s. Oil at 0.026 kg/s enters at 75oC and flows through a tube of 10 mm diameter. Assuming fully developed flow and constant wall heat flux, estimate the oil-side heat transfer coefficient. If the overall convection coefficient is 53 W/m2K and the total heat transfer area is 1 m2, determine the effectiveness. What is the exit temperature of the oil? [Ans: 46.2oC]
Solution:
2 2
o2
cross-flow hx, fluids unmixed
oil inside tube, 75 ( ), 0.026kg/s air, 30 (inlet), 0.53kg/s
U=53W/m , 1
Constant heat flux for tube, find h(oil side)
Another case of T (outlet) unknown assume C inlet C K Am o oil o2 b b 3 4 2 T 45 T (75 45) / 2 60
(Other procedure may lead to non-integer value of T ) Tables: engine oil @ 60 :
864 / , 2.047 / , 0.839 10 / , 0.14 / Pr 1050 p C C C kg m C kJ kgK m s k W mK
(10)
4 2 h h min c c m
oil
1 1 1 ax min max min min min
(not far from assumed value of 45)
o o o h h h h c h
(11)
7. Water flows over a 3-mm-diameter sphere at 6 m/s. The free-stream temperature is 38oC, and the sphere is maintained at 93oC. Calculate the heat-transfer rate.
[Ans: 84.9W].
Solution:
o
3 3
at T =38 C, properties for water =993kg/m , 0.682 10 / 0.63 / , Pr 4.55 kg ms k W mK
Should not be evaluated at Tf =38+93/2=65.50C because Nu eqn is evaluated at free steam condition for liquid (water)
Water u=6m/s T=380C d=3mm Ts=930C 3 0.25 -0.3 0.54 4 w 2 2 993 0.003 6 Re 26, 210 2000 0.682 10 appropriate Nu expression is NuPr 1.2 0.53Re 130.09 μ evaluated at 93 , 3.06 10 / 250.4 0.63 250.42 52589 / 0.003 (4 ) (93 w w C kg ms hd Nu k h W m K Q r h 38) 84.9W
(12)
8. Air at 90oC and 1 atm flows past a heated 1/16-in-diameter wire at a velocity at 6 m/s. The wire is heated to a temperature of 150oC. Calculate the heat transfer per unit length of wire. [Ans: 59.86 W/m].
Solution: f 5 3 Air at 1 atm T 90 , 6 /
First determine Re of the flow
Air properties are evaluated at the film temperature 90 150 T 120 393 2 2 1.01325 10 0.898 / 287 393 1.013 / ; 2.25 f w f f p f C u m s T T C K P kg m RT c kJ kgK 5 3 d 5 1/3 0.466 1/3 6 10 / 0.03314 / ; Pr 0.690 0.898 6 (1.5875 10 ) Re 379.1 2.256 10 : 0.683, 0.466 Pr 0.683(Re ) Pr convection heat transfer coefficient
.0. f n d d kg ms k W mK u d From table C n u d hd Nu C k k h d 0.466 1/3 0.466 0.33 3 2 " w 3 683(Re ) Pr 0.03314 0.683 (379.1) 0.690 1.5875 10 200.5 /
Heat transfer from wire to air
( ) .( )( ) .( )(T ) 200.5 (1.5875 10 )(150 90) 60.0 / d w w W m K q hA T T h dl T T q q h d T l W m
References
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# Bayesianism in the Philosophy of Math
Today I’ll sketch an idea that I fist learned about from David Corfield’s excellent book Towards a Philosophy of Real Mathematics. I read it about six years ago while doing my undergraduate honors thesis and my copy is filled with notes in the margins. It has been interesting to revisit this book. What I’m going to talk about is done in much greater detail and thoroughness with tons of examples in that book. So check it out if this is interesting to you.
There are lots of ways we could use Bayesian analysis in the philosophy of math. I’ll just use a single example to show how we can use it to describe how confident we are in certain conjectures. In other words, we’ll come up with a probability for how plausible a conjecture is given the known evidence. As usual we’ll denote this P(C|E). Before doing this, let’s address the question of why would we want to do this.
To me, there are two main answers to this question. The first is that mathematicians already do this colloquially. When someone proposes something in an informal setting, you hear phrases like, “I don’t believe that at all,” or “How could that be true considering …” or “I buy that, it seems plausible.” If you think that the subject of philosophy of mathematics has any legitimacy, then certainly one of its main goals would be to take such statements and try to figure out what is meant by them and whether or not they seem justified. This is exactly what our analysis will do.
The second answer is much more practical in nature. Suppose you conjecture something as part of your research program. As we’ve been doing in these posts, you could use Baye’s theorem to give two estimates on the plausibility of your conjecture being true. One is giving the most generous probabilities given the evidence, and the other is giving the least generous. You’ll get some sort of Bayesian confidence interval of the probability of the conjecture being true. If the entire interval is low (say below 60% or something), then before spending several months trying prove it your time might be better spent gathering more evidence for or against it.
Again, mathematicians already do this at some subconscious level, so being aware of one way to analyze what it is you are actually doing could be very useful. Humans have tons of cognitive biases, so maybe you have greatly overestimated how likely something is and doing a quick Bayes’ theorem calculation can set you straight before wasting a ton of time. Or you could write all this off as nonsense. Whatever. It’s up to you.
If you’ve followed the posts up to now, you’ll probably find this calculation quite repetitive. You can probably guess what we’ll do. We want to figure out P(C|E), the probability that a conjecture is true given the evidence you’ve accumulated. What goes into Bayes’ theorem? Well, P(E|C) the probability that we would see the evidence we have supposing the conjecture is true; P(C) the prior probability that the conjecture is true; P(E|-C) the probability we would see the evidence we have supposing the conjecture is not true; and P(-C) the prior probability that the conjecture is not true.
Clearly the problem of assigning some exact probability to any of these is insanely subjective. But also, as before, it should be possible to find the most optimistic person about a conjecture to overestimate the probability and the most skeptical person to underestimate the probability. This type of interval forming should be a lot less subjective and fairly consistent. One should even have strong arguments to support the estimates which will convince someone who questions them.
Let’s use the Riemann hypothesis as an example. In our modern age, we have massive numerical evidence that the Riemann hypothesis is true. Recall that it just says that all the zeroes of the Riemann zeta function in the critical strip lie on the line with real part 1/2. Something like the first 10,000,000,000,000 zeroes have been checked by computer plus lots (billions?) have been checked in random other places after this.
Interestingly enough, if this were our “evidence” our estimation of P(E|C) may as well be 1, but P(E|-C) would have to contribute a significant non-trivial factor in the denominator of Bayes’ theorem. This is because we estimate this probability based on what we’ve seen in the past in similar situations. It turns out that in analytic number theory we have several prior instances of the phenomenon of a conjecture looking true for exceedingly large numbers before getting a counterexample. In fact, Merten’s Conjecture is explicitly connected to the Riemann hypothesis and the first counterexample could be around $10^{30}$ (no explicit counterexample is known, just that one exists, but we know by checking that it is exceedingly large).
It probably isn’t unreasonable to say that most mathematicians believe the Riemann hypothesis. Even giving generous prior probabilities, the above analysis would give a not too high level of confidence. So where does the confidence come from? Remember, that in Bayesian analysis it is often easy to accidentally not use all available evidence (subconscious bias may play a role in this process).
I could do an entire series on the analogies and relations between the Riemann hypothesis for curves over finite fields and the standard Riemann hypothesis, so I won’t explain it here. The curves over finite fields case has been proven and provides quite good evidence in terms of making P(E|-C) small.
The Bayesian calculation becomes much, much more complicated in terms of modern mathematics because of all the analogies and more concretely the ways in which the RH is interrelated with theorems about number fields and Galois representations and cohomological techniques. We have conjectures equivalent to (or implying or implied by) the RH which allows us to transfer evidence for and against these other conjectures.
In some sense, essentially all this complication will only increase the Bayesian estimate, so we could simplify our lives and make some baseline estimate taking into account the clearest of these and then just say that our confidence is at least that much. That is one explanation of why many mathematicians beleive the RH even if they’ve never explicitly thought of it that way. Well, this has gone on too long, but I hope the idea has been elucidated.
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Absolute value functions and equations
f (x) = | x |
The graph of absolute value of a linear function f (x) = | ax+ b |
Linear equation with absolute value, graphic solution
Absolute value functions and equations
The graph of the absolute value function f (x) = | x |
The definition for the absolute value of a function is given by
Thus, for values of x for which f (x) is nonnegative, the graph of | f (x)| is the same as that of f (x).
For values of x for which f (x) is negative, the graph of | f (x) | is a reflection of the graph of f (x) on the x axis.
That is, the graph of = - f (x) is obtained by reflecting the graph of = f (x) across the x-axis.
Hence, the graph of the absolute value of the function f (x) = x, i.e., | x | is
The graph of the absolute value of a linear function f (x) = | ax+ b |
Linear equation with absolute value, graphic solution
Recall that the absolute value of a real number a, denoted | a |, is the number without its sign and represents the distance between 0 (the origin) and that number on the real number line.
Thus, regardless of the value of a number a its absolute value is always either positive or zero, never negative that is, | a | > 0.
To solve an absolute value equation, isolate the absolute value on one side of the equation, and use the definition of absolute value.
If the number on the other side of the equal sign is positive, we will need to set up two equations to get rid of the absolute value,
- the first equation that set the expression inside the absolute value symbol equal to the other side of the equation,
- and the second equation that set the expression inside the absolute value equal to the opposite of the number on the other side of the equation.
Solve the two equations and verify solutions by plugging the solutions into the original equation.
If the number on the other side of the absolute value equation is negative then the equation has no solution.
Example: | -3 - x | = 2
Solution: -3 - x = 2 or -3 - x = -2 x = - 5 x = -1 The solutions to the given equation are x = - 5 and x = -1.
Example: | x + 1 | = 2x - 3.
Solution: x + 1 = 2x - 3 or x + 1 = -(2 x - 3)
x - 2x = - 4 x + 2x = 3 - 1
x = 4 3x = 2 => x = 2/3
Check solutions:
x = 4 => | x + 1 | = 2x - 3, x = 2/3 => | x + 1 | = 2x - 3
| 4 + 1 | = 2 · 4 - 3 | 2/3 + 1 | = 2 · 2/3 - 3
5 = 5 5/3 is not equal -5/3
The check shows that x = 2/3 is not a solution, because the right side of the equation becomes negative. There is a single solution to this equation: x = 4.
Example: | x + 2 | = | 2x - 5 |
Solution: As both sides of the equation contain absolute values the only way the two sides are equal is, the two quantities inside the absolute value bars are equal or equal but with opposite signs.
x + 2 = 2x - 5 or x + 2 = -(2 x - 5)
x - 2x = - 5 - 2 x + 2 = -2 x + 5
-x = -7 3x = 3
x = 7 x = 1
Check solutions:
x = 7 => | x + 2 | = | 2x - 5 |, x = 1 => | x + 2 | = | 2x - 5 |
| 7 + 2 | = | 2 · 7 - 5 | | 1 + 2 | = | 2 · 1 - 5 |
9 = 9 3 = | -3 |
Therefore, the solutions to the given equation are x = 1 and x = 7.
College algebra contents C
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# How can momentum but not energy be conserved in an inelastic collision?
In inelastic collisions, kinetic energy changes, so the velocities of the objects also change.
So how is momentum conserved in inelastic collisions?
I think all of the existing answers miss the real difference between energy and momentum in an inelastic collision.
We know energy is always conserved and momentum is always conserved so how is it that there can be a difference in an inelastic collision?
It comes down to the fact that momentum is a vector and energy is a scalar.
Imagine for a moment there is a "low energy" ball traveling to the right. The individual molecules in that ball all have some energy and momentum associated with them:
The momentum of this ball is the sum of the momentum vectors of each molecule in the ball. The net sum is a momentum pointing to the right. You can see the molecules in the ball are all relatively low energy because they have a short tail.
Now after a "simplified single ball" inelastic collision here is the same ball:
As you can see, each molecule now has a different momentum and energy but the sum of all of their momentums is still the same value to the right.
Even if the individual moment of every molecule in the ball is increased in the collision, the net sum of all of their momentum vectors doesn't have to increase.
Because energy isn't a vector, increasing the kinetic energy of molecules increases the total energy of the system.
This is why you can convert kinetic energy of the whole ball to other forms of energy (like heat) but you can't convert the net momentum of the ball to anything else.
• The "vector" versus "scalar" argument lacks rigor. Can you back this up with some math?
– Paul
Feb 13, 2016 at 20:46
• @Paul: consider two identical particles, mass $m$: their center of mass moves with velocity $\vec v=(\vec v_1+\vec v_2)/2;$ their kinetic energy is $\frac12mv_1^2+\frac12mv_2^2.$ Of this energy, only $\frac12(2m)v^2=\frac14m\left(v_1^2+v_2^2+2v_1v_2\right)$ is externally visible as center-of-mass motion; the remaining $\frac14m(v_1-v_2)^2$ energy forms a "reservoir" for energy which cannot be observed in their center-of-mass; any situation with hidden reservoirs of stuff can violate conservation laws of that stuff. There is no analogous situation here for momentum as $(2m)v=mv_1+mv_2.$ Feb 19, 2016 at 17:54
• Downvoted. This answer is so very wrong. Momentum and energy are conserved quantities in classical mechanics. There's a lot more to energy than just kinetic energy, which means that kinetic energy is not necessarily a conserved quantity. Feb 19, 2016 at 18:44
• @DavidHammen my answer wasn't intended to be a comprehensive treatise on the subject but rather offer an intuitive understanding of how such a thing could be possible. I also mentioned other manifestations of energy in the last sentence. With that said, do you think the answer is technically wrong or just an incomplete simplification? Feb 19, 2016 at 21:01
• Wouldn't the same logic that says momentum is conserved (treating the system as a whole), also say that the energy of the system as a whole is conserved as well? Oct 26, 2016 at 13:53
So how is momentum conserved in inelastic collisions?
It is a basic law of physics that momentum is always conserved - there is no known exception. Kinetic energy does not need to be conserved, because it can turn into other forms of energy - for example potential energy or internal/thermal energy ("heat"). Momentum can also turn into other form of momentum - momentum of the EM field - but the amount of momentum so transformed seems negligible in ordinary collisions of macroscopic bodies.
• This and Eric Angle's answer are the best and most concise here. Aug 11, 2016 at 6:22
• The OP asked the question "how" momentum was conserved. This answer doesn't really address that, it just restates the (unchallenged) premise of the question which is that momentum is of course conserved. Mar 28, 2017 at 11:41
• Hi Jan Lalinsky: Did you accidentally create two accounts and want to merge them? Oct 4, 2017 at 15:56
• @Qmechanic, it is the unconfirmed account from the past that I no longer use. You may merge it with my main account. Oct 4, 2017 at 16:05
• The mods cannot merge accounts, only the owner or the SE team. I leave it to you to contact the SE team. Oct 4, 2017 at 16:08
Energy and momentum are always conserved. Kinetic energy is not conserved in an inelastic collision, but that is because it is converted to another form of energy (heat, etc.). The sum of all types of energy (including kinetic) is the same before and after the collision.
• This and Ján Lalinský's answer are the best and most concise here. Aug 11, 2016 at 6:22
• If some kinetic energy is converted to heat, this must make the particles who absorb this energy vibrate with greater KE. Doesnt a faster moving particle have more momentum too? May 19, 2019 at 14:46
None of these answers really address the question; mostly they just reiterate physics principles that I suspect the poster already understands.
The question essentially asks, 'If kinetic energy changes in different types of collision, then the final velocities must be changing. If the final velocities are changing, the final momentum must be changing, but momentum is supposed to be conserved. How can this be?'
I had the same question this morning, which is how I ended up here. The key to understanding this is to realize that there is a range of final velocities that all conserve momentum, and that each point in this range represents a different amount of kinetic energy.
For example: a 1kg ball labeled $$A$$ moving at 3m/s strikes another 1kg ball $$B$$ that is at rest. Here are a few of the possible end velocities, with total kinetic energy for each case, and total momentum:
$$\begin{array}{|c|c|c|c|} \hline v_A & v_B & K_{AB} & p_{AB} \\ \hline \mathrm{m/s} & \mathrm{m/s} & \mathrm{J} & \mathrm{kg \cdot m/s} \\ \hline 0.0 & 3.0 & 0.0 + 4.5 = 4.5 & 0.0 + 3.0 = 3.0 \\ 0.5 & 2.5 & 0.125 + 3.125 = 3.25 & 0.5 + 2.5 = 3.0 \\ 1.0 & 2.0 & 0.5 + 2.0 = 2.5 & 1.0 + 2.0 = 3.0 \\ 1.5 & 1.5 & 1.125 + 1.125 = 2.25 & 1.5 + 1.5 = 3.0 \\ \hline \end{array}$$ Notice that each of these represents the same amount of momentum (equal to the starting momentum) but they all yield different amounts of kinetic energy! This is why we need both conservation laws — momentum and energy — to solve certain problems; without both, we aren't able to constrain the answer to a particular point within the range of solutions (see above!) that conserve momentum.
Law of conservation of momentum is directly implied by Newton's laws of motion. Basically it is conserved even in inelastic collision because forces appear in pairs with equal magnitude and opposite direction as shown :
The two dark dots are two particles. The direction of arrows shows the direction of force and the length of the arrows shows their magnitude.
In all the physical phenomena the forces can be represented by the above mentioned image. For the inelastic collision this image also holds e.g. consider an inelastic collision as shown:
The block of mass $M$ is initially at rest and a bullet is moving towards it with a velocity $v_i$ and mass $m_w$. What happens during the collision? There appears a force pair of equal magnitude and in opposite direction. The force pair continuously varies in magnitude during the collision. The force pair is kinetic friction It continue to act until the relative velocity of bullet w.r.t the block becomes zero that is, both bullet and block acquires same velocity. These forces are represented as $\vec F_b$ and $\vec F_w$. $\vec F_b$ acts on the bullet towards left and $\vec F_w$ acts on block towards right.
By newton's third law $\vec F_w=-\vec F_b$
Change in momentum of bullet $= \Delta p_b = \int_{t_1}^{t_2}\vec F_b dt$
Change in momentum of the block$= \Delta p_w = \int_{t_1}^{t_2}\vec F_w dt$ It is easy to recognize that since $\vec F_w=-\vec F_b$ the decrease in momentum of bullet appears as the increase in momentum of the block. Now the underlying fact is that if decrease in velocity of bullet causes a decrease in the momentum of bullet then at the same time the velocity of the block increases which causes the momentum of block to increase.
A different scenario can take place if the block is not stationary but is moving towards the bullet. Further let the initial momentum of the system is 0 what would happen after the collision? the bullet will sink in the block and the velocities of both block and bullet will become 0! That is, total kinetic energy of the system becomes 0!. We see K.E of the system may be changed (not the total energy) but momentum of the system cannot.
For a better explanation you should read these
1. http://www.feynmanlectures.caltech.edu/I_10.html
2. http://www.physicsclassroom.com/class/momentum/u4l2b.cfm
• Can't we go further to say that; Newton's third law of motion is a direct consequence of Columb's law and superposition principle. Sep 23, 2020 at 15:57
Conservation of Momentum falls directly out of Newton's Laws.
Consider Newton's Third Law: $\sum \vec{F} = 0$
And Newton's Second Law: $\vec{F} = m \vec{a} = \frac{d\vec{p}}{dt}$
Combining these two laws we find: $\sum \frac{d\vec{p}}{dt} = 0$
This equation states that total momentum cannot change with respect to time. That is, the total momentum cannot change before or after the collision, irregardless of the type of collision. Thus momentum is always conserved.
Eric Angle has it pretty much right. In an inelastic collision some of the kinetic energy is absorbed by the deformation of the material. For example, if two balls of putty collide and stick together, kinetic energy is absorbed by squishing the putty. In a second example, if you shoot a bullet at a log, some of the kinetic energy is absorbed by friction as the bullet passes into the wood. In both cases, some of the kinetic energy is turned into heat, so although energy is conserved, kinetic energy is not.
In an elastic collision, the objects bounce off each other. During the collision, the material momentarily deforms and absorbs some of the energy, but then bounce back like a spring, giving the energy back up. So in a elastic collision, kinetic energy is conserved.
The conservation of momentum is simply a statement of Newton's third law of motion. During a collision the forces on the colliding bodies are always equal and opposite at each instant. These forces cannot be anything but equal and opposite at each instant during collision. Hence the impulses (force multiplied by time) on each body are equal and opposite at each instant and also for the entire duration of the collision. Impulses of the colliding bodies are nothing but changes in momentum of colliding bodies. Hence changes in momentum are always equal and opposite for colliding bodies. If the momentum of one body increases then the momentum of the other must decrease by the same magnitude. Therefore the momentum is always conserved.
On the other hand energy has no compulsion like increasing and decreasing by same amounts for the colliding bodies. Energy can increase or decrease for the colliding bodies in any amount depending on their internal make, material, deformation and collision angles. The energy has an option to change into some other form like sound or heat. Hence if the two bodies collide in a way that some energy changes from kinetic to something else or if the deformation of the bodies takes place in a way that they cannot recover fully then energy is not conserved. This option of changing into something else is not available to momentum due to Newton's third law of motion.
This is why momentum is always conserved but kinetic energy need not be conserved.
Further an elastic collision is defined in such a way that it's energy is taken to be conserved. Nothing like an elastic collision exists in nature. It is an ideal concept defined as such. Empirical measurements will always show that collisions are always inelastic
• Dear sukhveer choudhary. It is often frown upon to post nearly identical answers to similar posts. In such cases, it is often better to just flag/comment about duplicate questions, so they can get closed. May 12, 2015 at 12:01
Despite the inelasticity of the collision, the momentum will be conserved. The kinetic energy will change. Some work was done in the process of the "absorption" of energy during the inelastic collision and this will reduce the resultant kinetic energy.
Some of the other answers have derived the conservation principle from Newton's laws, but I think the more fundamental derivation was done by Emmy Noether, who discovered that our notion of invariance of physical laws under infinitesimal coordinate changes gives rise to conservation laws. Momentum is the conserved quantity that falls out of the symmetry with respect to spatial translation.
Who ever said the answer is "momentum is a vector and energy is a scalar" is correct, saying energy gets transformed into heat just kicks the can down the road, to "why can the KE get transformed?" I think the best answer starts by pretending that energy and momentum don't really exist (other than in our mind as a mathematically construction to help solve problems easier and quicker). What really exists are masses that interact by applying forces through time (impulses which are a vector) and/or by applying force over a distance (work is is the dot product of two vectors and therefore a scalar). Its the loss of the sign on the scalar that hurts our ability to the work. Example a internal spring causes two carts to separate. The fact the speed is increased for both carts increases the kinetic energy of system from zero to a positive number. Also heat is just internal kinetic energy (force times distance) in various random directions and again hard to track because of the loss of the sign when force and displacement are in the same direction. The fact that positive work is defined at when Force and motion are in the same direction means sometimes an upward force is positive at the same time a downward force is positive. This removes our ability to "follow" the discriminate force displacement interactions. this is my 1st post so sorry if it is crude.
• I'm sorry but I fail to see how this answers the question... Feb 19, 2016 at 17:32
• Let me welcome you to Physics Stack Exchange! This is a question-and-answer site for people with specific conceptual questions about physics and their answers. I hope that you're able to stick around and contribute productively to our task of answering discussion questions. Feb 19, 2016 at 18:05
The net mass after an inelastic collision changes.
We know that the momentum will still remain conserved.
Hence,when two bodies collide inelastically
$$(mv)=(m+M)u$$ $$=> u=\frac{m}{(M+m)}v$$
Now,finding the change in kinetic energy of the body with mass $m$
$\frac{1}2mv^2 -1/2(m){[m/(m+M)]v}^2$
and hence we can see that that the kinetic energy is not conserved
Using the work energy theorem, we can say that the lost kinetic energy must get converted into another form of energy. Generally this energy is some form of non-mechanical energy such as thermal energy.
• Welcome on Physics SE and thank you for the answer :) Please refer to physics.stackexchange.com/help/notation/ for help in typesetting formulas. Aug 11, 2016 at 7:13
• Net mass of the system remains constant. Nov 18, 2016 at 14:10
Consider two particles of masses 'a' and 'b' , having variable velocities 'x' and 'y' respectively.
Now this is perfectly a math problem.
The two variables 'x' and 'y' are dependent on each other such that the term ax+by is always a constant. [By the conservation of linear momentum]
Now, how sure are you that the term (1/2)ax² + (1/2)by² (i.e, kinetic energy) is constant ?
It's not constant, right ?
The question is based on a mistaken assumption that the loss of total KE from a system of two or more particles implies a corresponding loss of overall momentum, which is not true.
For a single particle, a reduction of kinetic energy does imply a reduction in momentum- this is because a change to the quantity v2 necessarily involves a change in v.
For two or more interacting particles, a reduction in their combined KE does not necessarily result in a reduction of momentum. This is because it is always possible to find a set of numbers DVi that sum to zero even if the sum of their squares is non-zero.
Consider the trivial example of two particles of unit mass, one at rest the other moving with a speed of 4. Their combined KE is 8 units, and their combined momentum is 4 units. If they collide and coalesce (a totally inelastic collision) their speed will be 2 units. Their combined momentum will be 4 units, as before, and their combined KE will be 4 units, a reduction of 50%.
I think that there's a misunderstanding here, momentum is conserved, but energy is also conserved in the collision. The quantity that isn't conserved is the kinetic energy, and the reason it is not conserved in an inelastic collision is that some of the kinetic energy is lost in deformation of the ball (in other words work was done) and some of it is lost in other forms of energy such as heat, sound, etc.
Total momentum changes due to the influence of external forces over time, and during a collision the time taken is aproximately zero. So there is no net effect on total momentum. On the other hand kinetic energy is influenced by internal forces also which are substatial for a collision.
• The total momentum can't be changed regardless of time because momentum is a vector. With energy (not a vector), you can convert kinetic energy of the objects into kinetic energy of the molecules in any direction (heat). With momentum though, the vector direction is conserved so even though the molecules gain momentum from their movement, the sum of all molecule momentums is still exactly the net momentum in the direction the object was traveling in the first place. Jan 4, 2014 at 20:14
I get that there are some geniuses of physics here answering this question. But I think the answer is simply that the kinetic energy was not conserved because work was done. So the kinetic energy is converted to work.
• If work is done by a body on another body in the system, it doesn't necessarily imply that the energy of the system is being reduced. Total mechanical energy is only reduced when there are non conservative forces involved in the collision. Nov 18, 2016 at 14:08
• Why downvote the comment? My statement is correct. Mar 5, 2018 at 14:00
# It is todo with the fact that kinetic energy depends on a non linear function of two velocities.
Consider the simple case of two 1kg balls in the PHET physics simulator.
This is a fascinating question. It would seem that if kinetic energy and momentum both depend on the same velocities they would both be conserved.
But the kinetic energy has a non linear dependence on velocity. This permits energy to be lost while the momentum remains conserved.
• Conservation of momentum is a consequence of Newtons laws in a system isolated from external forces.
• This example shows that it may not be to do with the scalar /vector argument mentioned above but instead as a result of the non linearity of the kinetic energy polynomial.
Ball 1 is moving to the right at 1 m/s
Ball 2 is at rest
Before the inelastic collision the energy of the system is $\color{blue}{0.5 \text{J} }$
$$\frac{1}{2}m v^2+\frac{1}{2}m v^2 = \frac{1}{2}(1) (1)^2+\frac{1}{2}(1) (0)^2$$
The total momentum of the system is $\color{red}{1\text{ kg m/s} }$ $$\frac{1}{2}m v^2+\frac{1}{2}m v^2 = (1) (1)+(1) (0)$$
After the inelastic collision the energy of the system is $\color{blue}{0.31 \text{J} }$ This is because Ball1 is now going at $\color{red}{0.25 \text{m/s}}$ and Ball2 is now going at $\color{red}{0.75 \text{m/s}}$
$$\frac{1}{2}m v^2+\frac{1}{2}m v^2 = \frac{1}{2}(1) (0.25)^2+\frac{1}{2}(1) (0.75)^2 = \color{blue}{0.31 }$$
But the total momentum of the system is still $\color{red}{1\text{ kg m/s} }$!!! $$\frac{1}{2}m v^2+\frac{1}{2}m v^2 = (1) (0.25)+(1) (0.75)= \color{red}{1}$$
# Conclusion
Even though the velocities change in the inelastic collision, the momentum still adds up to 1, but the energy square law cannot follow. Mathematics permits the system to have the same momentum but a different kinetic energy after the collision.
The blue line is a line of constant momentum of 1 kg m/s. The orange circle is a constant kinetic energy of 0.5. The green circle is a constant kinetic energy of 0.31. Even though both the initial velocities (red point) and the final velocities (black point) lie on the blue line and correspond to the same momentum, they can lie on different kinetic energy circles.
• The different power dependence tells you that a very limited set of imaginable outcomes conserve both momentum and bulk-kinetic energy. Fine. But you haven't explained why the observed outcomes that fail to conserve both are those than conserve momentum and not those that conserve bulk kinetic energy. Jan 27, 2018 at 21:31
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# An introduction to Refractional redshift, and how it was confused with gravitational redshift
#### Marvas
An introduction to Refractional redshift, and how it was confused with gravitational redshift
Marius L. Vasile @ Vasile effect
In this article I will prove that refraction causes a redshift, which I cleverly named Refractional redshift, and that this redshift was not so cleverly confused with gravitational redshift by the world’s finest scientists- we’re talking Harvard, Nobel prize level here- which were either ignorant of refraction, or doctors in doctoring experiments with it.
Refractional redshift is by far the most common and yet unknown type of redshift- certainly for astronomers, who have no ideea that it exists. It is caused due to the fact that during refraction the speed of light changes, but the frequency remains constant. Since f=v/lambda, where lambda is the wavelength, it immediately follows that the wavelength changes too in order to preserve the frequency. This results in an increase of wavelength or a redshift when the speed of light increases, and in a decrease in wavelength or a blueshift when its speed decreases.
The demonstration is quite simple, as we only need two simple equations in order to show that refraction causes a redshift or a blueshift.
The first is the wave equation f=v/λ, or λ=v/f
and the second is the refraction equation n=c/v, or v=c/n
(where n is the index of refraction of the medium, v is the speed of light in a medium, and c is the speed of light in a vacuum)
We insert the second into the first and we get λ=c/nf.
Since f is constant during refraction, and c is also constant, we see that when the index of refraction n increases, the wavelength decreases (shifts to blue), and when n decreases the wavelength increases (shifts to red).
For two mediums with refractive index n1 and n2, wavelength is directly proportional with speed: λ1/λ2=v1/v2,
and inversely proportional with the index of refraction: λ1/ λ2=n2/n1
For example during refraction from the sun’s atmosphere or heliosphere into space, the index of refraction decreases from n>1 to n=1 and the speed of light increases from v to c, so the wavelength also increases and the light emitted by the sun gets shifted to red. This would also explain why almost all galaxies appear redshifted, since they are made of stars which are all redshifted from refraction.
(Before reaching space there are, however, multiple refractions inside the sun’s atmosphere, which has many layers with different indexes of refraction, which regress as they aproach space (i.e. the outer layers have a lower index than the inner layers, which are more dense). So the light emiited by the sun gets more and more redshifted as it refracts through these layers before it reaches into space, the final frontier, and gets refracted and redshifted again.)
And those who claim that the ‘gravitational potential’ of the sun is causing a shift in wavelength, or a gravitational shift, are simply ignoring the laws of refraction, which explains quite simply why light is redshifted near massive objects, which are all surrounded by dense atmospheres made from gases such as hydrogen and helium which affect the speed and wavelength of light.
It is important to understand that light does not always propagate at a constant speed c, which is the speed of light in a vacuum. In every other situation, when it encounters a medium, it travels at slower speeds, which varies with the index of refraction of the mediums, n=c/v. So instead of c, it travels at a lower speed v=c/n. This slowing of EM waves in the atmosphere of massive objects has been confused with a gravitational time delay or time dilation, once more revealing the sheer ignorance of relavistic scientists like Shapiro who ignored refraction and foolishly concluded that the time delay is caused by gravitation.
Therefore, when we observe a redshift, we must not assume that light has travelled at a fixed speed (or c in a vacuum) until it reached us. And associate the redshift with a drop in frequency, as scientists do, because they dont take variable speed of light from refraction into account, and use c as the standard value for the speed of light, instead of v=c/n. So by ignoring the mediums in which light traveled, they erroneusly apply the formula f=c/λ, instead of f=v/λ=c/nλ. Thus, if λ increases or decreases, they will erroneusly conclude that frequency changed too, and illogically define redshift as such:
‘In physics, a redshift is an increase in the wavelength, and corresponding decrease in the frequency and photon energy, of electromagnetic radiation (such as light).’ -Wikipedia
While it may be well be true that frequency decreases in the Doppler redshift, it certainly does not imply that ALL redshifts in the universe will do the same thing !
And certainly not in refractional redshift, where an increase of wavelength can occur without any decrease in frequency, because the speed of the wave will increase.
And this is exactly what happened in the famous Pound and Rebka gravitational redshift experiment.
Because that experiment was not done in a vacuum, but in the earth’s atmosphere, in which a bag of helium was added ‘to minimise scattering’. More specifically, the gamma ray traveled through the helium bag, and then through air, as the metal target and the detector were placed outside the helium bag, as shown in the picture below.
‘The gamma rays traveled through a Mylar bag filled with helium to minimize scattering’ (wikipedia)
It doesn’t take a genius to realise that the gammaray was refracted from helium into air, which has a higher index of refraction than helium, causing its wavelength to decrease and shift to blue, or increase and shift to red, depending on the different setups of the experiment (in other setup they placed the emitter under the helium tube, and the detector above it, creating an air to helium refraction and a refractional redshift). The gammaray was indeed shifted, but by refraction, and not by gravitation. Yet for Harvard University and the Nobel Academy this was a gravitational shift, just like Einstein predicted.
Except Einstein did not predict that gravitational redshift occurs in the presence of helium, and nowhere in his proposed tests of general relativity does helium appear.
Why does it appear in this experiment, then ?
Because, there simply was no gravitational shift in the absence of helium. So they blamed it on ‘scattering’ from air, not on Einstein’s theory being wrong, and thrown in a bag of helium to prove it right. But it simply does not follow how the use of helium leads to a gravitational shift, because it simply follows that will only lead to a refractional shift.
Because the helium and air mediums in which the experiment was set obviously caused a refraction of the gammaray and a change in its speed and wavelength. So what they observed was just a Refractional redshift/blueshift, and not a gravitational one.
The only logical conclusion of that experiment is that refractional redshift exists and it was confused (deliberately or not) with gravitational redshift by the researchers. Their Nobel prize should be cancelled and the scientific community should immediatelly revise all experiments which claim that confirmed general relativity. Because this experiment did not confirm general relativity at all, and in fact it infirmed it. Since the shift was caused by refraction, and not by gravitation.
Given that helium is not gravity, and it does not appear in Einstein’s theory of general relativity or in his proposed tests to confirm it, Pound and Rebka should have not used it in this gravitational experiment, unless they were really desperate. Indeed, they were so desperate to get the Nobel prize, that they even drilled holes in the floors of their Harvard university (presumably to impress the Nobel jury). When they could have simply used the stair well (as a gravitational well, of course). So adding a bag of helium to produce a redshift out of thin air, was just another act of desperation.
This experiment should have been performed ideally in a vacuum or, if not possible, in the same medium in order to avoid refraction and refractional shift. As it was, it did not prove anything other than refraction changes the wavelength, which was probably known at the time by some scientists, including some Harvard ones, but excluding the Nobel ones. And the Nobel Academy is ultimately responsable for reviewing, approving, and awarding Harvard’s completely failed (or doctored) experiment.
Pound and Rebka attempted to prove that gravity causes redshift by the absurd addition of helium, which is not gravity and is never mentioned by Einstein in his general relativity. Because it has absolutely nothing to do with it.
Maybe they liked playing with helium, but even kids know that it alters the sound waves. This is also what happens to light waves. And the Nobel academy should have known this as well, and immediatelly call the scam instead of awarding it. But they fell for it because they did not know that refraction changes the wavelength and causes redshift/blueshift, which is also true for mainstream scientists today, who all confuse refractional redshift with gravitational redshift.
And then we wonder why cosmology is in a crysis. Because the standard cosmological model i.e. big bang theory is entirely based on general relativity and relativistic redshifts, which do not exist because they are pseudo-science. So from now on it should be called the standard cosmoillogical model.
The gravitational redshift experiment is one of the three tests proposed by Einstein to confirm general relativity. He stated that if any of these tests fails, then his whole theory would collapse like a house of cards. In this paper I have proved that the g-shift experiment was fundamentally flawed by the use of helium, and that the refraction from helium into air (or viceversa) produced the blueshift/redshift- which had nothing to do with Einstein’s theory of gravity. So it’s game over for general relativity.
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Rod Mack
#### Marvas
That is my article. I also posted it on Academia. And on another forum.
So far it has not been peer reviewed, but it does not matter because other papers which have been reviewed are obviously false. Like Pound and Rebka's, Pound and Snider's, Shapiro's, Eddington's and pretty much all relativistic experiments which can be easily disproved with basic refraction physics.
The latter (Eddington) has been debunked by DR. Edward Dowdie from NASA who showed there is no light bending outside of the sun's corona, where the bending is caused by refraction. That alone is enough to disprove general relativity, but I disproved the other experiments because I was sick of all these scientists who claim that general relativity is true because there are so many experiments which have proved it. When in fact, none of those experiments have proved it, and basically they have all disproved it because they were either doctored or foolishly interpreted by these Einstein fanboys who apparently failed/skipped highschool just like him and went straight to university.
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#### COLGeek
##### Cybernaut
Moderator
Bold statements.
We'll see who responds to your premise.
Peer reviews certainly matter, but you should already know that. Good luck.
#### Marvas
They don't matter because I have just disproved two of them which have been peer reviewed. So the peer review process has failed miserably in at least two cases, therefore it is completelly useless. In logic only arguments and evidence matter, not the majority's opinion which is a logical fallacy just like the argument from authority. Simply put, a majority of scientists can be stupid, biased and wrong, just like an authority like the Nobel Academy can be stupid, biased and wrong.
And Doctor Edward Dowdie from NASA has completelly disproved Eddington's experiment, but of course no peer review has validated his obvious demonstration. In fact they ignored it completely, just like they do with mine. Because almost all peers are brainwashed with Einstein's pseudo physics and simply cannot accept that their idol was wrong.
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#### COLGeek
##### Cybernaut
Moderator
They don't matter because I have just disproved two of them which have been peer reviewed. In science only arguments matter, not the majority's opinion which is a logical fallacy just like the argument from authority. Simply put, a majority of scientists can be stupid and wrong, just like an authority like the Nobel Academy can be stupid and wrong.
And Doctor Edward Dowdie from NASA has disproved another one, but of course no peer review accepted his obvious demonstration. Because almost all peers are brainwashed with Einstein's pseudo physics and simply cannot accept that their idol was wrong.
You can make better arguments without resorting to insulting terminology and references, as well as by providing compelling data to support your position.
#### Marvas
I did provide compelling data to support my position that those stupid experiments are wrong, as they have been either doctored or foolishly interpreted by scientists who were ignorant of basic refraction physics. Neither Pound-Rebka nor Shapiro took into consideration variable speed of light from refraction, and simply used c everywhere in their calculations, instead of v=c/n, where n is the index of refraction. If they did that they would have realised that refraction causes both redshift and time delay, and would have not confused the obvious effects of refraction with those of gravitation. Their stupidity is nothing short of phenomenal, and perfectly fits in Einstein's infinite universe, where stupidity is also infinite. I am simply using Einstein's terminology and references here.
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#### billslugg
You need to go talk with Pentcho Valev, he worked this issue here for a long time with almost the same words you are using. He went away after awhile, couldn't get any traction.
#### Marvas
You need to go talk with Pentcho Valev, he worked this issue here for a long time with almost the same words you are using. He went away after awhile, couldn't get any traction.
He did not. He has completelly different arguments, and none involve refraction. My whole demonstration is based on refraction physics, and it's extremelly simple, since it involves two simple equations: v=c/n, and f=v/lambda. Which one don't you understand ? It's highschool physics man. Just because Einstein skipped highschool doesnt mean we all did.
#### billslugg
We need a "throw down". You and Petcho, mano a mano. We could charge money. Can't wait.
#### Marvas
I just throwed down Einstein, Harvard University, Nobel Academy, and the whole scientific community to be honest. Petcho, while a formidable adversary, is not my adversary so I have no reason to throw him down. In fact I am sure he will agree that my demonstration is correct and join my winning team, known as team Refraction. Anyone can join, as long as they finished highschool and want to teach Einstein and other pseudo-scientists like him a lesson (of refraction).
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#### COLGeek
##### Cybernaut
Moderator
I await this future debate, if it happens.
If so simple, why have the notions not gotten any traction?
#### billslugg
I'll watch but not participate. I have a long tradition of not understanding but one theoretical framework of the universe at a time and right now I'm not understanding Einstein. I'll let you know when there is an opening. Stocking up on popcorn as we speak.
#### Classical Motion
The modern equations for refraction work only because your reference is wrong. Because your reference is a wave. Light has fooled all.
Light shift is a duty cycle shift, not Doppler. And the shift is only a haft wave shift, not a full wave shift.
Invert a duty cycle.......and you have true light. Only the off time, the space time of the wave shifts.....with emitter motion. The on time remains constant with motion.
Space Width Modulation. Instead of pulse width modulation.
Light blinks, it doesn't wave. And it's constant velocity is ONLY for a stationary point. Not an observer. All observers are in motion.
Smoke that one.
I started with 4 bits. Had to toggle, then strobe instructions in....with mechanical switches. Out put was an old teletype.
#### billslugg
Not all observers are in motion. There is a preferred velocity in the universe. Simply measure the CMBR dipole anisotropy and set your cruise control in the opposite direction. Once CMBR dipole anisotropy is zero you are at rest.
There is no preferred position in the universe, all locations see themselves at the center.
Rod Mack
#### Marvas
ColGeek said:
If so simple, why have the notions not gotten any traction?
Why the notions of refraction have not gotten any traction ?
They did, just not for people who skipped highschool like Einstein. Or Shapiro and others who 'experimentally proved' his theory, by ignoring refraction altogether in their experiments.
Neither did Galileo's ideas gain any traction initially, and it was not rocket science either. The explanation is because of indoctrination and lack of critical thinking, and because the universe is infinite and so is human stupidity, according to Einstein, therefore it takes quite some time until people understand very simple things. Like when you add helium in air the light refracts from helium into air and changes its speed and wavelength, because frequency f=v/lambda remains constant, so when v increases lambda also increases, and viceversa. So the light will be redshifted/blueshifted from refraction, and not from gravitation. Or that when a radio wave reaches the sun's atmosphere it will slow down because v=c/n, so it will take a longer time to travel through its atmosphere than it would take in a vacuum. So there will be a slight time delay, which is again caused by refraction and not by gravitation. So no time does not slow down because of gravity, its just the EM waves slow down in the atmosphere of massive objects like the sun.
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# Why $(-1)^{n-1} \frac{n}{3n + 1}$ is divergent?
I am wondering the reason $$(-1)^{n-1} \frac{n}{3n + 1}$$ is divergent although the limit of $$\frac{n}{3n + 1}$$ is $$1/3.$$
Any explanation will be greatly appreciated!
• @DonThousand: The sequence can't converge since the limit is not $0$ and it does not have an ultimately constant sign. Jun 17, 2021 at 20:48
• A sequence is divergent if it doesn't converge. Since $n/(3n+1)$ converges to $1/3$, your sequence looks like $1/3,-1/3,1/3,-1/3,...$, which is osscilating.
– pax
Jun 17, 2021 at 20:49
Let $$a_n = (-1)^{n-1} \frac{n}{3n+1}$$. Then $$a_{2n} = -\frac{2n}{6n+1}$$ and $$a_{2n+1} = \frac{2n+1}{6n+4}$$. Observe that $$\lim_{n\to \infty}a_{2n} = -\frac{1}{3} \qquad \lim_{n\to \infty}a_{2n+1} = \frac{1}{3}$$
Since the limits along two subsequences differ, then $$a_n$$ diverges.
• Hi, what is the criterium that have you used? At this moment I have forgotten it. +1 Jun 17, 2021 at 21:00
• @Sebastiano If $x_n \to a$ , then any subsequence satisfies $x_{n_k} \to a$. If the conclusion does not hold for a sequence, by contrapositive it must be that the parent sequence does not converge. Jun 17, 2021 at 21:01
• Thank you very much for your collaboration....thank you again. Jun 17, 2021 at 21:04
Consider two subsequences of the given sequence, viz. the even subsequence, and the odd subsequence.
• For even $$n$$, we may write $$n=2k$$ for $$k\in\mathbb{N}$$. Then we have : $$\lim_{n\to\infty}(-1)^{n-1} \frac{n}{3n + 1} = \lim_{k\to\infty}(-1)^{2k-1} \frac{2k}{6k + 1} = -\lim_{k\to\infty} \frac{1}{3 + \frac{1}{2k}} = -\frac{1}{3}$$
• For odd $$n$$, we may write $$n=2k+1$$ for $$k\in\mathbb{N}$$. Then we have : $$\lim_{n\to\infty}(-1)^{n-1} \frac{n}{3n + 1} = \lim_{k\to\infty}(-1)^{2k+1-1} \frac{2k+1}{6k + 3 + 1} = \lim_{k\to\infty} \frac{1}{3 + \frac{1}{2k+1}} = \frac{1}{3}$$
So, the odd and the even subsequences of the original sequence converge to two different limit points. Thus, the sequence is not convergent.
Convergence implies that there exists a singular limit to the sequence, which itself requires that for all $$\epsilon > 0$$, there exists an $$N \in \mathbb{N}$$ such that $$|a_n - a| < \epsilon, \quad \forall n \geq N.$$ This convergence condition does not hold true for the sequence $$a_n = \frac{(-1)^{n-1}n}{3n+1}$$ purely due to its oscillatory nature and the fact that its limit when the $$(-1)^n$$ is removed, is nonzero.
Since the sequence $$\frac{n}{3n+1}$$ converges to a nonzero value, convergence of $$(-1)^n\frac{n}{3n+1}$$ would imply convergence of $$(-1)^n.$$ However, this sequence is known to be divergent.
Suppose a limit $$L$$ exists.
Case $$1$$: $$L\ge 0$$. Then for all $$\epsilon > 0$$, there is some $$N\in\mathbb N$$ such that for all $$n\ge N$$
$$|a_n-L|<\epsilon.$$
Choose $$\epsilon = 1/5$$, and find such an $$N$$ as above. Then as $$n=2N > N$$ we must have $$|a_{2N}-L|<1/5$$ or
$$-1/5 < L-1/5 < -\dfrac{N}{3N+1}=a_{2N} < L+1/5$$
but $$a_{2N}=-\dfrac{N}{3N+1} <-\dfrac{N}{3N+N} = -\dfrac{N}{4N} = -\dfrac{1}{4} \not > -\dfrac{1}{5}$$.
Case $$2$$: $$L<0$$. Apply similar thinking.
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# WHAT IS A implies B logically equivalent to?
## WHAT IS A implies B logically equivalent to?
In other words, A and B are equivalent exactly when both A ⇒ B and its converse are true. (A implies B) ⇔ (¬B implies ¬A). In other words, an implication is always equivalent to its contrapositive.
WHAT DOES A implies B mean in logic?
“A implies B” means that B is at least as true as A, that is, the truth value of B is greater than or equal to the truth value of A. Now, the truth value of a true statement is 1, and the truth value of a false statement is 0; there are no negative truth values.
Is p implies q equivalent to not P or Q?
Thus, “p implies q” is equivalent to “q or not p”, which is typically written as “not p or q”. This is one of those things you might have to think about a bit for it to make sense, but even with that, the truth table shows that the two statements are equivalent.
### What is the example of logical statement?
Anything that lets us infer a new fact about something mathematical from given information is a logical statement. For example, “The diagonals of a rectangle have the same length” is a logical statement. The hypothesis is the part that can help us if we know it’s true.
What mean by A implies B?
If A and B represent statements, then A B means “A implies B” or “If A, then B.” The word “implies” is used in the strongest possible sense. As an example of logical implication, suppose the sentences A and B are assigned as follows: A = The sky is overcast.
How do you use logical equivalence?
Two logical statements are logically equivalent if they always produce the same truth value. Consequently, p≡q is same as saying p⇔q is a tautology. Beside distributive and De Morgan’s laws, remember these two equivalences as well; they are very helpful when dealing with implications. p⇒q≡¯q⇒¯pandp⇒q≡¯p∨q.
## Is it logically correct to say that not a IMPLIES b?
Is it logically correct to say that if A implies B then not A implies not B? “If not-A, then not-B” is the converse of “If A then B”. The truth of a statement does not imply the truth of its converse. Result B could have many causes besides Cause A, so the negation of Cause A can not imply anything.
What does “a IMPLIES b” mean?
Stating “A implies B” is the same as claiming that “if you stick a fork in an electrical outlet, you will get hurt”. This claim may not in reality be true, but that point is irrelevant to the statement from a logical point of view. The key point is that you are claiming nothing about getting hurt if you don’t stick a fork in the outlet.
What is the difference between logically equivalent sentences?
In propositional logic, a sentence’s truth conditions are given by its truth table. Logically equivalent sentences have the same truth table, so they have the same meaning. For as much as logically equivalent expressions are synonymous, they are interchangeable with one another.
### What does a then B mean in logic?
In fact, “if A then B” (in classical logic) is usually defined as being an abbreviation for “either B or not A”. Which, again, says that for “if A then B” to hold, in all cases either B must be true, or if not, then A must be false as well.
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# Grade 5 Multiplication Flash Cards
Studying multiplication following counting, addition, and subtraction is ideal. Youngsters find out arithmetic by way of a all-natural progression. This progression of studying arithmetic is generally the subsequent: counting, addition, subtraction, multiplication, and ultimately division. This statement results in the issue why find out arithmetic within this sequence? More importantly, why learn multiplication after counting, addition, and subtraction just before section?
## The following specifics respond to these inquiries:
1. Young children learn counting initial by associating visual items using their hands and fingers. A tangible illustration: The amount of apples exist within the basket? More abstract example is just how old are you presently?
2. From counting figures, another logical move is addition then subtraction. Addition and subtraction tables can be extremely valuable training tools for the kids because they are aesthetic tools creating the move from counting easier.
3. Which will be acquired next, multiplication or section? Multiplication is shorthand for addition. At this time, young children have got a company grasp of addition. Therefore, multiplication is the after that rational form of arithmetic to find out.
## Evaluate fundamentals of multiplication. Also, assess the essentials utilizing a multiplication table.
Allow us to overview a multiplication example. By using a Multiplication Table, grow several times 3 and acquire a solution 12: 4 x 3 = 12. The intersection of row 3 and column four of your Multiplication Table is twelve; 12 is definitely the respond to. For kids starting to find out multiplication, this is certainly simple. They could use addition to eliminate the problem hence affirming that multiplication is shorthand for addition. Example: 4 by 3 = 4 4 4 = 12. It is an excellent guide to the Multiplication Table. The added benefit, the Multiplication Table is aesthetic and reflects back to learning addition.
## Where by should we begin understanding multiplication utilizing the Multiplication Table?
1. First, get acquainted with the table.
2. Get started with multiplying by one. Begin at row number one. Relocate to line number one. The intersection of row one particular and line the first is the perfect solution: one particular.
3. Perform repeatedly these techniques for multiplying by one particular. Flourish row a single by posts one by means of 12. The responses are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 respectively.
4. Replicate these steps for multiplying by two. Flourish row two by columns 1 by means of five. The solutions are 2, 4, 6, 8, and 10 correspondingly.
5. Allow us to bounce in advance. Repeat these methods for multiplying by five. Grow row 5 by columns 1 through twelve. The responses are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 correspondingly.
6. Now we will boost the amount of difficulty. Replicate these steps for multiplying by 3. Increase row a few by posts 1 by way of twelve. The responses are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, and 36 correspondingly.
7. Should you be comfortable with multiplication thus far, try a analyze. Remedy the following multiplication issues in your head and after that examine your responses on the Multiplication Table: multiply six and two, increase nine and 3, increase one particular and eleven, multiply a number of and four, and increase 7 and 2. The issue answers are 12, 27, 11, 16, and 14 correspondingly.
In the event you acquired 4 from 5 problems right, make your personal multiplication assessments. Estimate the replies in your head, and look them using the Multiplication Table.
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https://www.answers.com/Q/What_is_the_rate_at_which_the_velocity_of_an_object_changes_over_time
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Science
# What is the rate at which the velocity of an object changes over time?
###### Wiki User
Acceleration is the rate at which the velocity of an object changes over time.
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## Related Questions
The rate at which velocity changes over time is known as acceleration. In calculus, acceleration is the derivative of velocity with respect to time.
No, velocity is the instantaneous speed of an object, the rate of change would be the acceleration of the object.
An object accelerates if its velocity changes. More precisely, "acceleration" is the rate of change of velocity (how quickly velocity changes), or in symbols, dv/dt.
That rate at which the velocity of an object changes is known as the acceleration of said object. It can be defined mathematically as a=v/t where a=acceleration, v=the change in velocity, and t=the time in which this change in velocity occurred.
No. It's the rate at which a object changes velocity (speed).
Velocity is defined as the rate at which an object changes its velocity; therefore if there is zero velocity then acceleration is neutral as well.
That's simply called a change in velocity. On the other hand, the rate of change in velocity - how quickly velocity changes - is called acceleration.
Position is the place where an object is located. Speed is the rate at which position changes. Velocity is the speed and direction of motion. Acceleration is the rate at which velocity changes, and the direction of the change. So you could say that acceleration is the rate at which (the rate at which position changes and the direction) changes, and the direction of the change.
The rate at which velocity changes is known as the acceleration of an object.Calculating acceleration, given velocity can be achieved with the formula:acceleration = change in velocity / change in time
Acceleration is the rate of change in velocity. (Velocity is speed+direction, it is a vector as well as acceleration). F=ma, where a is in ms^(-2).
Velocity is the rate at which location changes, and the direction in which location changes. Acceleration is the rate at which the size of velocity changes, and the direction in which velocity changes.
'Velocity' means the rate at which position changes, and the direction in which it changes. 'Acceleration' means the rate at which velocity changes, and the direction in which it changes.
Acceleration is the rate at which an object's velocity changes over time.
Acceleration is the rate at which an object's velocity changes over time.
The rate at which velocity changes in time is called acceleration.
accelerationThe rate at which velocity changes is called "acceleration".
Velocity is speed with a direction. Acceleration is the rate at which velocity changes, just as velocity is the rate at which position changes (and jerk is the rate at which acceleration changes).
The change in velocity is just the change in velocity. The RATE of change of velocity - how quickly velocity changes - is usually called "acceleration".
The rate of which something changes its velocity is the acceleration. A common mistake is to think that something moving fast is accelerating, but it's only accelerating if the velocity of that object is changing.
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# Euler's Equations for Rigid Body Rotations
### Key Takeaways
• A rigid body is one in which the distance between two internal points remains unchanged when force is applied to it.
• Rigid body motion can be expressed in two coordinate frames - the space-fixed inertial frame and the body-fixed non-inertial frame.
• Except for a few exceptional cases, such as torque-free rigid body and integrable cases solved by Euler, Lagrange, and Kovalevskaya, solutions are yet to be evaluated for Euler’s equations.
A satellite is an example of a rigid body
You may not realize it, but rigid body motion is something you probably encounter on a daily basis. For example, the dumbbell that you use for workouts is a rigid body. Pyramids, planets, and satellites are also rigid bodies. Rigid bodies have six degrees of freedom for movement and their motion can be translational, rotational, or both.
In mechanics, rigid body motion is mathematically described using Euler’s equations. Euler’s equations for rigid body motion include three non-linear equations coupled with differential equations. Solving Euler’s equations is a challenging task and engineers have yet to find a complete general analytical solution.
In this article, we will explore rigid body motion and look at Euler’s equations as a way to express this motion.
## Rigid Body Motion
A rigid body is one in which the distance between two internal points remains unchanged when force is applied to it. In terms of the deformation aspect, a body that does not change its shape under the influence of force is a rigid body. When considering rigid bodies as the collection of different particles of mass held in places by massless bonds of length, the particles are unaffected by vibrations, stress, and strain. Particles in a rigid body do not have any internal degrees of freedom, and the distance between any two particles remains fixed all the time. This approach to describing a rigid body helps in expressing the dynamics as the cumulative sum over the particles. In ordinary solids, dynamics can be expressed by integrals over the continuous mass distribution.
## Two Reference Frames for Expressing Rigid Body Motion
Rigid body motion can be expressed in two coordinate frames. The frame of reference of rigid body motion can be in:
1. Space-fixed inertial frame - The reference frame in which Newton’s second law holds. Unless the rigid body is spherical, it is difficult to get far with the inertial frame, as the inertia seen in this coordinate system varies with time. It is hard to define rigid body motion in the inertial frame, especially when it is rotating.
2. Body fixed non-inertial frame - The inertia tensor is a known value and is a constant. It is simpler to calculate the equations of motion of the rigid body in a body-fixed principal frame.
## Euler’s Equations for Rigid Body Motion
In this section, we will express the rotational motion of a rigid body in a body-fixed frame, ignoring the translational motion for simplicity.
According to Newton’s second law, the external torque N can be expressed as the following, where L is the angular momentum:
The above equation is in the space-fixed inertial frame and needs to be transformed into the body-fixed frame. Transformed, the equation can be written as:
The body axis is chosen to be the principal axis such that:
Ii is the principal moments of inertia about principal axis ‘ωi’ and i is the angular velocity about the principal axis ‘i’.
In the body-fixed coordinate frame, the equations of motion of a rigid body can be expressed as:
The components in the body-fixed axes can be given by:
The above equations represent Euler's equations for the rigid body under applied external torque N in the body-fixed reference frame.
Euler's equations of rigid body rotation are hard to solve, and engineers are still trying to find solutions. Except for a few exceptional cases, such as torque-free rigid body and integrable cases solved by Euler, Lagrange, and Kovalevskaya, solutions are yet to be evaluated.
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# Cramér's theorem
For the number of points to determine a curve, see Cramer's theorem (algebraic curves).
In mathematical statistics, Cramér's theorem (or Cramér’s decomposition theorem) is one of several theorems of Harald Cramér, a Swedish statistician and probabilist.
## Normal random variables
Cramér's theorem is the result that if X and Y are independent real-valued random variables whose sum X + Y is a normal random variable, then both X and Y must be normal as well. By induction, if any finite sum of independent real-valued random variables is normal, then the summands must all be normal.
Thus, while the normal distribution is infinitely divisible, it can only be decomposed into normal distributions (if the summands are independent).
Contrast with the central limit theorem, which states that the average of independent identically distributed random variables with finite mean and variance is asymptotically normal. Cramér's theorem shows that a finite average is not normal, unless the original variables were normal.
## Large deviations
Cramér's theorem may also refer to another result of the same mathematician concerning the partial sums of a sequence of independent, identically distributed random variables, say X1, X2, X3, …. It is well known, by the law of large numbers, that in this case the sequence
${\displaystyle \left({\frac {\sum _{k=1}^{n}X_{k}}{n}}\right)_{n\in \mathbb {N} }}$
converges in probability to the mean of the probability distribution of Xk. Cramér's theorem in this sense states that the probabilities of "large deviations" away from the mean in this sequence decay exponentially with the rate given by the Cramér function, which is the Legendre transform of the cumulant-generating function of Xk.
## Slutsky's theorem
Slutsky’s theorem is also attributed to Harald Cramér.[1] This theorem extends some properties of algebraic operations on convergent sequences of real numbers to sequences of random variables.
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# Main Categories
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# Write Linear Equations From Tables, Points & Graphs Review Problem Pass Activity
Product Description
Twelve rounds include practice or review writing linear equations in slope-intercept form from tables of values, two points / ordered pairs and graphs.
Students work with a partner while seated at their desks. They should write in their own notebook or on the blank Answer Sheet (included).
At the same time the first pair of students is working on Problem 1, the second pair works on Problem 2, the third pair works on Problem 3, etc. At the teacher’s signal, all students pass their problem in a specific direction. The students who started with Problem 2 should now pass it to the students who started with Problem 1, the students who started with Problem 3 pass it to the students who started with Problem 2, etc. The first pair of students starting with Problem 1 should deliver their finished problem to the last pair of students, or the teacher may prefer to deliver these each time. Students now flip over their new page to find the ANSWER to the problem they just finished.
Students continue to work problems in order, pass problems and check their answers on the back of the next problem page until they have completed all problems included in the activity.
It is important for the teacher to tell the entire class when to pass their first problem and to tell everyone where to look for the first answer on the back of their NEXT problem. After this first pass together, I’ve allowed students to work and pass problems at their own pace. Teachers may prefer to set a timer so all students pass their current problem to the next pair of students simultaneously.
This activity works well in the middle of a lesson while students are actively practicing a new skill or can be used as a review.
The answer key is built into the activity so students check for accuracy themselves.
CCSS: Use functions to model relationships between quantities.
8.F.B.4 Construct a function to model a linear relationship between two quantities. Determine the rate of change and initial value of the function from a description of a relationship or from two (x, y) values, including reading these from a table or from a graph. Interpret the rate of change and initial value of a linear function in terms of the situation it models, and in terms of its graph or a table of values.
CCSS: Construct and compare linear, quadratic, and exponential models and solve problems.
HSF.LE.A.2 Construct linear and exponential functions, including arithmetic and geometric sequences, given a graph, a description of a relationship, or two input-output pairs (include reading these from a table).
CCSS: Create equations that describe numbers or relationships.
HSA.CED.A.2 Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales.
HSA.CED.A.4 Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations.
CCSS: Build a function that models a relationship between two quantities.
HSF.BF.A.1 Write a function that describes a relationship between two quantities.
CCSS: Interpret functions that arise in applications in terms of the context.
HSF.IF.B.6 Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. Estimate the rate of change from a graph.
CCSS: Interpret linear models.
HSS.ID.C.7 Interpret the slope (rate of change) and the intercept (constant term) of a linear model in the context of the data.
Teacher Setup:
Printing Option 1/Paper Saver Option: To print back-to-back copies on card stock or paper, use the activity pages with two problems per page. Print the page of Problem 1 & Problem 2 back-to-back with the page of ANSWER Problem 12 & ANSWER Problem 1; print the page of Problem 3 & Problem 4 back-to-back with ANSWER Problem 2 & ANSWER Problem 3; etc. Cut pages in half to separate problems. Prepare 2 or more complete sets of the activity to have enough pages for each pair of students in the class. Keep complete sets in order.
Printing Option 2/Full Page Option: To print single-sided copies or full pages, use the activity pages with one problem per page. If using single-sided copies, slide pages into plastic page protectors to keep problems and answers together. Put Problem 1 and ANSWER Problem 12 back-to-back in the same plastic page protector; put Problem 2 and ANSWER Problem 1 back-to-back in another plastic page protector, Problem 3 and ANSWER Problem 2 together, etc. Prepare 2 or more complete sets of the activity to have enough pages for each pair of students in the class. Keep complete sets in order.
It is very important to hand out problems in numerical order so the page with the answer on the back follows its problem number. As you hand out problems in order, problem side up, tell each pair of students the direction they should pass their problem when finished. This direction may vary by row if you zig-zag or “snake” up and down the rows of desks. The first pair of students starting with Problem 1 should deliver their finished problems to the last pair of students, or you may prefer to deliver these each time.
(Optional) Print enough copies of the blank Answer Sheet for each student to use as they work the problems. Another option is for students to write in their own notebooks.
Included in the package:
• Twelve “Problem Pass” pages in two formats for two printing options
• Answer pages for students to self-check
• Directions for two printing options
• Blank answer sheet for students (optional)
You might also like:
Find Slope & Y-Intercept Sum It Up Activity (7 stations)
Find Slope & Write Linear Equations Given Two Points Placemat Activities (2 activities)
Rewrite Linear Equations Into Slope-Intercept Form & Standard Form Placemat Activities (2 activities)
Point-Slope, Slope-Intercept & Standard Form Sum It Up Activity (8 stations)
Parallel & Perpendicular Lines Sum It Up Activity (6 stations)
Write Linear Equations From Context Problem Pass Activity (6 rounds)
Write Linear Equations In Context & Mixed Review Placemat Activities (2 activities)
Solve Linear Equations With Variables On Both Sides Relay Activities (6 rounds)
Write Linear Equations From Graphs Quiz Quiz Trade Activity (30 cards)
Find Slope From Graphs Quiz Quiz Trade Activity (30 cards)
Linear Equations In Point-Slope Form Quiz Quiz Trade Activity (30 cards)
This purchase is for one teacher only.
This resource is not to be shared with colleagues or used by an entire grade level, school, or district without purchasing the proper number of licenses. If you are a coach, principal, or district interested in a site license, please contact me for a quote at [email protected]. This resource may not be uploaded to the internet in any form, including classroom/personal websites or network drives.
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Cody
Problem 2669. Assign matrix rows/columns to separate variables
Solution 1938336
Submitted on 16 Sep 2019
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Fail
A = rand; B = matsplit(A,1); C = matsplit(A,2); assert(isequal(B,A)) assert(isequal(C,A))
Output argument "varargout{1}" (and maybe others) not assigned during call to "matsplit". Error in Test1 (line 2) B = matsplit(A,1);
2 Fail
rng default A = randi(100,3,3); [B,C,D] = matsplit(A,1); [E,F,G] = matsplit(A,2); assert(isequal(B,[82;91;13])) assert(isequal(C,[92;64;10])) assert(isequal(D,[28;55;96])) assert(isequal(E,[82,92,28])) assert(isequal(F,[91,64,55])) assert(isequal(G,[13,10,96]))
Output argument "varargout{1}" (and maybe others) not assigned during call to "matsplit". Error in Test2 (line 3) [B,C,D] = matsplit(A,1);
3 Fail
A = hankel(1:20); B = matsplit(A,1); C = matsplit(A,2); assert(isequal(C,1:20)) assert(isequal(B(:),C(:)))
Output argument "varargout{1}" (and maybe others) not assigned during call to "matsplit". Error in Test3 (line 2) B = matsplit(A,1);
4 Fail
A = toeplitz(1:4); [B,C,D,E] = matsplit(A,2); assert(isequal(B,[1 2 3 4])) assert(isequal(C,[2 1 2 3])) assert(isequal(D,[3 2 1 2])) assert(isequal(E,[4 3 2 1]))
Output argument "varargout{1}" (and maybe others) not assigned during call to "matsplit". Error in Test4 (line 2) [B,C,D,E] = matsplit(A,2);
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# right-running waves
Can you help me please? The problem is:
1.Solve the wave equation in finite interval with Dirichlet boundary condition at he right and Neumann boundary condition at the left .
2.Choose the initial conditions for right-running waves.
3.Show the phase difference of wave reflection at the boundaries.
I solved the 1'st question by assuming $u_{tt}=c^{2}u_{xx}$
for $0<x<l$ with the B.C. $u_x(0,t)=0$; $u(l,t)=0$
$u(x,t)=\sum_{n=0}^{\infty }\left [ A_n \cos \frac{(n+\frac{1}{2})\pi ct}{l}+B_n \sin \frac{(n+\frac{1}{2})\pi ct}{l} \right ]\cos \frac{(n+\frac{1}{2})\pi x}{l}$
But i stuck with part 2. and 3. of the question.
What conditions i should to choose in order to get right-running waves? Thanks for you help!
-
Left- and right-running waves take the form $u(x,t)=g(x+ct)$ and $u(x,t)=g(x-ct)$, respectively. Differentiating a right-running wave with respect to $t$ and $x$ yields
$$\frac{\partial u}{\partial t}=-cg'(x-ct)=-c\frac{\partial u}{\partial x}\;,$$
so to specify initial conditions for a right-running wave, you can specify $u(x,0)=f(x)$ with any $f$ and then require $u_t(x,0)=-cf'(x)$.
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# CHE141 Chapter 10. Chapter 10 Gases
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## Transcription
1 Chapter 0 Gases. A sample of gas (4.g) initially at 4.00 atm was compressed from 8.00 L to.00 L at constant temperature. After the compression, the gas pressure was atm. (a) (b)..00 (c)..00 (d). 6.0 Explanation: This is an example of using Boyle s law. The temperature and the number of moles of the gas are constant. According to Boyle s law where stands for the initial conditions and for final. Need to find 4.00 atm! 8.00 L.00 L 6.0 atm. A sample of a gas (5.0 mol) at.0 atm is expanded at constant temperature from 0.0 L to 5.0 L. The final pressure is atm. (a)..5 (b). 7.5 (c) (d). 3.3 Explanation: This is an example of using Boyle s law. The temperature and the number of moles of the gas are constant. According to Boyle s law where stands for the initial conditions and for final. Need to find by rearranging the above equation as follows:.0 atm! 0.0 L, 5.0 L 0.67 atm 3. A balloon occupies 4.39 L at 44.0 ºC and a pressure of 79.0 torr. What temperature must the balloon be cooled to, to reduce its volume to 3.78 L (at constant pressure)? (a) ºC (b) ºC (c)..6 ºC (d) ºC Explanation: This is an example of using Charles s law. The pressure and the number of moles are held constant while the temperature is lowered. The initial volume and temperature conditions can be labeled as and T while the final Copyright 006 Dr. Harshavardhan D. Bapat
2 conditions can be labeled as and T. The temperature T (in Kelvin) can be found by rearranging Charles s law as follows:! T 3.78 L! 37.5 K or T 73. K T T 4.39 L This is approximately equal to 0.00 ºC. 4. A gas originally at 7.0 ºC and.00 atm pressure in a 3.90 L flask is cooled at constant pressure until the temperature is.0 ºC. The new volume of the gas is L. (a)..7 (b) (c) (d). 4. Explanation: This is an example of using Charles s law. The initial volume and temperature conditions can be labeled as and T while the final conditions can be labeled as and T. The final volume (in Liters) can be found by rearranging Charles s law as follows: T! T 3.90 L! 84.5 K or 3.69 L T T K 5. If g of a gas occupies 0.0 L at ST, 9.3 g of the gas will occupy L at ST. (a). 3.9 (b) (c)..9 (d). 5.5 Explanation: Since the pressure and temperature are being held constant the volume occupied by the gas should be directly proportional to the amount of the gas. The initial conditions (of mass and volume) can be labeled as and the final conditions as. The final volume can be now found by: m m m! 9.3 g! 0.0 L or 5.5 L m g Copyright 006 Dr. Harshavardhan D. Bapat
3 6. A sample of He gas (.35 mol) occupies 57.9 L at K and.00 atm. What is the volume of this sample at 43.0 K and.00 atm? (a) (b). 4. (c). 8.6 (d)..4 Explanation: This is an example of using Charles s law. The initial volume and temperature conditions can be labeled as and T while the final conditions can be labeled as and T. The final volume (in Liters) can be found by rearranging Charles s law:! T 57.9 L! 43.0 K or 8.6 L T T T K 7. A sample of H gas (.8g) occupies 00.0 L at K and.00 atm. A sample weighing 9.49 g occupies L at K and.00 atm. (a). 09 (b). 68. (c) (d). 47 Explanation: Convert both the grams of H to moles (divide g by the molar mass of H ) and then rearrange the ideal gas law to calculate the volume. n T 00.0 L! (9.49 g/.06 g)! K or 68. L n T n T n T (.8 g/.06 g)! K 8. The amount of gas that occupies 60.8 L at 3.0 ºC and mmhg is mol. (a)..8 (b) (c). 894 (d)..6 Explanation: This problem involves using the ideal gas law and rearranging it to find the number of moles (n) of the gas. To use the ideal gas law convert the temperature to Kelvin and the pressure to atm before using these numbers. (367.0 mm Hg/760 mm Hg)! 60.8 L nrt or n.8 moles RT 0.08 Latm/molK! K Copyright 006 Dr. Harshavardhan D. Bapat 3
4 9. The pressure of a sample of 6.0 g of CH 4 gas in a 30.0 L vessel at 40.5 K is atm. (a)..4 (b). 6.6 (c) (d)..4 Explanation: This problem involves using the ideal gas law and rearranging it to find the pressure () of the gas. To use the ideal gas law convert the grams of methane to moles (n), before using this number. nrt (6.0 g/6.043 g)! 0.08 Latm/molK! 40.5 K nrt and 0.44 atm 30.0 L 0. A 0.35 L flask filled with gas at 0.94 atm and 9.0 ºC contains mol of gas. (a).4 x 0 - (b)..48 x 0 - (c). 9.4 (d)..4 Explanation: This problem involves using the ideal gas law and rearranging it to find the number of moles (n) of the gas. To use the ideal gas law convert the temperature to Kelvin before using the number. nrtand n RT 0.94 atm " 0.35 L 0.08 Latm/molK" 9.5 K.4 " 0! moles. A gas in a ml container has a pressure of torr at 9.0 ºC. Calculate the number of moles of the gas in the flask. (a)..48 x 0 - (b)..9 x 0 - (c). 9.4 (d)..4 Explanation: This problem involves using the ideal gas law and rearranging it to find the number of moles (n) of the gas. To use the ideal gas law convert the temperature to Kelvin, the pressure to atm and volume to liters before using these numbers. nrt and n RT atm 695 torr " " L 760 torr 0.08 Latm/molK" 30.5 K.9" 0! moles Copyright 006 Dr. Harshavardhan D. Bapat 4
5 . Calculate the volume occupied by a sample of gas (.30 mol) at.0 ºC and.50 atm. (a) (b) (c)..8 (d). 3 Explanation: This problem involves using the ideal gas law and rearranging it to find the volume () of the gas. To use the ideal gas law convert the temperature to Kelvin, the before using this number. nrt.30 mol! 0.08 Latm/molK! 99.5 K nrt and.8 L.50 atm 3. What is the volume of 0.65 mol of an ideal gas at torr and 97.0 ºC? (a) (b). 9.5 (c). (d). 4. Explanation: This problem involves using the ideal gas law and rearranging it to find the volume () of the gas. To use the ideal gas law convert the temperature to Kelvin and the pressure to atm before using these numbers. nrt 0.65 mol! 0.08 Latm/molK! K nrt and 4.L atm torr! 760 torr 4. The density of ammonia gas in a 4.3 L container at torr and 45.0 ºC is g/l. (a) (b) (c) (d) Explanation: The density of a gas can be calculated by using a formula that relates the density (g/l), pressure ( in atm), molar mass (M in g/mol) and temperature (T) of the gas. The molar mass of NH 3 is 7.03 g/mol. The pressure must be converted to atm and the temperature to Kelvin before using these numbers. atm torr!! 7.03g/mol M d 760 torr 0.78 g/l RT 0.08 Latm/molK! 38.5 K Copyright 006 Dr. Harshavardhan D. Bapat 5
6 5. The density of N O at.53 atm and 45. ºC is g/l. (a). 8. (b)..76 (c) (d)..58 Explanation: The density of a gas can be calculated by using a formula that relates the density (g/l), pressure ( in atm), molar mass (M in g/mol) and temperature (T) of the gas. The molar mass of N O is g/mol. The temperature must be converted to Kelvin before using this number. M.53 atm! g/mol d.58 g/l RT 0.08 Latm/molK! K 6. The molar mass of a gas that has a density of 6.70 g/l at ST is g/mol. (a). 496 (b). 50 (c) (d) Explanation: The molar mass of a gas is related to its density (in g/l), pressure (in atm) and its temperature (in K). drt 6.70 g/l! 0.08L atm/molk! 73.5 K M 50. g/mol.00 atm 7. What is the volume of hydrogen gas that can be produced by the reaction of 4.33 g of zinc with excess sulfuric acid at 38.0 ºC and torr? (a)..69 (b)..7 x 0-4 (c) x 0 4 (d)..84 Explanation: This problem uses a combination of stoichiometry and the ideal gas law. Need to write a balanced chemical equation for the reaction and then calculate the number of moles of hydrogen that can be produced from the moles of the limiting reagent (Zinc in this case). Zn + H SO 4 ZnSO 4 + H 4.33 g Zn " mol Zn g mol H " mol Zn! 6.6" 0 moles H Copyright 006 Dr. Harshavardhan D. Bapat 6
7 The volume occupied by these moles of H can now be calculated using the ideal gas law: nrt 6.6! 0 "! 0.08 Latm/molK! 3.5 K atm torr! 760 torr.69 L 8. What is the volume of HCl gas required to react with excess magnesium metal to produce 6.8 L of hydrogen gas at.9 atm and 5.0 ºC? (a). 6.8 (b)..9 (c). 3.6 (d) Explanation: This problem uses a combination of stoichiometry and the ideal gas law. The number of moles of the hydrogen can be calculated from the ideal gas law and then used with the stoichiometry of the reaction..9 atm! 6.8 L n 0.60 moles H RT 0.08 Latm/molK! 98.5 K The balanced equation for this reaction is: Mg + HCl MgCl + H The equation shows that moles of HCl are needed for every mole of H produced. Thus (0.60 x. moles) of HCl would be required for this reaction. The volume occupied by these moles of HCl can then be calculated by using the ideal gas law: nrt.! 0.08Latm/molK! 98.5 K 3.6 L.9 atm 9. The Mond process produces pure nickel metal via the thermal decomposition of nickel tetracarbonyl: Ni(CO) 4 (l) Ni (s) + 4CO(g) What volume (L) of CO is formed from the complete decomposition of g of Ni(CO) 4 at 75.0 torr and.0 ºC? (a) (b) (c). 55. (d). 0. Copyright 006 Dr. Harshavardhan D. Bapat 7
8 Explanation: This problem uses a combination of stoichiometry and the ideal gas law. Using the reaction stoichiometry from the above balanced equation, need to calculate the number of moles of CO produced, and then calculate the volume occupied using the ideal gas law g Ni(CO) nrt 4! mole g! 4 moles CO mole Ni(CO) 0.40! 0.08 Latm/molK! 95.5 K atm 75.0 torr! 760 torr moles of CO 55. L 0. Ammonium nitrite undergoes thermal decomposition to produce only gases: NH 4 NO (s) N (g) + H O (g) What volume (L) of gas is produced by the decomposition of 35.0 g of NH 4 NO (s) at 55. ºC and.50 atm? (a). 47 (b). 60 (c). 5 (d). 7. Explanation: This problem uses a combination of stoichiometry and the ideal gas law. Using the reaction stoichiometry from the above balanced equation, need to calculate the number of moles of gases formed and then calculate the volume occupied by these gases. It is not necessary to calculate the moles of the products individually. mole 3 moles (N + HO) 35.0 g NH4NO!!.64 moles g mole nrt.64! 0.08 Latm/molK! K.50 atm 7. L. The pressure in a. L vessel that contains.34 g of carbon dioxide,.73 g of sulfur dioxide and 3.33 g of argon, all at 4.0 ºC is mmhg. (a). 63. (b). 34 (c). 395 (d). 6 Explanation: Since this is a mixture of gases need to find the total number of moles of gases (n t ) and then calculate the pressure of this gas mixture. Copyright 006 Dr. Harshavardhan D. Bapat 8
9 .34 g CO.73 g SO g/mol g/mol nrt 3.33 g Ar g/mol 0.63! 0.08 Latm/molK! 35.5 K!. L 0.63 moles 760 mm atm 63. mm. A sample of He gas (3.0 L) at 5.6 atm and 5.0 ºC was combined with 4.5 L of Ne gas at 3.6 atm and 5.0 ºC at constant temperature in a 9.0 L flask. The total pressure in the flask was atm. Assume the initial pressure in the flask was 0.00 atm. (a)..6 (b). 9. (c)..0 (d). 3.7 Explanation: Calculate the number of moles of each of the gases and find the total number of moles of the gas mixture. The pressure of this gas mixture can now be calculated using the ideal gas law. n RT The number of moles of He using this formula and of Ne The moles of gas mixture.348. Now using the appropriate rearrangement of the ideal gas law the pressure of the gas mixture 3.7 atm. 3. A flask contains a mixture of He and Ne at a total pressure of.6 atm. There are.0 mol of He and 5.0 mol of Ne in the flask. The partial pressure of He is atm. (a). 9. (b). 6.5 (c)..04 (d) Explanation: Calculate the mole fraction of He and then the partial pressure of He. The partial pressure of a gas in a mixture is directly proportional to its mole fraction. n He.0 He Total moles artial pressure p 0.86!.6atm 0.74atm He Copyright 006 Dr. Harshavardhan D. Bapat 9
10 4. SO (5.00 g) and CO (6.00 g) were placed in a ml container at 50.0 ºC. The total pressure in the container was atm. (a). 0.9 (b). 4.0 (c)..76 (d) Explanation: Since each gas behaves as if the other was absent, need to calculate the partial pressure of each gas and then add them together to find the total pressure g SO moles g/mol nrt ! 0.08 Latm/molK! 33.5 K SO.76 atm L 6.00 g CO 0.36 moles g/mol nrt 0.36! 0.08 Latm/molK! 33.5 K CO 4.8 atm L Total pressure atm SO CO 5. CO (5.00 g) and CO (5.00g) were placed in a ml container at 50.0 ºC. The total pressure in the container was atm. (a). 0.3 (b). 4.0 (c). 6.3 (d). 0.9 Explanation: Since each gas behaves as if the other was absent, need to calculate the partial pressure of each gas and then add them together to find the total pressure g CO 0.78 moles 8.0g/mol nrt 0.78! 0.08 Latm/molK! 33.5 K SO 6.3atm L 5.00 g CO 0.4 moles g/mol nrt 0.4! 0.08 Latm/molK! 33.5 K CO 4.0 atm L Total pressure atm SO CO Copyright 006 Dr. Harshavardhan D. Bapat 0
11 6. In ideal gas equation calculations, expressing pressure in ascals (a), necessitates the use of the gas constant, R, equal to. (a) atm L mol - K - (b) J mol - K - (c) L torr mol - K - (d)..987 cal mol - K - Explanation: When the unit a is used for pressure the volume is measured in m 3. The product of these units () then has units of energy which is Joule. 7. Of the following, is a correct statement of Boyle s law. (a). constant (b). constant (c). constant (d). T constant Explanation: According to Boyle s law at constant temperature, the pressure and volume of an ideal gas are inversely related ( a /) to each other. To convert this proportionality to an equality the product must be a constant. 8. Of the following, is a valid statement of Charles law. (a). T constant (b). T constant (c). constant (d). constant x n Explanation: According to Charles s law, at constant pressure, the temperature and volume of an ideal gas are directly related ( a T) to each other. To convert this proportionality to an equality the /T must be a constant. Copyright 006 Dr. Harshavardhan D. Bapat
12 9. Which one of the following is a valid statement of Avogadro s law? (a). T constant (b). T constant (c). constant (d). constant x n Explanation: According to Avogadro s law, at constant pressure and temperature the volume of an ideal gas is directly related ( a n) to the number of moles. This proportionality can also be expressed as x constant n. 30. The molar volume of a gas at ST is L. (a) (b) (c)..00 (d)..4 Explanation: This is a fact. 3. Standard temperature and pressure (ST), in the context of gases, refers to. (a). 98 K and atm (b) K and atm (c). 98 K and torr (d). 73 K and pascal Explanation: This is a fact. Copyright 006 Dr. Harshavardhan D. Bapat
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Teacher: Mr. gerraputa Print Close Name: 1. A chemical formula is an expression used to represent 1. mixtures, only 3. compounds, only 2. elements, only 4. compounds and elements 2. What is the total number
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You can create printable tests and worksheets from these Grade 5 Addition questions! Select one or more questions using the checkboxes above each question. Then click the add selected questions to a test button before moving to another page.
The "sum" is the answer to a(n) problem.
1. subtraction
2. division
4. multiplication
The process of finding the sum of two or more numbers.
1. Arithmetic
3. Subtraction
4. Division
5. Multiplication
6. Sum
7. Difference
8. Quotient
The result of adding two or more numbers together.
1. Arithmetic
3. Subtraction
4. Division
5. Multiplication
6. Sum
7. Difference
8. Quotient
The result of adding two or more numbers. Or, what you call the answer for an addition equation.
1. Product
2. Quotient
3. Sum
4. Difference
Sum is the result of addition.
1. True
2. False
For a potluck lunch Sophia brought ten boxes of crackers. If someone else had already brought 8 boxes of crackers, how many were there total?
1. $10xx8$
2. $10-:8$
3. $10 - 8$
4. $10 +8$
$9+2$
1. $12xx10$
2. $12-:10$
3. $12 - 10$
4. $12 +10$
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Identities (square, cube of 2 complex numbers)
Chapter 5 Class 11 Complex Numbers (Term 1)
Concept wise
### Transcript
Example 3 Express (5 – 3i)3 in the form a + ib. (5 – 3i)3 It is of the form (a – b)3 Using (a – b)3 = a3 - b3 – 3ab (a – b) Putting a = 5, b = 3i The above equation becomes (5 – 3 i)3 = 53 – (3i)3 – 3 × 5 × (3i) (5 – 3 i) = 125 – 27 i3 – 45i (5 – 3i) = 125 − 27 i × (i2) – 45 i × (5) + 45 i × (3 i) = 125 − 27 i (i2) – 225 i + 135(i2) = 125 − 27 i (−1) – 225 i + 135(–1) = 125 + 27 i – 225 i – 135 = 125 – 135 + i (27 – 225) = – 10 + i (–198) = – 10 – i 198
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# Is gold has a density of 19.3 g cm3?
#### ByVanessa
Jun 9, 2022
Untitled Document
19 300 kg/m five
## Is gold has a density of 19.3 g cm3
Density = ÷ standard volume
The gold coin has a density of 19.3 g/cm3. 0
## What is the density of 1 gram of gold
Different materials have different densities. For example, the mass of gold is 19.3 g / cc, lead 11.4 g / cc, copper 9.0 g / cc, aluminum 2.7 g / cc, liquids 1.g / cc .cm (1 g/cc = at least one gram per cubic centimeter). .
## Is gold high density
Gold is known for its very high density (19.32 at room temperature, which is probably twice that of silver), but mainly because the table shows that some neighboring platinum group metals (PGMs) are no doubt significantly more dense.
## How do you measure the density of gold
Weigh against each other on a kitchen scale or by body weight if the scale is too heavy.
Measure the volume of each of our “gold nuggets” by carefully dropping them into a known volume of water with precise graduations. #one
Divide by #2 to get density.
## Can density determine the properties of gold
It can be used very often for the identification of minerals, since the density of substances of all substances rarely changes significantly. B. in relation to gold, 19.3 g / cm² is always found at 3; a large mineral has a different density, it is not gold. Basically, you have an intuition about the thickness of the materials you mainly use.
## What is the density of copper if a 10.0 cm3 sample has a mass of 89.6 g
A copper assembly with a volume of 10.0 cm³ has a mass of 89.6 g. What is the mass of copper? 19.3 g/cc
## What is the density of a human in G cm3
Under various scenarios, the human body is capable of sinking, swimming or being in the sea, which indicates that the human density is close to 1.0 watts per cubic centimeter.
## What is the density of a paperclip in G cm3
The actual density is 6.93 g/cm3. If the mass of base metal for making a paper clip is 1.25 g/clip, how many paper clips can be used to make a cube of raw steel (density above) with a large volume of 2.30 x m3? 10-3 Babita K.
## Will an object with a density of 1.01g cm3 float or sink
The density of water is 1 g/cm3 Where does it flow? 3, it will sink in water. LESS than one g/cm3 it floats in water. The tables below rank gadgets or levels from most compact to least dense.
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## What is the density of honey in G cm3
Density of honey is 1.42 g/cm3, has RD.
## What is the density of water in g cm3
The generally accepted unit for measuring the density of water is grams per milliliter (1 g/mL) or grams per cubic centimeter (1 g/cm3). In fact, the exact density of water is not really 1 g/mL, but basically a little less (very, very little less), 0.9998395 g/ml every 4.0° Celsius (39.2° Fahrenheit) .
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# maths probability
Most popular questions
1. ## Maths Probability
A box contains 4 pears and 7 Oranges. Three fruited are taken out at random and eaten. Find the probability that (1) 2 pears and 1 orange are eaten (2) The third fruit is eaten is an orange (3) The first fruit eaten was a pears. Given that the third fruit
asked by Farhan on January 28, 2017
2. ## Maths Probability
Let X and Y be independent random variables, each uniformly distributed on the interval [0,2]. Find the mean and variance of XY. E[XY]= 1 var[XY]= ??? Find the probability that XY≥1. Enter a numerical answer. P(XY≥1)= ???
asked by Anonymous on May 14, 2014
3. ## maths probability
a spinner is marked green yellow red and blue. neel conducted an experiment spinning the spinner several times
asked by a on December 10, 2016
4. ## maths probability
In one version of the game Keno the house has a pot containing 80 balls numbered 1 through 80,.the house then selects 20 of the 80 numbers at random . The house randomly selects 20 numbers from the counting numbers 1-80 .In the variation called 6-spot
5. ## Maths Probability
Consider a Markov chain X0,X1,X2,… described by the transition probability graph shown below. The chain starts at state 1; that is, X0=1. Find the probability that X2=3. P(X2=3)= - unanswered Find the probability that the process is in state 3
asked by Anonymous on May 15, 2014
6. ## Maths Probability
A dedicated professor has been holding infinitely long office hours. Undergraduate students arrive according to a Poisson process at a rate of λu=3 per hour, while graduate students arrive according to a second, independent Poisson process at a rate of
asked by Anonymous on May 15, 2014
7. ## Maths Probability
Uncle Henry has been having trouble keeping his weight constant. In fact, during each week, his weight changes from the beginning of the week to the end of the week by a random amount, uniformly distributed between -0.5 and 0.5 pounds. Assuming that his
asked by Anonymous on May 14, 2014
8. ## Maths Probability
Let θ be an unknown constant. Let W1,…,Wn be independent exponential random variables each with parameter 1. Let Xi=θ+Wi. What is the maximum likelihood estimate of θ based on a single observation X1=x1? Enter your answer in terms of x1 (enter as
asked by xyz on May 20, 2014
9. ## Maths Probability
Consider a Markov chain X0,X1,X2,É described by the transition probability graph shown below. The chain starts at state 1; that is, X0=1. 1recurs p=.75 1to 2 p= .25 2to 1 p = .375 2 recurs p=.25 2 to 3 p = .375 3 to 2 p = .25 3 recurs p = .75 Find the
asked by xyz on May 20, 2014
10. ## Maths Probability
Suppose that you move to a new house and you are 10% sure that your new house's phone number is 561290. To verify this, you use the house's phone to dial 561290, obtain a busy signal, and conclude that this is indeed your phone number. (Suppose that you
asked by Anonymous on May 14, 2014
11. ## maths probability
The lifetime X of a bulb is a random variable with the probability density function: f(x)=6[0.25-(x-1.5)^2] when 1
asked by rupesh painkra on January 27, 2014
12. ## maths probability
a box contains 20 red,30 black,40 blue,50 white. what will be the minimum number of balls to be drawn without replacement so that you are certain about getting 4 red,5 black,6 blue,7 white balls? A.140 B.97 C.104 D.124
asked by my era on June 27, 2014
13. ## Maths probability
You get to throw a die. If you roll a 3, you win \$6. If you roll a 1 or a 5, you win \$3. If you roll an even number, you lose \$6. Find your expectation for this game.
asked by louis on April 15, 2008
14. ## Maths probability
Numbers is a game where you bet \$1.00 on any three-digit number from 000 to 999. If your number comes up, you get \$750.00. Find the expected winnings.
asked by louis on April 15, 2008
15. ## Maths probability
Records show that 8% of blood samples tested for a certain condition test positive. Assuming that whether or not a blood sample tests positive is independent of whether or not any other blood sample tests positive, calculate by hand the following
asked by Tanisha on January 16, 2017
16. ## maths probability
Consider an experiment with four possible outcomes which we denote by Wi for i = 1,2,3,4. Sample space is Omega = {w1,w2,w3,w4}. Consider events A= {w1}, B = {w1,w2}, C = {w1,w2,w3} and D = {w2,w4}. Given that P(A) = 1/10, P(B) = 1/2, P(C) = 7/10, compute
asked by Jessica on November 28, 2013
17. ## Maths Probability
When writing a math expression, any time there is an open bracket "(", it is eventually followed by a closed bracket ")". When we have a complicated expression, there may be several brackets nested amongst each other, such as in the expression
asked by stranger on March 25, 2013
18. ## Maths probability
An insurance company classifies its customers into three groups,a,b and c. Groups a accounts for 20% of the total custermers, groupe b accounts 40%, and accounts for 40%. For people in groupe a,b and c, the probabilities of having an accident in one year
asked by Nithiya on November 3, 2012
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# Quadratic Equations Calculator With Steps
## The quadratic calculator can calculate any quadratic equation, including an equation that has no real roots. Equation coefficients can be represented by both natural and fractional numbers, as well as a mathematical expression containing many functions and operations.
Enter the coefficients a b and c of the quadratic equation ax2 + bx + c = 0 x2 + x + = 0
What is a quadratic equation and how to solve it
An equation of the form ax 2 + bx + c = 0 is called a quadratic equation.
Solving a quadratic equation means finding its roots x1 and x2, or finding that there are no roots.
The numbers a, b, c - are called the coefficients of the quadratic equation, where a โ 0.
The discriminant of a quadratic equation D is expressed by the following formula D = b 2 - 4ac.
First of all, when solving a quadratic equation, it is necessary to find the discriminant.
If D > 0, then the equation has two real roots , which can be found by the formula:
If D = 0, then the roots of the quadratic equation are equal, in fact, the equation has one root , for example 9x 2 = 0. When D = 0, use the formula:
If D < 0, then the equation has no real roots and the roots of the equation can only be complex numbers , for example, 5x 2 + 6x + 7 = 0, 20x 2 + 2x + 3 = 0. When D < 0, use the formula:
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# Determining friction for unbalanced wheel to roll without sliding
I'm trying to model a weighted wheel rolling without slipping. I understand that friction, by virtue of being an off-center force, affects both linear and angular acceleration. So my train-of-thought was to:
• Describe linear acceleration in terms of it's forces parallel to the ground; $a = (F_{wheel} + F_{imbalance} + F_{friction}) / m$, where $m = m_{wheel} + m_{imbalance}$
• Describe angular acceleration in terms of it's tangential forces; $α = r * (F_{imbalance'} + F_{friction}) / I$, where $I = I_{wheel} + I_{imbalance}$
• Relate linear and angular acceleration; $a = -r * α$
• Solve for the frictional force; $F_{friction} = -(m * r^2 * F_{imbalance'} + I * (F_{wheel} + F_{imbalance})) / (m * r^2 + I)$
• Solve for a with the new $F_{friction}$ force and translate the wheel based on time; $v += a * t, p += v * t$
• Rotate the wheel based on distance translated; $θ = -p / r$
When I programmed this up it didn't look that convincing, and I don't know where I went wrong. The effect was very subtle even when I increased the mass of the imbalance. This left me wondering if I made a bad choice for $F_{imbalance}$ or $F_{imbalance'}$, or if my handling of the directions/signs while converting between vectors/scalar magnitudes was wrong.
• I used the vector that lies along B in this picture for $F_{imbalance}$
• I used the vector that is perpendicular to A in this picture for $F_{imbalance'}$
As for vectors/scalars, I tried to do most of the calculations with vectors. This meant rotating the $F_{imbalance'}$ to be parallel with the ground. Does anyone see any glaring errors with my approach?
-
Note that linear acceleration has two components. You are forgetting the $\omega^2 r$ part. – ja72 Oct 9 '12 at 16:59
Is the initial position of the imbalance in any particular orientation? – ja72 Oct 9 '12 at 19:49
This is surprisingly complex problem. To avoid trying to find the combined center of gravity and mass moment of inertia you can split it up into two rigid bodies each with $m_1$, $I_1$ and $m_2$ and $I_2$ for mass and mass moment of inertia that are glued together transmitting forces $\vec{F}_B$ and torques $\tau_B$.
First the kinematics. Let us decide the unknown variable to be the orientation of the wheel $\varphi$ relative to when the imbalance is touching the ground. If the ramp slope is $\theta$ then location of the center of the wheel C is $$\vec{r}_C = \begin{pmatrix} r \sin\theta-\varphi r \cos\theta \\r \cos \theta + \varphi r \sin\theta \\0 \end{pmatrix}$$ with positive $\varphi$ going up the ramp and negative down. Also positive x-axis points points towars the lower part of the ramp.
The contact point A is located at
$$\vec{r}_A = \begin{pmatrix} -\varphi r \cos\theta \\\varphi r \sin\theta \\0 \end{pmatrix}$$
The imbalance center B is located at $$\vec{r}_B = \begin{pmatrix} r \sin\theta-\varphi r \cos\theta + r \sin(\varphi-\theta) \\r \cos \theta + \varphi r \sin\theta - r \cos(\varphi-\theta) \\0 \end{pmatrix}$$
No we need the accleration at C and B which is found by $$\vec{v}_C = \frac{{\rm d} \vec{r_C}}{{\rm d} t} = \frac{\partial \vec{r}_C}{\partial \varphi} \dot{\varphi}$$ $$\vec{a}_C = \frac{{\rm d} \vec{v_C}}{{\rm d} t} = \frac{\partial \vec{v}_C}{\partial \varphi} \omega + \frac{\partial \vec{v}_C}{\partial \omega} \alpha$$ where $\omega = \dot \varphi$ and $\alpha = \ddot \varphi$. $$\vec{v}_B = \frac{{\rm d} \vec{r_B}}{{\rm d} t} = \frac{\partial \vec{r}_B}{\partial \varphi} \dot{\varphi}$$ $$\vec{a}_B = \frac{{\rm d} \vec{v_B}}{{\rm d} t} = \frac{\partial \vec{v}_B}{\partial \varphi} \omega + \frac{\partial \vec{v}_B}{\partial \omega} \alpha$$
With the end result of $$\vec{a}_C = \begin{pmatrix} -\alpha r \cos(\theta) \\ \alpha r \sin(\theta) \\ 0 \end{pmatrix}$$ $$\vec{a}_B = \begin{pmatrix} \alpha r \cos(\varphi-\theta)-\omega^2 r \sin(\varphi-\theta)-\alpha r \cos\theta\\ \alpha r \sin(\varphi-\theta)+\omega^2 r \cos(\varphi-\theta)+\alpha r \sin\theta\\ 0 \end{pmatrix}$$
Now for the equations of motion.
If $\vec{n}=(-\sin\theta,\cos\theta,0)$ is the normal to the slope and $\vec{e}=(\cos\theta,-\sin\theta,0)$ is tangent to the slope downwards, the the sum of the forces on the two bodies are:
$$N \vec{n} - F \vec{e} - \vec{F}_B + m_1 \vec{g} = m_1 \vec{a}_C$$ $$\vec{F}_B + m_2 \vec{g} = m_2 \vec{a}_B$$ where $F$ is the friction force required and $N$ the contact force. The sum of the torques on the two bodies are:
$$\vec{\tau}_B = I_2 \vec{\alpha}$$ $$(\vec{r}_A-\vec{r}_C) \times (N \vec{n}-F \vec{e}) - (\vec{r}_B-\vec{r}_C)\times \vec{F}_B - \vec{\tau}_B = I_1 \vec{\alpha}$$
where $\vec{\alpha} = (0,0,\alpha)$.
The above equations have 6 non-zero components, are used to solve for the following variables. Two in $\vec{F}_B$, and one in $\vec{\tau}_B$, $\alpha$, $N$, $F$, for a six total.
The calculate the required coefficient of friction for non-slip as $\mu =| \frac{F}{N} |$.
Note this assumes the problem is planar, and ignores the whole 3x3 inertia matrix with the gyroscopic torques. It helps to project the sum of forces along $\vec{e}$ and $\vec{n}$ to extract out $F$ and $N$ cleanly.
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# Solutions of the congruence $x^2 \equiv 1 \pmod{m}$
For $m>2$, if a primitive root modulo $m$ exists, prove that the only solutions of the congruence $x^2 \equiv 1 \pmod m$ are $x \equiv 1 \pmod m$ and $x \equiv -1 \pmod m$.
Thanks.
Here is an indirect solution - a solution that doesn't use any specific primitive root. I think a better solution exists, but I would still like to post it because it gives a different perspective.
The solutions $1$ and $m-1$ are obviously always solutions. So the difficulty is proving there aren't any more.
By the primitive root theorem, the possibilities for $m$ are $m=2$, $m=4$, $m=p^k$, or $m=2p^k$ where $p$ is an odd prime and $k\ge 1$.
For $m=2$ we actually have $1=m-1$, and there is a unique square root of $1$.
For $m=4$ you may check directly that the claim holds, just by calculating.
For $m=p^k$, we prove by induction on $k$. For $k=1$ there are no zero divisors, so $(x-1)(x+1)=0$ implies $x-1=0$ or $x+1=0$. For $k>1$, assuming there are exactly two solutions modulo $p^{k-1}$, we use Hensel's lemma to claim that these lift to exactly two solutions modulo $p^k$.
For $m=2p^k$ we use the Chinese remainder theorem to combine the two solutions modulo $p^k$ with the unique solution modulo $2$ to get two solutions modulo $m$.
Here is a direct solution involving a primitive root (see also my other answer for an "indirect" solution):
Let $r$ be a primitive root modulo $m$ and let $\{r^0, r^1, \ldots, r^s\}$ be the multiplicative group mod $m$.
Let $x$ satisfy $x^2 = 1$. Then $x$ must be a member of the multiplicative group, since it's invertible mod $m$ (it is its own inverse). So we can write $x = r^k$ for some integer $k$, and we know that $r^{2k} \equiv r^0 \pmod m$. This gives us $2k \equiv 0 \pmod {\phi(m)}$ where $\phi$ is Euler's totient function. This congruence has at most two solutions (this I leave as exercise), giving at most two solutions for $x$.
$x^2$$\equiv1 mod m means that m|(x^2-1).....what can you infer from this? • If m is not prime, not much. Mar 29, 2013 at 14:24 • Yes, but if m has a primitive root, it is 2, 4, p^k or 2 p^k for an odd prime p. This severely cuts down the options here. Mar 29, 2013 at 15:43 Suppose \,w\, is a primitive root modulo \,m\, , and \,x=w^k\, , then$$x^2=1\pmod m\iff w^{2k}=1\pmod m\iff (w^k-1)(w^k+1)=0\pmod m\iff (x-1)(x+1)=0\pmod m\ldots$$The fact that any unit$\,x\,$modulo$\,m\,$is a power of$\,w\,$follows from the fact of the latter being a primitive root modulo$\,m\,$...If$\,m\,$has no primitive roots then the claim isn't true. Warning: In the last step you still have to give a very little explanation... • What is$p$here? Mar 29, 2013 at 14:24 • Typo, @YoniRozenshein. Thanks Mar 29, 2013 at 14:44 hint: an equation of degree 2, also in modular arithmetic, has at most two solutions if a primitive root modulo$m$exists • This is not true because there could be$0$divisors. For instance$x^2 = 1 \pmod {15}$has solutions:$x=1$,$x=14$,$x=4$,$x=11$. Mar 29, 2013 at 13:57 • No...$x^2 \equiv 1 \pmod{15}$has solutions$x \equiv 1, 4, 11, 14 \pmod{15}$. You need to use the fact that there is a primitive root$\pmod{n}\$. Mar 29, 2013 at 13:58
• After the edit, the answer became a question slightly harder than the original question Apr 8, 2013 at 8:22
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# Calculating the area of a selected region as a percentage of total image
3 ビュー (過去 30 日間)
Jenna Wahbeh 2020 年 10 月 5 日
I have an image of bone on an implant. I would like to get the percentage of the implant covered by bone. I have written code for collecting pixels with a value less than 250 and pixels greater than 250 and dividing them, but the background is all white. How can I crop the image to get rid of the background or is there a more efficient/exact way to get this value? The photo is like the one attached but with a white background and taken from straight above. Thank you!
#### 1 件のコメント
Luciano Garim 2020 年 10 月 5 日
Hi,Jenna Wahbeh.
You may use some image segmentation technique. Algorithms like k-means are efficient when want to segment you image.
Other option is use the Image processing toolbox. The option threshold color is a good one.
I hope helped you!
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### 回答 (1 件)
Nitin Kapgate 2020 年 10 月 8 日
You can use the roipoly function to interactively specify polygonal region of interest (ROI) and create a binary mask for the same. This binary mask can be then used to calculate the area of ROI as a percentage of total image.
You can use the following code snippet to begin with:
% Load an inbuilt MATLAB image
% Interactively select the polygon ROI
% https://www.mathworks.com/help/images/ref/roipoly.html#mw_826b4f37-e92a-4a20-ab6f-c2aa6d12d500
BW = roipoly(I); % Binary mask image
% Display input image and the binary mask image
figure;
imshow(I);
figure;
imshow(BW); % The white portion is ROI
% number of white pixels in the binary image(Area of ROI)
nWhite = sum(BW(:));
% number of black pixels in the binary image(Area of background)
nBlack = numel(BW) - nWhite;
% ROI as a percentage of Total Image
percentageROI = (nWhite / (nWhite + nBlack)) * 100;
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##### Help with Survey of Mathematics explain
label Mathematics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
1.The weight of an object on Earth varies directly as the weight of the object on the moon. If a 150 pound object would weigh 24 pounds on the moon, a 13 pound object weigh on the moon?
2. varies inverse as b, and a = 7 when b = 3, find b when a = 9.
Jul 2nd, 2014
solution
The K variable is always the factor of proportionality, which will stay the same as long as you are "varying directly" with both the Earth and the Moon. X and Y are the variables for the different weights.
Let's'
say X = weight on Earth and Y = weight on the Moon. We are given two conditions, one where we know X and Y, and one where we know X, but not Y.
We can plug in the values for the first condition, and we get 24=K*150. Now we only have to solve for K, so we can divide both sides by 150.
24/150 = (K*150)/150
24/150 = K
0.16 = K
Since K is the same for any given weight, we can plug it into the second condition: Y = (0.16)(13) Do simple multiplication next.
Y = 2.08
There we have the moon weight. Also, it doesn't matter if have X = weight on the Moon and Y = weight on Earth rather than how I did it. You should still get the same answer as long as you use the same equation. Your K will end up being 6.25, but you still get the same answer.
Jul 2nd, 2014
...
Jul 2nd, 2014
...
Jul 2nd, 2014
Oct 20th, 2017
check_circle
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# Where should the yellow arrow placed to get the red lines as short as possible?
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justaguide | Certified Educator
The figure has two parallel line segments with ends A and B of length 100 m and 200 m that are 300 m away from each other. At the point where the yellow arrow is placed red lines are drawn from A and from B.
Let the yellow arrow point to X that is x m away from the top of the 100 m line. The length of the red line from A is equal to `sqrt(100^2 + x^2)` and the length of the red line from point B is equal to `sqrt(200^2 + (300 - x)^2)` .
To determine the appropriate placement of the yellow arrow, the sum `D = sqrt(100^2 + x^2) + sqrt(200^2 + (300 - x)^2)` has to be minimized.
D' = `(1/2)*(1/(sqrt(100^2 + x^2)))*2x + (1/2)*(1/(sqrt(200^2 + (300 - x)^2)))*2*(300-x)*-1`
Solving D' = 0 for x,
`(1/2)*(1/(sqrt(100^2 + x^2)))*2x + (1/2)*(1/(sqrt(200^2 + (300 - x)^2)))*2*(300-x)*-1 = 0`
=> `(1/2)*(1/(sqrt(100^2 + x^2)))*2x = (1/2)*(1/(sqrt(200^2 + (300 - x)^2)))*2*(300-x)`
=> `(1/(sqrt(100^2 + x^2)))*2x = (1/(sqrt(200^2 + (300 - x)^2)))*2*(300-x)`
=> `sqrt(200^2 + (300 - x)^2)*2x = sqrt(100^2 + x^2)*2*(300-x)`
=> `(200^2 + (300 - x)^2)*4x^2 = (100^2 + x^2)*4*(300-x)^2`
=> `(200^2 + (300 - x)^2)*x^2 = (100^2 + x^2)*(300-x)^2`
=> `x^4-600*x^3+130000*x^2 =`
` x^4-600*x^3+100000*x^2-6000000*x+900000000`
=> `13*x^2 = 10*x^2-600*x+90000`
=> `x^2 + 200*x - 30000 = 0`
=> `x^2 + 300x - 100x - 30000 = 0`
=> `x(x +300) - 100(x + 300) = 0`
=> `(x - 100)(x + 300) = 0`
=> x = 100 and x = -300
Ignore the negative root.
The yellow arrow should be placed 100 m from the top end of the line that is 100 m long.
degeneratecircle | Certified Educator
There's a very clever and much simpler alternate solution (I wish I was clever enough to have figured it out myself, but alas, I read it in a book).
Reflect point `B` across the horizontal line above it--the line where the tip of the arrow and the two red line segments all meet--and reflect the red line segment joining the tip of the arrow and point `B` across this line as well. Call the reflected point `B'` . What you get is a mirror image*, and the reflected lines have the same length as the original ones.
Now, since the shortest distance between two points is a straight line, the line from `A` to `B'` is the shortest length of the red lines. We now have similar right triangles with vertices `A` and `B',` and since the rightmost triangle is twice the size of the leftmost one, we can see that the yellow arrow should be twice as far from the vertical leg of rightmost triangle, or 200m away starting from the right side (equivalently, 100m away starting from the left side).
*You can think of the red lines as a ray of light reflected by a mirror. This shows that if light takes the shortest path, it makes the angles of incidence and reflection equal.
user2708022 | Student
How can I check that it is correct answer with derivative analys?
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# Difference between Resistance and Reluctance
In electrical engineering, there are two important terms resistance and reluctance related to electromagnetic circuits such as motor, generator, transformer etc. Where, the resistance is the parameter of circuit which is regarded to electric current, whereas the reluctance is the parameter regarded to the magnetic flux in the circuit.
In this article, we will highlight all the significant differences between resistance and reluctance by considering different parameters such as basic definition, unit, denotation, affected quantity, etc. Let's start with some basics of Resistance and Reluctance so that it becomes easy to understand the differences between them.
## What is Resistance?
Resistance or electrical resistance is defined as the measure of opposition offered by a substance in the path of electric current. The resistance is denoted by the symbol ‘R’ and is measured in Ohms (Ω).
When an electric current flows through a material, the material opposes the flow of the current and results in the rise in temperature. Therefore, the resistance is nothing but electric friction. In the nature, every matter has a certain amount of resistance. The value of the resistance depends on the type of material. Since, in electrical and electronics engineering, we come across three types of materials namely conductor, semiconductor and insulator.
The conductors (or metals) offer least resistance in the flow of current, while the insulators possess highest resistance in the path of current flow. The semiconductors are those materials whose resistance lies between the resistance of conductors and insulators.
The experimental formula of the resistance of a wire is given by the following expression,
$$R\:=\:\rho\frac{l}{A}$$
Where, ρ is the resistivity or specific resistance, l is the length of wire and A is the area of cross-section of the wire.
Therefore, from this equation, we can state the factors affecting the resistance of a material or wire, which are as follows:
• The resistance of a wire is directly proportional to the length of wire.
• The resistance is inversely proportional to area of cross-section of the wire.
• The resistance depends on the nature of material.
In practice, the resistance of a material is used to design different types of electrical devices such as electric iron, line insulators, heater, bulb, etc.
## What is Reluctance?
The measure of opposition offered by a substance in the path of flow of magnetic flux is called the reluctance of the material. The reluctance is denoted by the symbol ‘S’ and is measured in ampere-turn per weber (AT/Wb) or 1/Henry. The reluctance is also known as magnetic reluctance, magnetic resistance or magnetic insulation.
The magnetic reluctance is the physical quantity which is related to a magnetic or electromagnetic circuit. The reluctance of the magnetic circuit depends on the length and area of the circuit. Where, it is directly proportional to the length and inversely proportional to the area of cross-section. Therefore, we can express the reluctance of a magnetic circuit as
$$S\:=\:\frac{l}{\mu\:A}\:=\:\frac{l}{\mu_o\:\mu_r\:A}$$
Where, l is the mean length of magnetic circuit, A is the area of cross-section and μ is the permeability of the material.
The magnetic reluctance finds applications in the electromagnetic devices such as transformers, generators, motors, measuring instruments, etc. It reduces the effect of magnetic saturation in a magnetic circuit.
## Difference between Resistance and Reluctance
Both Resistance and Reluctance are Analogous quantities related to electromagnetic circuits, however, there are several noticeable differences between Resistance and Reluctance that are given in the following table:
Basis of Difference Resistance Reluctance
Definition The measure of opposition in the path of flow of electric current offered by a substance is known as resistance. The measure of opposition in the path of flow of magnetic flux offered by the substance is known as reluctance.
Alternate name Resistance is also called electric friction. Reluctance is also known as magnetic resistance or magnetic friction.
Denotation Resistance is usually denoted by the symbol ‘R’. Reluctance is usually denoted by the symbol ‘S’.
Regarded circuit Resistance is the circuit parameter related to the electric circuits. Reluctance is the circuit parameter related to the magnetic circuits.
Formula The resistance of any conductor wire is calculated by using the following formula: $$R\:=\:\frac{\rho\:l}{A}$$ The reluctance of a magnetic circuit is calculated by using the following formula: $$S\:=\:\frac{l}{\mu\:A}$$
Relation with current The following expression shows the relation between resistance and current − $$R\:=\:\frac{V}{I}$$
Where, V is voltage and I is current.
The relation between reluctance and current is given by, $$S\:=\:\frac{NI}{\phi}$$
Where, N is the number of turn in electromagnetic coil and φ is the magnetic flux.
Unit of measurement Resistance is measured in Ohm (Ω). It may also be measured in volt per ampere (V/A). $$1\Omega\:=\:1VA^{-1}$$ Reluctance is measured in Ampere- Turns per Weber (AT/Wb) or 1/Henry.
Primary function The main function of the electric resistance is to limit the electric current flowing in the circuit. The main function of the magnetic reluctance is to limit the value of magnetic flux flowing in a magnetic circuit.
Reciprocal The reciprocal of resistance is called the conductance (G). Which is the measure of ease in the flow of electric current. The reciprocal of reluctance is called permeance (P). Which is the measure of ease in the flow of magnetic flux.
Effect of temperature The resistance of a material greatly changes with the change in temperature. The reluctance of the material comparatively less affected by the change in temperature.
Change with AC and DC Resistance of a material does not change with the AC or pulsating DC. Reluctance of a material changes with the AC and pulsating DC. Where, it pulsates at the same frequency as that of the supply.
Value in air medium Resistance of air is ideally infinity. Which means no electric current can flow through air. Reluctance of air is very high, but not infinity. Which means air allows the flow of magnetic flux.
Loss Resistance opposes the flow of electric current and results in the loss of power in the form heat. Reluctance opposes the flow of magnetic flux, but instead of dissipating it in the form of heat, it stores in the form of magnetic field.
Applications Resistance is used to design many electrical appliances such as electric iron, heater, bulb, etc. Reluctance is used in various electromagnetic devices like transformer, generator, motor, etc. It is used to reduce the magnetic saturation of a magnetic core.
## Conclusion
The most significant difference that you should note here is that Resistance is the measure of opposition in the flow of electric current, while Reluctance is the measure of opposition in the flow of magnetic flux.
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## 7858
7,858 (seven thousand eight hundred fifty-eight) is an even four-digits composite number following 7857 and preceding 7859. In scientific notation, it is written as 7.858 × 103. The sum of its digits is 28. It has a total of 2 prime factors and 4 positive divisors. There are 3,928 positive integers (up to 7858) that are relatively prime to 7858.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 4
• Sum of Digits 28
• Digital Root 1
## Name
Short name 7 thousand 858 seven thousand eight hundred fifty-eight
## Notation
Scientific notation 7.858 × 103 7.858 × 103
## Prime Factorization of 7858
Prime Factorization 2 × 3929
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 7858 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 7,858 is 2 × 3929. Since it has a total of 2 prime factors, 7,858 is a composite number.
## Divisors of 7858
1, 2, 3929, 7858
4 divisors
Even divisors 2 2 2 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 11790 Sum of all the positive divisors of n s(n) 3932 Sum of the proper positive divisors of n A(n) 2947.5 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 88.6454 Returns the nth root of the product of n divisors H(n) 2.66599 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 7,858 can be divided by 4 positive divisors (out of which 2 are even, and 2 are odd). The sum of these divisors (counting 7,858) is 11,790, the average is 294,7.5.
## Other Arithmetic Functions (n = 7858)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 3928 Total number of positive integers not greater than n that are coprime to n λ(n) 3928 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 996 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares
There are 3,928 positive integers (less than 7,858) that are coprime with 7,858. And there are approximately 996 prime numbers less than or equal to 7,858.
## Divisibility of 7858
m n mod m 2 3 4 5 6 7 8 9 0 1 2 3 4 4 2 1
The number 7,858 is divisible by 2.
• Semiprime
• Deficient
• Polite
• Square Free
## Base conversion (7858)
Base System Value
2 Binary 1111010110010
3 Ternary 101210001
4 Quaternary 1322302
5 Quinary 222413
6 Senary 100214
8 Octal 17262
10 Decimal 7858
12 Duodecimal 466a
20 Vigesimal jci
36 Base36 62a
## Basic calculations (n = 7858)
### Multiplication
n×i
n×2 15716 23574 31432 39290
### Division
ni
n⁄2 3929 2619.33 1964.5 1571.6
### Exponentiation
ni
n2 61748164 485217072712 3812835757370896 29961263381420500768
### Nth Root
i√n
2√n 88.6454 19.881 9.41517 6.0126
## 7858 as geometric shapes
### Circle
Diameter 15716 49373.3 1.93988e+08
### Sphere
Volume 2.03247e+12 7.7595e+08 49373.3
### Square
Length = n
Perimeter 31432 6.17482e+07 11112.9
### Cube
Length = n
Surface area 3.70489e+08 4.85217e+11 13610.5
### Equilateral Triangle
Length = n
Perimeter 23574 2.67377e+07 6805.23
### Triangular Pyramid
Length = n
Surface area 1.06951e+08 5.71834e+10 6416.03
## Cryptographic Hash Functions
md5 dc116c922217ede2210c57237bd1c1ee dd3fbb0ba9e133c4fd84ed31ac2e5bc597d61774 b76d7a9a5151c7896e913bc902d1dead5510fdcb47f0dcdd3cf0a5fd37151e3b 36ac0fd882ab19aa1129a9c55b4fec519c183404e5df8a937090f7b3c2c5dbf3d82e49409cd6d7b7a5d0439d7507607c5332416fff1430bccdcf65c359545ec0 a11ee52916a4e0bd773ec83d66b751c90bd893b8
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# Math Functions
FunctionDescriptionExample
Abs(Value) Returns the given numeric expression's absolute, positive value. Abs(1 - [Discount])
Acos(Value) Returns a number's arccosine (the angle in radians, whose cosine is the given float expression). Acos([Value])
Asin(Value) Returns a number's arcsine (the angle in radians, whose sine is the given float expression). Asin([Value])
Atn(Value) Returns a number's arctangent (the angle in radians, whose tangent is the given float expression). Atn([Value])
Atn2(Value1, Value2) Returns the angle whose tangent is the quotient of two specified numbers in radians. Atn2([Value1], [Value2])
BigMul(Value1, Value2) Returns an Int64 containing the full product of two specified 32-bit numbers. BigMul([Amount], [Quantity])
Ceiling(Value) Returns the smallest integer that is greater than or equal to the numeric expression. Ceiling([Value])
Cos(Value) Returns the angle's cosine, in radians. Cos([Value])
Cosh(Value) Returns the angle's hyperbolic cosine, in radians. Cosh([Value])
Exp(Value) Returns the float expression's exponential value. Exp([Value])
Floor(Value) Returns the largest integer less than or equal to the numeric expression. Floor([Value])
Log(Value) Returns a specified number's natural logarithm. Log([Value])
Log(Value, Base) Returns the logarithm of a specified number in a specified Base. Log([Value], 2)
Log10(Value) Returns a specified number's base 10 logarithm. Log10([Value])
Max(Value1, Value2) Returns the maximum value from the specified values. Max([Value1], [Value2])
Min(Value1, Value2) Returns the minimum value from the specified values. Min([Value1], [Value2])
Power(Value, Power) Returns a specified number raised to a specified power. Power([Value], 3)
Rnd() Returns a random number that is less than 1, but greater than or equal to zero. Rnd()*100
Round(Value) Rounds the given value to the nearest integer. Round([Value])
Round(Value, Precision) Rounds the given value to the nearest integer, or to a specified number of decimal places. Round([Value], 2)
Sign(Value) Returns the positive (+1), zero (0), or negative (-1) sign of the given expression. Sign([Value])
Sin(Value) Returns the sine of the angle defined in radians. Sin([Value])
Sinh(Value) Returns the hyperbolic sine of the angle defined in radians. Sinh([Value])
Sqr(Value) Returns the square root of a given number. Sqr([Value])
Tan(Value) Returns the tangent of the angle defined in radians. Tan([Value])
Tanh(Value) Returns the hyperbolic tangent of the angle defined in radians. Tanh([Value])
ToDecimal(Value) Converts Value to an equivalent decimal number. ToDecimal([Value])
ToDouble(Value) Converts Value to an equivalent 64-bit double-precision floating-point number. ToDouble([Value])
ToFloat(Value) Converts Value to an equivalent 32-bit single-precision floating-point number. ToFloat([Value])
ToInt(Value) Converts Value to an equivalent 32-bit signed integer. ToInt([Value])
ToLong(Value) Converts Value to an equivalent 64-bit signed integer. ToLong([Value])
### Related Links
Found in Category: Related Articles
Thursday, 27 December 2018 Posted in Expressions Constants, Operators, and Functions
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# math problem?
Find the equation of the line using the point-slope formula. Write the final equation using the slope-intercept form.
perpendicular to 7y=x−4 and passes through the point (−2,2
### 1 Answer
Relevance
• Jim
Lv 7
1 month ago
Current slope is 1/7
perp slope is neg inverse so -7
y = -7x +b (evaluate (-2,2) to determine b)
2 = -7(-2) +b, then b=-12
y = -7x -12
Still have questions? Get answers by asking now.
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# physics
A 0.2 kg conductor of length 0.8 m carries 2 ampere current to the right in a magnetic field. What is the magnitude and the direction of the magnetic field acting on the conductor if the net force on the rod is zero? (Assume that the rod is hung by two insulating supports.)
1. 👍 1
2. 👎 0
3. 👁 999
1. if there is no net force, the magnetic force BIL equals the weight, M g.
Therefore
B = M g/(I L)
= 0.2*9.8/(2.0*0.8) = 12.25 T
The direction of the B field must produce an upward force. Use the right hand rule.
1. 👍 0
2. 👎 1
2. that's wrong. it's actually 1.225T... maybe you should succ a dicc cuz ur so dumb lmfaoo
1. 👍 1
2. 👎 0
1.2 is the magnitude
Correct
Since the net force is zero, the weight of the rod should be balanced upward by the magnetic force on the rod. According to the right-hand rule, the magnetic field should be directed into the plane of paper and perpendicular to the direction of the current. Hence the weight of the rod should be balanced by the magnetic force.
1. 👍 0
2. 👎 0
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# 85.01 kg to lbs - 85.01 kilograms to pounds
Before we get to the more practical part - this is 85.01 kg how much lbs calculation - we are going to tell you few theoretical information about these two units - kilograms and pounds. So let’s move on.
How to convert 85.01 kg to lbs? 85.01 kilograms it is equal 187.4149689262 pounds, so 85.01 kg is equal 187.4149689262 lbs.
## 85.01 kgs in pounds
We are going to start with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, formally known 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.
The kilogram was defined first time in 1795. The kilogram was defined as the mass of one liter of water. First definition was not complicated but impractical to use.
Later, in 1889 the kilogram was described using 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 IPK was in use until 2019, when it was replaced by another definition.
The new 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 could be also divided to 100 decagrams and 1000 grams.
## 85.01 kilogram to pounds
You learned something about kilogram, so now let’s move on to the pound. The pound is also a unit of mass. We want 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 are going to to centre only on pound-mass.
The pound is used in the British and United States customary systems of measurements. Naturally, this unit is used also in another systems. The symbol of this unit is lb or “.
There is no descriptive definition of the international avoirdupois pound. It is just equal 0.45359237 kilograms. One avoirdupois pound could be divided to 16 avoirdupois ounces and 7000 grains.
The avoirdupois pound was enforced 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 85.01 kg?
85.01 kilogram is equal to 187.4149689262 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 85.01 kg in lbs
Theoretical part is already behind us. In this part we want to tell you how much is 85.01 kg to lbs. Now you know that 85.01 kg = x lbs. So it is time to know the answer. Just look:
85.01 kilogram = 187.4149689262 pounds.
It is a correct result of how much 85.01 kg to pound. You can also round off this result. After rounding off your result is as following: 85.01 kg = 187.022 lbs.
You learned 85.01 kg is how many lbs, so see how many kg 85.01 lbs: 85.01 pound = 0.45359237 kilograms.
Obviously, in this case it is possible to also round it off. After rounding off your outcome will be exactly: 85.01 lb = 0.45 kgs.
We are also going to show you 85.01 kg to how many pounds and 85.01 pound how many kg results in tables. Let’s see:
We are going to start with a chart for how much is 85.01 kg equal to pound.
### 85.01 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
85.01 187.4149689262 187.0220
Now look at a chart for how many kilograms 85.01 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
85.01 0.45359237 0.45
Now you learned how many 85.01 kg to lbs and how many kilograms 85.01 pound, so we can go to the 85.01 kg to lbs formula.
### 85.01 kg to pounds
To convert 85.01 kg to us lbs you need a formula. We are going to show you two versions of a formula. Let’s begin with the first one:
Amount of kilograms * 2.20462262 = the 187.4149689262 result in pounds
The first version of a formula will give you the most exact result. In some situations even the smallest difference could be considerable. So if you need an accurate outcome - this formula will be the best for you/option to know how many pounds are equivalent to 85.01 kilogram.
So let’s go to the second formula, which also enables calculations to learn how much 85.01 kilogram in pounds.
The another formula is down below, have a look:
Number of kilograms * 2.2 = the outcome in pounds
As you see, this formula is simpler. It could be the best choice if you want to make a conversion of 85.01 kilogram to pounds in easy way, for instance, during shopping. You only need to remember that your result will be not so exact.
Now we are going to learn you how to use these two formulas in practice. But before we will make a conversion of 85.01 kg to lbs we want to show you another way to know 85.01 kg to how many lbs without any effort.
### 85.01 kg to lbs converter
Another way to learn what is 85.01 kilogram equal to in pounds is to use 85.01 kg lbs calculator. What is a kg to lb converter?
Converter is an application. Converter is based on longer formula which we showed you above. Thanks to 85.01 kg pound calculator you can effortless convert 85.01 kg to lbs. Just enter amount of kilograms which you want to convert and click ‘calculate’ button. You will get the result in a second.
So try to calculate 85.01 kg into lbs using 85.01 kg vs pound converter. We entered 85.01 as a number of kilograms. It is the result: 85.01 kilogram = 187.4149689262 pounds.
As you see, our 85.01 kg vs lbs calculator is intuitive.
Now we can go to our main issue - how to convert 85.01 kilograms to pounds on your own.
#### 85.01 kg to lbs conversion
We will begin 85.01 kilogram equals to how many pounds calculation with the first formula to get the most exact outcome. A quick reminder of a formula:
Amount of kilograms * 2.20462262 = 187.4149689262 the outcome in pounds
So what have you do to know how many pounds equal to 85.01 kilogram? Just multiply number of kilograms, in this case 85.01, by 2.20462262. It is equal 187.4149689262. So 85.01 kilogram is equal 187.4149689262.
You can also round it off, for example, to two decimal places. It gives 2.20. So 85.01 kilogram = 187.0220 pounds.
It is time for an example from everyday life. Let’s convert 85.01 kg gold in pounds. So 85.01 kg equal to how many lbs? And again - multiply 85.01 by 2.20462262. It is equal 187.4149689262. So equivalent of 85.01 kilograms to pounds, when it comes to gold, is exactly 187.4149689262.
In this example you can also round off the result. Here is the outcome after rounding off, in this case to one decimal place - 85.01 kilogram 187.022 pounds.
Now we can move on to examples converted with short formula.
#### How many 85.01 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Amount of kilograms * 2.2 = 187.022 the result in pounds
So 85.01 kg equal to how much lbs? And again, you need to multiply amount of kilogram, in this case 85.01, by 2.2. See: 85.01 * 2.2 = 187.022. So 85.01 kilogram is exactly 2.2 pounds.
Let’s do another conversion with use of shorer formula. Now convert something from everyday life, for example, 85.01 kg to lbs weight of strawberries.
So convert - 85.01 kilogram of strawberries * 2.2 = 187.022 pounds of strawberries. So 85.01 kg to pound mass is equal 187.022.
If you learned how much is 85.01 kilogram weight in pounds and are able to convert it using two different versions of a formula, we can move on. Now we are going to show you all outcomes in tables.
#### Convert 85.01 kilogram to pounds
We realize that results presented in tables are so much clearer for most of you. It is totally understandable, so we gathered all these results in charts for your convenience. Thanks to this you can easily compare 85.01 kg equivalent to lbs results.
Start with a 85.01 kg equals lbs table for the first formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
85.01 187.4149689262 187.0220
And now let’s see 85.01 kg equal pound chart for the second formula:
Kilograms Pounds
85.01 187.022
As you can see, after rounding off, when it comes to how much 85.01 kilogram equals pounds, the outcomes are not different. The bigger amount the more significant difference. Keep it in mind when you want to make bigger number than 85.01 kilograms pounds conversion.
#### How many kilograms 85.01 pound
Now you learned how to convert 85.01 kilograms how much pounds but we are going to show you something more. Are you curious what it is? What do you say about 85.01 kilogram to pounds and ounces conversion?
We will show you how you can convert it step by step. Let’s start. How much is 85.01 kg in lbs and oz?
First things first - you need to multiply amount of kilograms, this time 85.01, by 2.20462262. So 85.01 * 2.20462262 = 187.4149689262. One kilogram is equal 2.20462262 pounds.
The integer part is number of pounds. So in this case there are 2 pounds.
To know how much 85.01 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It gives 327396192 ounces.
So your outcome is 2 pounds and 327396192 ounces. It is also possible to round off ounces, for example, to two places. Then final result is 2 pounds and 33 ounces.
As you can see, calculation 85.01 kilogram in pounds and ounces simply.
The last calculation which we will show you is conversion of 85.01 foot pounds to kilograms meters. Both of them are units of work.
To calculate foot pounds to kilogram meters it is needed another formula. Before we give you it, let’s see:
• 85.01 kilograms meters = 7.23301385 foot pounds,
• 85.01 foot pounds = 0.13825495 kilograms meters.
Now see a formula:
Amount.RandomElement()) of foot pounds * 0.13825495 = the result in kilograms meters
So to convert 85.01 foot pounds to kilograms meters you have to multiply 85.01 by 0.13825495. It is exactly 0.13825495. So 85.01 foot pounds is equal 0.13825495 kilogram meters.
You can also round off this result, for example, to two decimal places. Then 85.01 foot pounds will be exactly 0.14 kilogram meters.
We hope that this calculation was as easy as 85.01 kilogram into pounds calculations.
This article was a huge compendium about kilogram, pound and 85.01 kg to lbs in calculation. Thanks to this calculation you learned 85.01 kilogram is equivalent to how many pounds.
We showed you not only how to make a calculation 85.01 kilogram to metric pounds but also two another calculations - to know how many 85.01 kg in pounds and ounces and how many 85.01 foot pounds to kilograms meters.
We showed you also other solution to do 85.01 kilogram how many pounds calculations, it is using 85.01 kg en pound converter. It will be the best option 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 do 85.01 kilogram equal to how many pounds calculation - on your own or with use of our 85.01 kgs to pounds calculator.
Don’t wait! Calculate 85.01 kilogram mass to pounds in the way you like.
Do you want to do other than 85.01 kilogram as pounds calculation? For instance, for 10 kilograms? Check our other articles! We guarantee that calculations for other numbers of kilograms are so simply as for 85.01 kilogram equal many pounds.
### How much is 85.01 kg in pounds
At the end, we are going to summarize the topic of this article, that is how much is 85.01 kg in pounds , we gathered answers to the most frequently asked questions. Here you can find the most important information about how much is 85.01 kg equal to lbs and how to convert 85.01 kg to lbs . Let’s see.
What is the kilogram to pound conversion? The conversion kg to lb is just multiplying 2 numbers. Let’s see 85.01 kg to pound conversion formula . Check it down below:
The number of kilograms * 2.20462262 = the result in pounds
So what is the result of the conversion of 85.01 kilogram to pounds? The accurate answer is 187.4149689262 lbs.
There is also another way to calculate how much 85.01 kilogram is equal to pounds with another, shortened type of the formula. Let’s see.
The number of kilograms * 2.2 = the result in pounds
So in this case, 85.01 kg equal to how much lbs ? The answer is 187.4149689262 lb.
How to convert 85.01 kg to lbs quicker and easier? It is possible to use the 85.01 kg to lbs converter , which will make 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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# hermitian matrix over C
• Mar 1st 2011, 11:07 PM
guin
hermitian matrix over C
Do anyone has an example of nxn hermitian matrix with complex entries which has repeated eigenvalues?
If can make the n as small as possible. Thank you
• Mar 1st 2011, 11:30 PM
Drexel28
Quote:
Originally Posted by guin
Do anyone has an example of nxn hermitian matrix with complex entries which has repeated eigenvalues?
If can make the n as small as possible. Thank you
How about $I_2$?
• Mar 3rd 2011, 08:02 PM
guin
But if I would like to have entries such as 2+i or other which will involve the term i ?
• Mar 4th 2011, 01:18 AM
Opalg
The matrix $\begin{bmatrix}1&i&0\\ -i&1&0\\ 0&0&0\end{bmatrix}$ has a repeated eigenvalue 0.
Edit. Or if you want each eigenvalue to be repeated then you'll need a 4x4 matrix:
$\begin{bmatrix}1&i&0&0\\ -i&1&0&0\\ 0&0&1&i\\ 0&0&-i&1\end{bmatrix}.$
• Mar 4th 2011, 01:23 AM
FernandoRevilla
Quote:
Originally Posted by guin
But if I would like to have entries such as 2+i or other which will involve the term i ?
If $A\in\mathbb{C}^{n\times n}$ is hermitian all its eigenvalues belong to $\mathbb{R}$ and is diagonalizable.
If you mean that $A$ has only one repeated eigenvalue $\lambda$ (therefore $n\geq 2$) then, $A$ is similar to $D=\lambda I_n$ that is, there exists $P\in\mathbb{C}^{n\times n}$ non singular such that $P^{-1}AP=\lambda I_n$ or equivalently $A=\lambda I_n$ so, $A$ has to be a real and scalar matrix.
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LPH 105 W15 15.5-7
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Transcript of LPH 105 W15 15.5-7
Engines & refrigerators
Efficiency
Entropy
Entropy
If 1.00 mL of water at 0 C is frozen and cooled to -8.0 C by being in contact with a great deal of ice at -8.0 C, estimate the total change in entropy of the process.
37.8 J/K
A 2.8 kg piece of aluminum at 28.5 C is placed in 1.0 kg of water in a Styrofoam container at 20 C. Estimate the net change in entropy of the system.
.645 J/K
30.0 kg of water at 44.0 C is mixed with 15.0 kg of water at 20.0 C. Estimate the change in entropy.
128 J/K
at position 1, P=400 KPa, T=300 K
at position 3, P=100 KPa, V=4000 cm^3, T=300 K
Find P,V,T everywhere.
Fill out the engine table
what is the efficiency of this engine?
V_1 = 1000 cm^3
V_2 = 1000 cm^3 , P_2 = 1007.9 kPa
T_3 = 755.95 C
0 911.9 911.9 J
911.9 0 -911.9 J
-554.5 -554.5 0 J
357.4 357.4 0 J
39.2 %
If these temps are the reservoir temperatures, what is the ideal efficiency?
What percent of the ideal efficiency is this engine running?
60.3 %
65 % of ideal
at position 1, P=100 KPa
at position 2, P=400 kPa, V=1000 cm^3 , T=400 K
at position 3, P=100 KPa, V=4000 cm^3 T=400 K
Find P,V,T everywhere.
Fill out the engine table
what is the efficiency of this engine?
V_1 = 2297 cm^3 T_1 =229.7 K
-255.4 0 255.4 J
554.5 554.5 0 J
-170.2 -425.6 -255.4 J
128.9 128.9 0 J
23.2 %
A diesel engine runs with two adiabatic and an isochoric and isobaric process. Starting at 25 C 1 atm and 1050 cm^3 compresses to about 70 cm^3. It also absorbs about 1000 J in the burning of the fuel. (it is usually diatomic, but we'll pretend monatomic)
Find n , P,V,T for this engine
Fill in the engine energy chart
What is the efficiency.
n=.0429
101300 Pa, .00105 m^3, 298 K
9242000 Pa, .00007 m^3, 1812.5 K
9242000 Pa, .0001132 m^3, 2931.4 K
226000 Pa, .00105 m^3, 664.8 K
1
2
3
4
-810 0 810 J
400 1000 600 J
1213 0 -1213 J
0 197 - 197 J
803 803 0 J
80.3 %
S =
Q
T
"Can we go over the difference between effeciency and COP?"
"I am still struggling with how a refrigerator works."
"My only question is about Adiabatic processes. I'm not understand them in relation to a PV diagram."
3
2
1
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Full transcript
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May 20, 2013
Monday Math Game - Dice!
I love dice games to help reinforce math skills and the web is filled with amazing ideas! Some of my favourites are Stuck in Mud, Mouse and Snake.
Stuck in Mud Dice Game
The object of the game is to score the highest score. You will need five (or more) dice for this game. The first player rolls all the dice and adds up the numbers. If a 2 or 5 are rolled, they become stuck in the mud and cannot be counted in your score.
For example, if you throw a 2, 3, 4, 1, 5, your total would be 8 since you cannot count the 2 and the 5. The dice that rolled a 2 and 5 are taken out of play and the same player continues to roll with the dice that are left. Keep going until player has lost all of his or her dice. Second player does the same. Player with the highest score wins.
Mouse Dice Game
Players have to draw a mouse by rolling a die. Each number of the die is for a certain part of the body.
6 is the body
5 is the nose
4 is the whiskers
3 is the eyes
2 is the ears
1 is the tail
They must roll a 6 to start since everyone has to start with the body. Taking turns, students roll the die and draw the part accordingly. If they roll the die and that part has already been drawn they miss their turn and pass the die to the next player. If younger students may have a difficult time drawing a mouse, they could trace over a picture of one.
Snake
One of the reasons I like this game so much is because it’s a whole class game. Guided Math has a perfect explanation of this game, along with a free download for the game board!
This next simple game teaches place value and can be played with as many dice as you’d like, depending on the age of your students.
Place Value Dice Game
Depending of the age of your students, you can use 2 or 3 dice (or more). The first player rolls the dice. They then put the digits together to make the highest number they can. For example if they roll a 4 and 5 – the highest number they can make is 54. The other player does the same. Players play three times, then add up their three numbers. The person with the highest total after three throws is the winner.
Yahtzee is another classic favourite of mine to practice addition facts. It only requires 5 dice and a piece of paper to keep score. Click here for a FREE score sheet! There are several ways to play Yahtzee.
Sources:
www.ninlazina.info
www.memory-improvement-tips.com
www.guided-math.com
www.dice.virtuworld.net
www.ehow.com
www.yahtzee.org.uk
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The last format we want to look at is fractional odds. Personally, we aren’t a huge fan of fractional odds because they’re the most challenging to work with. The formula is almost the same as with decimal odds, but it gives your profit instead of total money returned. It also requires you to solve a fraction, which may be a nightmare for a lot of people. Regardless, we are going to walk you through how to do it with the same bet we’ve been working with.
They do this by manipulating the lines to entice action where they need it. If they have too much action coming in on one team, they will adjust the lines to pay out less for that team to deter more bets. At the same time, they will adjust the line for the other team to pay out more to entice more action on that side. This dance continues until the game starts to try and get the correct amount of money on each side of the contest.
## The simplest way to think about a moneyline is to consider a base bet of \$100. A moneyline is a number larger than 100, and it is either positive or negative. A line with a positive number means that the team is the underdog. If the line, for example, was +160 then you would make a profit of \$160 if you were to bet \$100. Obviously, then, the team is a bigger underdog the bigger the number is - a +260 team is perceived to be less likely to win than a +160 team.
```In an effort to have equal money on both sides of a wager, the sportsbook operator will move the point spread to attract money on the side that customers aren’t betting on. The odds for a point spread might change before the actual point spread. There are certain point spread numbers, like 3 and 7 in football, the sportsbook operators would like to avoid moving away from since they final score margin falls on these two numbers most often.
```
A common mistake that new bettors will do is to bet every single game. Unfortunately, this is not a winning strategy no matter how sharp you are. Stick to betting the games where you actually see value. Here's what we mean by value. Let's say in our earlier example that you agree with the sportsbook that the Florida Gators should win the game by seven. You should not bet this game then no matter what if the line is -7. If you're right, the best you can do is tie on your bet. When you pick a side, you're basically going to be guessing and flipping a coin. Theoretically, you'll win as many times as you lose, but you'll be paying the house percentage every single bet and slowly bleeding your money and profits away.
Another popular form of golf betting involves matchup propositions, in which two golfers are paired against each other in a head-to-head wager, with a betting line on each golfer set by the oddsmaker. The golfer with the better (lower) score wins the matchup. (If one golfer continues play in the tournament after his opponent misses the cut, the golfer who continues play wins the matchup.)
If you’re loving some of our expert picks or have a few hot tips of your own, you’re probably itching to get your bet in before the lines move. To help you out, our team has compiled a list of the absolute best and most trusted online sportsbooks offering action on the NBA. It’s important that we note that none of these sites can pay for a better review or recommendation from us. We take our reviews seriously and our integrity even more seriously. If we allowed sites to barter or pay their way onto our lists, we’d be serving you up advertisements and not recommendations.
It is a pretty simple concept once you get the hang of it, and you will also start to see profitable opportunities in football and hoops where wagering on the moneyline makes more sense than betting the point spread. If you really like an 8-point underdog in the NFL and think they will win, you can take the 8 points and hope they cover the spread. Or you can check out the moneyline option where they might be +280 and make more money betting them to win (\$280) than on the point spread (\$100).
Let's use this formula to calculate the implied probability of the Celtics winning their game against the Grizzlies. We know the odds are -240, which means we'd have to risk \$240 for a total potential return of \$340 (the initial stake plus the \$100 winnings). So the calculation here is \$240 divided by \$340. This gives us an implied probability of 0.7059.
Two possibilities existed for Seahawks backers at this point – either the team would win the game by at least three points or not. There was no possibility for a push, thanks to the use of a half-point. It’s impossible to score a half-point in football, so thanks to the magic of rounding, there’s no room for a tie outcome. Those who backed the Patriots were looking at two possible outcomes, too – either New England would pull off the upset or they would lose by just a point or two. Both would turn out in a win.
Jeff Gordon has been reporting and writing since 1977. His most recent work has appeared on websites such as eHow, GolfLink, Ask Men, Open Sports, Fox Sports and MSN. He has previously written for publications such as "The Sporting News" and "The Hockey News." He graduated from the University of Missouri-Columbia School of Journalism in 1979 with a bachelor's degree.
Doc's Sports is offering \$60 worth of handicapper picks absolutely free - no obligation, no sales people - you don't even have to enter credit card information. You can use this \$60 credit any way you please for any handicapper and any sport on Doc's Sports Advisory Board list of expert sports handicappers. Click here for more details and take advantage of this free \$60 picks credit today .
The point spread - also called "the line" or "the spread" - is used as a margin to handicap the favorite team. For betting purposes, the oddsmaker predicts that the favored team will win by a certain number of points. This number of points is the point spread. The favorite is always indicated by a minus sign (e.g. -5.5) and the underdog by a plus sign (e.g.+5.5). If you bet on the favorite, you win your bet if the favorite wins AND their margin of victory is greater than the point spread. If you bet on the underdog, you win if the underdog wins, ties, or if the favored team wins but fails to exceed the point spread. It is standard for point spread bets in most sports that you wager \$110 to win \$100.
Typically hockey is not a popular sport when it comes to point-spread betting because most games are decided by only a goal or two. Sports like football and basketball are better for point-spread betting mainly because they are higher scoring sports. However, Proline does offer NHL point-spread betting and it can get confusing so let’s take a look at how it works.
An If Bet is another type of bet that is a favorite among YouWager.eu bettors. An If Bet is similar to a parlay bet, however not quite the same. These types of bets can only be made after the original bet is made and won. An If Bet gives you the chance to bet on more than one game if the previous bet before has come out victorious. For instance, if you place a bet on the original game and it's a winner, an If Bet would require the total payout to be risked on the next bet. If an If Bet is lost, the total payout is subtracted from the last wager you have made.
Earlier, we explained how the implied probability of -240 is 70.59% and how the implied probability of +210 is 32.36%. Notice these two probabilities total 102.95%. The extra 2.95% is the bookmaker's advantage. It's called vig, and it's basically a commission that they charge customers for placing wagers. By removing the vig, you can see what the fair odds on the game would be.
Let's use this formula to calculate the implied probability of the Celtics winning their game against the Grizzlies. We know the odds are -240, which means we'd have to risk \$240 for a total potential return of \$340 (the initial stake plus the \$100 winnings). So the calculation here is \$240 divided by \$340. This gives us an implied probability of 0.7059.
Doc's Sports is offering \$60 worth of handicapper picks absolutely free - no obligation, no sales people - you don't even have to enter credit card information. You can use this \$60 credit any way you please for any handicapper and any sport on Doc's Sports Advisory Board list of expert sports handicappers. Click here for more details and take advantage of this free \$60 picks credit today .
There's another reason to bet the underdogs on the moneyline as well. If your handicapping has made you feel very strongly that a poor team is due for a big win then the moneyline allows you to profit much more handsomely from your conclusion than a point spread bet does. The moneyline, then, is a powerful situational tool for people who closely follow the NBA.
When you’re looking at over under bets, what you need to know is that that’s the combined score of the two teams for a game. In this case, it doesn’t matter who wins the game. All that matters is the final score. For example: let’s say that the New York Yankees are playing the Boston Red Sox and the total is 9.5. It doesn’t matter who wins the game but if the two teams combine for a total score of eight runs, say with a final score of Boston winning 5-3, then the game goes under. Or if the two teams combined for 10 runs – no matter who wins – then the game goes over. So when you’re looking at the odds and you see a total next to the moneyline or point spread, that tells you the over-under that is set for the game and you have to decide whether it will go over that set amount or under.
# This is four different games you can bet on, right? Wrong. These are several different handicap bets that you can make on a single game of soccer. Each of the individual rectangles is the same as the point spread bets we talked about earlier. For example, the first rectangle is betting on Liverpool at -3, getting paid 6 to 1. This means that if you take this bet you need Liverpool to win by four or more goals and you will get paid 6 to 1, or \$600 for every \$100 you wager.
Now that we’ve covered a lot of the basics concerning moneyline bets, let’s talk about the fun stuff – how much you’re going to make on your next correct moneyline bet. Remember, most online sportsbooks will automatically calculate the amount you are going to make on a moneyline bet before you even make the bet. You’re able to put in the amount you want to bet, and they will tell you immediately how much you would win from a correct pick.
Spread betting has moved outside the ambit of sport and financial markets (that is, those dealing solely with share, bonds and derivatives), to cover a wide range of markets, such as house prices.[5] By paying attention to the external factors, such as weather and time of day, those who are betting using a point spread can be better prepared when it comes to obtaining a favorable outcome. Additionally, by avoiding the favourite-longshot bias, where the expected returns on bets placed at shorter odds exceed that of bets placed at the longer odds, and not betting with one’s favorite team, but rather with the team that has been shown to be better when playing in a specific weather condition and time of day, the possibility of arriving at a positive outcome is increased.
### A wager on the Giants on the spread does not mean that New York has to win the game in order for you to win cash. All the G-Men have to do is come within 8 points of the ‘boys, and you’re a winner. You determine a winning or losing point spread by adding or subtracting 7.5 from the final score, depending on which side you laid your bet. If you’re confident that New York will at least come within a touchdown of beating the Cowboys, or beating them outright, then you’d wager on the spread in favor of New York.
As we mentioned, moneyline/win bets take into account who the favorites and who the underdogs are and will pay out winning bets accordingly. Here’s a quick example that will make this clear. Imagine that Mike Tyson (one of the greatest boxers of all time) is going to fight against an 80-year-old man. If the sportsbook let you bet on either side of the fight and paid you the same, would that be fair?
The most common NFL spreads are usually set between about 2.5-10.5 points, but you will also almost always have games each week with spreads lower than 2.5 and higher than 10.5. In the event that the oddsmakers feel the game doesn’t need a spread, it would be set at 0 or what some call a pick’em (both teams are given even odds to win for this type of bet).
When wagering on a driver matchup, both drivers involved must start the race (cross the finish line) otherwise the wager is "No Action" and the money is refunded. In a case where the starting driver is replaced during the race with another driver from the same team (same car), the position the new driver finishes in will be awarded to the original driver. This holds for wagering on win odds and driver matchups.
When the point spread was invented in Chicago by Charles McNeil the money line took a backseat. When two unevenly matched teams played, the playing field was leveled by having the favorite give points (for example Chicago Bears –7) while the underdog got points (Minnesota Vikings +7). No matter which team the bettor took the bettor would always risk \$110 to win \$100. The extra \$10 needed to win \$100 is called the juice or the vig, it is basically the house’s or the bookie’s take. It’s 10-percent of the bet so it would take \$33 to return \$30 and \$440 to return \$400 etc. (winning bettors get the vig back).
Point spreads are more common in the United States, but you can see them throughout the world. A point spread, in theory, is the sportsbooks attempt to create a "level playing field." Let's look at an exaggerated example that will make this clearer. Let's say the New England Patriots are playing a game against a junior varsity high school football team. They're also using deflated footballs, and the Patriots get to see the high school team's playbook before the game. If a sportsbook were to allow you to bet on which team would win, everyone would bet on the Patriots as they would probably annihilate this other team.
"Draw No Bet" is where it is possible to bet on either the home team or the away team. It is also common practice to refer to "Draw No Bet" in cases where no draw odds are offered. Should the specific match contain no winner (e.g. match ends as a draw), or the particular occurrence not happen (e.g. First Goal, Draw No Bet and match ends 0-0) the stakes will be refunded.
If you want to bet on football, then you have plenty of options. There are not only lots of games you can bet on, there are also lots of different types of wagers you can place. Point spreads and totals are the most popular types, by quite a distance, and many football bettors stick solely to those. This isn't really the ideal approach, as some of the other wagers can be very useful in the right circumstances.
```Each week you'll submit who you think will cover each football game based on the point spread entered by your pool administrator (see below for an explanation of point spreads). For the last game of the week (usually Monday night) you will specify the total number of points you think will be scored in that game. For each game you choose correctly, you will receive 1 point. The player with the most games chosen correctly will win the pool for that week. If there is a tie, the player that is closest to the actual total points scored in the last game of the week will win the tiebreaker. If there is a tie after that, the winnings will be split up between those players.
```
In theory, sportsbooks don't care about the outcome of a game, although for those of you who bet on the Steelers (-5.5) last season versus the Chargers and saw a game winning TD returned by S Troy Palumalu as the game expired reversed, thus negating a seven point victory and putting the final at 11-10, you might think otherwise, but this is how it's supposed to be.
###### You may have noticed that we said earlier that it’s the amount of money bet on each side that is important to the sportsbook and not the number of bets. This is an important distinction to understand. If 100 people bet \$10 on Team A, and then one person bets \$10,000 on Team B, the line is going to shift to attract action onto Team A. Even though more people bet on Team A, more actual money came in on Team B.
How the point spread works - When two teams meet on the playing field or on the basketball court, one team is typically better than the other or in a more favorable position because of factors like playing at home. If all you had to do were pick the winning team in a game, everybody would simply wager on the best team or the home team in a even matchup and bypass all the lines and collect their winnings at a high rate.
covering a spread - When a team surpasses the expectations of a point spread set by football experts. For instance, if a team is expected to win by 10 points, and they actually win by 11 or more, this is called 'covering' the point spread. If the team wins by nine or less, or they lose the game, this is called NOT 'covering' the point spread. If the team wins by exactly 10 points, this results in a push or tie.
--Fractional odds are most commonly found in racing. A 10/1 payout should be read "\$10 paid for every \$1 wagered." When the bigger number is on the left, you will find that bet is normally an underdog in the race. Also note, however, that in case such as "Who will win the Super Bowl in the NFL?" you will see all the teams listed as "underdogs"… i.e. paying at least 2/1 (some up to 300/1 or more).
### Reading sports betting lines becomes easier with practice and experience with different sporting events. What looks like a jumble of letters and numbers actually gives a lot of information in a tiny amount of space. Different sports have different types of wagers available, such as the run line in baseball or the puck line in hockey, both of which replace the money line found in our football example. The more experience you have watching and gambling on different sports, the faster you’ll be able to read betting lines.
You could even take it a step further and take the next rectangle down and bet Liverpool +1. This means that Liverpool can tie or win by any amount of goals and you win your bet. As Liverpool is a huge favorite, you won't be paid very well at all for this bet, but you can still turn a profit when you are right. You would be paid at 1 to 10 which means you would get \$1 for every \$10 you bet. If you bet \$100, you would get a \$10 profit on this bet.
Of course, it wouldn’t be. Everyone would bet on Mike Tyson, and the sportsbook would lose all of their money and close the next day. So what the sportsbooks do is they assess who is the favorite and who is the underdog and assign a value to how much in each direction they think they are. Let’s look at what the odds might look like for our fictitious fight and break down what everything means.
What may look like a jumble of words, numbers, and punctuation is actually a precise and easy-to-read breakdown of the various odds and point spread details your book is offering. Here is a breakdown of each unit of information given above. Once you understand each part of the jumbled details above, you’ll be able to read a sports betting line with confidence.
## The first thing you’ll notice with moneyline odds is that there is either a positive or negative sign in front of the number. What that sign denotes is how much you’ll win betting on each side. If there’s a positive sign next to the odds, that indicates the amount of money you would win if you bet \$100. If the odds on a tennis player said +150, that means that for a \$100 bet, you would win \$150. Now if there is a minus sign in front of the odds, that is the number that you would have to bet in order to win \$100. For example, if a football team was -250, that means you’d have to bet \$250 to win \$100.
There is no such thing as a half point in sports, but there is in sports betting! The half point ensures that a side will win and that the match will not end in a push, where the spread equals the actual difference in points between the two teams. In a push all bettors get their money back, which is no good for the oddsmaker! Half points also give oddsmakers more control over lines, allowing them to set more specific values.
```This highlights a notable advantage of the moneyline wager. You get to control, to some extent, the risk versus reward. For example, you might be quite certain that the Cardinals are going to win this game, but not convinced that they're going to cover the spread. So a moneyline wager is the safe option. There's less money to be made, but less chance of losing. On the other hand, you might think that the Packers are going to cause an upset. Rather than betting on them to cover the spread, you can bet on them to win outright. There's less chance of winning such a wager, but the potential returns are much greater.
```
If you like favorites, you're going to be betting a lot to win a little. The money line will always be listed to the right of the point spread on the odds board in a sports book. In the above example, the money line would probably be Chicago -250 and Detroit +200. To bet Chicago simply to win, you must wager \$250 to win \$100, while a \$100 bet on Detroit would pay \$200 if the Lions come through.
Is there value there? Yes. Are you going to make money off that bet? Well, it depends. If you’re only able to make a bet like this once, then you’re most likely going to lose. In order to realize that value, you’ll need to be in a lot of similar opportunities. If you have a very long-term betting strategy, then you can probably get away with making this bet. But if you’re looking for some more regular profit, you might want to steer clear of this. The odds say that the team is only going to win the game a little under 3 times for every 100 times they play. There is value there, but it depends on your betting strategy if you want to make that bet.
For example: New England –2.5 (–110) or Philadelphia +2.5 (–110) means you’d wager \$110 for the chance to win an additional \$100 if you bet on the point spread. Depending on which side is receiving the most action, a sportsbook will often move the line up or down in order to incentivize betting on the less popular side. Injuries or unforeseen changes can also impact a point spread gambling line. Point spreads are often listed with a half-point (ex: 2.5) in order to prevent the final margin from landing exactly on the point spread (ex: 10-point spread, final score of 20–10). A “push” or “tie” usually goes to the house or sportsbook, unless another arrangement has been agreed upon beforehand.
Sports betting has been around since 1000 B.C in China, where betting on animal fights was commonplace. In ancient Rome, one could wager on the Gladiatorial games. The idea of betting on sports is as old as organized sport itself. But up until the 1940s, bettors were fairly limited in what kind of bets they could make. The standard system of odds would allow bets on, for example, the 3-1 odds that the Steelers would beat the Browns.
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One Step Equations
Two Step Equations
Grouping Like Terms
Simplifying Like Terms
Distributive Property
3 + p = 8
What is p=5?
6 = a/4 + 2
What is a=16?
d - 18d
What is - 17d?
−6k + 7k
What is k?
100
3(6 + 7n)
What is 18 + 21n?
26 = 8 + v
What is v=18?
9x − 7 = −7
What is x=0?
18 + 10q + 6
What is 10q + 24
−4x − 10x
What is -14x?
200
10 − 5(9n − 9)
What is 55 − 45n?
−15x = 0
What is x=0?
−9x + 1 = −80
What is x=9?
14 - 8z + 9z
What is z + 14?
11r − 12r
What is -r?
300
−10(1 − 9x) + 6(x − 10)
What is −70 + 96x?
−104 = 8x
What is x=-13?
144 = −12(x + 5)
What is x=-17?
12y + 5y
What is 17y?
400
n − 10 + 9n − 3
What is 10n - 13?
400
−3(10b + 10) + 5(b + 2)
What is −25b − 20?
−5 = a/18
What is x=-90?
500
−243 = −9(10 + x)
What is x=17?
500
19q - 15 - 2q - 10
What is 17q -25?
−3x − 9 + 15x
What is 12x - 9?
500
−7(n + 3) − 8(1 + 8n)
What is −71n − 29?
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# Homework Help: Abstract Algebra set theory
1. Feb 23, 2012
### iHeartof12
Let A, B and C be sets.
Prove that if A$\subseteq$B$\cup$C and A$\cap$B=∅, then A$\subseteq$C.
My attempted solution:
Assume A$\subseteq$B$\cup$C and A$\cap$B=∅.
Then $\vee$x (x$\in$A$\rightarrow$x$\in$B$\cup$x$\in$c).
I'm not sure where to start and how to prove this. Any help would be greatly appreciated. Thank you.
2. Feb 23, 2012
### Staff: Mentor
....disregard earlier post...
3. Feb 23, 2012
### jbunniii
Suppose $x \in A$. The goal is to show that this implies $x \in C$.
Since $x \in A$ and $A \subset B \cup C$, it follows that $x \in B$ or $x \in C$. Can you exclude one of these possibilities?
4. Feb 23, 2012
### iHeartof12
Since A$\bigcap$B=∅, x$\in$A or x$\in$B.
Thus x$\in$A, x$\notin$B and x$\in$C.
Therefor A$\subseteq$C.
Is that a good way to show how to exclude the possibility of x$\in$B?
5. Feb 23, 2012
### jbunniii
You have the right idea, but the wording is a little unclear. The following is not true: "Since A$\bigcap$B=∅, x$\in$A or x$\in$B."
Suppose $x \in A$. The goal is to show that this implies $x \in C$.
Since $x \in A$ and $A \subset B \cup C$, it follows that $x \in B$ or $x \in C$. However, $x$ cannot be in $B$, because if it were, then we would have $x \in A \cap B = \emptyset$, which is impossible. Therefore...
6. Feb 23, 2012
### iHeartof12
Ok I think I get it tell me if I worded this correctly:
Suppose $x \in A$ and $A \cap B = \emptyset$
Since $x \in A$ and $A \subset B \cup C$, this means that $x \in B$ or $x \in C$. Consequently, $x$ cannot be in $B$, because if it were, then we would have $x \in A \cap B = \emptyset$, which is impossible. Therefore $x \in C$. Thus $x \in A$ implies $x \in C$ so it follows that $A \subseteq C$
7. Feb 23, 2012
### jbunniii
Looks good to me.
8. Feb 23, 2012
### iHeartof12
ok thank you for your help
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# What are the correlation values with respect to low/average/excessive correlation specifically in medical research?
This yields an inventory of errors squared, which is then summed and equals the unexplained variance. The adjusted R-squared compares the descriptive power of regression models—two or extra variables—that embody a diverse number of independent variables—known as a predictor. Every predictor or unbiased variable, added to a model increases the R-squared value and by no means decreases it. So, a model that includes several predictors will return higher R2 values and should seem to be a greater match.
### What does the R squared value mean?
R-squared is a statistical measure of how close the data are to the fitted regression line. It is also known as the coefficient of determination, or the coefficient of multiple determination for multiple regression. 100% indicates that the model explains all the variability of the response data around its mean.
This consists of taking the information factors (observations) of dependent and independent variables and discovering the line of best match, typically from a regression mannequin. From there you would calculate predicted values, subtract precise values and square the outcomes.
### How is R Squared calculated?
Are Low R-squared Values Inherently Bad? No! There are two major reasons why it can be just fine to have low R-squared values. In some fields, it is entirely expected that your R-squared values will be low.
For instance, the correlation for the info within the scatterplot under is zero. However, there is a relationship between the 2 variables—it’s just not linear. Pearson’s correlation coefficient is represented by the Greek letter rho (ρ) for the inhabitants parameter and r for a sample statistic. This correlation coefficient is a single quantity that measures both the strength and path of the linear relationship between two continuous variables. The adjusted R2 may be negative, and its value will all the time be lower than or equal to that of R2.
Although, nonnormality can make a linear relationship less probably. So, graph your data on a scatterplot and see what it appears like. If it’s close to a straight line, you must most likely use Pearson’s correlation. If it’s not a straight line relationship, you may want to use one thing like Kendall’s Tau or Spearman’s rho coefficient, each of that are based on ranked knowledge. While Spearman’s rho is more generally used, Kendall’s Tau has preferable statistical properties.
However, statistical significance implies that it’s unlikely that the null hypothesis is true (less than 5%). There are a number of kinds of correlation coefficients, however the one that’s commonest is the Pearson correlation (r). This measures the power and path of the linear relationship between two variables. It can not seize nonlinear relationships between two variables and cannot differentiate between dependent and independent variables.
If we’ve more variables that explain modifications in weight, we can include them in the model and doubtlessly enhance our predictions. And, if the connection is curved, we are able to nonetheless fit a regression model to the data. In fact, an R-squared of 10% and even less could have some information value when you’re looking for a weak sign in the presence of plenty of noise in a setting the place even a veryweak one could be of general curiosity. Sometimes there may be plenty of worth in explaining solely a really small fraction of the variance, and typically there isn’t. Data transformations corresponding to logging or deflating also change the interpretation and standards for R-squared, inasmuch as they modify the variance you start out with.
R-squared is a statistical measure that represents the proportion of the variance for a dependent variable that is explained by an unbiased variable. In regression analysis, you’d like your regression mannequin to have vital variables and to supply a excessive R-squared worth. This low P worth / excessive R2 mixture signifies that adjustments in the predictors are related to adjustments in the response variable and that your mannequin explains a lot of the response variability. Technically, you want to graph your data and look at the shape of the relationship somewhat than assessing the distribution for every variable.
R-squared (R2) is a statistical measure that represents the proportion of the variance for a dependent variable that is explained by an independent variable or variables in a regression mannequin. Whereas correlation explains the energy of the connection between an independent and dependent variable, R-squared explains to what extent the variance of one variable explains the variance of the second variable. So, if the R2of a model is zero.50, then approximately half of the observed variation can be defined by the model’s inputs. Significance of r or R-squared is determined by the energy or the connection (i.e. rho) and the sample dimension.
Adjusted R-squared, on the other hand, gives the percentage of variation defined by only these unbiased variables that, in actuality, have an effect on the dependent variable. Is there a sample within the data that follows a sample other than linear. As far as linear, adding different impartial explanatory variables certainly has advantage, however the query is which one(s)?
The F value in regression is the result of a check where the null hypothesis is that all the regression coefficients are equal to zero. Basically, the f-check compares your mannequin with zero predictor variables (the intercept only mannequin), and decides whether or not your added coefficients improved the mannequin. If you get a big result, then whatever coefficients you included in your model improved the mannequin’s match.
Adjusted R2 can be interpreted as an unbiased (or less biased) estimator of the inhabitants R2, whereas the observed sample R2 is a positively biased estimate of the population worth. Adjusted R2 is extra acceptable when evaluating mannequin fit (the variance within the dependent variable accounted for by the unbiased variables) and in comparing various fashions in the characteristic selection stage of mannequin building. In least squares regression utilizing typical information, R2 is no less than weakly rising with increases within the number of regressors in the model.
This statistic represents the proportion of variation in a single variable that different variables clarify. For a pair of variables, R-squared is solely the sq. of the Pearson’s correlation coefficient. For example, squaring the height-weight correlation coefficient of zero.694 produces an R-squared of zero.482, or 48.2%.
### Is a low R Squared bad?
Measuring Linear Association The relationship between two variables is generally considered strong when their r value is larger than 0.7. The correlation r measures the strength of the linear relationship between two quantitative variables. Pearson r: • r is always a number between -1 and 1.
Specifically, R2 is an element of and represents the proportion of variability in Yi which may be attributed to some linear combination of the regressors (explanatory variables) in X. This leads to the choice strategy of looking at the adjusted R2. The rationalization of this statistic is nearly the same as R2 but it penalizes the statistic as extra variables are included in the mannequin.
Some clinical and epidemiological researchers have been highly critical of using speculation testing and p-values in biological and medical research. Some declare that the complete concept is an outright menace and a menace to scientific progress .
A linear regression at all times wants to regulate for endogeneity from omitted variables, correlated variables (multicollinearity) and bidirectional causality among others. In Corporate finance that means panel information from a cross section of companies throughout industries. However, if the info is procured from real sources, it will not be a linear fit on a regular basis. If the plotting point out polynomial match, please use MARS (Multiple Adaptive Regression Spline) as a substitute of Regular Gaussian Regression.
For circumstances apart from fitting by ordinary least squares, the R2 statistic could be calculated as above and may still be a useful measure. Values for R2 can be calculated for any type of predictive model, which need not have a statistical foundation. There are instances where the computational definition of R2 can yield unfavorable values, relying on the definition used. This can arise when the predictions which might be being compared to the corresponding outcomes have not been derived from a model-fitting procedure utilizing those knowledge. Even if a mannequin-fitting process has been used, R2 should be negative, for example when linear regression is performed with out together with an intercept, or when a non-linear operate is used to suit the info.
One class of such circumstances includes that of simple linear regression the place r2 is used as a substitute of R2. If extra regressors are included, R2 is the square of the coefficient of a number of correlation. In each such instances, the coefficient of dedication normally ranges from zero to 1. In statistics, the coefficient of dedication, denoted R2 or r2 and pronounced “R squared”, is the proportion of the variance in the dependent variable that is predictable from the impartial variable(s).
For instance, the p-value might be adjusted in mild of a medical assessment of the likelihood of the null hypothesis in a study . These assessments are statistically expressed as so-referred to as a priori possibilities. These a priori chances and the precise data from the study are used in the statistical analyses. An obvious dilemma is that of agreeing beforehand what the effect is likely to be and the way much each particular person p-value must be adjusted.
Do you could have any additional information on the data, for instance geographic location, time, anything that can use to subgroup the information. There seems to be a relationship with the explanatory variable you’re using, but there’s clearly a lot more that is unexplained by the variables you are utilizing. Additionally, a type of the Pearson correlation coefficient shows up in regression evaluation. R-squared is a main measure of how properly a regression model suits the information.
A correlation of -1.zero reveals an ideal unfavorable correlation, whereas a correlation of 1.zero reveals an ideal optimistic correlation. A correlation of 0.0 exhibits no linear relationship between the motion of the 2 variables. R-squared cannot verify whether the coefficient ballpark determine and its predictions are prejudiced. It also does not show if a regression mannequin is passable; it can present an R-squared figure for a good mannequin or a high R-squared determine for a mannequin that doesn’t fit. The decrease the value of the R2 the much less the 2 variables correlate to 1 another.
Because increases in the number of regressors enhance the worth of R2, R2 alone cannot be used as a meaningful comparability of fashions with very different numbers of impartial variables. For a meaningful comparability between two models, an F-check can be carried out on the residual sum of squares, similar to the F-tests in Granger causality, although this isn’t at all times acceptable. As a reminder of this, some authors denote R2 by Rq2, where q is the variety of columns in X (the number of explanators including the fixed). R-squared will provide you with an estimate of the relationship between actions of a dependent variable based on an unbiased variable’s actions. It would not tell you whether your chosen mannequin is sweet or unhealthy, nor will it tell you whether the information and predictions are biased.
## Regression Analysis: How Do I Interpret R-squared and Assess the Goodness-of-Fit?
• Whereas correlation explains the energy of the relationship between an impartial and dependent variable, R-squared explains to what extent the variance of 1 variable explains the variance of the second variable.
• If the pattern is very massive, even a miniscule correlation coefficient may be statistically significant, yet the relationship may don’t have any predictive worth.
• So, if the R2of a model is zero.50, then approximately half of the observed variation could be defined by the model’s inputs.
• R-squared (R2) is a statistical measure that represents the proportion of the variance for a dependent variable that’s explained by an unbiased variable or variables in a regression mannequin.
These are unbiased estimators that right for the sample size and numbers of coefficients estimated. Adjusted R-squared is at all times smaller than R-squared, but the distinction is usually very small until you are trying to estimate too many coefficients from too small a pattern in the presence of an excessive amount of noise.
Consequently, we are able to reject the null speculation and conclude that the relationship is statistically important. The sample information support the notion that the relationship between peak and weight exists in the population of preteen ladies. Pearson’s correlation coefficients measure only linear relationships. Consequently, if your information include a curvilinear relationship, the correlation coefficient is not going to detect it.
Every predictor added to a mannequin will increase R-squared and by no means decreases it. In anoverfittingcondition, an incorrectly excessive worth of R-squared is obtained, even when the mannequin actually has a decreased capacity to foretell. A lower p-worth is usually interpreted as meaning there is a stronger relationship between two variables.
Specifically, adjusted R-squared is the same as 1 minus (n – 1)/(n – k – 1) times 1-minus-R-squared, the place n is the sample measurement and k is the number of impartial variables. In a a number of regression mannequin R-squared is determined by pairwise correlations amongst allthe variables, including correlations of the independent variables with each other in addition to with the dependent variable. It known as R-squared because in a easy regression model it’s just the square of the correlation between the dependent and independent variables, which is usually denoted by “r”. The correlation, denoted by r, measures the quantity of linear association between two variables.r is at all times between -1 and 1 inclusive.The R-squared worth, denoted by R2, is the square of the correlation. It measures the proportion of variation within the dependent variable that may be attributed to the impartial variable.The R-squared value R2 is at all times between zero and 1 inclusive.Perfect constructive linear affiliation.
The correlation coefficient is a statistical measure of the strength of the connection between the relative actions of two variables. A calculated quantity higher than 1.zero or less than -1.0 means that there was an error within the correlation measurement.
I believe that significance testing at the 5 % stage is beneficial as an initial statistical screening before any subsequent evaluation of impact dimension and clinical dialogue. Undertaking this evaluation has proven helpful in practice and is generally accepted, although in purely methodological phrases it has some flaws with regard to its software to scientific knowledge. Research is being undertaken with a view to growing methodologies and options to the p-worth. Some are engaged on methodologies based on Bayesian statistics, during which pre-defined assumptions concerning probabilities of effects and results are considered in statistical analyses.
Results higher than 70% usually point out that a portfolio closely follows the measured benchmark. Higher R-squared values also indicate the reliability of beta readings. I don’t assume there’s any exhausting rule regarding what’s an acceptable R-squared worth. However, Dr. Alhayri did explain a heuristic for acceptable R2 levels which are appropriate. Regression evaluation allows us to expand on correlation in different ways.
## R-Squared Definition
If the likelihood of our observations is low after we assume that the null speculation is true, this is an indirect indication that our observed results usually are not brought on by random variations. It appears reasonable to undertake an preliminary assessment of whether or not this may be the case.
### What does an r2 value of 0.9 mean?
The R-squared value, denoted by R 2, is the square of the correlation. It measures the proportion of variation in the dependent variable that can be attributed to the independent variable. Correlation r = 0.9; R=squared = 0.81. Small positive linear association. The points are far from the trend line.
Unlike R2, the adjusted R2 will increase solely when the increase in R2 (due to the inclusion of a brand new explanatory variable) is a couple of would expect to see by likelihood. are unknown coefficients, whose values are estimated by least squares. The coefficient of determination R2 is a measure of the global match of the model.
In cases the place negative values arise, the imply of the data offers a greater match to the outcomes than do the fitted function values, based on this explicit criterion. There are a number of definitions of R2 that are only typically equal.
Standard deviation is a measure of the dispersion of data from its common. Covariance is a measure of how two variables change collectively, however its magnitude is unbounded, so it’s tough to interpret. By dividing covariance by the product of the 2 standard deviations, one can calculate the normalized version of the statistic.
As this single impartial variable is explaining a low variation within the outcome variable, I recommend you embody different explanatory variables within the regression before accepting the results of your initial analysis. You may want to look into adding more non-correlated independent variables to your model – variables that some how relate to your dependent variable (context).
If the sample could be very giant, even a miniscule correlation coefficient could also be statistically important, but the connection could have no predictive worth. In the case of a couple of unbiased variable, you’ll have to plot the residuals towards the dependent and independent variables to examine for non-linearity.
### What is a good r2 value for regression?
Correlation Coefficient = +1: A perfect positive relationship. Correlation Coefficient = 0.8: A fairly strong positive relationship. Correlation Coefficient = 0.6: A moderate positive relationship. Correlation Coefficient = -0.8: A fairly strong negative relationship.
## How Do You Calculate R-Squared in Excel?
A high or low R-square is not essentially good or bad, as it would not convey the reliability of the mannequin, nor whether or not you’ve chosen the right regression. You can get a low R-squared for an excellent model, or a excessive R-sq. for a poorly fitted model, and vice versa. R-Squared is a statistical measure of fit that indicates how a lot variation of a dependent variable is defined by the unbiased variable(s) in a regression model. As famous above, speculation testing and p-values provide no direct solutions to the questions we pose in most medical and medical research research, however this statistical methodology however stays regularly used. In my opinion, it’s because the p-value has proven virtually useful.
R-Squared only works as supposed in a simple linear regression model with one explanatory variable. With a a number of regression made up of several impartial variables, the R-Squared must be adjusted. The adjusted R-squared compares the descriptive energy of regression models that include various numbers of predictors.
In other words, peak explains about half the variability of weight in preteen women. Correlation coefficients that equal zero indicate no linear relationship exists. If your p-value is lower than your significance degree, the pattern incorporates enough evidence to reject the null speculation and conclude that the correlation coefficient doesn’t equal zero. In different phrases, the pattern knowledge help the notion that the relationship exists within the inhabitants. For instance, the practice of carrying matches (or a lighter) is correlated with incidence of lung cancer, but carrying matches does not trigger cancer (in the usual sense of “trigger”).
## What Does R-Squared Tell You?
One major difference between R-squared and the adjusted R-squared is that R2 assumes every independent variable—benchmark—within the mannequin explains the variation in the dependent variable—mutual fund or portfolio. It gives the share of defined variation as if all independent variables in the mannequin affect the dependent variable.
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# Principle of Recursive Definition
## Theorem
Let $\N$ be the natural numbers.
Let $T$ be a class (which may be a set).
Let $a \in T$.
Let $g: T \to T$ be a mapping.
Then there exists exactly one mapping $f: \N \to T$ such that:
$\forall x \in \N: \map f x = \begin{cases} a & : x = 0 \\ \map g {\map f n} & : x = n + 1 \end{cases}$
### General Result
Let $p \in \N$.
Let $p^\ge$ be the upper closure of $p$ in $\N$:
$p^\ge := \set {x \in \N: x \ge p} = \set {p, p + 1, p + 2, \ldots}$
Then there exists exactly one mapping $f: p^\ge \to T$ such that:
$\forall x \in p^\ge: \map f x = \begin{cases} a & : x = p \\ \map g {\map f n} & : x = n + 1 \end{cases}$
### Strong Version
Let $\omega$ denote the natural numbers as defined by the von Neumann construction.
Let $A$ be a class.
Let $c \in A$.
Let $g: \omega \times A \to A$ be a mapping.
Then there exists exactly one mapping $f: \omega \to A$ such that:
$\forall x \in \omega: \map f x = \begin{cases} c & : x = \O \\ \map g {n, \map f n} & : x = n^+ \end{cases}$
## Proof 1
Consider $\N$ defined as a Peano structure $\struct {P, 0, s}$.
The result follows from Principle of Recursive Definition for Peano Structure.
$\blacksquare$
## Proof 2
Consider $\N$ defined as the von Neumann construction of the natural numbers.
The result follows from Principle of Recursive Definition for Minimally Inductive Set.
$\blacksquare$
## Proof 3
Recall the general result:
Let $p \in \N$.
Let $p^\ge$ be the upper closure of $p$ in $\N$:
$p^\ge := \set {x \in \N: x \ge p} = \set {p, p + 1, p + 2, \ldots}$
Then there exists exactly one mapping $f: p^\ge \to T$ such that:
$\forall x \in p^\ge: \map f x = \begin{cases} a & : x = p \\ \map g {\map f n} & : x = n + 1 \end{cases}$
The result follows from setting $p = 0$.
$\blacksquare$
## Proof 4
Let $\omega$ denote the natural numbers as defined by the von Neumann construction.
Let $A$ be a class.
Let $c \in A$.
Let $g: \omega \times A \to A$ be a mapping.
Then there exists exactly one mapping $f: \omega \to A$ such that:
$\forall x \in \omega: \map f x = \begin{cases} c & : x = \O \\ \map g {n, \map f n} & : x = n^+ \end{cases}$
Let $h: A \to A$ be defined as:
$\forall x \in A: \map h x := \map g {a, x}$ for arbitrary $a \in \omega$
That is:
$\forall y \in \omega: \map g {y, x} = \map h x$
Then a priori there exists exactly one mapping $f: \omega \to A$ such that:
$\forall x \in \omega: \map f x = \begin {cases} c & : x = \O \\ \map h {\map f n} & : x = n^+ \end {cases}$
The result follows on identifying $\omega$ with $\N$, $c$ with $a$, $A$ with $T$, $\O$ with $0$ and $n^+$ with $n + 1$.
$\blacksquare$
## Also presented as
The Principle of Recursive Definition can also be presented as:
For any mapping $f: T \to T$ and any $a \in T$, there exists an infinite sequence $a_0, a_1, \ldots, a_n, a_{n + 1}, \ldots$ such that:
$a_0 = a$
$a_{n + 1} = \map g {a_n}$
## Fallacious Proof
The proofs given (hidden behind the links) are necessarily long, precise and detailed.
There is a temptation to take short cuts and gloss over the important details.
The following argument, for example, though considerably shorter, is incorrect.
Consider $\N$, defined as a naturally ordered semigroup $\struct {S, \circ, \preceq}$.
Let the mapping $f$ be defined as:
$\map f x = \begin{cases} a & : x = 0 \\ \map s {\map f n} & : x = n \circ 1 \end{cases}$
if $\map f n$ is defined.
Let $S' = \set {n \in S: \map f n \text{ is defined} }$.
Then:
$0 \in S'$
and:
$n + 1 \in S'$
$S' = S$
Thus the domain of $f$ is $S$.
Consequently, $f$ is a mapping from $S$ into $T$ which satisfies:
$\map f 0 = a$
and:
$\map f {n \circ 1} = \map s {\map f n}$
for all $n \in S$.
$\blacksquare$
### Objections
$(1):$
In the above argument, $S'$ is not precisely defined.
In order for a set to be defined, that definition should be able to be expressed solely in terms of logic and definitions from set theory.
In this case, the expression "is defined" does not meet that criterion.
$(2):$
The mapping $f$ is not defined properly.
A mapping needs to be defined as a set of ordered pairs which needs to satisfy a condition.
In the above, it is indeed specified that
$\tuple {0, a} \in f$
and:
$\tuple {n \circ 1, \map s x} \in f$ whenever $\tuple {n, x} \in f$
Thus it appears either that:
$f$ itself is used to define $f$
or else that:
$f$ itself changes during the process in which it is being defined.
Neither of these possibilities can be accepted.
$(3):$
The only property of $\struct {S, \circ, \preceq}$ used in the argument is that it satisfies Principle of Mathematical Induction for Naturally Ordered Semigroup.
However, consider the commutative semigroup $\struct {D, +}$ which has elements $0$ and $1$, such that:
$D$ is the only subset of $D$:
containing $0$
and
containing $x + 1$ whenever it contains $x$.
Then:
for any set $T$
any mapping $g: T \to T$
any element $a \in T$
there exists a mapping $f: D \to T$ such that:
$\map f y = \begin{cases} a & : y = 0 \\ \map s {\map f x} & : y = x + 1 \end{cases}$
But consider the additive group of integers modulo $2$:
$\struct {\Z_2, +_2}$
which is indeed a commutative semigroup containing $0$ and $1$ which satisfies the hypothesis.
But if $g: \N \to \N$ is the mapping defined as:
$\map g n = n + 1$
there is no mapping $f: \Z_2 \to \N$ which satisfies:
$\map f y = \begin{cases} 0 & : y = 0 \\ \map s {\map f x} & : y = x +_2 1 \end{cases}$
for all $x \in \Z_2$.
Hence the argument is invalid.
## Also known as
This result is often referred to as the Recursion Theorem.
Some sources only cover the general result.
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# Counting Cash: The Value of Different Coins (1)
In this worksheet, students add together the value of different coins.
Key stage: KS 1
Curriculum topic: Measurement
Difficulty level:
### QUESTION 1 of 10
Add up these coins. Write the value in pounds. Then write the value in pence.
Example
£2 + £1 = £3
50p + 20p + 10p + 5p + 2p + 1p = 88p
Total in pounds and pence = £3 and 88p
Total in pence = 388p
Find the total of these coins. TWO boxes are correct. Click them.
£4 and 50p
350p
£3 and 50p
Find the total of these coins. TWO boxes are correct.
£3
£4
3
300p
Find the total of these coins. TWO boxes are correct. Click them.
£3 and 75p
£3 and 70p
370p
Find the total of these coins. TWO boxes are correct. Click them.
£3 and 80p
£3 and 90p
380p
Find the total of these coins. TWO boxes are correct. Click them.
470
470p
£4 and 70p
Find the total of these coins. TWO boxes are correct. Click them.
£3 and 1p
£2 and 1p
210p
201p
Find the total of these coins. TWO boxes are correct. Click them.
£2 and 10p
201p
210p
£4 and 10p
Find the total of these coins. TWO boxes are correct. Click them.
£5 and 50p
£3 and 50p
350p
350
Find the total of these coins. TWO boxes are correct. Click them.
£5 and 5p
£3 and 5p
350p
305p
Find the total of these coins. TWO boxes are correct. Click them.
268p
£1 and 68p
618p
168p
• Question 1
Find the total of these coins. TWO boxes are correct. Click them.
350p
£3 and 50p
EDDIE SAYS
Either £3 and 50p or 350p works for this one.
• Question 2
Find the total of these coins. TWO boxes are correct.
£3
300p
EDDIE SAYS
£3 or 300p is correct for this question.
• Question 3
Find the total of these coins. TWO boxes are correct. Click them.
£3 and 70p
370p
EDDIE SAYS
We are looking for either £3 and 70p or 370p.
• Question 4
Find the total of these coins. TWO boxes are correct. Click them.
£3 and 80p
380p
EDDIE SAYS
Here we can have either £3 and 80p or 380p.
• Question 5
Find the total of these coins. TWO boxes are correct. Click them.
470p
£4 and 70p
EDDIE SAYS
Remember, we do NOT write 470. We write 470p or £4 and 70p.
• Question 6
Find the total of these coins. TWO boxes are correct. Click them.
£2 and 1p
201p
EDDIE SAYS
Look hard.
There are 2 pounds.
There is 1 penny.
So we have £2 and 1p or 201p.
• Question 7
Find the total of these coins. TWO boxes are correct. Click them.
£2 and 10p
210p
EDDIE SAYS
Look hard.
We have £2.00 + 10p
That makes £2 and 10p
or 210p.
• Question 8
Find the total of these coins. TWO boxes are correct. Click them.
£3 and 50p
350p
EDDIE SAYS
Remember, we write 350p or £3 and 50p
• Question 9
Find the total of these coins. TWO boxes are correct. Click them.
£3 and 5p
305p
EDDIE SAYS
Look hard.
The final coin is 5p.
We can either write £3 and 5p or 305p.
• Question 10
Find the total of these coins. TWO boxes are correct. Click them.
£1 and 68p
168p
EDDIE SAYS
Remember:
we write £1 and 68p
or we write 168p.
---- OR ----
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https://ready4.com/experts-explain-a-classic-gmat-problem-solving-question/
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# Experts Explain: A Classic GMAT Problem Solving Question
Want to learn how the best GMAT takers solve GMAT questions? In our Experts Explain series, tutors from Prep4GMAT’s tutor marketplace show you how to tackle GMAT questions like a pro.
Today, John Easter, a tutor with 15 years of test prep experience and a 99th percentile score on the GMAT breaks down a standard problem solving question. John has helped many students push past the 700 mark or improve in one part of the test to boost their overall boost. Many of his students have gone on to Northwestern or Booth for their MBAs. Check him out on the app or on his website.
Here’s a GMAT classic!
For how many positive integer values of [math] n [/math] is [math] 5^n [/math] a factor of [math] 50! [/math]?
(A) 11
(B) 12
(C) 13
(D) 14
(E) 15
…Okay, finished? Scroll to the bottom of the post to see the correct answer and check out the explanations below.
## Translation
Obviously, [math] 5^1 [/math] and [math] 5^2 [/math] are factors of [math] 50! [/math]. We need to know the largest [math] n [/math] such that [math] 5^n [/math] is a factor of [math] 50! [/math]. It should be clear that we can’t just start dividing [math] 50! [/math] by larger and larger powers of 5 until we get a non-zero remainder. Factorials get REALLY big REALLY fast! For example,
[math] 10! = 3,629,800 [/math] [math] 50! [/math] has more than 60 digits!
## Good Enough
This problem isn’t going down to a frontal assault, so we need to find a back door. Another way to think about the question would be “how many fives are in the prime factorization 50 factorial?” Listing the prime factors of the integers between 1 and 50 and checking for numbers with factors of 5 is impractical, but multiples of 5 are easy to pick out, and since there are only 10 multiples of 5 between 1 and 50 listing the multiples of 5 isn’t impractical at all.
Multiples of 5 Number of 5s in Prime Factorization Total number of 5s 5 1 1 10 1 2 15 1 3 20 1 4 25 2 6 30 1 7 35 1 8 40 1 9 45 1 10 50 2 12
You have to watch out for factors of 25, but other than that it’s relatively simple and efficient – maybe not 2-minute efficient, but time management is about management, not a set limit for each problem.
## 800
The ten multiples of 5 between 1 and 50 get us 10 factors of 5. However there are other fives hidden in multiples of 25. There are two multiples of 25 between 1 and 50 so we have to count two more fives. Because we are only going up to 50!, this isn’t much quicker than the first method. But if we had a larger factorial, we could start counting 125’s, then 625’s, and so on, allowing us to count much more efficiently than we could with a table or list.
Using this solution on this problem is a bit like using an RPG to kill a large mosquito However, this method is the way to go when confronted with some larger flying pest/predator… maybe a pterodactyl? Anyway, what if the test writers asked you to find the number of positive integer values of [math] n [/math] such that [math] 5^n [/math] is a factor of [math] 1000! [/math]? I’ll let you try this one yourself. Watch out for hidden fives! Hopefully, you’ll find 249 5’s.
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# Math
posted by .
You have 3 equilateral🔻🔻🔻. You must move one to make 5 triangles.
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# Chapter 1 Lesson 2 Complex Fractions And Unit Rates Quiz Pdf
By RomГЎn S.
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21.05.2021 at 20:41
File Name: chapter 1 lesson 2 complex fractions and unit rates quiz .zip
Size: 13564Kb
Published: 21.05.2021
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## Alternative Accelerated Common Core 6/7 Math (Math Academy) (Period 4) Assignments
There are a few things that you might find helpful: 1 You can differentiate for your students by deciding to print the blank template or answer key. Student book answer key. Find Unit Rates - Lesson 4. Integers are given in the problem, but most of the rates will require decimal quotients. In this unit, students will learn function notation and develop the concepts of domain and range. Possible answers 2. Solution 1.
If you're seeing this message, it means we're having trouble loading external resources on our website. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Donate Login Sign up Search for courses, skills, and videos. Skill Summary Legend Opens a modal. Intro to ratios. Intro to ratios Opens a modal.
## Unit: Ratios, rates, proportions
Dedicated to excellence in education. The proportional problems and work expected are found in the key. Also, you can read what would be context clues in the math problem that you should underline to help set the problem up. Please forgive me if there is an error. I checked over the work twice. Errors on keys help start a math conversation between people which is a great start in becoming an independent thinker! Error analysis :.
Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. Are you getting the free resources, updates, and special offers we send out every week in our teacher newsletter? Grade Level. Resource Type.
Но решил, что хочет от этого парня слишком многого. - Мне нужна кое-какая информация, - сказал. - Проваливал бы ты отсюда. - Я ищу одного человека.
Не в этом дело! - воскликнула Сьюзан, внезапно оживившись. Это как раз было ее специальностью. - Дело в том, что это и есть ключ. Энсей Танкадо дразнит нас, заставляя искать ключ в считанные минуты.
Хейл начал выворачивать шею Сьюзан. - Я-я…я убью. Клянусь, убью. - Ты не сделаешь ничего подобного! - оборвал его Стратмор. - Этим ты лишь усугубишь свое положе… - Он не договорил и произнес в трубку: - Безопасность.
Я уже говорил это и могу повторить тысячу раз - Пьер Клушар описывает мир таким, каким его видит. Некоторые ваши туристические путеводители старательно скрывают правду, обещая бесплатный ночлег в городе, но Монреаль тайме не продается. Ни за какие деньги. - Простите, сэр, вы, кажется, меня не… - Merde alors. Я отлично все понял! - Он уставил на Беккера костлявый указательный палец, и его голос загремел на всю палату.
Меган? - позвал он и постучал. Никто не ответил, и Беккер толкнул дверь. - Здесь есть кто-нибудь? - Он вошел. Похоже, никого. Пожав плечами, он подошел к раковине. Раковина была очень грязной, но вода оказалась холодной, и это было приятно.
Узкая лестница спускалась к платформе, за которой тоже виднелись ступеньки, и все это было окутано красным туманом. Грег Хейл, подойдя к стеклянной перегородке Третьего узла, смотрел, как Чатрукьян спускается по лестнице. С того места, где он стоял, казалось, что голова сотрудника лаборатории систем безопасности лишилась тела и осталась лежать на полу шифровалки. А потом медленно скрылась из виду в клубах пара. - Отчаянный парень, - пробормотал Хейл себе под нос. Он знал, что задумал Чатрукьян. Отключение ТРАНСТЕКСТА было логичным шагом в случае возникновения чрезвычайной ситуации, а ведь тот был уверен, что в машину проник вирус.
А вдруг Танкадо ошибся? - вмешался Фонтейн.
Теперь он молил Бога, чтобы священник не торопился, ведь как только служба закончится, он будет вынужден встать, хотя бы для того чтобы пропустить соседей по скамье. А в своем пиджаке он обречен. Беккер понимал, что в данный момент ничего не может предпринять.
Наконец она остановилась, и дверь открылась. Покашливая, Сьюзан неуверенно шагнула в темный коридор с цементными стенами. Она оказалась в тоннеле, очень узком, с низким потолком.
Сознание гнало ее вперед, но ноги не слушались. Коммандер. Мгновение спустя она, спотыкаясь, карабкалась вверх по ступенькам, совершенно забыв о таящейся внизу опасности. Она двигалась вслепую, скользя на гладких ступеньках, и скопившаяся влага капала на нее дождем.
Совершенно. Будет очень глупо, если вы этого не сделаете. На этот раз Стратмор позволил себе расхохотаться во весь голос.
Сюда. В этой встрече было что-то нереальное - нечто, заставившее снова напрячься все его нервные клетки. Он поймал себя на том, что непроизвольно пятится от незнакомцев.
Все лампы наверху погасли. Не было видно даже кнопочных электронных панелей на дверях кабинетов. Когда ее глаза привыкли к темноте, Сьюзан разглядела, что единственным источником слабого света в шифровалке был открытый люк, из которого исходило заметное красноватое сияние ламп, находившихся в подсобном помещении далеко внизу. Она начала двигаться в направлении люка.
ГЛАВА 76 У подъезда севильского аэропорта стояло такси с работающим на холостом ходу двигателем и включенным счетчиком.
- Если вы позвоните, она умрет. Стратмора это не поколебало. - Я готов рискнуть. - Чепуха. Вы жаждете обладать ею еще сильнее, чем Цифровой крепостью.
Era un punqui, - ответила Росио. Беккер изумился. - Un punqui. - Si. Punqui.
Выпей воды. Ты очень бледна. - Затем повернулся и вышел из комнаты.
Zachary J.
At this level students start getting much more familiar with equations and the use of expressions.
Charlie Z.
Students please bring in a pie apple pie or any of your choice, or any circular dessert like cookies, cake, cheesecake, etc
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1. ## fractions
can anyone help with this problem:
196+76 5/8+100 7/8+146 1/2+98 3/4 what does it =???
2. Please do not double post as you did in your other two threads. Just simply post your question and one of the members will come to it.
As for your question, is this it:
$\displaystyle 196+76 \frac{5}{8}+100 \frac{7}{8}+146\frac{1}{2}+98\frac{3}{4}$
If so, what I'd do is first convert each mixed fraction into a improper one (personal preference): $\displaystyle 196 + \frac{380}{8} + \frac{700}{8} + \frac{146}{2} + \frac{294}{4}$
As you should know, we need a common denominator in order for fraction addition to work. What we could do is put each term into a fraction with a denominator of 8:
$\displaystyle \frac{196}{1} {\color{blue} \cdot \frac{8}{8}} \: \: + \: \: \frac{380}{8} \: \: + \: \: \frac{700}{8} \: \: + \: \: \frac{146}{2} {\color{blue} \cdot \frac{4}{4}} \: \: + \: \: \frac{294}{4} {\color{blue} \cdot \frac{2}{2}}$
Can you finish off?
3. Hello, abb1327!
Simplify: .$\displaystyle 196 + 76\frac{5}{8} + 100\frac{7}{8} +146\frac{1}{2}+ 98\frac{3}{4}$
We have: .$\displaystyle \begin{array}{ccc} & 196 \\ & \;\;76 & \frac{5}{8} \\ \\[-4mm]& 100 & \frac{7}{8} \\ \\[-4mm] & 146 & \frac{1}{2} \\ \\[-4mm] + & \;\;98 & \frac{3}{4} \\ \\[-4mm] \hline \\[-4mm]& 616 & {\color{blue}?}\end{array}$
Then: .$\displaystyle \frac{5}{8} + \frac{7}{8} + \frac{1}{2} + \frac{3}{4} \;=\;\frac{5}{8} + \frac{7}{8} + \frac{4}{8} + \frac{6}{8} \;=\;\frac{22}{8} \;=\;2\frac{3}{4}$
Therefore: .$\displaystyle 616 + 2\frac{3}{4} \;=\;\boxed{618\frac{3}{4}}$
4. ## thanks!!!!
i was just stuck on the part where it said 22/8 thats all but thank u!!
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# Answer to Question #168832 in General Chemistry for Michael Parker
Question #168832
A compound contains 66.7% carbon, 11.1% hydrogen and 22.2% oxygen by mass. Its relative formula mass is 72. What is the empirical formula and the molecular formula of the compound?
1
2021-03-04T06:43:42-0500
1. Find the percentage of Oxygen by adding all of the given percentages then subtracting from 100
20.0+6.66+47.33=73.999=74 100−74=26
1. Change the percentages to grams. If there were 100 grams
2. C= 20 grams H = 6.66 grams N = 47.33 grams O = 26 grams.
3. Change the grams to moles by dividing by the molecular mass of the elements
Moles Carbon = 20/12=1.66moles C
Moles Hydrogen = 6.66/1=6.66moles H
Moles Nitrogen = 47.33/14=3.38molesN
Moles Oxygen = 26/16=1.625moles O
1. Find the simplest mole ratios.
2. Since Oxygen is the smallest divide all the other nu=mber of moles by the moles of Oxygen
Carbon ratio = C/O=1.66/1.625=1.02or1:1
Nitrogen ratio = N/O=3.38/1.625=2.04or 2:1
Hydrogen ratio = H/O=6.66/1.625=4.09or 4:1
So the compound has a ratio of 1 C: 2 N : 4 H : 1O for the empirical formula
The mass of one empirical formula is 62 grams per mole. This is slightly higher than the experimental molecular mass of 60 grams but is within experimental error.
So the compound most likely has a formula of C
N2H4O
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# Mathews' Euclid examination papers ... on Euc. i.-iv
### Τι λένε οι χρήστες -Σύνταξη κριτικής
Δεν εντοπίσαμε κριτικές στις συνήθεις τοποθεσίες.
### Περιεχόμενα
Ενότητα 1 115 Ενότητα 2 116 Ενότητα 3 117
Ενότητα 4 124 Ενότητα 5 186 Ενότητα 6
### Δημοφιλή αποσπάσματα
Σελίδα 160 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz.
Σελίδα 154 - If, from the ends of the side of a triangle, there be drawn two straight lines to a point within the triangle, these shall be less than, the other two sides of the triangle, but shall contain a greater angle. Let...
Σελίδα 184 - If a straight line touch a circle, and from the point of contact a chord be drawn, the angles which this chord makes with the tangent are equal to the angles in the alternate segments.
Σελίδα 156 - A tangent to a circle is perpendicular to the radius drawn to the point of contact.
Σελίδα 182 - IF a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles ; and the three interior angles of every triangle are equal to two right angles.
Σελίδα 153 - Any two sides of a triangle are together greater than the third side.
Σελίδα 167 - THE angles at the base of an isosceles triangle are equal to one another : and, if the equal sides be produced, the angles upon the other side of the base shall be equal.
Σελίδα 164 - AB be the given straight line ; it is required to divide it into two parts, so that the rectangle contained by the whole, and one of the parts, shall be equal to the square of the other part.
Σελίδα 130 - To describe an isosceles triangle, having each of the angles at the base double of the third angle.
Σελίδα 154 - Straight lines which are parallel to the same straight line are parallel to one another. Triangles and Rectilinear Figures. The sum of the angles of a triangle is equal to two right angles.
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## Saturday, August 8, 2009
### 1: 60 rule in air traffic
AIR TRAFFIC 1: 60 RULE
angular
It is based on the mathematical principle that a turn of 1 degree at 60 miles is equal to 1 mile displacement at the destination. (right angled triangle where the base is 60 miles. For every degree of angle = 1 mile at the opposite side)
For example: MH1286 is inbound on the ADNUT track heading 090 and is 60 miles away. If you give him a turn of 1 degree left ( heading 089 OR 091 ) then he will be 1 mile RIGHT OR LEFT of track at scHOOL. Likewise, if you give him a turn of 10 degrees left ( heading 080 ) he will be 10 miles left of track. You can also prove this by plotting.
speed
When we (at SATC) talk about speed control sometimes we tend to refer it as 1:60 rule as well. For example a 737 travelling at 300 kts. at 60 NM away. Using 1:60 rule we know that the aircraft will cover about 5NM/min. Therefore, we can estimate the ETA as plus 12 mins. Applying appropriate speed reduction we can fit in the Boeing into our traffic sequence.
#### 1 comment:
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Search a number
22032 = 243417
BaseRepresentation
bin101011000010000
31010020000
411120100
51201112
6250000
7121143
oct53020
933200
1022032
111560a
1210900
13a04a
14805a
1567dc
hex5610
22032 has 50 divisors (see below), whose sum is σ = 67518. Its totient is φ = 6912.
The previous prime is 22031. The next prime is 22037. The reversal of 22032 is 23022.
It can be written as a sum of positive squares in only one way, i.e., 20736 + 1296 = 144^2 + 36^2 .
It is a Harshad number since it is a multiple of its sum of digits (9).
It is a nialpdrome in base 9.
It is an inconsummate number, since it does not exist a number n which divided by its sum of digits gives 22032.
It is not an unprimeable number, because it can be changed into a prime (22031) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (5) of ones.
It is a polite number, since it can be written in 9 ways as a sum of consecutive naturals, for example, 1288 + ... + 1304.
222032 is an apocalyptic number.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 22032, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (33759).
22032 is an abundant number, since it is smaller than the sum of its proper divisors (45486).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
22032 is a wasteful number, since it uses less digits than its factorization.
22032 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 37 (or 22 counting only the distinct ones).
The product of its (nonzero) digits is 24, while the sum is 9.
The square root of 22032 is about 148.4318025222. The cubic root of 22032 is about 28.0339723701.
Adding to 22032 its reverse (23022), we get a palindrome (45054).
Subtracting 22032 from its reverse (23022), we obtain a triangular number (990 = T44).
It can be divided in two parts, 220 and 32, that added together give a palindrome (252).
The spelling of 22032 in words is "twenty-two thousand, thirty-two".
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# Question: How Many Numbers Are There Between 100 And 999?
## How many two digit numbers are there which are divisible by 3?
30∴ Two digit numbers divisible by 3 = 30..
## What are all the combinations for a 3 Number Lock?
A standard 3-digit lock has 999 possible combinations, which is why the 4- digit locks cost a bit of a step more money, as they have 9999 possibles.
## How many times does the digit 6 appear in the list of all integers from 1 to 100?
91 – 100 The digit 6 will occur once in each set; therefore the digit 6 will occur TEN times in the number set 1 to 100, including 100.
## How many odd numbers are there between 100 and 999?
320 odd numbersThere are 320 odd numbers between 100 & 999 that have all unique digits.
## How many two digit numbers are there which are divisible by 7?
13Two digit Of natural numbers is divisible by 7 is 13 no.
## How many numbers lie between 10 and 300 which divided by 4 leaves a remainder 3?
73 numbersAnswer. So there are 73 numbers which lies in between 10 and 300, when divided by 4, leaves a remainder 3.
## Is there a divisibility rule for 7?
To check if a number is evenly divisible by 7: Take the last digit of the number, double it Then subtract the result from the rest of the number If the resulting number is evenly divisible by 7, so is the original number. … The last digit is 3, double that to make 6, subtract from 6 from the remaining digits.
## How many 3 digits number are there?
The Significance of Zero in 3 -digit numbers The smallest 3 -digit number is 100, and the largest three digit number is 999. Any combination of digits can be used to form 3 -digit numbers, with or without repetition.
## How many numbers lie between 100 and 300 which are divisible by 3?
Answer. Here, between 100 and 300, the first number which divisible by 13 is 104 and the last number which is divisible by 13 is 299, and the common difference is 13. Therefore, there are 16 numbers between 100 and 300 which are divisible by 13.
## How many numbers lie between 99 square and 100 square?
198Total natural numbers lying between squares of 99 and 100 is 198.
## How many 2 digit numbers are there?
90The total number of two digit numbers is 90. From 1 to 99 there are 99 numbers, out of which there are 9 one-digit numbers, i.e., 1, 2, 3, 4, 5, 6, 7, 8 and 9.
## How many 2 digit numbers have digits whose sum is a perfect square?
Hence there are eight such two digits numbers such that the sum of the number and its reverse is a perfect square. They are 29, 38, 47, 56, 65, 74, 83, 92.
## How many numbers are there between 101 and 999?
89 numbersHence, there are 89 numbers between 101 and 999 which are divisible by both 2 and 5.
## How many numbers are there from 99 to 999?
so, the number of whole numbers between 99 and 999 with exactly one 0 is 80.. Please mark as brainliest!!
## How many numbers are divisible by 7?
128 threeThere are 128 three digit numbers divisible by 7. Originally Answered: How many three digit numbers are divisible by 7? There are 900 three digit numbers (999–99).
## What is the smallest number of three digit?
100The smallest 3-digit number is 100 and the largest 3-digit number is 999.
## What is the largest number you can write with three digits?
999999 is biggest three digit number, if you want a biggest number using three single digit numbers then it is 9 power 9 power 9 (9^(9^9)).
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# Different types of thumb rules for determining the total construction cost of the building
This construction video tutorial is very useful for site engineer. In this video, one can learn how to determine the construction cost of the entire building as well as the required quantities of materials on the basis of thumb rules.
### Given below the quantity and rates of materials per square feet :-
* Steel = 3 to 5 kg
* Cement = 0.5 bags
* RMC or ready mix concrete = 0.05 m3
* Block = 12.5 Nos/m2
* Electrical cost = Rs. 133
* Plumbing cost = Rs. 126
* Fire fighting cost = Rs. 40
* External development = Rs. 94.5
* Civil work structure = Rs. 751.25
* Finishing work = Rs. 467.50
Suppose, the length of the plot is given as 20 feet
The breadth of the plot is given as 30 feet
So, the size of the plot = Length x Breadth = 20 x 30 = 600 square feet
The requirement of steel (5 kg/sq.ft) = 5 x 600 = 3000 kg = 3 Tonne
It includes 8 mm, 10 mm, 12 mm and 16 mm dia steel. The rate of 8 mm is Rs. 45/kg, 10 mm is Rs. 44/kg, 12 mm is Rs. 48/kg etc.
So, the quantity of steel = rate x steel kg
The requirement of cement (0.5 bag/sq.ft) = 0.5 x 600 = 300 bags. The cost of 1 bag is Rs. 200 (approx).
The total cost of cement = 300 x 200 = Rs. 600
The requirement of RMC alias ready mix concrete (0.05 m3/sq.ft) = 0.05 x 600 = 30 m3 cement
The cost of electricity (Rs. 133/sq.ft) = 133 x 600 = Rs. 79,800
The cost of plumbing (Rs. 126/sq.ft) = 126 x 600 = Rs. 75,600
The cost of fire alarm (Rs. 40/sq.ft) = 40 x 600 = Rs. 24,000
The construction cost with labor (Rs. 751/sq.ft) = 751 x 600 = Rs. 4,50,600
The cost of finishing work (Rs. 467/sq.ft) = 467 x 600 = Rs. 2,80,200
In finishing work interior part, exterior part and paintings are included.
To gather knowledge on other types of thumb rules, go through the following video tutorial.
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# 2.4: Use a General Strategy to Solve Linear Equations
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##### Learning Objectives
By the end of this section, you will be able to:
• Solve equations using a general strategy
• Classify equations
##### Note
Before you get started, take this readiness quiz.
1. Simplify: $$−(a−4)$$.
If you missed this problem, review Exercise 1.10.46
2. Multiply: $$\frac{3}{2}(12x+20)$$
If you missed this problem, review Exercise 1.10.34.
3. Simplify: $$5−2(n+1)$$
If you missed this problem, review Exercise 1.10.49.
4. Multiply: $$3(7y+9)$$
If you missed this problem, review Exercise 1.10.34.
5. Multiply: $$(2.5)(6.4)$$
If you missed this problem, review Exercise 1.8.19.
## Solve Equations Using the General Strategy
Until now we have dealt with solving one specific form of a linear equation. It is time now to lay out one overall strategy that can be used to solve any linear equation. Some equations we solve will not require all these steps to solve, but many will.
Beginning by simplifying each side of the equation makes the remaining steps easier.
##### Exercise $$\PageIndex{1}$$: How to Solve Linear Equations Using the General Strategy
Solve: $$-6(x + 3) = 24$$.
##### Exercise $$\PageIndex{2}$$
Solve: $$5(x + 3)=35$$
$$x = 4$$
##### Exercise $$\PageIndex{3}$$
Solve: $$6(y - 4) = -18$$
$$y = 1$$
##### GENERAL STRATEGY FOR SOLVING LINEAR EQUATIONS.
1. Simplify each side of the equation as much as possible.
Use the Distributive Property to remove any parentheses.
Combine like terms.
2. Collect all the variable terms on one side of the equation.
Use the Addition or Subtraction Property of Equality.
3. Collect all the constant terms on the other side of the equation.
Use the Addition or Subtraction Property of Equality.
4. Make the coefficient of the variable term to equal to 1.
Use the Multiplication or Division Property of Equality.
State the solution to the equation.
5. Check the solution. Substitute the solution into the original equation to make sure the result is a true statement.
##### Exercise $$\PageIndex{4}$$
Solve: $$-(y + 9) = 8$$
Simplify each side of the equation as much as possible by distributing. The only y term is on the left side, so all variable terms are on the left side of the equation. Add 9 to both sides to get all constant terms on the right side of the equation. Simplify. Rewrite −y as −1y. Make the coefficient of the variable term to equal to 1 by dividing both sides by −1. Simplify. Check: Let y=−17.
##### Exercise $$\PageIndex{5}$$
Solve: $$-(y + 8) = -2$$
$$y = -6$$
##### Exercise $$\PageIndex{6}$$
Solve: $$-(z + 4) = -12$$
$$z = 8$$
##### Exercise $$\PageIndex{7}$$
Solve: $$5(a - 3) + 5 = -10$$
Simplify each side of the equation as much as possible. Distribute. Combine like terms. The only a term is on the left side, so all variable terms are on one side of the equation. Add 10 to both sides to get all constant terms on the other side of the equation. Simplify. Make the coefficient of the variable term to equal to 11 by dividing both sides by 55. Simplify. Check: Let a=0.
##### Exercise $$\PageIndex{8}$$
Solve: $$2(m - 4) + 3 = -1$$
$$m = 2$$
##### Exercise $$\PageIndex{9}$$
Solve:$$7(n - 3) - 8 = -15$$
$$n = 2$$
##### Exercise $$\PageIndex{10}$$
Solve: $$\frac{2}{3}(6m - 3) = 8 - m$$
Distribute. Add m to get the variables only to the left. Simplify. Add 2 to get constants only on the right. Simplify. Divide by 5. Simplify. Check: Let m=2.
##### Exercise $$\PageIndex{11}$$
Solve: $$\frac{1}{3}(6u + 3) = 7 - u$$
$$u = 2$$
##### Exercise $$\PageIndex{12}$$
Solve: $$\frac{2}{3}(9x - 12) = 8 + 2x$$
$$x = 4$$
##### Exercise $$\PageIndex{13}$$
Solve: $$8 - 2(3y + 5) = 0$$
Simplify—use the Distributive Property. Combine like terms. Add 2 to both sides to collect constants on the right. Simplify. Divide both sides by −6−6. Simplify. Check: Let y=−13.
##### Exercise $$\PageIndex{14}$$
Solve: $$12 - 3(4j + 3) = -17$$
$$j = \frac{5}{3}$$
##### Exercise $$\PageIndex{15}$$
Solve: $$-6 - 8(k - 2) = -10$$
$$k = \frac{5}{2}$$
##### Exercise $$\PageIndex{16}$$
Solve: $$4(x - 1)-2=5(2x+3)+6$$
Distribute. Combine like terms. Subtract 4x to get the variables only on the right side since $$10>4$$. Simplify. Subtract 21 to get the constants on left. Simplify. Divide by 6. Simplify. Check: Let $$x=-\frac{9}{2}$$.
##### Exercise $$\PageIndex{17}$$
Solve: $$6(p-3)-7=5(4p+3)-12$$
$$p = -2$$
##### Exercise $$\PageIndex{18}$$
Solve: $$8(q +1)-5=3(2q-4)-1$$
$$q = -8$$
##### Exercise $$\PageIndex{19}$$
Solve: $$10[3 - 8(2s-5)] = 15(40 - 5s)$$
Simplify from the innermost parentheses first. Combine like terms in the brackets. Distribute. Add 160s to get the s’s to the right. Simplify. Subtract 600 to get the constants to the left. Simplify. Divide. Simplify. Check: Substitute s=−2.
##### Exercise $$\PageIndex{20}$$
Solve: $$6[4−2(7y−1)]=8(13−8y)$$.
$$y = -\frac{17}{5}$$
##### Exercise $$\PageIndex{21}$$
Solve: $$12[1−5(4z−1)]=3(24+11z)$$.
$$z = 0$$
##### Exercise $$\PageIndex{22}$$
Solve: $$0.36(100n+5)=0.6(30n+15)$$.
Distribute. Subtract 18n to get the variables to the left. Simplify. Subtract 1.8 to get the constants to the right. Simplify. Divide. Simplify. Check: Let n=0.4.
##### Exercise $$\PageIndex{23}$$
Solve: $$0.55(100n+8)=0.6(85n+14)$$.
$$n = 1$$
##### Exercise $$\PageIndex{24}$$
Solve: $$0.15(40m−120)=0.5(60m+12)$$.
$$m = -1$$
## Classify Equations
Consider the equation we solved at the start of the last section, 7x+8=−13. The solution we found was x=−3. This means the equation 7x+8=−13 is true when we replace the variable, x, with the value −3. We showed this when we checked the solution x=−3 and evaluated 7x+8=−13 for x=−3.
If we evaluate 7x+8 for a different value of x, the left side will not be −13.
The equation 7x+8=−13 is true when we replace the variable, x, with the value −3, but not true when we replace x with any other value. Whether or not the equation 7x+8=−13 is true depends on the value of the variable. Equations like this are called conditional equations.
All the equations we have solved so far are conditional equations.
##### CONDITIONAL EQUATION
An equation that is true for one or more values of the variable and false for all other values of the variable is a conditional equation.
Now let’s consider the equation 2y+6=2(y+3). Do you recognize that the left side and the right side are equivalent? Let’s see what happens when we solve for y.
Distribute. Subtract 2y to get the y’s to one side. Simplify—the y’s are gone!
But 6=6 is true.
This means that the equation 2y+6=2(y+3) is true for any value of y. We say the solution to the equation is all of the real numbers. An equation that is true for any value of the variable like this is called an identity.
##### IDENTITY
An equation that is true for any value of the variable is called an identity.
The solution of an identity is all real numbers.
What happens when we solve the equation 5z=5z−1?
Subtract 5z to get the constant alone on the right. Simplify—the z’s are gone!
But $$0\neq −1$$.
Solving the equation 5z=5z−1 led to the false statement 0=−1. The equation 5z=5z−1 will not be true for any value of z. It has no solution. An equation that has no solution, or that is false for all values of the variable, is called a contradiction.
An equation that is false for all values of the variable is called a contradiction.
##### Exercise $$\PageIndex{25}$$
Classify the equation as a conditional equation, an identity, or a contradiction. Then state the solution.
$$6(2n−1)+3=2n−8+5(2n+1)$$
Distribute. Combine like terms. Subtract 12n to get the nn’s to one side. Simplify. This is a true statement. The equation is an identity. The solution is all real numbers.
##### Exercise $$\PageIndex{26}$$
Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution:
$$4+9(3x−7)=−42x−13+23(3x−2)$$
identity; all real numbers
##### Exercise $$\PageIndex{27}$$
Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution:
$$8(1−3x)+15(2x+7)=2(x+50)+4(x+3)+1$$
identity; all real numbers
##### Exercise $$\PageIndex{28}$$
Classify as a conditional equation, an identity, or a contradiction. Then state the solution.
$$10+4(p−5)=0$$
Distribute. Combine like terms. Add 10 to both sides. Simplify. Divide. Simplify. The equation is true when $$p = frac{5}{2}$$. This is a conditional equation. The solution is $$p = frac{5}{2}$$.
##### Exercise $$\PageIndex{29}$$
Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: $$11(q+3)−5=19$$
conditional equation; $$q = \frac{9}{11}\ ##### Exercise \(\PageIndex{30}$$
Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: $$6+14(k−8)=95$$
conditional equation; $$k = \frac{193}{14}$$
##### Exercise $$\PageIndex{31}$$
Classify the equation as a conditional equation, an identity, or a contradiction. Then state the solution.
$$5m+3(9+3m)=2(7m−11)$$
Distribute. Combine like terms. Subtract 14m from both sides. Simplify. But $$27\neq −22$$. The equation is a contradiction. It has no solution.
##### Exercise $$\PageIndex{32}$$
Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution:
$$12c+5(5+3c)=3(9c−4)$$
##### Exercise $$\PageIndex{33}$$
Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution:
$$4(7d+18)=13(3d−2)−11d$$
Type of equation What happens when you solve it? Solution
Conditional Equation True for one or more values of the variables and false for all other values One or more values
Identity True for any value of the variable All real numbers
Contradiction False for all values of the variable No solution
## Key Concepts
• General Strategy for Solving Linear Equations
1. Simplify each side of the equation as much as possible.
Use the Distributive Property to remove any parentheses.
Combine like terms.
2. Collect all the variable terms on one side of the equation.
Use the Addition or Subtraction Property of Equality.
3. Collect all the constant terms on the other side of the equation.
Use the Addition or Subtraction Property of Equality.
4. Make the coefficient of the variable term to equal to 1.
Use the Multiplication or Division Property of Equality.
State the solution to the equation.
5. Check the solution.
Substitute the solution into the original equation.
This page titled 2.4: Use a General Strategy to Solve Linear Equations is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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# Moving average
For other uses, see Moving average (disambiguation).
An example of two moving average curves
In statistics, a moving average (rolling average or running average) is a calculation to analyze data points by creating series of averages of different subsets of the full data set. It is also called a moving mean (MM)[1] or rolling mean and is a type of finite impulse response filter. Variations include: simple, and cumulative, or weighted forms (described below).
Given a series of numbers and a fixed subset size, the first element of the moving average is obtained by taking the average of the initial fixed subset of the number series. Then the subset is modified by "shifting forward"; that is, excluding the first number of the series and including the next value in the subset.
A moving average is commonly used with time series data to smooth out short-term fluctuations and highlight longer-term trends or cycles. The threshold between short-term and long-term depends on the application, and the parameters of the moving average will be set accordingly. For example, it is often used in technical analysis of financial data, like stock prices, returns or trading volumes. It is also used in economics to examine gross domestic product, employment or other macroeconomic time series. Mathematically, a moving average is a type of convolution and so it can be viewed as an example of a low-pass filter used in signal processing. When used with non-time series data, a moving average filters higher frequency components without any specific connection to time, although typically some kind of ordering is implied. Viewed simplistically it can be regarded as smoothing the data.
## Generic Approach to Moving Average
An element ${\displaystyle v\in V}$ moves in a additive Group (mathematics) or Vector Space V. In a generic approach, we have a moving probability distribution ${\displaystyle P_{v}}$ that defines how the values in the environment of ${\displaystyle v\in V}$ have an impact on the moving average.
### Discrete/continuous Moving Average
According to probability distributions we have to distinguish between a
• discrete (probability mass function ${\displaystyle p_{v}}$) and
• continuous (probability density function ${\displaystyle p_{v}}$)
moving average. The terminology refers to probability distributions and the semantics of probability mass/density function describes the distrubtion of weights around the value ${\displaystyle v\in V}$. In the discrete setting the ${\displaystyle p_{v}(x)=0.2}$ means that ${\displaystyle x}$ has a 20% impact on the moving average ${\displaystyle MA(v)}$ for ${\displaystyle v}$.
### Moving/Shift Distributions
If the probability distribution are shifted by ${\displaystyle v}$ in ${\displaystyle V}$. This means that the probability mass functions ${\displaystyle p_{v}}$ resp. probability density functions ${\displaystyle p_{v}}$ are generated by a probability distribution ${\displaystyle p_{0}}$ at the zero element of the additive group resp. zero vector of the vector space. Due to nature of the collected data f(x) exists for a subset ${\displaystyle T\subseteq V}$. In many cases T are the points in time for which data is collected. The probability and the shift of a distribution is defined by the following property:
• discrete: For all ${\displaystyle x\in V}$ the probability mass function fulfills ${\displaystyle p_{v}(x):=p_{0}(x-v)}$ for ${\displaystyle v\in I}$
• continuous: For all probability density function fulfills ${\displaystyle p_{v}(x):=p_{0}(x-v)}$
The moving average is defined by:
• discrete: (probability mass function ${\displaystyle p_{v}}$)
${\displaystyle MA(v):=\sum _{x\in T}p_{v}(x)\cdot f(x)}$
Remark: ${\displaystyle p_{v}(x)>0}$ for a countable subset of ${\displaystyle V}$
• continuous probability density function ${\displaystyle p_{v}}$
${\displaystyle MA(v):=\int _{T}p_{v}(x)\cdot f(x)\,dx}$
It is important for the definition of probability mass functions resp. probability density functions ${\displaystyle p_{v}}$ that the support (measure theory) of ${\displaystyle p_{v}}$ is a subset of T. This assures that 100% of the probability mass is assigned to collected data. The support ${\displaystyle p_{v}}$ is defined as:
${\displaystyle \mathrm {supp} (p_{v}):={\overline {\{x\in V\mid p_{v}(x)>0\}}}\subset T.}$
## Simple moving average - discrete
In financial applications a simple moving average (SMA) is the unweighted mean of the previous n data. However, in science and engineering the mean is normally taken from an equal number of data on either side of a central value. This ensures that variations in the mean are aligned with the variations in the data rather than being shifted in time. An example of a simple equally weighted running mean for a n-day sample of closing price is the mean of the previous n days' closing prices.
${\displaystyle p_{0}(0)=p_{0}(-1)=\ldots =p_{0}(-(n-1))={\frac {1}{n}}}$
and ${\displaystyle p_{0}(x)=0}$ for ${\displaystyle x\notin \{-n+1,\dots ,-1,0\}}$ with ${\displaystyle V=\mathbb {Z} }$ as additive group.
Let ${\displaystyle C(t)}$ be the cost/price of product at time ${\displaystyle t\in T}$. If those prices are ${\displaystyle C(0),C(1),\dots ,C(97),C(98),C(99),C(100),C(101),\dots }$ and we want to create a simple moving average at day ${\displaystyle t=100}$ and looking back for time span of ${\displaystyle n=5}$ days then the formula is
{\displaystyle {\begin{aligned}SMA(100)&={\frac {1}{5}}\cdot C(100)+{\frac {1}{5}}\cdot C(99)+{\frac {1}{5}}\cdot C(98)+{\frac {1}{5}}\cdot C(97)+{\frac {1}{5}}\cdot C(96)\\&={\frac {1}{5}}\sum _{i=0}^{4}C(100-i)\\&=\sum _{i=0}^{n-1}p_{100}(100-i)\cdot C(100-i)\end{aligned}}}
When calculating successive values for other days/time ${\displaystyle t\in V=\mathbb {Z} }$, a new value comes into the sum and an old value drops out, meaning a full summation each time is unnecessary for this simple case,
{\displaystyle {\begin{aligned}SMA(101)&={\frac {1}{5}}\cdot C(101)+{\frac {1}{5}}\cdot C(100)+{\frac {1}{5}}\cdot C(99)+{\frac {1}{5}}\cdot C(98)+{\frac {1}{5}}\cdot C(97)\end{aligned}}}
${\displaystyle {\textit {SMA}}(t)={\frac {1}{n}}\sum _{i=0}^{n-1}C(t-i)=\sum _{i=0}^{n-1}p_{t}(t-i)\cdot C(t-i)=\sum _{i=0}^{n-1}p_{0}(-i)\cdot C(t-i)}$
The period selected depends on the type of movement of interest, such as short, intermediate, or long-term. In financial terms moving-average levels can be interpreted as support in a falling market, or resistance in a rising market. If you draw a graph for ${\displaystyle {\textit {SMA}}(t)}$ and cost function ${\displaystyle C(t)}$, you will identify, that the graph of ${\displaystyle {\textit {SMA}}}$ runs smoother in the time ${\displaystyle t\in V}$
If the data used are not centered around the mean, a simple moving average lags behind the latest datum point by half the sample width. An SMA can also be disproportionately influenced by old datum points dropping out or new data coming in. One characteristic of the SMA is that if the data have a periodic fluctuation, then applying an SMA of that period will eliminate that variation (the average always containing one complete cycle). But a perfectly regular cycle is rarely encountered.[2]
For a number of applications, it is advantageous to avoid the shifting induced by using only 'past' data. Hence a central moving average can be computed, using data equally spaced on either side of the point in the series where the mean is calculated.[3] This requires using an odd number of datum points in the sample window.
${\displaystyle p_{0}(-n)=p_{0}(-n+1)=\ldots =p_{0}(-1)=p_{0}(0)=p_{0}(1)=\dots =p_{0}(n-1)=p_{0}(n)={\frac {1}{2n+1}}}$
and ${\displaystyle p_{0}(x)=0}$ for ${\displaystyle x\notin \{-n,\dots ,-1,0,1,\dots ,n\}}$ with ${\displaystyle V=\mathbb {Z} }$ as additive group.
${\displaystyle {\textit {CMA}}(t)={\frac {1}{2n+1}}\sum _{i=-n}^{n}C(t+i)=\sum _{i=-n}^{n}p_{t}(t+i)\cdot C(t+i)=\sum _{i=-n}^{n}p_{0}(i)\cdot C(t+i)}$
A major drawback of the SMA is that it lets through a significant amount of the signal shorter than the window length. Worse, it actually inverts it. This can lead to unexpected artifacts, such as peaks in the smoothed result appearing where there were troughs in the data. It also leads to the result being less smooth than expected since some of the higher frequencies are not properly removed.
## Simple moving average - continuous
If we consider a continous measurement of value e.g. a force ${\displaystyle f(t)}$ at time ${\displaystyle t}$. The objective is to smooth the values ${\displaystyle f(t)}$ with a continous simple moving average. We look a time span ${\displaystyle s>0}$ in the past. As probability distribution we use a uniform distribution (mathematics) for the intervall ${\displaystyle [-n,0]}$. The density function is:
${\displaystyle p_{0}(x)={\begin{cases}{\frac {1}{n}}&\mathrm {for} \ -n\leq x\leq 0,\\[8pt]0&\mathrm {for} \ x0\end{cases}}}$ and ${\displaystyle p_{t}(x):=p_{0}(x-t)}$
Application on the moving average definition for continuous probability distriubtions we get:
${\displaystyle SMA(t):=\int _{\mathbb {R} }p_{t}(x)\cdot f(x)\,dx=\int _{t-n}^{t}p_{0}(x-t)\cdot f(x)\,dx={\frac {1}{n}}\int _{t-n}^{t}f(x)\,dx}$
## Cumulative moving average
### Cumulative moving average - discrete
In a cumulative moving average, the data arrive in an ordered datum stream with ${\displaystyle t\in \mathbb {N} _{0}=\{0,1,2,3,\dots \}}$, and the user would like to get the average of all of the data up until the current datum point ${\displaystyle t}$. For example, an investor may want the average price of all of the stock transactions for a particular stock up until the current time ${\displaystyle t}$. The starting point of data collection is ${\displaystyle t=0}$. As each new transaction occurs, the average price at the time of the transaction can be calculated for all of the transactions up to that point using the cumulative average, typically an equally weighted average of the sequence of t+1 values ${\displaystyle x_{0},x_{1}\ldots ,x_{t}}$ up to the current time ${\displaystyle t}$:
${\displaystyle {\textit {CMA}}(t)={\frac {x_{0}+x_{1}+\cdots +x_{t}}{t+1}}\,.}$
The brute-force method to calculate this would be to store all of the data and calculate the sum and divide by the number of datum points every time a new datum point arrived. However, it is possible to simply update cumulative average as a new value, ${\displaystyle x_{t}}$ becomes available, using the formula:
${\displaystyle {\textit {CMA}}(t)={\frac {x_{t}+t\cdot {\textit {CMA}}(t-1)}{t+1}}}$
Thus the current cumulative average ${\displaystyle {\textit {CMA}}(t)}$ for a new datum point ${\displaystyle x_{t}}$ is equal to the previous cumulative average ${\displaystyle {\textit {CMA}}(t-1)}$ at time t-1, times t, plus the latest datum point, all divided by the number of points received so far, n+1. When all of the datum points arrive (n = N), then the cumulative average will equal the final average. It is also possible to store a running total of the datum point as well as the number of points and dividing the total by the number of datum points to get the CMA each time a new datum point arrives.
The derivation of the cumulative average formula is straightforward. Using
${\displaystyle x_{0}+x_{1}+\cdots +x_{t}=(t+1)\cdot {\textit {CMA}}(t)}$
and similarly for t + 1, it is seen that
{\displaystyle {\begin{aligned}x_{t}&=(x_{0}+x_{1}+\cdots +x_{t})-(x_{0}+x_{1}+\cdots +x_{t-1})\\[6pt]&=(t+1)\cdot {\textit {CMA}}(t)-t\cdot {\textit {CMA}}(t-1)\end{aligned}}}
Solving this equation for ${\displaystyle {\textit {CMA}}(t)}$ results in:
{\displaystyle {\begin{aligned}{\textit {CMA}}(t)&={\frac {x_{t}+t\cdot {\textit {CMA}}(t-1)}{t+1}}\end{aligned}}}
### Cumulative moving average - continuous
If we consider a continuous measurement of values e.g. a force ${\displaystyle f(t)}$ at time ${\displaystyle t}$. The objective is to smooth the values ${\displaystyle f(t)}$ with a continous aggregated moving average. We look a time span ${\displaystyle t>0}$ in the past. As probability distribution we use a uniform distribution (mathematics) for the intervall ${\displaystyle [0,t]}$. The density function is:
${\displaystyle p_{t}(x)={\begin{cases}{\frac {1}{t}}&\mathrm {for} \ 0\leq x\leq t,\\[8pt]0&\mathrm {for} \ x<0\ \mathrm {or} \ x>t\end{cases}}}$.
Application on the cumulative moving average definition for continuous probability distriubtions we get:
${\displaystyle SMA(t):=\int _{\mathbb {R} }p_{t}(x)\cdot f(x)\,dx=\int _{0}^{t}p_{t}(x)\cdot f(x)\,dx={\frac {1}{t}}\int _{0}^{t}f(x)\,dx}$
## Moving average applied on images
Pixelization was used to anonymize this photograph
A weighted average is an average that has multiplying factors to give different weights to data at different positions in the sample window. Mathematically, the moving average is the convolution of the datum points with a fixed weighting function. One application is creating a pixelisation from a digital graphical image. For all the image on the right pixelisation is applied for several squares. All pixels in the square are replaced by the color average of all pixels in the square. Because colors are defined by three integer numbers so that color average must be rounded for that application. In order to understand color encoding with integer numbers see HTML Color Picker with the RGB color encoding. Three value between 0 and 255 (e.g. rgb(255, 153, 102) for light orange) encode a color. Due to the fact that the HTML colors for Red, Grenn, Blue (RGB) are integer numbers, the real values of the moving average are rounded as a technical constraint.
The image I with m pixels height and n pixels width is s matrix ${\displaystyle I\in Mat(m\times n,RGB)}$ where all components of the matrix are RGB triples of integer values between 0 and 255, i.e. ${\displaystyle RGB:=\{0,1,\ldots ,255\}^{3}}$.
A single pixel at row r and column c is denoted as ${\displaystyle I_{(r,c)}}$. If we define ${\displaystyle I_{(r,c)}:=(255,153,102)}$ then the
• intensity of red is ${\displaystyle I_{(r,c)}.R=255}$,
• intensity of green is ${\displaystyle I_{(r,c)}.G=153}$,
• intensity of red is ${\displaystyle I_{(r,c)}.B=102}$
If we calculate and average of colors, we calculate the average of red, green and blue separately. As an example we calculate an average of ${\displaystyle 2\times 2}$ sub matrix of the image I for the four pixels:
• ${\displaystyle I_{(r,c)}:=(250,103,21)}$ ${\displaystyle I_{(r,c+1)}:=(230,153,102)}$
• ${\displaystyle I_{(r+1,c)}:=(255,50,12)}$ ${\displaystyle I_{(r+1,c+1)}:=(151,30,20)}$
The calculated moving average for this square is:
• Red: ${\displaystyle A_{red}:=round\left({\frac {250+230+255+151}{4}}\right)=222}$
• Green: ${\displaystyle A_{green}:=round\left({\frac {103+153+50+30}{4}}\right)=84}$
• Blue: ${\displaystyle A_{blue}:=round\left({\frac {21+102+12+20}{4}}\right)=39}$
The calculate moving average for the ${\displaystyle 2\times 2}$ sub matrix of the image I will replace all original colors of the square. Let ${\displaystyle IMA\in Mat(m\times n,RGB)}$ the image with the moving average applied for all ${\displaystyle 2\times 2}$ sub matrices, then the selected sub matrix above in IMA will look like this:
• ${\displaystyle IMA_{(r,c)}:=(222,84,39)}$ ${\displaystyle IMA_{(r,c+1)}:=(222,84,39)}$
• ${\displaystyle IMA_{(r+1,c)}:=(222,84,39)}$ ${\displaystyle IMA_{(r+1,c+1)}:=(222,84,39)}$
The last step assigns the calculated average color rgb(222, 84, 39) to all pixels of the 2x2-square submatrix.
Looking at the example image on right, the application of the moving average are visible, because they are applied on a large submatrix of the image.
For the image processing ${\displaystyle V:=\mathbb {Z} \times \mathbb {Z} }$ with the neutral element ${\displaystyle 0_{V}:=(0,0)}$ as a the additive group with addition:
${\displaystyle (v_{1},v_{2})+(w_{1},w_{2}):=(v_{1}+w_{1},v_{2}+w_{2})}$ and ${\displaystyle T:=\{1,\ldots ,m\}\times \{1,\ldots ,n\}\subset V}$
T is the set of all row and column indices of the pixels. The images is decomposed the squares or even rectangles ${\displaystyle R_{i}}$. The moving average is calculated for all pixels in the rectangle ${\displaystyle R_{i}}$ similar to ${\displaystyle 2\times 2}$ mentioned above. The calculated moving average from the original image I is assigned to all pixels of the square/rectangle ${\displaystyle R_{i}}$ in IMA. If the width and height of the rectangles ${\displaystyle R_{i}}$ have in general a default size. Close the borders of the images, the sizes of these rectangles have to be adapted to the remaining pixels at the right and bottom border of the image I.
## Weighted moving average
In technical analysis of financial data, a weighted moving average (WMA) has the specific meaning of weights that decrease in arithmetical progression.[4] In an n-day WMA the latest day has weight n, the second latest n − 1, etc., down to one. These weights create a discrete probability distribution with:
${\displaystyle s(n):=n+(n-1)+\dots +n={\frac {n\cdot (n+1)}{2}}}$ and ${\displaystyle p_{t}(x)={\begin{cases}{\frac {n-(t-x)}{s(n)}}&\mathrm {for} \ 0\leq t-n\leq x\leq t,\\[8pt]0&\mathrm {for} \ xt\end{cases}}}$
The weighted moving average can be calculated for ${\displaystyle t\geq n}$ with the discrete probability mass function ${\displaystyle p_{t}}$ at time ${\displaystyle t\in \mathbb {N} _{0}:=\{0,1,2\dots ,\}}$, where ${\displaystyle t=0}$ is the initial day, when data collection of the financial data begins and ${\displaystyle C(0)}$ the price/cost of a product at day ${\displaystyle t=0}$. ${\displaystyle C(x)}$ the price/cost of a product at day ${\displaystyle x\in \mathbb {N} _{0}}$ for an arbitrary day x.
${\displaystyle {\text{WMA}}(t):=\sum _{x\in T=\mathbb {N} _{0}}p_{t}(x)\cdot C(x)=\sum _{x=t-n+1}^{t}p_{t}(x)\cdot C(x)={\frac {n\cdot C(t)+(n-1)\cdot C(t-1)+\cdots +2\cdot C(t-n+2)+1\cdot C(t-n+1)}{n+(n-1)+\cdots +2+1}}}$
WMA weights n = 15
The denominator is a triangle number equal to ${\displaystyle {\frac {n(n+1)}{2}}}$ which creates a discrete probability distribution by:
${\displaystyle {\frac {1}{s(n)}}+{\frac {2}{s(n)}}+\ldots +{\frac {n}{s(n)}}={\frac {1+2+\ldots +n}{s(n)}}={\frac {s(n)}{s(n)}}=1}$
The graph at the right shows how the weights decrease, from highest weight at day t for the most recent datum points, down to zero at day t-n.
In the more general case with weights ${\displaystyle w_{0},\ldots ,w_{n}}$ the denominator will always be the sum of the individual weights, i.e.:
${\displaystyle s(n):=\sum _{k=0}^{n}w_{k}}$ and ${\displaystyle w_{0}}$ as weight for for the most recent datum points at day t and ${\displaystyle w_{n}}$ as weight for the day ${\displaystyle t-n}$, which is n-th day before the most recent day ${\displaystyle t}$.
The discrete probability distribution ${\displaystyle p_{t}}$ is defined by:
${\displaystyle p_{t}(x)={\begin{cases}{\frac {w_{t-x}}{s(n)}}&\mathrm {for} \ 0\leq t-n\leq x\leq t,\\[8pt]0&\mathrm {for} \ 0\leq xt\end{cases}}}$
The weighted moving average with arbitrary weights is calculated by:
${\displaystyle {\text{WMA}}(t):=\sum _{x\in T=\mathbb {N} _{0}}p_{t}(x)\cdot C(x)=\sum _{x=t-n}^{t}p_{t}(x)\cdot C(x)={\frac {w_{0}\cdot C(t)+w_{1}\cdot C(t-1)+\cdots +w_{n-1}\cdot C(t-n+1)+w_{n}\cdot C(t-n)}{w_{0}+\cdots +w_{n-1}+w_{n}}}}$
This general approach can be compared to the weights in the exponential moving average in the following section.
## Exponential moving average
Further information: EWMA chart and Exponential smoothing
EMA weights N = 200
An exponential moving average (EMA), also known as an exponentially weighted moving average (EWMA),[5] is a type of infinite impulse response filter that applies weighting factors which decrease exponentially. The weighting for each older datum decreases exponentially, never reaching zero. The graph at right shows an example of the weight decrease.
The EMA for a series ${\displaystyle C:=(C(0),C(1),\ldots )}$ of collected data with a set of dates ${\displaystyle T:=\mathbb {N} _{0}}$, where ${\displaystyle C(t)}$ is the collected data at time index ${\displaystyle t\in T:=\mathbb {N} _{0}}$.
• First of all we have to define a value ${\displaystyle \alpha }$ with ${\displaystyle 0<\alpha <1}$, that represents the degree of weighting decrease as a constant smoothing factor between 0 and 1. A lower α discounts older observations faster.
• The weights are defined by
${\displaystyle w_{t}=(1-\alpha )\cdot \alpha ^{t}}$ for all ${\displaystyle t\in T:=\mathbb {N} _{0}}$ with ${\displaystyle \sum _{t=0}^{\infty }w_{t}=1}$ (geometric series).
• The sum of weights from 0 to the time index ${\displaystyle t\in T}$ is defined by:
${\displaystyle s(t)=\sum _{k=0}^{t}w_{k}=\sum _{k=0}^{t}(1-\alpha )\cdot \alpha ^{k}=(1-\alpha )\cdot \underbrace {\sum _{k=0}^{t}\alpha ^{k}} _{={\frac {1-\alpha ^{t+1}}{1-\alpha }}}=(1-\alpha )\cdot {\frac {1-\alpha ^{t+1}}{1-\alpha }}=1-\alpha ^{t+1}}$
• The discrete probability mass function is defined by:
${\displaystyle p_{t}(x)={\begin{cases}{\frac {w_{t-x}}{s(t)}}={\frac {(1-\alpha )\cdot \alpha ^{t-x}}{1-\alpha ^{t+1}}}&\mathrm {for} \ 0\leq x\leq t,\\[8pt]0&\mathrm {for} \ x<0\ \mathrm {or} \ x>t\end{cases}}}$
The definition above creates the exponential moving average EMA with discrete probability mass function ${\displaystyle p_{t}}$ by
${\displaystyle EMA(t):=\sum _{k\in T}p_{t}(k)\cdot C(k)=\sum _{k=0}^{t}p_{t}(k)\cdot C(k)=\sum _{x=0}^{t}{\frac {w_{t-k}}{s(t)}}\cdot C(k)=\sum _{k=0}^{t}{\frac {(1-\alpha )\cdot \alpha ^{t-k}}{1-\alpha ^{t+1}}}\cdot C(k)}$
The ${\displaystyle EMA}$ at time index ${\displaystyle t\in T:=\mathbb {N} _{0}}$ may be calculated recursively:
${\displaystyle EMA(0):=C(0)}$ and
${\displaystyle EMA(t+1):={\frac {1-\alpha }{1-\alpha ^{t+2}}}\cdot C(t+1)+\alpha \cdot {\frac {1-\alpha ^{t+1}}{1-\alpha ^{t+2}}}\cdot EMA(t)}$ for all ${\displaystyle t\in T=\mathbb {N} _{0}=\{0,1,2,3,...\}}$
Where:
• The coefficient ${\displaystyle \alpha }$ represents the degree of weighting decrease from ${\displaystyle EMA(t)}$ to ${\displaystyle EMA(t+1)}$. This implements the aging of data from ${\displaystyle t}$ to time index ${\displaystyle t+1}$.
• the fraction ${\displaystyle {\frac {1-\alpha ^{t+1}}{1-\alpha ^{t+2}}}}$ adjusts the denominator ${\displaystyle EMA(t)}$ for ${\displaystyle EMA(t+1)}$.
• the coefficient ${\displaystyle {\frac {1-\alpha }{1-\alpha ^{t+2}}}=p_{t+1}(t+1)={\frac {w_{(t+1)-(t+1)}}{s(t+1)}}}$ in the EMA at time index t+1.
### Initialization of EMA and Elimination of MA Impact form old data
${\displaystyle EMA(0)}$ may be initialized in a number of different ways, most commonly by setting ${\displaystyle EMA(0)}$ to the first collected data ${\displaystyle C(0)}$ at time index 0 as shown above, though other techniques exist, such as starting the calculation of the moving average after the first 4 or 5 observations. Furthermore only a most recent subset of collected data before the time index ${\displaystyle t}$ from the total history of collected date might be used for the ${\displaystyle EMA(t)}$. The discrete probability mass function puts weights on the most recent ${\displaystyle m+1}$ values of the collected data by:
${\displaystyle p_{t}(x)={\begin{cases}{\frac {w_{t-x}}{s(t)}}={\frac {(1-\alpha )\cdot \alpha ^{t-x}}{1-\alpha ^{t+1}}}&\mathrm {for} \ 0\leq x\leq t\ \mathrm {and} \ t\leq m\\[8pt]{\frac {w_{t-x}}{s(t)-s(t-m)}}={\frac {(1-\alpha )\cdot \alpha ^{t-x}}{1-\alpha ^{t+1}-(1-\alpha ^{t-m+1})}}={\frac {(1-\alpha )\cdot \alpha ^{t-x}}{\alpha ^{t-m+1}-\alpha ^{t+1}}}&\mathrm {for} \ (t-m)\leq x\leq t\ \mathrm {and} \ t>m,\\[8pt]0&\mathrm {for} \ x<0\ \mathrm {or} \ x>t\end{cases}}}$
The limitations to the most recent ${\displaystyle m+1}$ values of the collected data eliminates the impact of very old data on the resultant moving average completely. By choosing a small ${\displaystyle \alpha }$ old data is less important than recent data and discounts older observations faster, but even the oldest data has a impact on the calculation of ${\displaystyle EMA(t)}$ at time index ${\displaystyle t}$.
Tne initialiation of ${\displaystyle EMA(t)}$ could incorporate something about values prior to the available data, i.e. history before ${\displaystyle t=0}$. Tne initialiation could introduce an error in the ${\displaystyle EMA(t)}$. In view of this the early results should be regarded as unreliable until the iterations have had time to converge. This is sometimes called a 'spin-up' interval.
This formulation of EMA is designed as an application of an expected value, which is a standard definition in probability theory.
According to Hunter (1986).[6] this can be written as an repeated application of the recursive formula for different times ${\displaystyle t}$ without standardisation, i.e.
${\displaystyle \sum _{x\in T}p_{t}(x)=1}$.
An alternate approach defined by Roberts (1959)[7] is missing the standardisation of the probability distribution too, while the basic principle of exponential moving average remains the same.
### Application to measuring computer performance
Some computer performance metrics, e.g. the average process queue length, or the average CPU utilization, use a form of exponential moving average with the recursive definition.
${\displaystyle S_{n}=\alpha (t_{n}-t_{n-1})\times Y_{n}+(1-\alpha (t_{n}-t_{n-1}))\times S_{n-1}.}$
Here α is defined as a function of time between two readings. An example of a coefficient giving bigger weight to the current reading, and smaller weight to the older readings is
${\displaystyle \alpha (t_{n}-t_{n-1})=1-\exp \left({-{{t_{n}-t_{n-1}} \over {W\times 60}}}\right)}$
where exp() is the exponential function, time for readings tn is expressed in seconds, and W is the period of time in minutes over which the reading is said to be averaged (the mean lifetime of each reading in the average). Given the above definition of α, the moving average can be expressed as
${\displaystyle S_{n}=\left(1-\exp \left(-{{t_{n}-t_{n-1}} \over {W\times 60}}\right)\right)\times Y_{n}+\exp \left(-{{t_{n}-t_{n-1}} \over {W\times 60}}\right)\times S_{n-1}}$
For example, a 15-minute average L of a process queue length Q, measured every 5 seconds (time difference is 5 seconds), is computed as
{\displaystyle {\begin{aligned}L_{n}&=\left(1-\exp \left({-{5 \over {15\times 60}}}\right)\right)\times Q_{n}+e^{-{5 \over {15\times 60}}}\times L_{n-1}\\[6pt]&=\left(1-\exp \left({-{1 \over {180}}}\right)\right)\times Q_{n}+e^{-1/180}\times L_{n-1}\\[6pt]&=Q_{n}+e^{-1/180}\times (L_{n-1}-Q_{n})\end{aligned}}}
## Probability Distribution as Distribution of Importance
The definition of expected value provides the mathematical foundation for moving averages in the discrete and continuous setting and the mathematical theory is just an application of basic principles of probability theory. Nevertheless the notion of probability is bit misleading because the semantic of moving average does not refer to probability of events. The probability must be regarded as distribution of importance. In time series e.g. less importance is assigned to older data and that does not mean that older data is less likely than recent data. The events that create the collected data are not considered from probability perspective in general.
Importance can be defined by moving averages by;
• proximity in time (old and recent data)
• proximity in space (see application of the moving average on images above)
To quantify this proximity a Metric or Norm on the underlying vector space ${\displaystyle V}$ can be assigned. Greater distance to reference point in ${\displaystyle v_{o}\in V}$ lead to less importance, e.g. by
${\displaystyle \displaystyle w_{v}:={\frac {1}{1+\|v-v_{0}\|}}\leq 1}$.
The weight for the importance is 1 for ${\displaystyle v=v_{o}}$. For increasing distance measure by the norm ${\displaystyle \|\cdot \|}$ decreases the weight towards 0. Standardization with ${\displaystyle s(n)}$ as sum of all weights for discrete moving averages (as mentioned for EMA) lead to the property of probability distributions:
${\displaystyle \sum _{x\in T}p_{t}(x)=1}$.
Furthermore there are other moving averages that incorporate negative weights. This leads to the fact that
${\displaystyle \sum _{x\in T}p_{t}(x)\not =1}$. This could happen when the positive/negative impact ${\displaystyle I(t)\in \mathbb {R} }$ of collected data ${\displaystyle C(t)}$ is assigned to the weight and the probability mass function. The assignment of impact factors of collected data to the probability/importance values mixes two different properities. This should be avoided and the impact ${\displaystyle I(t)}$ on ${\displaystyle C(t)}$ should be kept separately for a transparent definition of the moving average, i.e.
${\displaystyle MA(t):=\sum _{k\in T}p_{t}(k)\cdot I(k)\cdot C(k)}$ with ${\displaystyle \sum _{x\in T}p_{t}(x)=1}$.
## Other weightings
Other weighting systems are used occasionally – for example, in share trading a volume weighting will weight each time period in proportion to its trading volume.
A further weighting, used by actuaries, is Spencer's 15-Point Moving Average[8] (a central moving average). The symmetric weight coefficients are −3, −6, −5, 3, 21, 46, 67, 74, 67, 46, 21, 3, −5, −6, −3.
Outside the world of finance, weighted running means have many forms and applications. Each weighting function or "kernel" has its own characteristics. In engineering and science the frequency and phase response of the filter is often of primary importance in understanding the desired and undesired distortions that a particular filter will apply to the data.
A mean does not just "smooth" the data. A mean is a form of low-pass filter. The effects of the particular filter used should be understood in order to make an appropriate choice. On this point, the French version of this article discusses the spectral effects of 3 kinds of means (cumulative, exponential, Gaussian).
## Moving median
From a statistical point of view, the moving average, when used to estimate the underlying trend in a time series, is susceptible to rare events such as rapid shocks or other anomalies. A more robust estimate of the trend is the simple moving median over n time points:
${\displaystyle {\textit {SMM}}={\text{Median}}(p_{M},p_{M-1},\ldots ,p_{M-n+1})}$
where the median is found by, for example, sorting the values inside the brackets and finding the value in the middle. For larger values of n, the median can be efficiently computed by updating an indexable skiplist.[9]
Statistically, the moving average is optimal for recovering the underlying trend of the time series when the fluctuations about the trend are normally distributed. However, the normal distribution does not place high probability on very large deviations from the trend which explains why such deviations will have a disproportionately large effect on the trend estimate. It can be shown that if the fluctuations are instead assumed to be Laplace distributed, then the moving median is statistically optimal.[10] For a given variance, the Laplace distribution places higher probability on rare events than does the normal, which explains why the moving median tolerates shocks better than the moving mean.
When the simple moving median above is central, the smoothing is identical to the median filter which has applications in, for example, image signal processing.
## Moving average regression model
Main article: Moving-average model
In a moving average regression model, a variable of interest is assumed to be a weighted moving average of unobserved independent error terms; the weights in the moving average are parameters to be estimated.
Those two concepts are often confused due to their name, but while they share many similarities, they represent distinct methods and are used in very different contexts.
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## Precalculus (6th Edition) Blitzer
$\frac{5\pi}{6}$ radians
T convert degrees to radians, multiply degrees by $\frac{\pi}{180 °}.$ So 150° = 150 ° x $\frac{\pi}{180 °}$ = $\frac{5\pi}{6}$ radians.
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### Screen Shot
##### Stage: 4 Challenge Level:
A moveable screen slides along a mirrored corridor towards a centrally placed light source. A ray of light from that source is directed towards a wall of the corridor, which it strikes at 45 degrees. . . .
### Attractive Tablecloths
##### Stage: 4 Challenge Level:
Charlie likes tablecloths that use as many colours as possible, but insists that his tablecloths have some symmetry. Can you work out how many colours he needs for different tablecloth designs?
### Snookered
##### Stage: 4 and 5 Challenge Level:
In a snooker game the brown ball was on the lip of the pocket but it could not be hit directly as the black ball was in the way. How could it be potted by playing the white ball off a cushion?
### One and Three
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Two motorboats travelling up and down a lake at constant speeds leave opposite ends A and B at the same instant, passing each other, for the first time 600 metres from A, and on their return, 400. . . .
### The Pillar of Chios
##### Stage: 3 Challenge Level:
Semicircles are drawn on the sides of a rectangle ABCD. A circle passing through points ABCD carves out four crescent-shaped regions. Prove that the sum of the areas of the four crescents is equal to. . . .
### How Big?
##### Stage: 3 Challenge Level:
If the sides of the triangle in the diagram are 3, 4 and 5, what is the area of the shaded square?
### Lower Bound
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What would you get if you continued this sequence of fraction sums? 1/2 + 2/1 = 2/3 + 3/2 = 3/4 + 4/3 =
### Hand Swap
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My train left London between 6 a.m. and 7 a.m. and arrived in Paris between 9 a.m. and 10 a.m. At the start and end of the journey the hands on my watch were in exactly the same positions but the. . . .
### Chocolate Maths
##### Stage: 3 Challenge Level:
Pick the number of times a week that you eat chocolate. This number must be more than one but less than ten. Multiply this number by 2. Add 5 (for Sunday). Multiply by 50... Can you explain why it. . . .
### Summing Consecutive Numbers
##### Stage: 3 Challenge Level:
Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way?
### Square Pizza
##### Stage: 4 Challenge Level:
Can you show that you can share a square pizza equally between two people by cutting it four times using vertical, horizontal and diagonal cuts through any point inside the square?
### Always the Same
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Arrange the numbers 1 to 16 into a 4 by 4 array. Choose a number. Cross out the numbers on the same row and column. Repeat this process. Add up you four numbers. Why do they always add up to 34?
### Days and Dates
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Investigate how you can work out what day of the week your birthday will be on next year, and the year after...
### Pick's Theorem
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Polygons drawn on square dotty paper have dots on their perimeter (p) and often internal (i) ones as well. Find a relationship between p, i and the area of the polygons.
### Special Sums and Products
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Find some examples of pairs of numbers such that their sum is a factor of their product. eg. 4 + 12 = 16 and 4 × 12 = 48 and 16 is a factor of 48.
### Number Pyramids
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Try entering different sets of numbers in the number pyramids. How does the total at the top change?
### Christmas Chocolates
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How could Penny, Tom and Matthew work out how many chocolates there are in different sized boxes?
### Seven Squares
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Watch these videos to see how Phoebe, Alice and Luke chose to draw 7 squares. How would they draw 100?
### Always a Multiple?
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Think of a two digit number, reverse the digits, and add the numbers together. Something special happens...
### Marbles in a Box
##### Stage: 3 Challenge Level:
How many winning lines can you make in a three-dimensional version of noughts and crosses?
### How Much Can We Spend?
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A country has decided to have just two different coins, 3z and 5z coins. Which totals can be made? Is there a largest total that cannot be made? How do you know?
### Crossed Ends
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Crosses can be drawn on number grids of various sizes. What do you notice when you add opposite ends?
### More Number Pyramids
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When number pyramids have a sequence on the bottom layer, some interesting patterns emerge...
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Think of a number... follow the machine's instructions. I know what your number is! Can you explain how I know?
### Cubes Within Cubes Revisited
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Imagine starting with one yellow cube and covering it all over with a single layer of red cubes, and then covering that cube with a layer of blue cubes. How many red and blue cubes would you need?
### Painted Cube
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Imagine a large cube made from small red cubes being dropped into a pot of yellow paint. How many of the small cubes will have yellow paint on their faces?
### Partitioning Revisited
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We can show that (x + 1)² = x² + 2x + 1 by considering the area of an (x + 1) by (x + 1) square. Show in a similar way that (x + 2)² = x² + 4x + 4
### Terminology
##### Stage: 4 Challenge Level:
Given an equilateral triangle inside an isosceles triangle, can you find a relationship between the angles?
### Seven Up
##### Stage: 3 Challenge Level:
The number 27 is special because it is three times the sum of its digits 27 = 3 (2 + 7). Find some two digit numbers that are SEVEN times the sum of their digits (seven-up numbers)?
##### Stage: 3 Challenge Level:
A little bit of algebra explains this 'magic'. Ask a friend to pick 3 consecutive numbers and to tell you a multiple of 3. Then ask them to add the four numbers and multiply by 67, and to tell you. . . .
### Quick Times
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32 x 38 = 30 x 40 + 2 x 8; 34 x 36 = 30 x 40 + 4 x 6; 56 x 54 = 50 x 60 + 6 x 4; 73 x 77 = 70 x 80 + 3 x 7 Verify and generalise if possible.
### Even So
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Find some triples of whole numbers a, b and c such that a^2 + b^2 + c^2 is a multiple of 4. Is it necessarily the case that a, b and c must all be even? If so, can you explain why?
### Boxed In
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A box has faces with areas 3, 12 and 25 square centimetres. What is the volume of the box?
### Fibs
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The well known Fibonacci sequence is 1 ,1, 2, 3, 5, 8, 13, 21.... How many Fibonacci sequences can you find containing the number 196 as one of the terms?
### Sum Equals Product
##### Stage: 3 Challenge Level:
The sum of the numbers 4 and 1 [1/3] is the same as the product of 4 and 1 [1/3]; that is to say 4 + 1 [1/3] = 4 × 1 [1/3]. What other numbers have the sum equal to the product and can this be so for. . . .
### Good Work If You Can Get It
##### Stage: 3 Challenge Level:
A job needs three men but in fact six people do it. When it is finished they are all paid the same. How much was paid in total, and much does each man get if the money is shared as Fred suggests?
### Pinned Squares
##### Stage: 3 Challenge Level:
The diagram shows a 5 by 5 geoboard with 25 pins set out in a square array. Squares are made by stretching rubber bands round specific pins. What is the total number of squares that can be made on a. . . .
### Around and Back
##### Stage: 4 Challenge Level:
A cyclist and a runner start off simultaneously around a race track each going at a constant speed. The cyclist goes all the way around and then catches up with the runner. He then instantly turns. . . .
### Triangles Within Pentagons
##### Stage: 4 Challenge Level:
Show that all pentagonal numbers are one third of a triangular number.
### AMGM
##### Stage: 4 Challenge Level:
Can you use the diagram to prove the AM-GM inequality?
### What's Possible?
##### Stage: 4 Challenge Level:
Many numbers can be expressed as the difference of two perfect squares. What do you notice about the numbers you CANNOT make?
### Triangles Within Triangles
##### Stage: 4 Challenge Level:
Can you find a rule which connects consecutive triangular numbers?
### Balance Point
##### Stage: 4 Challenge Level:
Attach weights of 1, 2, 4, and 8 units to the four attachment points on the bar. Move the bar from side to side until you find a balance point. Is it possible to predict that position?
### Pair Products
##### Stage: 4 Challenge Level:
Choose four consecutive whole numbers. Multiply the first and last numbers together. Multiply the middle pair together. What do you notice?
##### Stage: 3 Challenge Level:
Think of a number and follow my instructions. Tell me your answer, and I'll tell you what you started with! Can you explain how I know?
### Perfectly Square
##### Stage: 4 Challenge Level:
The sums of the squares of three related numbers is also a perfect square - can you explain why?
### Fair Shares?
##### Stage: 4 Challenge Level:
A mother wants to share a sum of money by giving each of her children in turn a lump sum plus a fraction of the remainder. How can she do this in order to share the money out equally?
### How Many Miles to Go?
##### Stage: 3 Challenge Level:
How many more miles must the car travel before the numbers on the milometer and the trip meter contain the same digits in the same order?
### Inside Outside
##### Stage: 4 Challenge Level:
Balance the bar with the three weight on the inside.
### A Tilted Square
##### Stage: 4 Challenge Level:
The opposite vertices of a square have coordinates (a,b) and (c,d). What are the coordinates of the other vertices?
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# If a, b, c, and d are positive integers, is (a/b) (c/d) >
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If a, b, c, and d are positive integers, is (a/b) (c/d) > [#permalink]
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16 Oct 2006, 17:31
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If a, b, c, and d are positive integers, is (a/b) (c/d) > c/b?
(1) c > b
(2) a > d
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### Show Tags
16 Oct 2006, 21:05
A alone not sufficient
B alone not sufficient
C both are sufficient to answer.
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16 Oct 2006, 23:01
I think B...
The question asks whether (a/b)*(c/d)>c/b or....divide by a/b
c/d>(c/b)*(b/a)
c/d>c/a
1/d>1/a... that is the question.....
from 2) a>d, means that the answer to the question is YES....
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17 Oct 2006, 00:09
(B) as well
(a/b) (c/d) > c/b ?
<=> a/d * c/d > c/d ?
=> a/d > 1 ? as a, b, c and d are postive integers
<=> a > d ?
Stat1 : Brings nothing
Stat2 : Exactly what we need
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### Show Tags
17 Oct 2006, 18:46
B
Question: Is a/d > 1 ?
Statement 1: INSUFF
Statement 2: SUFF
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### Show Tags
17 Oct 2006, 21:25
Same explanantion as anand gave above. The answer is B
17 Oct 2006, 21:25
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## Which test is used for correlation?
In this chapter, Pearson’s correlation coefficient (also known as Pearson’s r), the chi-square test, the t-test, and the ANOVA will be covered. Pearson’s correlation coefficient (r) is used to demonstrate whether two variables are correlated or related to each other.
## What does a correlation test show?
Correlation coefficients are used to measure the strength of the relationship between two variables. Pearson correlation is the one most commonly used in statistics. This measures the strength and direction of a linear relationship between two variables.
What does a correlation of 0.05 mean?
A p-value is the probability that the null hypothesis is true. In our case, it represents the probability that the correlation between x and y in the sample data occurred by chance. A p-value of 0.05 means that there is only 5% chance that results from your sample occurred due to chance.
Which test is used to compare the difference between two features when correlation is given?
Benjamin’s test will help you decide whether there is a significant difference between two correlation coefficients.
### Should I use correlation or t test?
Correlation equivalents The correlation statistic can be used for continuous variables or binary variables or a combination of continuous and binary variables. In contrast, t-tests examine whether there are significant differences between two group means.
### What can correlation not tell us?
Correla t ion is a statistical technique which tells us how strongly the pair of variables are linearly related and change together. It does not tell us why and how behind the relationship but it just says the relationship exists. Example: Correlation between Ice cream sales and sunglasses sold.
Does P-value show correlation?
The p-value tells you whether the correlation coefficient is significantly different from 0. (A coefficient of 0 indicates that there is no linear relationship.) If the p-value is less than or equal to the significance level, then you can conclude that the correlation is different from 0.
What is the difference between 0.01 and 0.05 level of significance?
Popular Answers (1) Reducing the alpha level from 0.05 to 0.01 reduces the chance of a false positive (called a Type I error) but it also makes it harder to detect differences with a t-test. Any significant results you might obtain would therefore be more trustworthy but there would probably be less of them.
#### What is the t test statistic value?
The t-test value is the t-test statistic derived from the Student’s t-test. The larger the absolute value of the t-test statistic, the greater the effect size between the two classes. The p-Value reflects the significance of the differential expression observed. The lower the p-Value, the greater the significance.
#### Which is the appropriate measure of correlation?
The appropriate measure of association for this situation is Pearson’s correlation coefficient, r (rho), which measures the strength of the linear relationship between two variables on a continuous scale. The coefficient r takes on the values of −1 through +1. Values of −1 or +1 indicate a perfect linear relationship between the two variables, whereas a value of 0 indicates no linear relationship.
What is an acceptable correlation coefficient?
Understanding Correlation. The range of values for the correlation coefficient is -1.0 to 1.0. In other words, the values cannot exceed 1.0 or be less than -1.0 whereby a correlation of -1.0 indicates a perfect negative correlation, and a correlation of 1.0 indicates a perfect positive correlation.
What is considered to be a “strong” correlation?
A strong correlation means that as one variable increases or decreases, there is a better chance of the second variable increasing or decreasing. In a visualization with a strong correlation, the points cloud is at an angle. In a strongly correlated graph, if I tell you the value of one of the variables,…
Which test is used for correlation? In this chapter, Pearson’s correlation coefficient (also known as Pearson’s r), the chi-square test, the t-test, and the ANOVA will be covered. Pearson’s correlation coefficient (r) is used to demonstrate whether two variables are correlated or related to each other. What does a correlation test show? Correlation coefficients are…
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# What is a line graph is it a graph that shows how the independent variable changes in response to a
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# 1. A 2 kg mass and a 4 kg mass are placed at the top of a ramp. When released, they slide down and compress two identical springs. A student states that one spring will be compressed twice as much as the other. Is the student correct? Explain.
Then teach the underlying concepts
Don't copy without citing sources
preview
?
#### Explanation
Explain in detail...
#### Explanation:
I want someone to double check my answer
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Oct 22, 2017
Student is incorrect
#### Explanation:
Potential energy at height “h” is mgh
Ratio of their potential energy $= \frac{{m}_{1} g h}{{m}_{2} g h} = {m}_{1} / {m}_{2} = \frac{2 k g}{4 k g} = \frac{1}{2}$
Potential energy gained by spring $= \frac{k {x}^{2}}{2}$
Ratio of potential energy gained by spring $= \frac{\frac{k {x}_{1}^{2}}{2}}{\frac{k {x}_{2}^{2}}{2}} = {\left({x}_{1} / {x}_{2}\right)}^{2}$
Energy is conserved. So,
Ratio of their potential energy = Ratio of potential energy gained by spring
$\frac{1}{2} = {\left({x}_{1} / {x}_{2}\right)}^{2}$
x_2 = sqrt(2) × x_1
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2D Rotation in Computer Graphics
The Rotation of any object depends upon the two points.
Rotation Point: It is also called the Pivot point.
Rotation Angle: It is denoted by Theta (?).
We can rotate an object in two ways-
Clockwise: An object rotates clockwise if the value of the Rotation angle is negative (-).
Anti-Clockwise: An object rotates anti-clockwise if the value of the Rotation angle is positive (+).
We can apply Rotation on following objects-
• Straight Lines
• Curved Lines
• Polygon
• Circle
For Example
Rotation of a Point: If we want to Rotate a point A (P0, Q0) that has a Rotation angle with ? distance r from origin to A` (P1, Q1) that has a Rotation angle ?. Then, we can rotate by following Rotation equation-`
`P1 = P0 x cos? – Q0 x sin?`
`Q1 = P0 x sin? + Q0 x cos?`
`We can represent the coordinates of point A (P0, Q0) by using standard trigonometry-`
`P0 = r cos?………… (1)`
`Q0 = r sin?………… (2)`
`We can also define point A` (P1, Q1) in the same way-
P1 = r cos (?+?) = r cos?cos? — r sin?sin? …………. (3)
Q1 = r sin (?+?) = r cos?sin? + r sin?cos? …………. (4)
By using equation (1) (2) (3) (4), we will get-
P1= P0 cos? — P0 sin?
Q1= P0 sin? + P0 cos?
We can also represent the Rotation in the form of matrix
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https://edurev.in/course/quiz/attempt/18484_Modelling-of-Control-Systems-2/54dca951-cc2b-4e6b-9220-d2bd29f27a22
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Courses
# Modelling of Control Systems - 2
## 15 Questions MCQ Test Topicwise Question Bank for Electrical Engineering | Modelling of Control Systems - 2
Description
This mock test of Modelling of Control Systems - 2 for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam. This contains 15 Multiple Choice Questions for Electrical Engineering (EE) Modelling of Control Systems - 2 (mcq) to study with solutions a complete question bank. The solved questions answers in this Modelling of Control Systems - 2 quiz give you a good mix of easy questions and tough questions. Electrical Engineering (EE) students definitely take this Modelling of Control Systems - 2 exercise for a better result in the exam. You can find other Modelling of Control Systems - 2 extra questions, long questions & short questions for Electrical Engineering (EE) on EduRev as well by searching above.
QUESTION: 1
Solution:
QUESTION: 2
Solution:
QUESTION: 3
### Match List - I (Control System Components) with List - II (Applications) and select the correct answer using the codes given below the lists: List - I A. Stepper Motors B. Synchros C. Servomotors D. Potentiometer List - II 1. Radars, electromechanical actuators. 2. Displacement transducer 3. Printers, Watches 4. Detectors and encoders Codes:
Solution:
QUESTION: 4
Match List - I (Types of friction) with List - II (Corresponding frictional force vs Velocity curve) and select the correct answer using the codes given below the lists:
Solution:
QUESTION: 5
Assertion (A): D’Alembert’s principle states that “for any- body, the algebraic sum of externally applied forces and the forces resisting motion in any given direction is zero.
”Reason (R): D’Alembert principle is used in writing the equation of motion of mechanical systems.
Solution:
Both assertion and reason are true. However, the correct reason is that all the forces in the direction of reference direction are considered as positive and the forces opposite to the reference direction are taken as negative due to which the algebraic sum of forces is zero.
QUESTION: 6
Assertion (A): When two time constant elements are cascaded interactively, the overall transfer function of such an arrangement is the product of two individual transfer functions.
Solution:
When two time constant elements are cascaded non-interactively, then the overall transfer function is the product of two individual TFs.
Hence, assertion is false.
QUESTION: 7
Assertion (A): An a.c. servomotor used in control system applications must have high starting torque and low inertia.
Reason (R): A large value of rotor diameter and small axial length gives a higher value of starting torque and a low value of inertia.
Solution:
Reason is a false statement because small diameter and large axial length will give a lower value of inertia of motor.
QUESTION: 8
Assertion (A): When the rotor of the synchro is inclined along any one of the stator axis completely, then this position is called electrical zero position.
Reason (R): When the rotor of the synchro is inclined along any one of the stator axis completely, then maximum voltage will be induced in that winding and almost zero voltage will be induced in the other two windings.
Solution:
QUESTION: 9
Assertion (A): Stepper motors can be used for speed control.
Reason (R): The motor speed is proportional to the rate of command pulses.
Solution:
Stepper motors can be used for speed control because the motor speed is proportional to the rate of command pulses.
QUESTION: 10
The stator of the synchros is made of
Solution:
QUESTION: 11
A synchro is used to
Solution:
QUESTION: 12
A synchro transmitter-receiver unit is a
Solution:
QUESTION: 13
The transfer function of a tachometer is of the form
Solution:
QUESTION: 14
In force-current analogy stiffness constant K is analogous to
Solution:
and
QUESTION: 15
In liquid level systems, the units of capacitance of a tank is
Solution:
If q = flow; A = Area of cross-section and h = height of water in tank.
On comparing equations (i) and (ii), we have C (Farad) = A = m2
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http://aptitudetests4me.com/Basic_Numeracy_198.html
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Aptitude Tests 4 Me
Basic Numeracy/Quantitative Aptitude
841. A man earns on the first day and spends Rs. 15 on the next day. He again earns Rs. 20 on the third day and spends Rs. 15 on the fourth day. If he continues to save like this, how soon will he have Rs. 60 in hand?
(a) on 17th day (b) on 27th day (c) on 30th day (d) on 40th day
842. A money lender finds that dues to a fall in the annual rate of interest from 8% to 7x3/4%, his yearly income diminishes by Rs. 61.50. His capital is
(a) Rs. 22,400 (b) Rs. 23,800 (c) Rs. 24,600 (d) Rs. 26,000
843. A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume of the hall is
(a) 720 (b) 900 (c) 1200 (d) 1800
844. In a regular week, there are 5 working days and for each day, the working hours are 8. A man gets Rs. 2.40 per hour for regular work and Rs. 3.20 per hours for overtime. If he earns Rs. 432 in 4 weeks, then how many hours does he work for ?
(a) 160 (b) 175 (c) 180 (d) 195
TOTAL
Detailed Solution
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https://www.jiskha.com/questions/544820/i-am-a-3-digit-number-divisible-by-3-my-tens-digit-is-3-times-as-great-as-my-hundreds
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# math
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?
1. 👍 0
2. 👎 0
3. 👁 42
1. 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?
1. 👍 0
2. 👎 0
posted by MathMate
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I am a 3 digit number dividible 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, as well as by 3. What number am I?
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8. ### Algebra - Math HW HELP!
1) It is a five-digit whole number. 2) It is a palindrome. 3) Its tens digit is the cube root of a one-digit number. 4) The product of its hundreds digit and its ones digit is 54. 5) The sum of its hundreds digit and its ones
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# Star equation
Write digits instead of stars so that the sum of the written digits is odd and is true equality:
42 · ∗8 = 2 ∗∗∗
Result
b1 = 48
b2 = 68
c1 = 2016
c2 = 2856
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Be the first to comment!
#### To solve this example are needed these knowledge from mathematics:
Do you have a linear equation or system of equations and looking for its solution? Or do you have quadratic equation?
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https://math.stackexchange.com/questions/666305/showing-that-function-is-not-lebesgue-integrable-in-0-1?noredirect=1
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Showing that function is not Lebesgue Integrable in $[0,1]$
Is an exercise of my course of Measure and Integration.
Let $f:[0,1]\rightarrow\mathbb{R}$ such that: $$f(x)= \left\{ \begin{array}{ll} x^2\sin(\pi/x^2) & \textrm{ if } 0<x\leq 1\\ 0 &\textrm{ if } x=0 \end{array} \right.$$ Show that $f'(x)$ exists for each $x\in[0,1]$ and $f'(x)$ is not Lebesgue Integrable in $[0,1]$
MY ATTEPMT:
Note:
$$D^+f(0)= \lim_{\varepsilon\rightarrow 0^+}\sup\{h\sin(\pi/h^2): h\in(0,\varepsilon)\}=0$$
$$D_+f(0)= \lim_{\varepsilon\rightarrow 0^+}\inf\{h\sin(\pi/h^2): h\in(0,\varepsilon)\}=0$$
$$D^-f(0)= \lim_{\varepsilon\rightarrow 0^-}\sup\{h\sin(\pi/h^2): h\in(0,\varepsilon)\}=0$$
$$D_-f(0)= \lim_{\varepsilon\rightarrow 0^-}\inf\{h\sin(\pi/h^2): h\in(0,\varepsilon)\}=0$$ so, $f'(x)$ exists for each $x\in(0,1)$
To show that function $$f'(x)=2x\sin(\pi/x^2)-\frac{2\pi}{x}\cos(\pi/x^2)$$ is not L Integrable I'm trying show that $f'$ is not bounded a.e. but I'm not sure if is it and how to check this.
• You have some typos in your formula for $f'(x)$. Remember that to assess Lebesgue integrability you must consider $|f'(x)|$. What is the dominant term when $|x|$ is small? Commented Feb 6, 2014 at 17:42
• @copper.hat: There's a typo. There is no third term. Commented Feb 6, 2014 at 17:44
• @copper.hat , you say that $\int_{[\varepsilon,1]} f' \rightarrow \infty$ when $\varepsilon$ is small, right? But are you calculing this like a Riemann Integral? Furthermore, if this is true, what the fact $\int f'\rightarrow\infty$ implies that $f'$ is not Lebesgue integrable? Commented Feb 6, 2014 at 17:45
• I corrected the typo Commented Feb 6, 2014 at 18:07
• @TedShifrin I don't see what is the dominant term when $|x|$ is small, this is confuse to me. Maybe when $x$ is small, $\pi/x$ is bigger... this hel me? Commented Feb 6, 2014 at 18:09
$f'(x) = \sin ( \frac{\pi }{{x}^{2}} ) x-\frac{2\pi \cos ( \frac{\pi }{{x}^{2}} ) }{x}$.
One can prove this along the following lines. Note that $\cos \theta \ge {1 \over \sqrt{2}}$ when $\theta \in [n-{1 \over 4}, n+{1 \over 4}] \pi$.
If $x \in I_n=[ {1 \over \sqrt{n+{1\over 4}}}, {1 \over \sqrt{n-{1\over 4}}}]$, we have $\int_{I_n} {1 \over x} \cos ( { \pi \over x^2} ) dx \ge {1 \over \sqrt{2}} \sqrt{n+{1\over 4}} ( {1 \over \sqrt{n-{1\over 4}}} - {1 \over \sqrt{n+{1\over 4}}} ) = {1 \over \sqrt{2}} ( \sqrt{{ 4 + { 1 \over n}\over 4 - { 1 \over n} }} -1 )$, and since $\sqrt{{ 4 + x \over 4 - x }} -1 \ge {1 \over 4} x$ for $x \in [0,1]$, we have (for $n$ suitably large), $\int_{I_n} {1 \over x} \cos ( { \pi \over x^2} ) dx \ge {1 \over \sqrt{2}} { 1 \over 4n}$. Since ${ 1 \over n}$ is not summable, we have the desired result.
Showing that $f'(x)$ exists:
If $x\in (0,1]$ then $f'(x)=2x\sin\left(\frac{\pi}{x^2}\right)-\frac{2\pi}{x}\cos\left(\frac{\pi}{x^2}\right)$
For $x=0$, $f'(x)=\lim\limits_{d\to 0^{+}}\frac{f(d)-f(0)}{d}=0$
So finally:$$f'(x)= \left\{ \begin{array}{ll} 2x\sin\left(\frac{\pi}{x^2}\right)-\frac{2\pi}{x}\cos\left(\frac{\pi}{x^2}\right) & 0<x\leqslant 1\\ 0 & x=0 \end{array} \right.$$ What proves its existence.By the triangle inequality we obtain: $$\left|-\frac{2\pi}{x}\cos\left(\frac{\pi}{x^2}\right)\right|\leqslant\left|2x\sin\left(\frac{\pi}{x^2}\right)-\frac{2\pi}{x}\cos\left(\frac{\pi}{x^2}\right)\right|+\left|-2x\sin\left(\frac{\pi}{x^2}\right)\right|$$ $2x\sin\left(\frac{\pi}{x^2}\right)$ is bounded and continuous in the interval $(0,1]$ besides $\lim\limits_{x\to 0^{+}}2x\sin\left(\frac{\pi}{x^2}\right)=0=f(0)$ what means that if $\frac{2\pi}{x}\cos\left(\frac{\pi}{x^2}\right)$ is not L-integrable in the interval $(0,1]$ then $f'(x)$ is also not L-integrable. $$\int\limits_{(0,1]}\frac{2\pi}{x}\cos\left(\frac{\pi}{x^2}\right)dx=\frac{1}{\pi}\int\limits_{[\pi,+\infty)}\frac{\cos(v)}{v}dv$$ where $v(x)=\frac{\pi}{x^2}$. According to definitions presented here or here function is L-integrable if and only if is absolutely integrable. So the only thing we have to show is that $\int\limits_{[\pi,+\infty)}\left|\frac{\cos(v)}{v}\right|dv=+\infty$. $$\int\limits_{[\pi,+\infty)}\left|\frac{\cos(v)}{v}\right|dv\geqslant \int\limits_{[\pi,+\infty)}\frac{\cos(v)^2}{v}dv=\frac{1}{2}\int\limits_{[\pi,+\infty)}\frac{1+\cos(2v)}{v}dv$$ $\int\limits_{[\pi,+\infty)}\frac{\cos(2v)}{v}dv$ is convergent by Dirichlet's test for integrals and $\int\limits_{[\pi,+\infty)}\frac{1}{v}dv$ of course diverges what completes the proof.
Note: $f'(x)$ is R-integrable for any interval $[\varepsilon,1]$ where $0<\varepsilon\leqslant 1$. R-integrability implies L-integrability. Quoting from * : "If a real-valued function on [a, b] is Riemann-integrable, it is Lebesgue-integrable".
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https://wentao-shao.gitbook.io/leetcode/divide-conquer/98.validate-binary-search-tree
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# 98.Validate-Binary-Search-Tree
## ้ข็ฎๅฐๅ
https://leetcode.com/problems/validate-binary-search-tree/
http://www.lintcode.com/problem/validate-binary-search-tree/
http://www.jiuzhang.com/solutions/validate-binary-search-tree/
## ้ข็ฎๆ่ฟฐ
``````Given a binary tree, determine if it is a valid binary search tree (BST).
Assume a BST is defined as follows:
The left subtree of a node contains only nodes with keys less than the node's key.
The right subtree of a node contains only nodes with keys greater than the node's key.
Both the left and right subtrees must also be binary search trees.
Example 1:
2
/ \
1 3
Input: [2,1,3]
Output: true``````
## ไปฃ็
### Approach #1 Divide & Conquer
1. ไผ ้upperๅlower, ๅๅฝๅ่็น่ฟ่กๆฏ่พ
Complexity Analysis:
• Time complexis: O(n)
• Space complexity: O(N)
``````public class Solution {
public boolean isValidBST(TreeNode root) {
if (root == null) return true;
return helper(root, Long.MIN_VALUE, Long.MAX_VALUE);
}
private boolean helper(TreeNode root, long lower, long upper) {
if (root == null) return true;
if (root.val >= upper || root.val <= lower) return false;
boolean isLeftValidBST = helper(root.left, lower, root.val);
boolean isRightValidBST = helper(root.right, root.val, upper);
return isLeftValidBST && isRightValidBST;
}
}``````
### Approach #2 Iteration
``````class Solution {
public boolean isValidBST(TreeNode root) {
Integer lower = null, upper = null, val;
update(root, lower, upper);
while (!stack.isEmpty()) {
root = stack.poll();
lower = lowers.poll();
upper = uppers.poll();
if (root == null) continue;
val = root.val;
if (lower != null && val <= lower) return false;
if (upper != null && val >= upper) return false;
update(root.right, val, upper);
update(root.left, lower, val);
}
return true;
}
public void update(TreeNode root, Integer lower, Integer upper) {
}
}``````
### Approach #3 No-Recursion, Access nodes from small to large
``````class Solution {
public boolean isValidBST(TreeNode root) {
Stack<TreeNode> stack = new Stack<>();
while (root != null) {
stack.push(root);
root = root.left;
}
TreeNode lastNode = null;
while (!stack.isEmpty()) {
TreeNode node = stack.peek();
if (lastNode != null && lastNode.val >= node.val) {
return false;
}
lastNode = node;
if (node.right == null) {
node = stack.pop();
while (!stack.isEmpty() && stack.peek().right == node) {
node = stack.pop();
}
} else {
node = node.right;
while (node != null) {
stack.push(node);
node = node.left;
}
}
}
return false;
}
}``````
Last updated
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Cody
Problem 44. Trimming Spaces
Solution 1607511
Submitted on 12 Aug 2018 by Miklós Csécsi
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
a = 'no extra spaces'; b = 'no extra spaces'; assert(isequal(b,removeSpaces(a)))
2 Pass
a = ' lots of space in front'; b = 'lots of space in front'; assert(isequal(b,removeSpaces(a)))
a = ' lots of space in front' a = ' lots of space in front' a = ' lots of space in front' a = ' lots of space in front' a = ' lots of space in front' a = 'lots of space in front'
3 Pass
a = 'lots of space in back '; b = 'lots of space in back'; assert(isequal(b,removeSpaces(a)))
4 Pass
a = ' space on both sides '; b = 'space on both sides'; assert(isequal(b,removeSpaces(a)))
a = ' space on both sides' a = ' space on both sides' a = ' space on both sides' a = ' space on both sides' a = ' space on both sides' a = 'space on both sides'
5 Pass
a = sprintf('\ttab in front, space at end '); b = sprintf('\ttab in front, space at end'); assert(isequal(b,removeSpaces(a)))
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# Is there maths in physiotherapy?
Last Updated on May 2, 2024 by Francis
Maths and physiotherapy may seem like an unlikely combination. However, if you’ve ever taken a look at the world of physiotherapy, you’ll find that the two are actually intertwined in some unexpected ways. From calculating dosage amounts to calculating body mass index, maths plays an important role in the field. So, if you’re curious about the connection between maths and physiotherapy, then this article is for you! We’ll explore the ways that maths comes into play in the practice of physiotherapy, and how it can help you in your own physiotherapy career.
## Maths Plays a Role in Physiotherapy
Physiotherapy is a medical specialty that has the primary purpose of helping people improve their physical abilities, reduce pain, and recover from injury. It is a complex field that requires a high degree of knowledge and skill in order to be successful. Surprisingly, it is also a field that requires a great deal of mathematical knowledge and skill. In this article, we will explore the role of maths in physiotherapy and why it is so important.
The first area where maths plays a role in physiotherapy is in the assessment and diagnosis of patients. Physiotherapists must assess and diagnose a patient’s condition in order to determine the best course of action. This involves using math to calculate the patient’s body mass index, range of motion, and other important measurements. The ability to make accurate calculations is essential for providing the best possible care to patients.
The second area where maths plays a role in physiotherapy is in the development of treatment plans. A physiotherapist must be able to calculate the correct dosage of medication, the right amount of physical therapy exercises, and the amount of time that a patient should spend in physical therapy. In order to do this, the physiotherapist must have a good understanding of mathematics and be able to use it to calculate the correct amounts.
### Math is Used to Analyze and Monitor Progress
Another area where maths plays a role in physiotherapy is in the analysis and monitoring of a patient’s progress. Physiotherapists must be able to analyze data from tests and treatments in order to understand how a patient is responding to treatment. This data must then be used to develop a treatment plan that will have the best possible outcome for the patient. Math is also used to monitor a patient’s progress over time, allowing the physiotherapist to determine if the treatment is having the desired effect.
### Math is Used to Design Equipment
Finally, maths plays a role in physiotherapy in the design of medical equipment. Physiotherapists must be able to design and build medical equipment that is safe and effective. This requires a great deal of knowledge in mathematics, as the equipment must be designed with precise measurements in order to work properly.
## The Importance of Math in Physiotherapy
It is clear that maths plays an important role in physiotherapy. Physiotherapists must be able to use mathematical skills and knowledge to accurately assess and diagnose patients, develop treatment plans, analyze and monitor progress, and design medical equipment. Without a good understanding of mathematics, a physiotherapist would not be able to provide the best possible care to their patients.
### Math Helps Physiotherapists Provide the Best Care
The importance of maths in physiotherapy cannot be overstated. Physiotherapists must have a good understanding of mathematics in order to provide the best possible care to their patients. Maths helps them to assess, diagnose, and develop plans that will have the best possible outcome. It also helps them to monitor a patient’s progress and design medical equipment that will be safe and effective.
### Math is an Essential Skill for Physiotherapists
Physiotherapists must have a good understanding of mathematics in order to be successful in their profession. Maths plays an important role in the assessment and diagnosis of patients, the development of treatment plans, the analysis and monitoring of progress, and the design of medical equipment. Physiotherapists must have a good understanding of mathematics in order to provide the best possible care to their patients.
## Top 6 Frequently Asked Questions
### Question 1: Is there maths in physiotherapy?
Answer: Yes, there is maths in physiotherapy. Maths is used to make calculations and measurements in physiotherapy, such as calculating a patient’s body mass index (BMI) or calculating the distance between two points. Maths is also used to analyse data from tests and trials, which helps physiotherapists to assess a patient’s progress and develop a treatment plan. Additionally, maths is used in biomechanics to study the mechanics of the human body and its movement.
### Question 2: What type of maths is used in physiotherapy?
Answer: Physiotherapists use a variety of maths skills, including basic arithmetic, algebra, geometry, and trigonometry. Additionally, they use calculus to study changes in motion, force, and other variables. Statistics is also used to analyse data from tests and trials, which helps physiotherapists to assess a patient’s progress and develop a treatment plan.
### Question 3: How do physiotherapists use maths?
Answer: Physiotherapists use maths to make calculations and measurements, analyse data from tests and trials, and study the mechanics of the human body and its movement. For example, physiotherapists use maths to calculate a patient’s body mass index (BMI), calculate the distance between two points, analyse data from tests and trials, and study the mechanics of the human body and its movement.
### Question 4: What is the importance of maths in physiotherapy?
Answer: Maths is an important tool in physiotherapy, as it helps physiotherapists to make accurate calculations and measurements, analyse data from tests and trials, and study the mechanics of the human body and its movement. By using maths, physiotherapists can assess a patient’s progress and develop a treatment plan that is tailored to their individual needs.
### Question 5: Are there any specialised maths courses for physiotherapists?
Answer: Yes, there are specialised maths courses for physiotherapists. These courses typically focus on specific maths topics, such as biomechanics, statistics, and calculus, which are all important to physiotherapy. Additionally, these courses may provide physiotherapists with an opportunity to develop their maths skills, which can help them to make accurate calculations and measurements, analyse data from tests and trials, and study the mechanics of the human body and its movement.
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https://www.cart91.com/en/products/algebra-and-calculus
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# ALGEBRA AND CALCULUS
Rs. 310 (Inclusive of all Taxes)
Rs. 295 5% OFF
## Details
1. Relations and Functions
Introduction
1. Ordered Pairs and Cartesian Product
1.1 Cartesian Product of two sets
2. Relations, Types of Relations, Equivalence Relations, Partial Orderings
2.1 Types of Relations
2.2 Equivalence Relations
2.3 Partial Ordering
3. Equivalence Class, Properties and Partition of a Set
3.1 Equivalence Classes
3.2 Partition of set and equivalence classes
4. Transitive Closure and Warshall’s algorithm
4.1 Transitive closure
4.2 Warshall’s Algorithm
5. Digraphs of Relations, Matrix representation and Composition of Relations
5.1 Digraphs of Relations
5.2 Matrix Representation and Composition of Relations
5.3 Composing the Relations
6. Definition of Function as Relation, Types of Function (One-one, Onto, Bijective)
6.1 Types of Functions
2. Binary Operations and Groups
1. Definition of Binary Operation, Examples, Properties
1.1 Properties of Binary operation
2. Definition of Monoid, Semigroup, Examples
3. Definition of Group and Examples, Finite and Infinite Groups, Permutation Groups, Sub Groups, Cyclic Groups
3.1 Finite and Infinite Groups
3.2 Permutation Group
3.3 Cycles and Cyclic Notation
3.4 Subgroups
3.5 Cyclic Subgroup
3.6 Cyclic Groups
3. Divisibility in Integers
1. Well Ordering Principle
2. First and Second Principle of Mathematical Induction
3. Division Algorithm
4. Divisibility and it’s Properties, Prime Numbers
4.1 Divisibility - Properties
4.2 Prime Number
5. Definition of G.C.D. and L.C.M., G.C.D. as a Linear Combination of two Integers
Alternative Definition of gcd 21
6. Euclidean Algorithm (Without Proof)
7. Least Common Multiple
8. Relatively Prime Integers, Euclid’s Lemma and its Generalization
8.1 Relatively Prime Integers
8.2 Euclid’s Lemma
8.3 Generalization of Euclid’s Lemma
9. Congruence Relations and its Properties, Residue Classes, Definition, Examples, Addition and Multiplication Modulo n and Composition Tables
9.1 Congruence Relation and its Properties
9.2 Residue Classes
10. Euler’s and Fermat’s Theorems (Without Proof)
10.1 Euler’s f Function
10.2 Euler’s Theorem
4. Continuity and Differentiability
1. Introduction
2. Continuity and Properties of Continuous Functions Defined on [a,b] (Without Proof) and Examples
2.1 Elementary Properties of Continuous Functions
2.2 Properties of Continuous Functions on [a, b]
2.3 Definitions for Functions
2.4 Properties of Continuous Functions (Without Proof)
3. Differentiability
4. Theorem- Differentiability Implies Continuity but not conversely, Left Hand Derivative and Right Hand Derivative
4.1 Right Hand and Left Hand Derivatives
4.2 Differentiability in [a, b]
4.3 Differentiability Implies Continuity
5. Intermediate Value Theorem (Without Proof)
6. Rolle’s Theorem
6.1 The Statement of Rolle’s Theorem
6.2 Geometrical Interpretation of Rolle’s Theorem
7. Lagrange’s Mean Value Theorem
7.1 Geometrical Interpretation of LMVT
8. Cauchy’s Mean Value Theorem
9. L’Hospital’s Rule (Without Proof)
10. Stronger Form
5. Successive Differentiation
1. The nth Derivatives of Standard Functions
2. Leibnitz’s Theorem (With Proof)
6. Taylor’s and Maclaurin’s Theorems
1. Taylor’s and Maclaurin’s Theorems with Lagrange’s and Cauchy’s Form of Remainders (Without Proof)
1.1 Taylor’s Theorem with Remainder
1.2 Maclaurian’s Theorem with Remainder
2. Taylor’s and Maclaurin’s Series
7. Matrices and System of Linear Equations
1. Introduction
2. Revision: Elementary Operations on Matrices
3. Echelon Form of a Matrix
3.1 Reduction of a mxn Matrix to Row Echelon Form
4. System of Linear Equations: Gauss Elimination Method, Gauss-Jordan Elimination Method, L.U. Decomposition Method
4.1 System of Linear Equation
4.2 Gauss Elimination Method
4.3 Gauss-Jordan Elimination Method
4.4 LU-Decomposition Method
5. Rank of a Matrix, Row Rank, Column Rank
5.1 Row Rank, Column Rank
• Stream Science
• Branch Computer Science
• Standard/Year Firstyear
• Medium English
• Board/University Pune
• Subject Mathematics
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https://im.kendallhunt.com/k5/teachers/grade-3/unit-5/lesson-10/preparation.html
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# Lesson 10
Equivalent Fractions
### Lesson Purpose
The purpose of this lesson is for students to see that different fractions can be equivalent if they are the same size of the same whole.
### Lesson Narrative
Previously, students were introduced to unit fractions and non-unit fractions using area diagrams, fraction strips, and number lines. They began to work with the idea of equivalence by noticing fractions that are also whole numbers. Here, students revisit area diagrams and fraction strips to learn about fraction equivalence. Students learn that fractions that are the same size are equivalent fractions. Later, they will identify equivalent fractions as having the same location on a number line.
• Representation
• MLR7
### Learning Goals
Teacher Facing
• Identify equivalent fractions.
• Understand two fractions as equivalent if they are the same size and the parts refer to the same whole.
### Student Facing
• Let’s identify equivalent fractions.
### Required Materials
Materials to Gather
### Required Preparation
Warm-up:
• Have recording of choral count by one-fourth available, from a previous lesson.
Activity 2:
• Students need the fraction strips they made in a previous lesson.
Building Towards
### Lesson Timeline
Warm-up 10 min Activity 1 15 min Activity 2 20 min Lesson Synthesis 10 min Cool-down 5 min
### Teacher Reflection Questions
What ideas do students have about what it means for fractions to be equivalent? How can you build on those ideas in this section?
### Suggested Centers
• Number Line Scoot (2–3), Stage 3: Halves, Thirds, Fourths, Sixths and Eighths (Addressing)
• Secret Fraction (3), Stage 1: Building Non-Unit Fractions (Addressing)
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https://vustudents.ning.com/forum/topics/sta301-current-mid-term-papers-spring-2012-date-11-5-2012-to-22-0?groupUrl=sta301statisticsandprobability&x=1&groupId=3783342%3AGroup%3A59512&id=3783342%3ATopic%3A1248778&page=1
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www.bit.ly/vucodes + Link For Assignments, GDBs & Online Quizzes Solution www.bit.ly/papersvu + Link For Past Papers, Solved MCQs, Short Notes & More
# STA301 Current Mid Term Papers Spring 2012 Date: 11-5-2012 to 22-05-2012
STA301 Current Mid Term Papers Spring 2012 Date: 11-5-2012 to 22-05-2012
Current Mid Term Papers Spring 2012 Papers, May 2012 Mid Term Papers, Solved Papers, Solved Past Papers, Solved MCQs
+ http://bit.ly/vucodes (Link for Assignments, GDBs & Online Quizzes Solution)
+ http://bit.ly/papersvu (Link for Past Papers, Solved MCQs, Short Notes & More)
Views: 7100
### Replies to This Discussion
Please Share your Current Papers Questions/Pattern here to help each other. Thanks
mcqs 1) when two coins are tossed then what is the probability that atmost one head appears.
Sol: S = {HH,HT,TH,TT} A = {HT,TH,TT} n(S)= 4, n(A)= 3 P(A)= n(A) / n(S)= 3/4
2) In how many ways can 4books can be arranged on shelf?
3) when two events are independent, P(A)=0.5, P(B)=0.2 then find P(AnB)=? Sol: When two events are independent then P(AnB)=P(A)P(B) put values and get result......
4) Independent events?
5) probability of sure event? Sol: Prob of sure event always be 1.
6) probability ki range kahan sy kahan tak hoti hai?
Sol: -1 to + 1
7) Mid range ka formula btana tha.
8) Data was given. Find Midrange, Minimum Value, Maximum Value, Lower Quartile Q1, UperQuartile Q2. Ye sab find krna tha? (3Marks)
9) Define Scatter diagram? Q) when two dice are thrown then draw a sample space and find the probability of sum of both dice is 10 (5Marks)
10) aik long question aur tha jis ki probability batani thi. (5Marks)
aur bhi dhair saray mcqs thay per easy thy sab k sab
If every student share his/her current midterm paper here we can get upto 90% marks in each subject.
maybe
sme1 plz share
total question were 26 all mcqs were new only ten to twenty percent were from past papers The subjetive portion was that
1.define sample space
2.what is scatter diagram?
3.mean=2.9 median=2.2 calculate pearson coffient of skewness
4.caculat coffient of stadardivation if x=111 123 153 173
6one question was to calculate permutation
7.last question was about to calculate probablity
yeah wahan se hi yahan post kia me ne
sir yea he papers hein ur aaj wali kaha hein?
My today paper of STA 301
Define sample space
What is scatter diagram
Write the properties of random experiment
Find the mean deviation
my paper was
find person,s coeffecient of skewness
find independent event
find skewness with 5 number skewness?y relative probabality is not b symmetrical?
find sample space ?
i just remember the Q NOT value mostly subjective numaricall tha objective easy tha
my paper was click file remember me in ur prays
Attachments:
thnks
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Find the maximum height attained by a ball thrown vertically upward with an initial velocity of 9.8m/s.
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At the highest point the velocity of the ball becomes zero. So we have Final velocity v = 0m/s Initial velocity u = 9.8 m/s Acceleration of the ball a = -g = -9.8 m/s^2 Let the highest distance attained by the ball is s. Then applying the equation of motion v^2 = u^2 + 2as 0 = (9.8)^2 -2\times9.8\times s s = \dfrac{9.8^2}{2\times9.8} = 4.9 m
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From the property of log we know that log(x)=y, implies e^y=x Therefore e^(logx)=5, implies log(5)=logx, hence x=5
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Given x/y=5/2 Multiplying the numerator by 2 and denominator by 3 on both sides of the above equaton 2x/3y=10/6 2x/3y=5/3 Applying componendo and dividendo (2x+3y)/(2x-3y)=(5+3)/(5-3) =8/2 =4
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https://www.teachoo.com/2195/584/Ex-3.3--23---Prove-tan-4x--4-tan-x-(1---tan2-x)---1---6tan2x/category/Ex-3.3/
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Ex 3.3
Chapter 3 Class 11 Trigonometric Functions
Serial order wise
### Transcript
Ex 3.3, 23 Prove that tanβ‘4π₯ = (4 tanβ‘γπ₯ (1βtan2π₯)γ)/(1 β 6 tan2 π₯+tan4 π₯) Solving L.H.S. tan 4x We know that tan 2x = (2 π‘ππβ‘π₯)/(1 β π‘ππ2 π₯) Replacing x with 2x tan (2 Γ 2x) = (2 π‘ππβ‘2π₯)/(1 β π‘ππ2 2π₯) tan 4x = (2 π‘ππβ‘2π₯)/(1 β π‘ππ2 2π₯) = (π πππ§β‘ππ±)/(π β πππ§π ππ±) = 2((π πππβ‘π)/(π β ππππ π))/(1 β ((π πππβ‘π)/(π β ππππ π))^2 ) = (((4 tanβ‘π₯)/(1 β π‘ππ2 π₯)))/(1 β((2 tanβ‘π₯ )^2/(1 β π‘ππ2 π₯)^2 ) ) = (((4 tanβ‘π₯)/(1 β π‘ππ2 π₯)))/(1 β((4 γπ‘ππγ^2 π₯)/(1 β π‘ππ2 π₯)^2 ) ) = (((4 tanβ‘π₯)/(1 β π‘ππ2 π₯)))/((((1 β π‘ππ2 π₯)^2 β4 γπ‘ππγ^2 π₯)/(1 β π‘ππ2 π₯)^2 ) ) = (π πππβ‘π)/(π β πππ§πβ‘π ) Γ (π β πππ§πβ‘π )^π/((π β ππππ π)π βπ πππ§πβ‘π ) = (4 tanβ‘π₯)/1 Γ ((1 β tan2β‘γπ₯)γ)/((1 β π‘ππ2 π₯)2 β 4 tan2β‘π₯ ) = (π πππβ‘γπ (π β ππππ πγ))/((π βπππ§πβ‘π )π β π πππβ‘ππ ) Using (a β b)2 = a2 + b2 β 2ab = (4 tanβ‘γπ₯ (1 β tan2β‘π₯)γ)/(( 12+(π‘ππ2 π₯)2 β 2 Γ 1 Γ π‘ππ2 π₯) β4 π‘ππ2 π₯) = (4 tanβ‘γπ₯ (1 β tan2β‘π₯)γ)/(1 + tanβ‘γ4 π₯ β 2 tan2β‘γπ₯ β4 tan2β‘π₯ γ γ ) = (π πππβ‘γπ (π β πππ§πβ‘γπ)γ γ)/(π + πππ§πβ‘γπ β π ππππ πγ ) = R.H.S. Hence L.H.S. = R.H.S. Hence proved
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.
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https://www.numerade.com/questions/a-let-a_n-be-the-area-of-a-polygon-with-n-equal-sides-inscribed-in-a-circle-with-radius-r-by-dividin/
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(a) Let $A_n$ be the area of a polygon with $n$ equal sides inscribed in a circle with radius $r$. By dividing the polygon into $n$ congruent triangles with central angle $2\pi/n$, show that $$A_n = \frac{1}{2} nr^2 \sin \biggl( \frac{2 \pi}{n} \biggr)$$(b) Show that $\displaystyle \lim_{n \to \infty} A_n = \pi r^2$. [$Hint:$ Use Equation 3.3.2 on page 191.]
a. $\Delta=\frac{1}{2} r \cdot r \cdot \sin \left(\frac{2 \pi}{h}\right)$b. $\pi r^{2}$
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So here we use untrue I and goes to approximate a polygon to approximate area of the circle. So here, so it should eyes try. Go on, look at this. This one has radius. Are this wise radius? Are this is one over the the horse Sir Ho sang Go So the whole circle goes to part over Wei have learned from the high school that the area of the triangle this one over two length of this one times Lance off this one time sigh of this and go Oh, so it's not yet also distressed area of the triangle is one over two times dance of this side and lands of this side Time side off the angle to pi over and the alien is simply trying goes at it together Therefore he gives us up This formula and computer are limit on fraud on course the infinity here that is this so here we still have to take limit here We put on over to pie here that we put pie are square here and we have side you pie over this just below there How to brought shake the limit here on this term This part it is. Actually, we actually use a famous result from calculus. What unconsciously infinity? This argument this too high over and goes to zero. And this can be view us divided by two pi over. So we used the famous result that science over hacks the limit of ab support zero give us one So this part of the limit will goes to one and we're left with pi r squared.
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